WEBVTT 00:00:00.000 --> 00:00:04.890 align:middle line:90% [SQUEAKING][RUSTLING][CLICKING] 00:00:04.890 --> 00:00:18.093 align:middle line:90% 00:00:18.093 --> 00:00:22.030 align:middle line:84% CASEY RODRIGUEZ: So today we're going 00:00:22.030 --> 00:00:31.390 align:middle line:84% to apply functional analysis to a simple Dirichlet 00:00:31.390 --> 00:00:33.940 align:middle line:84% problem, i.e., an Ordinary Differentiable 00:00:33.940 --> 00:00:39.580 align:middle line:84% Equation, on a line with conditions at the boundary. 00:00:39.580 --> 00:00:42.940 align:middle line:84% Typically, when you encounter ODEs for the first time, 00:00:42.940 --> 00:00:45.670 align:middle line:84% say, in your Ordinary Differentiable Equations class, 00:00:45.670 --> 00:00:48.140 align:middle line:84% you always have an equation you want to solve, 00:00:48.140 --> 00:00:53.090 align:middle line:84% and then you specify maybe the function 00:00:53.090 --> 00:00:54.770 align:middle line:84% at a point and its first derivative, 00:00:54.770 --> 00:00:56.840 align:middle line:84% if it's the second-order derivative or just 00:00:56.840 --> 00:00:59.960 align:middle line:84% your second-order differentiable equation or just the function 00:00:59.960 --> 00:01:01.227 align:middle line:90% evaluated at that point. 00:01:01.227 --> 00:01:02.810 align:middle line:84% But now this Dirichlet problem, you're 00:01:02.810 --> 00:01:05.663 align:middle line:84% going to be specifying the function at two endpoints. 00:01:05.663 --> 00:01:07.580 align:middle line:84% So what's the problem we're going to look at-- 00:01:07.580 --> 00:01:10.870 align:middle line:90% 00:01:10.870 --> 00:01:24.100 align:middle line:84% let V be a continuous real value function. 00:01:24.100 --> 00:01:25.650 align:middle line:90% And we will consider-- 00:01:25.650 --> 00:01:28.678 align:middle line:90% 00:01:28.678 --> 00:01:30.095 align:middle line:84% we'll call the Dirichlet problem-- 00:01:30.095 --> 00:01:36.480 align:middle line:90% 00:01:36.480 --> 00:01:42.000 align:middle line:84% minus u double prime of x, plus V of x. 00:01:42.000 --> 00:01:45.380 align:middle line:90% u of x equals f of x. 00:01:45.380 --> 00:01:51.120 align:middle line:90% 00:01:51.120 --> 00:01:56.970 align:middle line:84% And with boundary conditions, u of 0 equals 0. 00:01:56.970 --> 00:01:58.710 align:middle line:90% u of 1 equals 0. 00:01:58.710 --> 00:02:03.120 align:middle line:84% So what do we want to do with this? 00:02:03.120 --> 00:02:05.680 align:middle line:90% 00:02:05.680 --> 00:02:13.530 align:middle line:90% So the question is given. 00:02:13.530 --> 00:02:14.550 align:middle line:90% So what's the input? 00:02:14.550 --> 00:02:19.480 align:middle line:84% You may think of f as being kind of a force. 00:02:19.480 --> 00:02:22.020 align:middle line:84% So you're given a force and you would 00:02:22.020 --> 00:02:24.000 align:middle line:84% like to compute the solution or show 00:02:24.000 --> 00:02:27.360 align:middle line:84% there exists a solution on the interval 0, 00:02:27.360 --> 00:02:30.030 align:middle line:90% 1 satisfying the equation. 00:02:30.030 --> 00:02:31.750 align:middle line:84% And we'll say a classical solution, 00:02:31.750 --> 00:02:33.840 align:middle line:84% meaning it's twice continuously differentiable. 00:02:33.840 --> 00:02:37.020 align:middle line:84% So this equation makes sense at every x. 00:02:37.020 --> 00:02:49.540 align:middle line:84% So given f continuous function, does there 00:02:49.540 --> 00:03:02.010 align:middle line:90% exist a unique solution u of x. 00:03:02.010 --> 00:03:06.090 align:middle line:90% And in C2 0, 1, so the 2-- 00:03:06.090 --> 00:03:10.230 align:middle line:84% C2 meaning two continuous derivatives. 00:03:10.230 --> 00:03:12.750 align:middle line:84% C2 means two continuous derivatives. 00:03:12.750 --> 00:03:17.280 align:middle line:84% So does there exist a unique solution 00:03:17.280 --> 00:03:21.260 align:middle line:90% to the Dirichlet problem here? 00:03:21.260 --> 00:03:23.930 align:middle line:84% Does there exist a unique solution to this ODE 00:03:23.930 --> 00:03:26.480 align:middle line:84% with these boundary conditions at 0 and 1? 00:03:26.480 --> 00:03:31.430 align:middle line:84% And the purpose of this section is to apply functional analysis 00:03:31.430 --> 00:03:34.970 align:middle line:84% that we've done to say the answer is yes 00:03:34.970 --> 00:03:36.775 align:middle line:90% if V is non-negative. 00:03:36.775 --> 00:03:42.120 align:middle line:90% 00:03:42.120 --> 00:03:44.380 align:middle line:90% OK. 00:03:44.380 --> 00:03:48.880 align:middle line:84% If V is not necessarily non-negative, 00:03:48.880 --> 00:03:53.350 align:middle line:84% then the answer is it depends on f. 00:03:53.350 --> 00:03:55.870 align:middle line:90% 00:03:55.870 --> 00:03:59.920 align:middle line:84% All right, but given any f, does there exists a unique solution 00:03:59.920 --> 00:04:02.840 align:middle line:90% to the Dirichlet problem? 00:04:02.840 --> 00:04:04.960 align:middle line:84% When V is non-negative, the answer is yes. 00:04:04.960 --> 00:04:07.600 align:middle line:84% And this is what we're going to prove. 00:04:07.600 --> 00:04:09.580 align:middle line:84% This is the main goal of this section. 00:04:09.580 --> 00:04:13.350 align:middle line:90% 00:04:13.350 --> 00:04:17.170 align:middle line:84% And I'll state this answer as two theorems. 00:04:17.170 --> 00:04:20.200 align:middle line:84% One thing that there exists a unique solution. 00:04:20.200 --> 00:04:22.570 align:middle line:90% And the other will be-- 00:04:22.570 --> 00:04:26.978 align:middle line:84% or I should say that a solution to this is unique. 00:04:26.978 --> 00:04:29.020 align:middle line:84% And then the second part, the more involved part, 00:04:29.020 --> 00:04:32.290 align:middle line:84% will be that there does exist at least one solution 00:04:32.290 --> 00:04:35.742 align:middle line:84% to this problem, assuming V is non-negative. 00:04:35.742 --> 00:04:40.160 align:middle line:90% 00:04:40.160 --> 00:04:42.710 align:middle line:84% So the first thing we'll prove is 00:04:42.710 --> 00:04:46.715 align:middle line:84% that any solution to the Dirichlet problem is unique. 00:04:46.715 --> 00:04:53.470 align:middle line:90% 00:04:53.470 --> 00:05:01.480 align:middle line:84% f is a continuous function on 0, 1. 00:05:01.480 --> 00:05:09.410 align:middle line:84% u1, u2 are twice continuously differentiable functions that 00:05:09.410 --> 00:05:17.795 align:middle line:90% satisfy the Dirichlet problem. 00:05:17.795 --> 00:05:22.230 align:middle line:90% 00:05:22.230 --> 00:05:25.200 align:middle line:84% I should say we are now working under the assumption 00:05:25.200 --> 00:05:26.230 align:middle line:90% that V is non-negative. 00:05:26.230 --> 00:05:30.640 align:middle line:84% So I should also put that in here-- 00:05:30.640 --> 00:05:32.340 align:middle line:90% suppose V is non-negative. 00:05:32.340 --> 00:05:34.710 align:middle line:84% We won't always have to assume V is non-negative in some 00:05:34.710 --> 00:05:37.950 align:middle line:84% of these theorems that we'll prove about this problem 00:05:37.950 --> 00:05:39.270 align:middle line:90% or about certain operators. 00:05:39.270 --> 00:05:42.900 align:middle line:84% So it would be good if I specify when 00:05:42.900 --> 00:05:44.560 align:middle line:90% I'm assuming V is non-negative. 00:05:44.560 --> 00:05:49.830 align:middle line:84% So suppose V is non-negative and f is a continuous function 00:05:49.830 --> 00:05:52.500 align:middle line:84% and I have two solutions to the Dirichlet problem, 00:05:52.500 --> 00:05:57.580 align:middle line:90% then they must be the same. 00:05:57.580 --> 00:06:02.850 align:middle line:84% Now the proof I think I've given this as an assignment 00:06:02.850 --> 00:06:06.120 align:middle line:84% or as a problem in an assignment in past classes, 00:06:06.120 --> 00:06:11.400 align:middle line:84% but when in doubt integrate by parts. 00:06:11.400 --> 00:06:21.170 align:middle line:84% So by subtracting let u be u 1 minus u 2. 00:06:21.170 --> 00:06:25.220 align:middle line:84% Then u is a twice continuously differentiable function 00:06:25.220 --> 00:06:34.890 align:middle line:84% which satisfies u of x of V of x, times u of x equal 0. 00:06:34.890 --> 00:06:41.190 align:middle line:84% And u of 0 equals u of 1 equals 0. 00:06:41.190 --> 00:06:43.410 align:middle line:84% Now what we'd like to conclude is that u is 0. 00:06:43.410 --> 00:06:45.035 align:middle line:84% And how we're going to do this if we're 00:06:45.035 --> 00:06:47.820 align:middle line:84% going to integrate by parts. so then multiply this equation 00:06:47.820 --> 00:06:49.330 align:middle line:84% by something and integrate by parts. 00:06:49.330 --> 00:06:54.910 align:middle line:84% So we have 0 is equal to the integral from 0 00:06:54.910 --> 00:06:59.650 align:middle line:84% to 1 of minus u double prime of x, 00:06:59.650 --> 00:07:07.915 align:middle line:84% plus V of x, u of x times the complex conjugate of u 00:07:07.915 --> 00:07:10.330 align:middle line:90% of x, dx. 00:07:10.330 --> 00:07:13.010 align:middle line:90% 00:07:13.010 --> 00:07:16.716 align:middle line:84% Now this is equal to integral from 0 to 1, 00:07:16.716 --> 00:07:28.790 align:middle line:84% u prime of x times u of x, plus integral from 0 to 1, 00:07:28.790 --> 00:07:33.990 align:middle line:84% V of x times cancelled value of u of x squared. 00:07:33.990 --> 00:07:37.270 align:middle line:90% 00:07:37.270 --> 00:07:38.590 align:middle line:90% And now we integrate by part. 00:07:38.590 --> 00:07:40.290 align:middle line:84% So integration by part says that if I 00:07:40.290 --> 00:07:42.600 align:middle line:84% have a derivative on a function, I 00:07:42.600 --> 00:07:47.070 align:middle line:84% can move it to the other function, and then 00:07:47.070 --> 00:07:48.390 align:middle line:90% plus boundary terms. 00:07:48.390 --> 00:07:50.850 align:middle line:84% And changing this, minus sine to a cosine. 00:07:50.850 --> 00:07:54.980 align:middle line:84% So this is equal to u prime of x-- 00:07:54.980 --> 00:08:02.870 align:middle line:84% u bar of x evaluated at 0, 1, with this minus 00:08:02.870 --> 00:08:04.250 align:middle line:90% gives me a plus. 00:08:04.250 --> 00:08:06.260 align:middle line:90% And we go from 0 to 1. 00:08:06.260 --> 00:08:13.730 align:middle line:84% We find with x the derivative of complex conjugate of u dx 00:08:13.730 --> 00:08:21.245 align:middle line:90% plus V of x, d of x squared, dx. 00:08:21.245 --> 00:08:24.170 align:middle line:90% 00:08:24.170 --> 00:08:28.760 align:middle line:84% Now we use the boundary conditions that at 0 and at 1, 00:08:28.760 --> 00:08:29.900 align:middle line:90% u of x is 0. 00:08:29.900 --> 00:08:32.960 align:middle line:84% So its complex conjugate is 0, so please go away. 00:08:32.960 --> 00:08:36.500 align:middle line:84% And what I'm left with is integral from 0 to 1. 00:08:36.500 --> 00:08:41.570 align:middle line:84% And I'm not going to keep writing the x part. 00:08:41.570 --> 00:08:44.810 align:middle line:84% So integral from 0 to 1 of u prime squared, plus integral 00:08:44.810 --> 00:08:48.290 align:middle line:90% from 1 to 1, V times u prime-- 00:08:48.290 --> 00:08:51.900 align:middle line:84% or times the magnitude of u squared. 00:08:51.900 --> 00:08:57.110 align:middle line:84% Now since V is non-negative, V times the absolute value 00:08:57.110 --> 00:08:58.560 align:middle line:90% of u squared is non-negative. 00:08:58.560 --> 00:09:04.470 align:middle line:84% So this is certainly bigger than 0, 1, u prime squared. 00:09:04.470 --> 00:09:06.900 align:middle line:84% And remember, what I started off with was 0. 00:09:06.900 --> 00:09:08.760 align:middle line:84% So I showed 0 is bigger than or equal 00:09:08.760 --> 00:09:11.085 align:middle line:90% to this non-negative quantity. 00:09:11.085 --> 00:09:15.230 align:middle line:90% 00:09:15.230 --> 00:09:19.530 align:middle line:84% Now u is twice continuously differentiable 00:09:19.530 --> 00:09:23.370 align:middle line:84% so that u prime, that's a continuous function-- 00:09:23.370 --> 00:09:25.000 align:middle line:84% non-negative continuous function. 00:09:25.000 --> 00:09:35.990 align:middle line:84% So since this equals 0, I get that u prime is identically 00:09:35.990 --> 00:09:39.815 align:middle line:90% 0, which implies u is constant. 00:09:39.815 --> 00:09:42.980 align:middle line:90% 00:09:42.980 --> 00:09:46.730 align:middle line:84% And because u is constant and u of 0 equals 0, 00:09:46.730 --> 00:09:53.130 align:middle line:84% I get that u is identically 0, the 0 function, 00:09:53.130 --> 00:09:54.680 align:middle line:90% which is what I wanted to show. 00:09:54.680 --> 00:09:58.640 align:middle line:90% u equals 0 means u1 equals u2. 00:09:58.640 --> 00:10:03.550 align:middle line:90% So that's uniqueness. 00:10:03.550 --> 00:10:05.170 align:middle line:84% That's half of what we wanted to show, 00:10:05.170 --> 00:10:08.050 align:middle line:84% which was if V is non-negative, then there exists 00:10:08.050 --> 00:10:12.250 align:middle line:84% a unique solution to the Dirichlet problem given any f, 00:10:12.250 --> 00:10:16.370 align:middle line:84% any input to the right-hand side. 00:10:16.370 --> 00:10:19.870 align:middle line:84% Now we're going to turn to unique existence, which 00:10:19.870 --> 00:10:25.070 align:middle line:84% is more interesting and harder parts. 00:10:25.070 --> 00:10:31.360 align:middle line:84% So now we're going on to existence. 00:10:31.360 --> 00:10:34.850 align:middle line:90% 00:10:34.850 --> 00:10:38.050 align:middle line:84% So if you don't know how to do something, 00:10:38.050 --> 00:10:40.550 align:middle line:90% try and do something easier. 00:10:40.550 --> 00:10:44.720 align:middle line:84% So let's start off with the V equals 0 case. 00:10:44.720 --> 00:10:45.625 align:middle line:90% So we'll look at-- 00:10:45.625 --> 00:11:00.660 align:middle line:90% 00:11:00.660 --> 00:11:02.790 align:middle line:84% so we have the following theorem that 00:11:02.790 --> 00:11:08.070 align:middle line:84% says we can uniquely solve the Dirichlet problem when 00:11:08.070 --> 00:11:08.640 align:middle line:90% equals 0. 00:11:08.640 --> 00:11:10.650 align:middle line:84% So that just means we want to solve minus u 00:11:10.650 --> 00:11:13.980 align:middle line:84% double prime equals f of x with these boundary conditions. 00:11:13.980 --> 00:11:16.830 align:middle line:84% And we can actually write down explicitly the solution 00:11:16.830 --> 00:11:22.530 align:middle line:84% in terms of f as a compact self-adjoint operator, which 00:11:22.530 --> 00:11:24.430 align:middle line:84% I've alluded to maybe at a couple of points 00:11:24.430 --> 00:11:25.180 align:middle line:90% here in the class. 00:11:25.180 --> 00:11:36.370 align:middle line:84% So let K x, y be the function given by minus 1 times y here 00:11:36.370 --> 00:11:45.660 align:middle line:84% if y is less than or equal to x, is less than or equal to 1, 00:11:45.660 --> 00:11:47.400 align:middle line:90% and now switch. 00:11:47.400 --> 00:11:55.640 align:middle line:90% 00:11:55.640 --> 00:11:59.390 align:middle line:84% So it's piece-wise defined across the diagonal. 00:11:59.390 --> 00:12:02.300 align:middle line:84% And it's continuous across the diagonal when y equals x. 00:12:02.300 --> 00:12:03.450 align:middle line:90% I get the same thing. 00:12:03.450 --> 00:12:07.520 align:middle line:84% So this thing, in fact, is a continuous function 00:12:07.520 --> 00:12:10.004 align:middle line:90% on 0, 1 cross 0, 1. 00:12:10.004 --> 00:12:14.170 align:middle line:90% 00:12:14.170 --> 00:12:15.415 align:middle line:90% We have this function here. 00:12:15.415 --> 00:12:22.620 align:middle line:90% 00:12:22.620 --> 00:12:32.300 align:middle line:84% Define an operator, A f of x to be integral K of x, y, 00:12:32.300 --> 00:12:34.050 align:middle line:90% f of y, dy. 00:12:34.050 --> 00:12:41.310 align:middle line:90% 00:12:41.310 --> 00:12:43.740 align:middle line:84% And A is a bounded linear operator on L2. 00:12:43.740 --> 00:12:51.620 align:middle line:90% 00:12:51.620 --> 00:12:54.634 align:middle line:84% And in fact, it's a self-adjoint compact operator. 00:12:54.634 --> 00:13:05.520 align:middle line:90% 00:13:05.520 --> 00:13:07.070 align:middle line:90% Self-adjoint operator. 00:13:07.070 --> 00:13:10.550 align:middle line:90% 00:13:10.550 --> 00:13:16.160 align:middle line:84% And basically A times f is the inverse of u double prime, 00:13:16.160 --> 00:13:18.770 align:middle line:84% or A times f solves the Dirichlet problem. 00:13:18.770 --> 00:13:23.020 align:middle line:90% 00:13:23.020 --> 00:13:27.500 align:middle line:90% f is C 0, 1. 00:13:27.500 --> 00:13:42.860 align:middle line:84% And u equals Af is the peak solution to the Dirichlet 00:13:42.860 --> 00:13:44.723 align:middle line:90% problem when V equals 0. 00:13:44.723 --> 00:13:49.560 align:middle line:90% 00:13:49.560 --> 00:13:52.320 align:middle line:84% I'm not writing out all the argument of x, 00:13:52.320 --> 00:14:00.320 align:middle line:84% but it should be understood u of 0 equals u of 1 equals 0. 00:14:00.320 --> 00:14:03.170 align:middle line:84% So when V equals 0, we can write down the explicit solution 00:14:03.170 --> 00:14:06.230 align:middle line:84% in terms of this, what's called an integral operator 00:14:06.230 --> 00:14:08.300 align:middle line:84% because you take a function, multiply it 00:14:08.300 --> 00:14:10.970 align:middle line:84% by another function, and integrate. 00:14:10.970 --> 00:14:14.000 align:middle line:84% This thing here is usually referred to 00:14:14.000 --> 00:14:18.230 align:middle line:84% as the Green's function for this differentiable operator. 00:14:18.230 --> 00:14:20.510 align:middle line:84% And what this theorem says is that the solution 00:14:20.510 --> 00:14:24.620 align:middle line:84% to your differentiable equation, your Dirichlet problem, 00:14:24.620 --> 00:14:28.280 align:middle line:84% is given by an integral operator, which 00:14:28.280 --> 00:14:31.880 align:middle line:84% is a compact self-adjoint operator on L2. 00:14:31.880 --> 00:14:33.830 align:middle line:84% So it shouldn't come too much of a surprise 00:14:33.830 --> 00:14:38.360 align:middle line:84% that the solution to a Dirichlet problem u 00:14:38.360 --> 00:14:40.430 align:middle line:84% can be written as an integral operator, 00:14:40.430 --> 00:14:44.240 align:middle line:84% right, because by the fundamental theorem of calculus 00:14:44.240 --> 00:14:47.120 align:middle line:84% integration and differentiation are inverse of each other. 00:14:47.120 --> 00:15:02.960 align:middle line:90% 00:15:02.960 --> 00:15:14.520 align:middle line:84% So let's C be the sup over 0, 1 cross 00:15:14.520 --> 00:15:19.000 align:middle line:84% 0, 1 of the absolute value of K. So I'm not writing x, y. 00:15:19.000 --> 00:15:23.260 align:middle line:84% So this is the supremum of K over the square of 0, 00:15:23.260 --> 00:15:27.160 align:middle line:84% 1 across 0, 1, which is finite, since K is continuous. 00:15:27.160 --> 00:15:31.650 align:middle line:90% 00:15:31.650 --> 00:15:37.490 align:middle line:84% And by the Cauchy-Schwarz inequality we get that A times 00:15:37.490 --> 00:15:47.414 align:middle line:84% f of x, A f of x, is equal to the integral from 0, 1, K x, y, 00:15:47.414 --> 00:15:52.110 align:middle line:84% f of y dy, is less than or equal to-- now if I bring 00:15:52.110 --> 00:15:57.210 align:middle line:84% the absolute value inside and then bound K by C-- 00:15:57.210 --> 00:15:58.710 align:middle line:90% my name is Casey-- 00:15:58.710 --> 00:16:05.610 align:middle line:84% is bounded by C times the integral from 0, 1 of f of y 00:16:05.610 --> 00:16:06.270 align:middle line:90% dy. 00:16:06.270 --> 00:16:09.696 align:middle line:84% And now I apply Cauchy-Schwarz to this, 00:16:09.696 --> 00:16:12.720 align:middle line:84% this I can think of as this quantity times 1. 00:16:12.720 --> 00:16:23.660 align:middle line:84% So this is less than or equal to C times the integral 0, 1, 1 00:16:23.660 --> 00:16:26.160 align:middle line:90% squared raised to the 1/2 power. 00:16:26.160 --> 00:16:31.160 align:middle line:84% At certain points I will stop writing of x dx, or of y dy. 00:16:31.160 --> 00:16:38.690 align:middle line:84% And the meaning should be clear, integral 0, 1 f squared 1/2. 00:16:38.690 --> 00:16:44.330 align:middle line:84% And I get C times the L2 norm of f. 00:16:44.330 --> 00:16:45.470 align:middle line:90% So what have I done? 00:16:45.470 --> 00:16:49.300 align:middle line:84% I've bounded for every x in 0, 1, 00:16:49.300 --> 00:16:56.000 align:middle line:84% A f of x, by a constant times f and L2. 00:16:56.000 --> 00:17:10.730 align:middle line:84% And I also have, if I look A f of x minus A f of z, 00:17:10.730 --> 00:17:18.069 align:middle line:84% so I can just again by bringing the absolute value 00:17:18.069 --> 00:17:20.410 align:middle line:90% inside the integral, bounded. 00:17:20.410 --> 00:17:22.510 align:middle line:84% And so the difference here is going 00:17:22.510 --> 00:17:27.740 align:middle line:84% to be K of x, y minus K of x, z times f of y and integrate it. 00:17:27.740 --> 00:17:32.860 align:middle line:84% So what I get is that this is less than or equal to sup y 00:17:32.860 --> 00:17:48.242 align:middle line:84% in 0, 1, K of x, y minus K of z, y times the L2 norm of f. 00:17:48.242 --> 00:17:50.450 align:middle line:84% So I'm not going to give all the details because I've 00:17:50.450 --> 00:17:52.730 align:middle line:84% actually assigned this problem before, 00:17:52.730 --> 00:17:55.580 align:middle line:84% that if K is a continuous function, then 00:17:55.580 --> 00:18:00.320 align:middle line:84% this is a bounded compact operator on L2. 00:18:00.320 --> 00:18:04.642 align:middle line:84% So maybe you haven't done the exercises yet, but in any case 00:18:04.642 --> 00:18:07.100 align:middle line:84% I'll just sketch out from here that just based on these two 00:18:07.100 --> 00:18:25.190 align:middle line:84% estimates and the Arzela-Ascoli theorem, 00:18:25.190 --> 00:18:28.480 align:middle line:84% which gives you sufficient conditions 00:18:28.480 --> 00:18:32.770 align:middle line:84% that a sequence of continuous functions 00:18:32.770 --> 00:18:35.260 align:middle line:84% has a conversion subsequence in the space 00:18:35.260 --> 00:18:43.050 align:middle line:84% of continuous functions, you can conclude that A 00:18:43.050 --> 00:18:49.100 align:middle line:90% is the compact operator on L2. 00:18:49.100 --> 00:19:00.240 align:middle line:90% 00:19:00.240 --> 00:19:04.740 align:middle line:84% Now so that shows is A compact operator on L2. 00:19:04.740 --> 00:19:05.760 align:middle line:90% Y is itself adjoint. 00:19:05.760 --> 00:19:08.620 align:middle line:90% 00:19:08.620 --> 00:19:15.070 align:middle line:84% Let f, g be continuous functions on 0, 1. 00:19:15.070 --> 00:19:19.600 align:middle line:84% Then if I look at a times f paired with g, 00:19:19.600 --> 00:19:22.510 align:middle line:84% this is in the L2 pairing, this is 00:19:22.510 --> 00:19:29.930 align:middle line:84% equal to integral of 0, 1, integral of 0, 1, K of x, y, 00:19:29.930 --> 00:19:35.240 align:middle line:90% f of y dy, V of x dx. 00:19:35.240 --> 00:19:37.760 align:middle line:90% 00:19:37.760 --> 00:19:40.820 align:middle line:84% Now I have a double integral involving nothing 00:19:40.820 --> 00:19:42.060 align:middle line:90% but a continuous function. 00:19:42.060 --> 00:19:48.530 align:middle line:84% So I can apply Fubini's theorem to interchange the integration. 00:19:48.530 --> 00:19:55.010 align:middle line:84% So this becomes 0, 1, integral 0, 1. 00:19:55.010 --> 00:19:57.920 align:middle line:84% I can write it this way-- f of y. 00:19:57.920 --> 00:19:59.510 align:middle line:84% So I haven't done anything yet, I'm 00:19:59.510 --> 00:20:05.885 align:middle line:84% just distributing, bringing this gx inside 00:20:05.885 --> 00:20:07.260 align:middle line:84% and stating that this integral is 00:20:07.260 --> 00:20:12.030 align:middle line:84% equal to this double integral, dy dx. 00:20:12.030 --> 00:20:16.020 align:middle line:84% I guess I'm using Fubini's theorem there. 00:20:16.020 --> 00:20:18.360 align:middle line:84% And Fubini's theorem says if you're 00:20:18.360 --> 00:20:23.040 align:middle line:84% integrating continuous functions over a box, 00:20:23.040 --> 00:20:30.390 align:middle line:84% then it doesn't matter if you integrate dy first or dx first. 00:20:30.390 --> 00:20:33.630 align:middle line:90% So then I can just undo this. 00:20:33.630 --> 00:20:36.788 align:middle line:84% And, oh, the pairing should have a complex conjugate over g. 00:20:36.788 --> 00:20:41.120 align:middle line:90% 00:20:41.120 --> 00:20:47.150 align:middle line:84% And so if I just integrate dx first and kind of undo things, 00:20:47.150 --> 00:20:49.130 align:middle line:84% I can write this as the integral of f 00:20:49.130 --> 00:20:57.560 align:middle line:84% of y times the integral from 0, 1 of K of x, y, g of x dx. 00:20:57.560 --> 00:21:00.605 align:middle line:84% Now the complex conjugate over K, 00:21:00.605 --> 00:21:06.500 align:middle line:84% all of this complex conjugate, dy. 00:21:06.500 --> 00:21:09.230 align:middle line:84% Again, you can check that this is equal to this previous thing 00:21:09.230 --> 00:21:12.350 align:middle line:84% by Fubini, because then this iterated integral becomes 00:21:12.350 --> 00:21:15.530 align:middle line:84% just an integral dx dy, which again by Fubini 00:21:15.530 --> 00:21:17.360 align:middle line:84% doesn't matter what the order is. 00:21:17.360 --> 00:21:20.030 align:middle line:84% And the complex conjugate hits this complex conjugate. 00:21:20.030 --> 00:21:23.090 align:middle line:84% I get back K. This complex conjugate gets g, 00:21:23.090 --> 00:21:24.170 align:middle line:90% and I get that. 00:21:24.170 --> 00:21:27.320 align:middle line:90% 00:21:27.320 --> 00:21:31.630 align:middle line:84% So this says that Af paired with g 00:21:31.630 --> 00:21:47.590 align:middle line:84% is equal to f paired with B times g, where Bg of x now. 00:21:47.590 --> 00:21:52.330 align:middle line:84% So I got out a function of y integrating x. 00:21:52.330 --> 00:21:57.220 align:middle line:84% So if I switch the dummy variables, 00:21:57.220 --> 00:22:06.160 align:middle line:84% I can write this as K y, x, complex conjugate, dy dy. 00:22:06.160 --> 00:22:11.660 align:middle line:90% 00:22:11.660 --> 00:22:16.220 align:middle line:84% But now the thing to note is that K, what is K? 00:22:16.220 --> 00:22:19.530 align:middle line:90% So K is this function here. 00:22:19.530 --> 00:22:21.630 align:middle line:90% It's a real value function. 00:22:21.630 --> 00:22:26.270 align:middle line:84% So the complex conjugate just doesn't matter. 00:22:26.270 --> 00:22:27.090 align:middle line:90% It's real value. 00:22:27.090 --> 00:22:29.600 align:middle line:84% So the complex conjugate is equal to the original function. 00:22:29.600 --> 00:22:31.520 align:middle line:90% And it's symmetric in x and y. 00:22:31.520 --> 00:22:35.600 align:middle line:84% If I switch x and y, I get K x, y back. 00:22:35.600 --> 00:22:39.260 align:middle line:84% So K of y, x is equal to K of x, y. 00:22:39.260 --> 00:22:41.456 align:middle line:90% So this is equal to 0, 1. 00:22:41.456 --> 00:22:46.340 align:middle line:90% K of x, y, g of y dy. 00:22:46.340 --> 00:22:50.090 align:middle line:84% But this is just by definition equal to A times g. 00:22:50.090 --> 00:23:01.460 align:middle line:84% So we've proven that Af, g equals Af paired with Ag now 00:23:01.460 --> 00:23:08.210 align:middle line:84% for all f, g, which are continuous, which, remember, 00:23:08.210 --> 00:23:10.600 align:middle line:90% is a subset of 0, 1. 00:23:10.600 --> 00:23:12.670 align:middle line:84% But not just a subset, it's a dense subset. 00:23:12.670 --> 00:23:18.020 align:middle line:90% 00:23:18.020 --> 00:23:28.760 align:middle line:84% Since continuous functions on 0, 1 are dense in L2, 00:23:28.760 --> 00:23:32.990 align:middle line:84% and the density argument implies that this relation has to hold 00:23:32.990 --> 00:23:34.940 align:middle line:90% for every f and g in L2. 00:23:34.940 --> 00:23:38.456 align:middle line:90% 00:23:38.456 --> 00:23:47.690 align:middle line:84% It's Af, g equals f, Ag, not just 00:23:47.690 --> 00:23:52.350 align:middle line:84% for all continuous functions, but for all functions in L2. 00:23:52.350 --> 00:23:57.260 align:middle line:90% 00:23:57.260 --> 00:23:59.706 align:middle line:84% That proves that A is self-adjoint. 00:23:59.706 --> 00:24:11.430 align:middle line:90% 00:24:11.430 --> 00:24:13.140 align:middle line:84% All right, and so the last part, which 00:24:13.140 --> 00:24:16.830 align:middle line:84% is verifying that in fact defining u to be Af, 00:24:16.830 --> 00:24:19.260 align:middle line:84% gives you a twice continuously differentiable function 00:24:19.260 --> 00:24:22.650 align:middle line:84% which satisfies the Dirichlet problem when V equals 0. 00:24:22.650 --> 00:24:25.810 align:middle line:84% It just follows from direct computation. 00:24:25.810 --> 00:24:40.250 align:middle line:84% So f is in 0, 1, then if I define u of x to be Af of x, 00:24:40.250 --> 00:24:44.870 align:middle line:84% and I actually write out the integral 00:24:44.870 --> 00:24:51.180 align:middle line:84% over the various domains on how it's defined, 00:24:51.180 --> 00:24:57.837 align:middle line:84% this is first I pick up an integral 0 00:24:57.837 --> 00:25:09.600 align:middle line:84% to x of x minus 1 times y, f of y, dy, plus x times 0, 00:25:09.600 --> 00:25:15.980 align:middle line:90% 1 of y minus 1, dy. 00:25:15.980 --> 00:25:18.950 align:middle line:84% And now I can just apply the fundamental theorem of calculus 00:25:18.950 --> 00:25:29.510 align:middle line:84% to show that indeed by fundamental theorem of calculus 00:25:29.510 --> 00:25:33.020 align:middle line:84% that u is twice continuously differentiable. 00:25:33.020 --> 00:25:35.880 align:middle line:90% 00:25:35.880 --> 00:25:41.820 align:middle line:84% And minus u double prime gives me f. 00:25:41.820 --> 00:25:45.500 align:middle line:90% 00:25:45.500 --> 00:25:48.350 align:middle line:90% And u is given by Af. 00:25:48.350 --> 00:25:50.300 align:middle line:84% This is the unique solution to that problem 00:25:50.300 --> 00:25:52.910 align:middle line:84% because we've already proven that when V is non-negative, 00:25:52.910 --> 00:25:56.070 align:middle line:84% there exists only one solution to the Dirichlet problem. 00:25:56.070 --> 00:26:02.570 align:middle line:84% So in the case V equals 0, we can write down 00:26:02.570 --> 00:26:06.720 align:middle line:84% the explicit solution in terms of this integral operator, 00:26:06.720 --> 00:26:10.851 align:middle line:84% which on L2 is the self-adjoint contact operator. 00:26:10.851 --> 00:26:14.010 align:middle line:90% 00:26:14.010 --> 00:26:24.870 align:middle line:84% All right, so to solve the Dirichlet problem, 00:26:24.870 --> 00:26:39.105 align:middle line:84% so now what's the plan for V not equal to 0? 00:26:39.105 --> 00:26:42.890 align:middle line:90% 00:26:42.890 --> 00:26:45.780 align:middle line:90% So the plan is this-- 00:26:45.780 --> 00:26:53.580 align:middle line:84% that if I have minus u double prime plus V times u, 00:26:53.580 --> 00:26:58.050 align:middle line:84% so this is just kind of formal stuff, 00:26:58.050 --> 00:27:00.870 align:middle line:84% and then we'll actually prove rigorous statements 00:27:00.870 --> 00:27:03.780 align:middle line:84% and prove that there exists a solution in the end. 00:27:03.780 --> 00:27:07.580 align:middle line:84% We already have uniqueness, like I said. 00:27:07.580 --> 00:27:09.550 align:middle line:84% How do we solve this differentiable equation 00:27:09.550 --> 00:27:11.710 align:middle line:84% with, again, the boundary conditions which 00:27:11.710 --> 00:27:14.815 align:middle line:90% I'm not going to write down? 00:27:14.815 --> 00:27:17.738 align:middle line:90% 00:27:17.738 --> 00:27:19.155 align:middle line:84% So let me just write it down here. 00:27:19.155 --> 00:27:22.240 align:middle line:90% 00:27:22.240 --> 00:27:27.510 align:middle line:84% So that implies that minus u double prime equals f minus Vu. 00:27:27.510 --> 00:27:36.160 align:middle line:90% And so equals f plus minus Vu. 00:27:36.160 --> 00:27:39.400 align:middle line:84% Now think of this as just a fixed function g. 00:27:39.400 --> 00:27:47.730 align:middle line:84% And therefore by this existence and uniqueness result 00:27:47.730 --> 00:27:53.430 align:middle line:84% that we have for the V equals 0 case, so think of this as g. 00:27:53.430 --> 00:27:56.190 align:middle line:84% So now I have minus u double prime equals g. 00:27:56.190 --> 00:28:17.250 align:middle line:84% So this implies that u equals A applied to f minus Vu. 00:28:17.250 --> 00:28:19.440 align:middle line:84% So the unique solution to minus u double prime 00:28:19.440 --> 00:28:21.600 align:middle line:90% is equal to A applied to g. 00:28:21.600 --> 00:28:24.310 align:middle line:90% g is f minus V, so I have this. 00:28:24.310 --> 00:28:37.830 align:middle line:84% So I get-- by this if and only if here, which, 00:28:37.830 --> 00:28:42.390 align:middle line:84% if I distribute this through, is the same as saying the identity 00:28:42.390 --> 00:28:47.040 align:middle line:84% plus the operator given by A applied to the multiplication 00:28:47.040 --> 00:28:51.480 align:middle line:84% by V-- so when I write V here, I mean multiplication by V-- 00:28:51.480 --> 00:28:57.430 align:middle line:84% applied to u equals A applied to f. 00:28:57.430 --> 00:29:00.730 align:middle line:84% And now this is good because we've gotten rid 00:29:00.730 --> 00:29:03.070 align:middle line:90% of this differentiation. 00:29:03.070 --> 00:29:07.870 align:middle line:84% And now we're talking about solving an equation that 00:29:07.870 --> 00:29:10.120 align:middle line:90% involves bounded operators. 00:29:10.120 --> 00:29:12.610 align:middle line:84% The identity is a bounded operator. 00:29:12.610 --> 00:29:14.530 align:middle line:84% Multiplication by a continuous function 00:29:14.530 --> 00:29:16.450 align:middle line:90% is a bounded operator on L2. 00:29:16.450 --> 00:29:20.290 align:middle line:84% And A is a compact self-adjoint operator on L2. 00:29:20.290 --> 00:29:25.480 align:middle line:84% So now we're solving an equation involving bounded operators 00:29:25.480 --> 00:29:28.720 align:middle line:90% on the Hilbert space. 00:29:28.720 --> 00:29:32.590 align:middle line:84% What would make it even better is if this thing was-- 00:29:32.590 --> 00:29:34.480 align:middle line:84% so we already know it's a compact operator 00:29:34.480 --> 00:29:36.190 align:middle line:90% because A is compact. 00:29:36.190 --> 00:29:37.570 align:middle line:84% If we knew this was self-adjoint, 00:29:37.570 --> 00:29:39.290 align:middle line:84% but that doesn't necessarily hold because 00:29:39.290 --> 00:29:42.970 align:middle line:84% of the adjoint of AV will be V times A. 00:29:42.970 --> 00:29:45.320 align:middle line:90% So these don't exactly commute. 00:29:45.320 --> 00:29:48.370 align:middle line:84% But we can get around that and reduce ourselves 00:29:48.370 --> 00:29:53.050 align:middle line:84% to studying an equation that involves 00:29:53.050 --> 00:29:56.800 align:middle line:84% compact self-adjoint operators by a nice little trick. 00:29:56.800 --> 00:30:08.895 align:middle line:90% So write u as-- 00:30:08.895 --> 00:30:11.820 align:middle line:90% 00:30:11.820 --> 00:30:16.470 align:middle line:84% but I'll say A to the 1/2 V. A to the 1/2 meaning 00:30:16.470 --> 00:30:19.920 align:middle line:84% its square gives me A. The fact that such a thing exists 00:30:19.920 --> 00:30:24.300 align:middle line:84% is not clear right now, but we will show it, in fact, exists. 00:30:24.300 --> 00:30:25.410 align:middle line:90% Again, formal stuff. 00:30:25.410 --> 00:30:30.600 align:middle line:84% So write u as A to the 1/2 applied to V, where now V 00:30:30.600 --> 00:30:32.610 align:middle line:90% is the thing we solve for. 00:30:32.610 --> 00:30:37.560 align:middle line:84% And if we stick this into this equation here, 00:30:37.560 --> 00:30:42.770 align:middle line:84% we get A. We pull out an A to the 1/2 applied to I plus A 00:30:42.770 --> 00:30:52.450 align:middle line:84% to the 1/2 V, A to the 1/2, by to V equals Af. 00:30:52.450 --> 00:30:55.610 align:middle line:90% 00:30:55.610 --> 00:30:58.160 align:middle line:84% So this should be little v, not this capital V. 00:30:58.160 --> 00:31:03.650 align:middle line:84% And therefore if I can believe I can cancel 1/2 powers, 00:31:03.650 --> 00:31:06.815 align:middle line:84% I've reduced myself to study the equation 00:31:06.815 --> 00:31:12.120 align:middle line:84% A to the 1/2 V times A to the 1/2, 00:31:12.120 --> 00:31:16.410 align:middle line:84% applied to little v equals A to the 1/2 f. 00:31:16.410 --> 00:31:18.030 align:middle line:90% Remember, f is given. 00:31:18.030 --> 00:31:21.510 align:middle line:84% So whatever this thing is on the right side, 00:31:21.510 --> 00:31:22.780 align:middle line:90% we know that ahead of time. 00:31:22.780 --> 00:31:26.610 align:middle line:84% So our problem is to solve this equation here. 00:31:26.610 --> 00:31:30.360 align:middle line:90% Now what's the great thing? 00:31:30.360 --> 00:31:33.120 align:middle line:84% The great thing is is that because A 00:31:33.120 --> 00:31:38.760 align:middle line:84% is compact self-adjoint and in fact a non-negative operator, A 00:31:38.760 --> 00:31:43.170 align:middle line:84% to the 1/2 exists is also a compact operator 00:31:43.170 --> 00:31:44.520 align:middle line:90% and it's also self-adjoint. 00:31:44.520 --> 00:31:46.500 align:middle line:84% So then we have this self-adjoint operator 00:31:46.500 --> 00:31:49.480 align:middle line:84% on both sides of V. Both of them are compact. 00:31:49.480 --> 00:31:57.970 align:middle line:84% So this whole thing will be a compact self-adjoint operator. 00:31:57.970 --> 00:32:00.720 align:middle line:84% So if we want to be able to solve for V-- i.e., 00:32:00.720 --> 00:32:02.760 align:middle line:84% invert this thing on the left side-- 00:32:02.760 --> 00:32:06.367 align:middle line:84% this is just a compact self-adjoint operator 00:32:06.367 --> 00:32:07.950 align:middle line:84% plus the identity, which you can think 00:32:07.950 --> 00:32:13.170 align:middle line:84% of as a compact self-adjoint operator minus 1 00:32:13.170 --> 00:32:14.110 align:middle line:90% times the identity. 00:32:14.110 --> 00:32:18.738 align:middle line:84% So now that's an equation we know how to solve. 00:32:18.738 --> 00:32:20.280 align:middle line:84% We have the Fredholm alternative that 00:32:20.280 --> 00:32:22.950 align:middle line:84% says we can invert this operator if and only 00:32:22.950 --> 00:32:28.440 align:middle line:84% if this entire operator doesn't have a null space 00:32:28.440 --> 00:32:30.570 align:middle line:84% or doesn't have a non-trivial null space. 00:32:30.570 --> 00:32:37.690 align:middle line:84% So that's the plan is we will reduce ourselves to studying-- 00:32:37.690 --> 00:32:41.380 align:middle line:84% or we will prove that we can invert this operator here. 00:32:41.380 --> 00:32:44.320 align:middle line:84% First, we have to prove that we can 00:32:44.320 --> 00:32:48.460 align:middle line:84% find such that A to the 1/2, meaning an operator that's 00:32:48.460 --> 00:32:53.440 align:middle line:84% square gives me A, prove the properties we need, and then 00:32:53.440 --> 00:32:56.290 align:middle line:84% also show that this operator is convertible, 00:32:56.290 --> 00:33:03.810 align:middle line:84% define V as inverse of this times this, 00:33:03.810 --> 00:33:08.690 align:middle line:84% u as A to the 1/2 times V, and conclude 00:33:08.690 --> 00:33:11.112 align:middle line:90% that u solves our problem. 00:33:11.112 --> 00:33:12.320 align:middle line:90% So that's where we're headed. 00:33:12.320 --> 00:33:22.820 align:middle line:84% So now to get this plan off the ground, 00:33:22.820 --> 00:33:26.350 align:middle line:84% we need to show that we can come up 00:33:26.350 --> 00:33:31.600 align:middle line:84% with such a compact self-adjoint operator whose square gives me 00:33:31.600 --> 00:33:39.860 align:middle line:84% A. So as a first step in this direction, 00:33:39.860 --> 00:33:46.200 align:middle line:84% we are going to compute the spectrum of the operator 00:33:46.200 --> 00:33:49.890 align:middle line:84% A, which again is the inverse of this Dirichlet problem. 00:33:49.890 --> 00:33:55.660 align:middle line:90% 00:33:55.660 --> 00:33:58.680 align:middle line:84% So when f is continuous, it's the unique twice continuously 00:33:58.680 --> 00:34:02.210 align:middle line:84% differentiable function that's second derivative gives me f 00:34:02.210 --> 00:34:03.460 align:middle line:90% and 0 where at the end points. 00:34:03.460 --> 00:34:06.040 align:middle line:84% But in general, it's this integral operator. 00:34:06.040 --> 00:34:17.030 align:middle line:84% So first thing I want to prove is that null space of A 00:34:17.030 --> 00:34:20.584 align:middle line:84% is the zero vector or the zero function. 00:34:20.584 --> 00:34:27.659 align:middle line:90% 00:34:27.659 --> 00:34:30.620 align:middle line:84% And so it has no non-trivial null space. 00:34:30.620 --> 00:34:50.781 align:middle line:84% And the orthonormal eigenvectors for A 00:34:50.781 --> 00:35:06.740 align:middle line:84% are given by uk of x equals square root of 2, sine k 00:35:06.740 --> 00:35:08.440 align:middle line:90% pi of x. 00:35:08.440 --> 00:35:18.730 align:middle line:84% Here, k equals a natural number With associated eigenvalues, 00:35:18.730 --> 00:35:23.680 align:middle line:84% lambda k equals 1 over k squared pi squared. 00:35:23.680 --> 00:35:32.430 align:middle line:90% 00:35:32.430 --> 00:35:36.450 align:middle line:84% So let me just make a brief remark. 00:35:36.450 --> 00:35:38.760 align:middle line:84% We have via the spectral theorem that we proved 00:35:38.760 --> 00:35:41.310 align:middle line:84% last time that for a compact self-adjoint operator 00:35:41.310 --> 00:35:51.400 align:middle line:84% we can find or that the eigenvectors form 00:35:51.400 --> 00:35:57.244 align:middle line:84% an orthonormal basis for the range of the null space of A. 00:35:57.244 --> 00:36:00.280 align:middle line:84% And then to complete it to an orthonormal basis of all 00:36:00.280 --> 00:36:02.860 align:middle line:84% of the Hilbert space, in this case L2, 00:36:02.860 --> 00:36:04.990 align:middle line:84% we just need to take an orthonormal basis 00:36:04.990 --> 00:36:06.340 align:middle line:90% of the null space of A. 00:36:06.340 --> 00:36:08.950 align:middle line:84% But the null space of A is the zero vector. 00:36:08.950 --> 00:36:25.804 align:middle line:84% So by the spectral theorem for compact self-adjoint operators, 00:36:25.804 --> 00:36:31.680 align:middle line:84% we get that square root of 2 sine k pi 00:36:31.680 --> 00:36:35.260 align:middle line:90% x, k to a point infinity. 00:36:35.260 --> 00:36:49.470 align:middle line:84% So this is, orthonormal basis for L2 0, 1. 00:36:49.470 --> 00:36:52.970 align:middle line:90% 00:36:52.970 --> 00:36:56.030 align:middle line:84% So we use the spectral theorem to conclude that this 00:36:56.030 --> 00:36:58.130 align:middle line:90% is an orthonormal basis for L2. 00:36:58.130 --> 00:37:00.680 align:middle line:84% You can also prove it directly using 00:37:00.680 --> 00:37:03.970 align:middle line:84% what we know about E to the I in x 00:37:03.970 --> 00:37:09.630 align:middle line:84% being an orthonormal basis for minus pi to pi. 00:37:09.630 --> 00:37:11.970 align:middle line:84% Just by rescaling E to the I in pi 00:37:11.970 --> 00:37:15.690 align:middle line:84% x is an orthonormal basis for L2 to minus 1 to 1. 00:37:15.690 --> 00:37:17.280 align:middle line:84% Now we have functions here defined 00:37:17.280 --> 00:37:26.310 align:middle line:84% on 0, 1, which we can extend by odd parity to minus 1 to 1. 00:37:26.310 --> 00:37:29.760 align:middle line:84% And the only parts of the E to the I in pi 00:37:29.760 --> 00:37:35.910 align:middle line:84% x that any expansion for an odd function that come out only 00:37:35.910 --> 00:37:37.890 align:middle line:90% involve the sine-- 00:37:37.890 --> 00:37:39.700 align:middle line:90% these guys. 00:37:39.700 --> 00:37:43.470 align:middle line:84% So that's without knowing these are the eigenfunctions 00:37:43.470 --> 00:37:46.230 align:middle line:84% or eigenvectors for this operator A, 00:37:46.230 --> 00:37:48.780 align:middle line:84% you could also conclude that this is an orthonormal basis 00:37:48.780 --> 00:37:50.070 align:middle line:90% for L2 of 0, 1. 00:37:50.070 --> 00:37:53.820 align:middle line:90% 00:37:53.820 --> 00:38:02.350 align:middle line:84% All right, so to prove this theorem, 00:38:02.350 --> 00:38:06.340 align:middle line:84% so what we're going to do is to prove that the null space of A 00:38:06.340 --> 00:38:16.485 align:middle line:90% is trivial is we will show-- 00:38:16.485 --> 00:38:19.720 align:middle line:90% 00:38:19.720 --> 00:38:22.460 align:middle line:84% so you can go about it a couple of different ways. 00:38:22.460 --> 00:38:28.630 align:middle line:84% We'll show that the range of A is, in fact, dense in L2. 00:38:28.630 --> 00:38:34.950 align:middle line:90% 00:38:34.950 --> 00:38:38.020 align:middle line:84% So first thing we're showing is that the null space is trivial. 00:38:38.020 --> 00:38:53.710 align:middle line:84% So we'll show that the range of A closure equals all of L2. 00:38:53.710 --> 00:38:55.960 align:middle line:84% Remember, this is equal to the orthogonal complement 00:38:55.960 --> 00:38:56.810 align:middle line:90% of the null space. 00:38:56.810 --> 00:39:00.040 align:middle line:84% So if the orthogonal complement is the entire space, 00:39:00.040 --> 00:39:05.060 align:middle line:84% then that means the null space of A is just a trivial vector. 00:39:05.060 --> 00:39:12.600 align:middle line:84% So I need to be able to show that a dense subspace of L2 00:39:12.600 --> 00:39:29.710 align:middle line:84% can be solved for by A. So let u be polynomial on 0, 1, 00:39:29.710 --> 00:39:32.750 align:middle line:84% and f equals minus u double prime. 00:39:32.750 --> 00:39:36.542 align:middle line:90% 00:39:36.542 --> 00:39:37.750 align:middle line:90% Now by the previous theorem-- 00:39:37.750 --> 00:39:42.150 align:middle line:90% 00:39:42.150 --> 00:39:42.930 align:middle line:90% I should say-- 00:39:42.930 --> 00:39:50.180 align:middle line:90% 00:39:50.180 --> 00:39:59.280 align:middle line:84% [INAUDIBLE] apart-- with u of 0 equals u of 1 00:39:59.280 --> 00:40:13.790 align:middle line:84% equals 0 and f equals minus u double prime. 00:40:13.790 --> 00:40:23.800 align:middle line:84% Then by the previous theorem, A applied 00:40:23.800 --> 00:40:39.110 align:middle line:84% to f is the unique solution to the Dirichlet problem 00:40:39.110 --> 00:40:41.285 align:middle line:90% with that function V being 0. 00:40:41.285 --> 00:40:44.340 align:middle line:90% 00:40:44.340 --> 00:40:46.110 align:middle line:90% But remember, u, so f-- 00:40:46.110 --> 00:40:51.657 align:middle line:90% 00:40:51.657 --> 00:40:54.330 align:middle line:84% write it this way-- i.e., f applied 00:40:54.330 --> 00:40:58.540 align:middle line:84% to u double prime should give me f. 00:40:58.540 --> 00:41:04.365 align:middle line:84% And Af of 0 equals Af of 1 equals 0. 00:41:04.365 --> 00:41:08.050 align:middle line:90% 00:41:08.050 --> 00:41:12.040 align:middle line:84% But how do we define f? f is minus 00:41:12.040 --> 00:41:14.890 align:middle line:84% u double prime. u is a polynomial on 0, 1. 00:41:14.890 --> 00:41:16.090 align:middle line:90% That's 0 at the endpoints. 00:41:16.090 --> 00:41:19.090 align:middle line:84% And therefore I conclude that Af equals u. 00:41:19.090 --> 00:41:24.630 align:middle line:90% 00:41:24.630 --> 00:41:25.565 align:middle line:90% I hope this is clear. 00:41:25.565 --> 00:41:28.630 align:middle line:90% 00:41:28.630 --> 00:41:40.960 align:middle line:84% Now since a set of polynomials on 0, 00:41:40.960 --> 00:41:55.670 align:middle line:84% 1 are vanishing at x equals 0 and 1 are dense in L2. 00:41:55.670 --> 00:41:56.770 align:middle line:90% Now why is this? 00:41:56.770 --> 00:41:59.370 align:middle line:84% This is because we know that continuous functions that 00:41:59.370 --> 00:42:03.450 align:middle line:84% are vanishing at the two endpoints are dense in L2. 00:42:03.450 --> 00:42:09.180 align:middle line:84% And by the Weierstrass approximation theorem, 00:42:09.180 --> 00:42:10.920 align:middle line:84% every continuous function on 0, 1 00:42:10.920 --> 00:42:14.260 align:middle line:84% can be approximated by a polynomial. 00:42:14.260 --> 00:42:18.092 align:middle line:84% And it's not too difficult to convince yourself 00:42:18.092 --> 00:42:19.550 align:middle line:84% that if that's the case, then I can 00:42:19.550 --> 00:42:23.960 align:middle line:84% also approximate every continuous function 00:42:23.960 --> 00:42:25.850 align:middle line:84% that's vanishing at the two endpoints 00:42:25.850 --> 00:42:29.790 align:middle line:84% by a polynomial that's vanishing at the two endpoints. 00:42:29.790 --> 00:42:33.270 align:middle line:84% So since I can approximate every continuous function on 0, 00:42:33.270 --> 00:42:36.600 align:middle line:84% 1 vanishing at the endpoints by a polynomial vanishing 00:42:36.600 --> 00:42:40.560 align:middle line:84% at the endpoints, then those polynomials 00:42:40.560 --> 00:42:43.740 align:middle line:84% vanish at the endpoint are dense in L2 of 0, 1. 00:42:43.740 --> 00:42:46.200 align:middle line:84% So we'll just say here that this follows 00:42:46.200 --> 00:42:53.665 align:middle line:84% from, like I said, density of 0, 1. 00:42:53.665 --> 00:42:58.600 align:middle line:84% Now I'll put a 0 here, meaning it's 0 at the two endpoints, 00:42:58.600 --> 00:43:09.150 align:middle line:84% and Weierstrass approximation theorem. 00:43:09.150 --> 00:43:13.350 align:middle line:84% So we've been able to solve for every u that's 00:43:13.350 --> 00:43:15.330 align:middle line:90% dense in L2, right. 00:43:15.330 --> 00:43:18.840 align:middle line:84% So every polynomial that's 0 at the endpoints 00:43:18.840 --> 00:43:30.110 align:middle line:84% is in the range of A. And therefore, range of A 00:43:30.110 --> 00:43:38.990 align:middle line:84% contains a dense subset of L2 0, 1. 00:43:38.990 --> 00:43:52.870 align:middle line:84% And therefore, the closure has to be all of 0, 1. 00:43:52.870 --> 00:43:58.240 align:middle line:84% And then since null space of A is 00:43:58.240 --> 00:44:03.060 align:middle line:84% equal to the orthogonal complement-- 00:44:03.060 --> 00:44:06.090 align:middle line:84% or I should say the orthogonal complement of the null space 00:44:06.090 --> 00:44:09.060 align:middle line:90% of A is equal to the range-- 00:44:09.060 --> 00:44:14.140 align:middle line:84% the closure of the range, and this equals L2 of 0, 1, 00:44:14.140 --> 00:44:21.730 align:middle line:84% I conclude that the null space is just the trivial vector. 00:44:21.730 --> 00:44:23.050 align:middle line:90% So A has no null space. 00:44:23.050 --> 00:44:28.010 align:middle line:84% So that by the spectrum theorem and orthonormal 00:44:28.010 --> 00:44:32.090 align:middle line:84% basis for L2 of 0, 1 is given by the eigenvectors of A. 00:44:32.090 --> 00:44:34.460 align:middle line:84% And so now we'll prove the eigenvectors 00:44:34.460 --> 00:44:37.220 align:middle line:90% are given by this form. 00:44:37.220 --> 00:44:42.150 align:middle line:84% Now I'll just give you a brief sketch of this. 00:44:42.150 --> 00:44:45.230 align:middle line:84% So let's solve for the eigenvalues and eigenvectors. 00:44:45.230 --> 00:44:53.470 align:middle line:84% Suppose that lambda does not equals 0, 00:44:53.470 --> 00:44:59.590 align:middle line:84% u is an element of L2 that has unit length or unit 00:44:59.590 --> 00:45:06.670 align:middle line:84% norm, and A applied to u equals lambda times u. 00:45:06.670 --> 00:45:15.717 align:middle line:84% Now then-- let's see-- write that a minute ago. 00:45:15.717 --> 00:45:25.220 align:middle line:90% 00:45:25.220 --> 00:45:25.720 align:middle line:90% OK. 00:45:25.720 --> 00:45:28.440 align:middle line:90% 00:45:28.440 --> 00:45:33.030 align:middle line:84% Then u equals 1 over lambda times 00:45:33.030 --> 00:45:35.130 align:middle line:84% Au, which is fine, because lambda is non-zero, 00:45:35.130 --> 00:45:36.570 align:middle line:90% so I can divide by that. 00:45:36.570 --> 00:45:40.585 align:middle line:84% Now I didn't say this when I was discussing-- 00:45:40.585 --> 00:45:42.210 align:middle line:84% let's see if we still have it up there. 00:45:42.210 --> 00:45:45.120 align:middle line:90% 00:45:45.120 --> 00:45:47.723 align:middle line:90% We do, so we can talk about it. 00:45:47.723 --> 00:45:51.110 align:middle line:90% 00:45:51.110 --> 00:45:57.470 align:middle line:84% Now for any function in L2, by this estimate here, 00:45:57.470 --> 00:46:01.680 align:middle line:84% this also proves that A applied to f is a continuous function. 00:46:01.680 --> 00:46:06.050 align:middle line:84% So this also shows that A applied 00:46:06.050 --> 00:46:10.580 align:middle line:84% to f is, in fact, a continuous function. 00:46:10.580 --> 00:46:13.700 align:middle line:90% 00:46:13.700 --> 00:46:15.160 align:middle line:90% I hope that's clear. 00:46:15.160 --> 00:46:16.960 align:middle line:90% Because why is this so? 00:46:16.960 --> 00:46:19.420 align:middle line:84% I need to make this thing on the left-hand side 00:46:19.420 --> 00:46:22.280 align:middle line:84% small if x and z are close together. 00:46:22.280 --> 00:46:25.420 align:middle line:84% Now that only depends on some number, the L2 norm 00:46:25.420 --> 00:46:27.640 align:middle line:90% of f times this quantity here. 00:46:27.640 --> 00:46:33.610 align:middle line:84% K is a continuous function on 0, 1 cross 0, 1. 00:46:33.610 --> 00:46:38.050 align:middle line:84% So as long as I make x and z close, then x, y and z, 00:46:38.050 --> 00:46:39.580 align:middle line:90% y will be close. 00:46:39.580 --> 00:46:42.595 align:middle line:84% And since K is continuous, this quantity here will be small. 00:46:42.595 --> 00:46:46.220 align:middle line:90% 00:46:46.220 --> 00:46:48.490 align:middle line:84% And therefore, the thing that's smaller than that 00:46:48.490 --> 00:46:50.830 align:middle line:90% will be small. 00:46:50.830 --> 00:46:56.350 align:middle line:84% So that's why if I take a function which is in L2 00:46:56.350 --> 00:46:58.550 align:middle line:84% and hit it by A, I get a continuous function. 00:46:58.550 --> 00:47:03.910 align:middle line:84% So u equals 1 over lambda times this continuous function 00:47:03.910 --> 00:47:06.040 align:middle line:90% implies that u is continuous. 00:47:06.040 --> 00:47:08.380 align:middle line:84% But now we're going to feed that back in. 00:47:08.380 --> 00:47:12.700 align:middle line:84% Because if u is continuous, then A applied to u 00:47:12.700 --> 00:47:17.950 align:middle line:84% is twice continuously differentiable, 00:47:17.950 --> 00:47:23.020 align:middle line:84% which implies that u equals 1 over lambda Au 00:47:23.020 --> 00:47:26.440 align:middle line:84% is twice continuously differentiable. 00:47:26.440 --> 00:47:31.270 align:middle line:84% This is what's called a bootstrap argument, I guess. 00:47:31.270 --> 00:47:32.550 align:middle line:90% Is that the right word? 00:47:32.550 --> 00:47:35.910 align:middle line:90% 00:47:35.910 --> 00:47:40.350 align:middle line:84% Anyways, that doesn't sound right. 00:47:40.350 --> 00:47:43.350 align:middle line:90% 00:47:43.350 --> 00:47:57.110 align:middle line:84% In any case, so we conclude that a eigenfunction of A 00:47:57.110 --> 00:48:00.580 align:middle line:84% is twice continuously differentiable, right. 00:48:00.580 --> 00:48:07.910 align:middle line:84% So now u is a twice continuously differentiable function, 00:48:07.910 --> 00:48:10.370 align:middle line:84% which is equal to 1 over lambda times Au. 00:48:10.370 --> 00:48:17.630 align:middle line:90% 00:48:17.630 --> 00:48:21.890 align:middle line:84% So another way to write this as A 00:48:21.890 --> 00:48:28.390 align:middle line:84% applied into u over lambda, because A is a linear operator. 00:48:28.390 --> 00:48:31.770 align:middle line:84% Now A applied to something is a unique function 00:48:31.770 --> 00:48:34.140 align:middle line:84% whose second derivatives times minus 1 00:48:34.140 --> 00:48:35.610 align:middle line:90% gives me that thing inside. 00:48:35.610 --> 00:48:41.190 align:middle line:84% So I conclude that minus u double prime equals 1 00:48:41.190 --> 00:48:43.590 align:middle line:90% over lambda applied to u. 00:48:43.590 --> 00:48:48.060 align:middle line:84% And along with the boundary conditions, u of 0 00:48:48.060 --> 00:48:50.250 align:middle line:90% equals u of 1 equals 0. 00:48:50.250 --> 00:48:56.080 align:middle line:90% 00:48:56.080 --> 00:48:59.065 align:middle line:84% But now I know how to solve this equation. 00:48:59.065 --> 00:49:05.070 align:middle line:90% 00:49:05.070 --> 00:49:11.420 align:middle line:84% It has to be a superposition of two functions. 00:49:11.420 --> 00:49:12.110 align:middle line:90% And I get that. 00:49:12.110 --> 00:49:17.300 align:middle line:84% So u of x must be equal to A times sine, 1 00:49:17.300 --> 00:49:21.110 align:middle line:84% over square root of lambda times x, 00:49:21.110 --> 00:49:26.270 align:middle line:84% plus B times the cosine 1 over square root of lambda times x. 00:49:26.270 --> 00:49:35.660 align:middle line:84% And now since u of 0 has to give me 0, 00:49:35.660 --> 00:49:40.790 align:middle line:84% that tells me that B equals 0, which tells me 00:49:40.790 --> 00:49:47.060 align:middle line:84% u of x equals A times sine 1 over lambda x, 00:49:47.060 --> 00:49:53.448 align:middle line:84% and since u has unit length with A not equal to 0. 00:49:53.448 --> 00:49:57.670 align:middle line:84% And now u of 1 equals 0 implies that 1 00:49:57.670 --> 00:50:02.920 align:middle line:84% over square root of lambda has to be 00:50:02.920 --> 00:50:06.550 align:middle line:90% an integer multiple of pi. 00:50:06.550 --> 00:50:09.560 align:middle line:90% 00:50:09.560 --> 00:50:12.500 align:middle line:90% If u of x is to be non-zero. 00:50:12.500 --> 00:50:18.890 align:middle line:84% And therefore, u of x is equal to A times sine k 00:50:18.890 --> 00:50:20.780 align:middle line:90% pi x for some k. 00:50:20.780 --> 00:50:26.450 align:middle line:84% And the fact that we get square root of 2 00:50:26.450 --> 00:50:28.400 align:middle line:84% comes from the normalization condition. 00:50:28.400 --> 00:50:33.200 align:middle line:90% 00:50:33.200 --> 00:50:38.432 align:middle line:84% Implies that u of x equals square root of 2 sine k 00:50:38.432 --> 00:50:41.780 align:middle line:90% pi x for some k natural number. 00:50:41.780 --> 00:50:54.830 align:middle line:90% 00:50:54.830 --> 00:50:57.770 align:middle line:84% So square root of 2 times sine k pi 00:50:57.770 --> 00:51:01.040 align:middle line:84% x as k varies over the natural numbers 00:51:01.040 --> 00:51:06.450 align:middle line:84% gives me an orthonormal basis for L2 00:51:06.450 --> 00:51:09.720 align:middle line:84% that consists of eigenvectors of A. 00:51:09.720 --> 00:51:19.570 align:middle line:84% The eigenvalues are 1 over k squared pi squared. 00:51:19.570 --> 00:51:23.520 align:middle line:84% So I can think of A as simply multiplying each eigenvector 00:51:23.520 --> 00:51:25.390 align:middle line:90% by 1 over k squared pi squared. 00:51:25.390 --> 00:51:27.450 align:middle line:84% And if I want to define A to the 1/2, which 00:51:27.450 --> 00:51:31.590 align:middle line:84% is why I'm doing all this, then we'll 00:51:31.590 --> 00:51:36.150 align:middle line:84% define an operator so that it takes something 00:51:36.150 --> 00:51:41.100 align:middle line:84% in this orthonormal basis and simply multiply this by 1 00:51:41.100 --> 00:51:45.600 align:middle line:84% over k pi, which is half of what A would do, 00:51:45.600 --> 00:51:47.820 align:middle line:84% or half power of what A would do. 00:51:47.820 --> 00:51:50.808 align:middle line:84% This is also how one could define 00:51:50.808 --> 00:51:52.350 align:middle line:84% what's called the functional calculus 00:51:52.350 --> 00:51:57.510 align:middle line:84% for self-adjoint compact operators, which you can then 00:51:57.510 --> 00:52:00.280 align:middle line:84% extend to self-adjoint operators as well. 00:52:00.280 --> 00:52:02.520 align:middle line:90% So let's make this a definition. 00:52:02.520 --> 00:52:07.440 align:middle line:90% 00:52:07.440 --> 00:52:23.430 align:middle line:84% So if f is in L2 0, 1 with f given by this Fourier expansion 00:52:23.430 --> 00:52:29.820 align:middle line:84% with ck given by interval 0 to 1, 00:52:29.820 --> 00:52:33.390 align:middle line:84% f of x square root of 2, sine k pi 00:52:33.390 --> 00:52:47.440 align:middle line:84% x dx, and we define an operator which we call A to the 1/2, 00:52:47.440 --> 00:52:52.660 align:middle line:84% although I'm not saying that its square is A just yet. 00:52:52.660 --> 00:53:02.875 align:middle line:84% We define linear operator A by A f of x equals sum from k 00:53:02.875 --> 00:53:11.110 align:middle line:84% equals 1 to infinity of 1 over k pi ck square root of 2, 00:53:11.110 --> 00:53:13.030 align:middle line:90% sine k pi x. 00:53:13.030 --> 00:53:15.670 align:middle line:90% 00:53:15.670 --> 00:53:16.170 align:middle line:90% 1/2. 00:53:16.170 --> 00:53:23.380 align:middle line:90% 00:53:23.380 --> 00:53:27.160 align:middle line:84% So right now I just have this expression for 00:53:27.160 --> 00:53:31.450 align:middle line:84% given a function in L2 with this expansion in terms 00:53:31.450 --> 00:53:33.550 align:middle line:90% of this orthonormal basis. 00:53:33.550 --> 00:53:36.160 align:middle line:84% I do this to the coefficients, I take the coefficients 00:53:36.160 --> 00:53:40.360 align:middle line:84% and multiply them by 1 over k pi. 00:53:40.360 --> 00:53:48.970 align:middle line:84% And my claim is that A to the 1/2 00:53:48.970 --> 00:54:00.230 align:middle line:84% is a compact self-adjoint operator on L2. 00:54:00.230 --> 00:54:06.000 align:middle line:90% 00:54:06.000 --> 00:54:08.760 align:middle line:84% And this notation is not just for show. 00:54:08.760 --> 00:54:11.440 align:middle line:84% I take A to the 1/2 and compose it with itself, 00:54:11.440 --> 00:54:12.720 align:middle line:90% meaning I square it. 00:54:12.720 --> 00:54:15.240 align:middle line:84% I get the operator A, which remember 00:54:15.240 --> 00:54:25.690 align:middle line:84% was defined as this integral operator involving k, right. 00:54:25.690 --> 00:54:32.520 align:middle line:84% If you like, though, so let me just [INAUDIBLE].. 00:54:32.520 --> 00:54:37.830 align:middle line:84% Remember A, when it hits each of these sine k pi x's, spits out 00:54:37.830 --> 00:54:40.110 align:middle line:84% 1 over k squared times pi squared. 00:54:40.110 --> 00:54:45.240 align:middle line:84% So A applied to f would be this thing multiplied by 1 over k 00:54:45.240 --> 00:54:46.450 align:middle line:90% squared pi squared. 00:54:46.450 --> 00:54:48.300 align:middle line:84% And so it makes sense that A to the 1/2 00:54:48.300 --> 00:54:51.720 align:middle line:84% should be something that when I apply it again I get back A, 00:54:51.720 --> 00:54:57.708 align:middle line:84% since I have 1 over k squared pi squared when I hit the A, 00:54:57.708 --> 00:55:00.420 align:middle line:84% I multiply the coefficients by 1 over k pi, 00:55:00.420 --> 00:55:02.280 align:middle line:84% then this should give me what I want. 00:55:02.280 --> 00:55:08.310 align:middle line:84% And now it's just the process of confirming these facts 00:55:08.310 --> 00:55:11.400 align:middle line:90% that we need. 00:55:11.400 --> 00:55:14.670 align:middle line:84% But you may think of A as this integral operator. 00:55:14.670 --> 00:55:19.110 align:middle line:84% You can think of A as a solution operator for this Dirichlet 00:55:19.110 --> 00:55:20.010 align:middle line:90% problem. 00:55:20.010 --> 00:55:26.700 align:middle line:84% Or you may simply think of it as, I take a function f, expand 00:55:26.700 --> 00:55:31.440 align:middle line:84% it in terms of sine k pi x in orthonormal basis for L2 of 0, 00:55:31.440 --> 00:55:40.130 align:middle line:84% 1, and I simply multiply by 1 over k squared pi squared. 00:55:40.130 --> 00:55:46.610 align:middle line:84% How does this jive with what I just said about A also 00:55:46.610 --> 00:55:47.840 align:middle line:90% being the solution operator? 00:55:47.840 --> 00:55:51.170 align:middle line:84% Well, if I have A applied to this quantity, 00:55:51.170 --> 00:55:54.720 align:middle line:84% then I have sine k pi x over k squared pi squared. 00:55:54.720 --> 00:55:56.840 align:middle line:84% Now let's say I take two derivatives of that, 00:55:56.840 --> 00:55:59.180 align:middle line:84% then I get k squared pi squared over k 00:55:59.180 --> 00:56:00.710 align:middle line:90% squared pi squared, I get 1. 00:56:00.710 --> 00:56:03.990 align:middle line:84% So I get back out f for the minus sine. 00:56:03.990 --> 00:56:05.465 align:middle line:90% So all of those things jive. 00:56:05.465 --> 00:56:08.480 align:middle line:90% 00:56:08.480 --> 00:56:10.520 align:middle line:84% All right, so proof of this theorem. 00:56:10.520 --> 00:56:14.894 align:middle line:90% 00:56:14.894 --> 00:56:23.795 align:middle line:84% You write, say, two functions in L2 expanded in this basis. 00:56:23.795 --> 00:56:41.380 align:middle line:90% 00:56:41.380 --> 00:56:43.980 align:middle line:84% And so first thing I would like to show 00:56:43.980 --> 00:56:49.150 align:middle line:84% is that this operator is bounded, linear operator. 00:56:49.150 --> 00:56:56.050 align:middle line:84% So I want the L2 norm of fA to the 1/2 f squared. 00:56:56.050 --> 00:57:03.390 align:middle line:84% So this is the L2 norm squared of the function 00:57:03.390 --> 00:57:13.830 align:middle line:84% with coefficients given by k pi square root of 2 times 00:57:13.830 --> 00:57:18.230 align:middle line:90% k pi x, all squared. 00:57:18.230 --> 00:57:25.660 align:middle line:84% And now, so the L2 norm of this function given by this Fourier 00:57:25.660 --> 00:57:27.580 align:middle line:84% expansion here is just the sum of the squares 00:57:27.580 --> 00:57:30.850 align:middle line:84% of the coefficients appearing in front of the square root of 2 00:57:30.850 --> 00:57:32.740 align:middle line:90% times sine k pi x. 00:57:32.740 --> 00:57:37.300 align:middle line:84% So this is equal to, by Parseval's identity equals 00:57:37.300 --> 00:57:42.610 align:middle line:84% ck squared over k squared pi squared. 00:57:42.610 --> 00:57:47.680 align:middle line:84% And 1 over k squared pi squared as k goes from 1 to infinity 00:57:47.680 --> 00:57:52.690 align:middle line:84% is bounded by 1 over pi squared equals 1 00:57:52.690 --> 00:57:56.140 align:middle line:84% to infinity of ck squared equals 1 00:57:56.140 --> 00:58:02.400 align:middle line:84% over pi squared, again, by Parseval's identity, 00:58:02.400 --> 00:58:08.070 align:middle line:84% or a completeness of this eigenbasis 00:58:08.070 --> 00:58:12.130 align:middle line:84% that gives me back the L2 norm of that. 00:58:12.130 --> 00:58:13.780 align:middle line:90% So that proves boundedness. 00:58:13.780 --> 00:58:18.370 align:middle line:90% What about self-adjointness? 00:58:18.370 --> 00:58:23.500 align:middle line:84% A to the 1/2 f, but in take its inner product with g, 00:58:23.500 --> 00:58:28.300 align:middle line:84% again using how these things are defined 00:58:28.300 --> 00:58:32.410 align:middle line:84% and the fact that the inner product between two functions 00:58:32.410 --> 00:58:35.800 align:middle line:84% is given by just the little L2 pairing of the coefficients 00:58:35.800 --> 00:58:40.660 align:middle line:84% that appear in front of the square root of 2 sine k pi x. 00:58:40.660 --> 00:58:48.800 align:middle line:90% This is ck over k times pi, dk. 00:58:48.800 --> 00:58:51.985 align:middle line:84% And now we just move this over here. 00:58:51.985 --> 00:58:55.100 align:middle line:90% 00:58:55.100 --> 00:59:03.020 align:middle line:84% And these are real numbers, so I can move them over on the d 00:59:03.020 --> 00:59:04.910 align:middle line:84% without taking their complex conjugate. 00:59:04.910 --> 00:59:11.270 align:middle line:84% And that's just equal to f inner product A to the 1/2 g. 00:59:11.270 --> 00:59:15.203 align:middle line:84% So that proves that A is self-adjoint. 00:59:15.203 --> 00:59:21.490 align:middle line:90% 00:59:21.490 --> 00:59:30.520 align:middle line:84% And now finally, we show that A to the 1/2 squared gives me 00:59:30.520 --> 00:59:43.040 align:middle line:90% A. So let f be-- 00:59:43.040 --> 00:59:45.260 align:middle line:90% OK, so I rewrote that part. 00:59:45.260 --> 00:59:46.610 align:middle line:90% I don't need to write it again. 00:59:46.610 --> 00:59:55.610 align:middle line:84% If I look at A to the 1/2 f, this is by definition 00:59:55.610 --> 01:00:00.350 align:middle line:84% equal to A to the 1/2 applied to now the function 01:00:00.350 --> 01:00:03.470 align:middle line:84% given in this eigenbasis or this orthonormal basis 01:00:03.470 --> 01:00:11.880 align:middle line:84% by ck over k pi square root of 2 sine k pi x. 01:00:11.880 --> 01:00:14.640 align:middle line:84% Again, how do I apply A to the 1/2? 01:00:14.640 --> 01:00:18.450 align:middle line:84% I take the coefficients in front of the basis 01:00:18.450 --> 01:00:20.720 align:middle line:90% function and divide by k pi. 01:00:20.720 --> 01:00:26.340 align:middle line:84% So then I get k equals 1 to infinity of ck over k 01:00:26.340 --> 01:00:34.665 align:middle line:84% squared pi squared, square root of 2 sine k pi x. 01:00:34.665 --> 01:00:39.070 align:middle line:90% 01:00:39.070 --> 01:00:41.200 align:middle line:90% Let's leave it here. 01:00:41.200 --> 01:00:44.890 align:middle line:84% Now going from here to what we have next, since each of these 01:00:44.890 --> 01:00:53.110 align:middle line:84% is an eigenfunction of A with eigenvalue 1 over k 01:00:53.110 --> 01:00:55.120 align:middle line:84% squared pi squared, this is also equal 01:00:55.120 --> 01:01:03.160 align:middle line:84% to A applied to the square root of 2 sine k pi x. 01:01:03.160 --> 01:01:06.880 align:middle line:84% And now it's perfectly legitimate 01:01:06.880 --> 01:01:14.600 align:middle line:84% to pull this A out of the infinite sum, 01:01:14.600 --> 01:01:27.400 align:middle line:84% because this is a basis for L2, or really by [INAUDIBLE].. 01:01:27.400 --> 01:01:37.490 align:middle line:84% In any case, I can pull the A out and I get square root of 2, 01:01:37.490 --> 01:01:39.800 align:middle line:90% sine k pi x. 01:01:39.800 --> 01:01:51.610 align:middle line:90% 01:01:51.610 --> 01:01:52.420 align:middle line:90% Why can I do this? 01:01:52.420 --> 01:01:58.300 align:middle line:84% It's because so if I put a finite in here, then sum from k 01:01:58.300 --> 01:02:03.490 align:middle line:84% equals 1 to n converges in L2 norm to this quantity 01:02:03.490 --> 01:02:06.380 align:middle line:90% here with an infinity. 01:02:06.380 --> 01:02:12.400 align:middle line:84% So since A is a bounded linear operator, 01:02:12.400 --> 01:02:15.520 align:middle line:84% the limit as n goes to infinity of A applied to the finite sum 01:02:15.520 --> 01:02:17.500 align:middle line:84% is equal to A applied to the infinite sum. 01:02:17.500 --> 01:02:20.620 align:middle line:84% Now A applied to a finite sum is given by this, 01:02:20.620 --> 01:02:21.985 align:middle line:90% letting n go to infinity again. 01:02:21.985 --> 01:02:24.510 align:middle line:90% 01:02:24.510 --> 01:02:27.380 align:middle line:84% This we can take the A in and out. 01:02:27.380 --> 01:02:33.860 align:middle line:90% 01:02:33.860 --> 01:02:36.160 align:middle line:84% But this was just the expansion of f. 01:02:36.160 --> 01:02:37.540 align:middle line:90% So this is equal to Af. 01:02:37.540 --> 01:02:40.450 align:middle line:84% So we have proven that A to the 1/2 01:02:40.450 --> 01:02:44.440 align:middle line:84% squared, meaning A to the 1/2 applied to A to the 1/2 of f 01:02:44.440 --> 01:02:45.490 align:middle line:90% gives me back Af. 01:02:45.490 --> 01:02:51.110 align:middle line:90% 01:02:51.110 --> 01:02:53.390 align:middle line:84% Now let me give a brief sketch on why 01:02:53.390 --> 01:02:54.640 align:middle line:90% this is a compact operator. 01:02:54.640 --> 01:02:57.260 align:middle line:84% We've shown it's a bounded self-adjoint operator whose 01:02:57.260 --> 01:03:16.830 align:middle line:84% square gives me A. So A is a compact operator. 01:03:16.830 --> 01:03:26.090 align:middle line:84% We'll show that the image of the unit ball has equi-small tails. 01:03:26.090 --> 01:03:37.080 align:middle line:84% I want to say here is suffices to show that, say, A applied 01:03:37.080 --> 01:03:50.205 align:middle line:90% to f has equi-small tails. 01:03:50.205 --> 01:03:51.830 align:middle line:84% The fact that this has equi-small tails 01:03:51.830 --> 01:03:54.320 align:middle line:84% then implies that the closure also has equi-small tails. 01:03:54.320 --> 01:03:55.970 align:middle line:84% And therefore by this theorem we proved 01:03:55.970 --> 01:04:01.040 align:middle line:84% about characterizing compact subsets of Hilbert spaces, 01:04:01.040 --> 01:04:06.150 align:middle line:84% we conclude that the closure of this set is compact-- 01:04:06.150 --> 01:04:11.620 align:middle line:90% I should have 1/2, 1/2-- 01:04:11.620 --> 01:04:15.700 align:middle line:84% and therefore A to the 1/2 is a compact operator. 01:04:15.700 --> 01:04:32.090 align:middle line:84% So let epsilon be positive, and choose N as a number 01:04:32.090 --> 01:04:37.250 align:middle line:84% so that 1 over N squared is less than epsilon squared. 01:04:37.250 --> 01:04:39.560 align:middle line:84% We can do that because 1 over capital N squared 01:04:39.560 --> 01:04:41.760 align:middle line:84% goes to 0 as capital N goes to infinity. 01:04:41.760 --> 01:04:45.050 align:middle line:90% So I can always find such an N. 01:04:45.050 --> 01:04:47.150 align:middle line:84% And now we claim that this N is the one that 01:04:47.150 --> 01:04:52.480 align:middle line:84% works for showing this set has equi-small tails. 01:04:52.480 --> 01:05:01.950 align:middle line:84% Let f be in L2 with norm less than or equal to 1. 01:05:01.950 --> 01:05:04.620 align:middle line:90% 01:05:04.620 --> 01:05:10.770 align:middle line:84% And again, if I look at the sum of the Fourier coefficients 01:05:10.770 --> 01:05:16.080 align:middle line:84% given in this basis of square root 2 sine k pi x of A 01:05:16.080 --> 01:05:30.680 align:middle line:84% to the 1/2 f, this is equal to, by definition 01:05:30.680 --> 01:05:39.780 align:middle line:84% of how A operates, this is equal to simply the coefficient 01:05:39.780 --> 01:05:45.880 align:middle line:84% of f squared over k squared pi squared. 01:05:45.880 --> 01:05:48.490 align:middle line:84% And this is less than or equal to-- 01:05:48.490 --> 01:05:52.990 align:middle line:84% k is bigger than N, so this is less than or equal to 1 01:05:52.990 --> 01:05:55.280 align:middle line:84% over n squared with a pi squared. 01:05:55.280 --> 01:05:57.820 align:middle line:90% But that's something like 9. 01:05:57.820 --> 01:06:00.280 align:middle line:84% So I can leave it and still be less than 01:06:00.280 --> 01:06:04.450 align:middle line:90% or equal to, sum, sum, k. 01:06:04.450 --> 01:06:10.810 align:middle line:84% I can now take this to be all ck's. 01:06:10.810 --> 01:06:17.000 align:middle line:84% And this is equal to 1 over N squared, 2 squared, 01:06:17.000 --> 01:06:20.030 align:middle line:84% which is less than or equal to 1 over N squared. 01:06:20.030 --> 01:06:22.240 align:middle line:84% And then we chose an N so that 1 over N squared 01:06:22.240 --> 01:06:24.470 align:middle line:90% is less than epsilon squared. 01:06:24.470 --> 01:06:27.740 align:middle line:84% So that's less than epsilon squared, 01:06:27.740 --> 01:06:32.450 align:middle line:84% therefore showing that the set has equi-small tails. 01:06:32.450 --> 01:06:38.600 align:middle line:84% And therefore the closure of this set is compact. 01:06:38.600 --> 01:06:51.080 align:middle line:90% 01:06:51.080 --> 01:06:54.030 align:middle line:84% So we had this A to the 1/2 operator, 01:06:54.030 --> 01:06:57.620 align:middle line:84% which we needed to carry out this plan 01:06:57.620 --> 01:07:04.270 align:middle line:84% that we sketched at the beginning solving the Dirichlet 01:07:04.270 --> 01:07:05.150 align:middle line:90% problem. 01:07:05.150 --> 01:07:09.370 align:middle line:84% So now we just need to check that that operator I 01:07:09.370 --> 01:07:12.790 align:middle line:84% had before, which is A to the 1/2 composed 01:07:12.790 --> 01:07:15.490 align:middle line:84% with multiplication by B composed with A to the 1/2 01:07:15.490 --> 01:07:17.450 align:middle line:84% is a compact self-adjoint operator. 01:07:17.450 --> 01:07:21.098 align:middle line:84% So first, let me just state a theorem about multiplying 01:07:21.098 --> 01:07:22.140 align:middle line:90% by a continuous function. 01:07:22.140 --> 01:07:26.620 align:middle line:84% Let V be a real value continuous function. 01:07:26.620 --> 01:07:39.100 align:middle line:90% 01:07:39.100 --> 01:07:45.460 align:middle line:84% Define m sub v as in multiplication f 01:07:45.460 --> 01:07:52.740 align:middle line:84% of x to be this V of x f of x for f in L2. 01:07:52.740 --> 01:07:58.660 align:middle line:90% 01:07:58.660 --> 01:08:07.850 align:middle line:84% Then this is a bounded linear operator on L2, 01:08:07.850 --> 01:08:09.632 align:middle line:90% and it's self-adjoint. 01:08:09.632 --> 01:08:20.470 align:middle line:90% 01:08:20.470 --> 01:08:25.390 align:middle line:84% So I'm actually going to leave this as an exercise. 01:08:25.390 --> 01:08:29.350 align:middle line:90% It's not too difficult to prove. 01:08:29.350 --> 01:08:32.319 align:middle line:84% Again, just multiplying by this function, which 01:08:32.319 --> 01:08:35.982 align:middle line:84% is continuous and bounded gives you boundedness automatically. 01:08:35.982 --> 01:08:37.899 align:middle line:84% And the fact that its real value will give you 01:08:37.899 --> 01:08:39.069 align:middle line:90% the self-adjointness. 01:08:39.069 --> 01:08:41.979 align:middle line:90% 01:08:41.979 --> 01:08:48.235 align:middle line:84% So from this, you can conclude the following. 01:08:48.235 --> 01:08:59.890 align:middle line:84% Let V between a function be real value, 01:08:59.890 --> 01:09:04.859 align:middle line:84% then the operator T equals A to the 1/2 composed 01:09:04.859 --> 01:09:10.350 align:middle line:84% with multiplication by v, composed with A to the 1/2 01:09:10.350 --> 01:09:11.790 align:middle line:90% satisfies the following. 01:09:11.790 --> 01:09:19.180 align:middle line:90% 01:09:19.180 --> 01:09:32.149 align:middle line:84% One is T is a self-adjoint compact operator on L2. 01:09:32.149 --> 01:09:36.850 align:middle line:90% 01:09:36.850 --> 01:09:43.800 align:middle line:84% And one extra condition is T is bounded. 01:09:43.800 --> 01:09:46.500 align:middle line:84% So in fact T is a bounded operator 01:09:46.500 --> 01:09:54.120 align:middle line:90% from L2 to continuous functions. 01:09:54.120 --> 01:09:56.940 align:middle line:90% 01:09:56.940 --> 01:09:59.990 align:middle line:84% And if I take an L2 function and stick it into this function, 01:09:59.990 --> 01:10:03.200 align:middle line:84% I get out a continuous function in a bounded way. 01:10:03.200 --> 01:10:06.720 align:middle line:90% 01:10:06.720 --> 01:10:09.650 align:middle line:84% So one follows immediately from what 01:10:09.650 --> 01:10:18.850 align:middle line:84% we've proven already in this exercise of a theorem 01:10:18.850 --> 01:10:20.830 align:middle line:90% that I gave, right. 01:10:20.830 --> 01:10:24.610 align:middle line:84% Multiplication by V is a self-adjoint operator. 01:10:24.610 --> 01:10:26.620 align:middle line:84% A to the 1/2 is a self-adjoint operator. 01:10:26.620 --> 01:10:28.790 align:middle line:84% A to the 1/2 is a self-adjoint operator. 01:10:28.790 --> 01:10:31.810 align:middle line:84% So when I take the adjoint of this quantity, 01:10:31.810 --> 01:10:35.440 align:middle line:84% I will get the adjoint of A to the 1/2 in front, 01:10:35.440 --> 01:10:37.810 align:middle line:84% adjoint of mv, adjoint of A to the 1/2. 01:10:37.810 --> 01:10:40.340 align:middle line:84% They are equal to each other themselves. 01:10:40.340 --> 01:10:42.440 align:middle line:84% So I get out something self-adjoint. 01:10:42.440 --> 01:10:47.140 align:middle line:84% This is the composition of, if you like, 01:10:47.140 --> 01:10:49.310 align:middle line:84% a bounded operator with a compact operator. 01:10:49.310 --> 01:10:52.460 align:middle line:90% So I get a compact operator. 01:10:52.460 --> 01:10:54.700 align:middle line:84% So that proves it's a self-adjoint compact operator 01:10:54.700 --> 01:10:55.720 align:middle line:90% on L2. 01:10:55.720 --> 01:11:02.170 align:middle line:84% Why is this a bounded operator from L2 01:11:02.170 --> 01:11:03.350 align:middle line:90% to continuous functions? 01:11:03.350 --> 01:11:16.130 align:middle line:84% Well, it suffices to show that this A to the 1/2 01:11:16.130 --> 01:11:26.610 align:middle line:84% is a bounded operator from L2 to continuous functions. 01:11:26.610 --> 01:11:27.267 align:middle line:90% Why is that? 01:11:27.267 --> 01:11:29.100 align:middle line:84% Because if I take an L2 function and feed it 01:11:29.100 --> 01:11:32.038 align:middle line:84% into this operator T, then A to the 1/2 01:11:32.038 --> 01:11:33.330 align:middle line:90% would be a continuous function. 01:11:33.330 --> 01:11:36.690 align:middle line:84% Multiplication by V will be a continuous function, again, 01:11:36.690 --> 01:11:39.130 align:middle line:84% because multiplication by V is continuous. 01:11:39.130 --> 01:11:40.600 align:middle line:90% So V is continuous. 01:11:40.600 --> 01:11:43.080 align:middle line:84% And A to the 1/2 applied to a continuous function 01:11:43.080 --> 01:11:44.740 align:middle line:84% again spits out a continuous function. 01:11:44.740 --> 01:11:47.550 align:middle line:84% So suffices to show that A to the 1/2 01:11:47.550 --> 01:11:53.240 align:middle line:84% is a bounded linear operating from L2 to continuous function. 01:11:53.240 --> 01:12:07.610 align:middle line:84% So let f be given in a Fourier expansion 01:12:07.610 --> 01:12:16.316 align:middle line:84% in terms of the sine k pi x's, then 01:12:16.316 --> 01:12:20.920 align:middle line:84% if A to the 1/2 in f of x, this is 01:12:20.920 --> 01:12:36.180 align:middle line:84% equal to k pi square root of 2 sine k pi of x. 01:12:36.180 --> 01:12:38.610 align:middle line:84% And now to show that A to the 1/2 01:12:38.610 --> 01:12:40.360 align:middle line:84% applied to f is a continuous function, 01:12:40.360 --> 01:12:42.420 align:middle line:84% we'll apply the Weierstrass M-test right. 01:12:42.420 --> 01:12:47.260 align:middle line:84% So this is an infinite sum of continuous functions. 01:12:47.260 --> 01:12:48.840 align:middle line:84% So to apply the Weierstrass M-test, 01:12:48.840 --> 01:12:51.870 align:middle line:84% I have to say that I can bound this thing by something 01:12:51.870 --> 01:12:52.620 align:middle line:90% which is summable. 01:12:52.620 --> 01:12:56.950 align:middle line:90% 01:12:56.950 --> 01:12:59.540 align:middle line:90% So I have this. 01:12:59.540 --> 01:13:02.965 align:middle line:84% And then I also, if I take the absolute value of-- 01:13:02.965 --> 01:13:09.560 align:middle line:90% 01:13:09.560 --> 01:13:15.710 align:middle line:84% that's less than or equal to ck over k pi 01:13:15.710 --> 01:13:21.830 align:middle line:84% square root of 2, which is less than or equal to Tk over k. 01:13:21.830 --> 01:13:23.180 align:middle line:90% And I claim this is summable. 01:13:23.180 --> 01:13:26.186 align:middle line:90% 01:13:26.186 --> 01:13:29.330 align:middle line:84% And if I sum from k equal 1 to infinity, 01:13:29.330 --> 01:13:34.340 align:middle line:84% ck over k by Cauchy-Schwarz, this 01:13:34.340 --> 01:13:40.210 align:middle line:84% is less than or equal to sum over k, 1 over k squared, 01:13:40.210 --> 01:13:47.480 align:middle line:84% 1/2 sum k ck squared raised to the 1/2. 01:13:47.480 --> 01:13:50.080 align:middle line:84% And remember, this is just the L2 norm of f, 01:13:50.080 --> 01:13:53.420 align:middle line:84% because the sine k pi x's are in L2 basis. 01:13:53.420 --> 01:14:07.910 align:middle line:84% So that equals something like pi squared over 6. 01:14:07.910 --> 01:14:09.200 align:middle line:90% I don't know. 01:14:09.200 --> 01:14:15.741 align:middle line:84% Let's say it is, this thing here times L2 norm. 01:14:15.741 --> 01:14:20.460 align:middle line:90% 01:14:20.460 --> 01:14:26.700 align:middle line:84% So each of these continuous functions in this infinite sum 01:14:26.700 --> 01:14:30.130 align:middle line:90% is bounded by a constant. 01:14:30.130 --> 01:14:34.060 align:middle line:84% And those constants are summable by this computation. 01:14:34.060 --> 01:14:37.980 align:middle line:84% So thus, that implies A to the 1/2 f 01:14:37.980 --> 01:14:47.150 align:middle line:84% is continuous function by Weierstrass M-test. 01:14:47.150 --> 01:14:54.370 align:middle line:84% Not only that, this computation we did shows that A to the 1/2 01:14:54.370 --> 01:14:58.180 align:middle line:84% applied to f of x is less than or equal to pi 01:14:58.180 --> 01:15:03.790 align:middle line:84% squared over 6 raised to the 1/2 times the L2 norm in f. 01:15:03.790 --> 01:15:07.330 align:middle line:84% So it's a bounded linear operator from L2 01:15:07.330 --> 01:15:09.403 align:middle line:84% to the space of continuous functions. 01:15:09.403 --> 01:15:16.170 align:middle line:90% 01:15:16.170 --> 01:15:19.765 align:middle line:90% So where are we? 01:15:19.765 --> 01:15:28.350 align:middle line:84% We have all the pieces we need in place to solve 01:15:28.350 --> 01:15:34.910 align:middle line:90% our problem, Dirichlet problem. 01:15:34.910 --> 01:15:42.580 align:middle line:90% 01:15:42.580 --> 01:15:45.000 align:middle line:90% All the ingredients are ready. 01:15:45.000 --> 01:15:47.350 align:middle line:90% We just need to cook them. 01:15:47.350 --> 01:15:57.820 align:middle line:90% 01:15:57.820 --> 01:16:00.380 align:middle line:84% So I was mentioning the Weierstrass M-test. 01:16:00.380 --> 01:16:03.260 align:middle line:84% That should have been covered in 18.100. 01:16:03.260 --> 01:16:07.857 align:middle line:84% There's an infinite sum of continuous functions. 01:16:07.857 --> 01:16:09.440 align:middle line:84% And each of those continuous functions 01:16:09.440 --> 01:16:10.850 align:middle line:90% is bounded by a constant. 01:16:10.850 --> 01:16:13.160 align:middle line:84% And those constants are summable, 01:16:13.160 --> 01:16:17.140 align:middle line:84% then you get a continuous function out in the end. 01:16:17.140 --> 01:16:22.410 align:middle line:84% So now the theorem that concludes the existence 01:16:22.410 --> 01:16:36.220 align:middle line:84% part of the Dirichlet problem, let V C 0,1 continuous 01:16:36.220 --> 01:16:39.580 align:middle line:90% function be non-negative. 01:16:39.580 --> 01:16:42.800 align:middle line:90% 01:16:42.800 --> 01:16:45.030 align:middle line:84% So it's real value and non-negative. 01:16:45.030 --> 01:16:48.878 align:middle line:90% 01:16:48.878 --> 01:16:56.050 align:middle line:84% And let f be a continuous function. 01:16:56.050 --> 01:17:09.750 align:middle line:84% Then there exists a unique twice continuously differentiable 01:17:09.750 --> 01:17:19.630 align:middle line:84% function on 0, 1, solving the Dirichlet problem, 01:17:19.630 --> 01:17:24.790 align:middle line:84% minus u double prime, plus V multiplied by u 01:17:24.790 --> 01:17:33.670 align:middle line:84% equals f on 0, 1 and boundary conditions u of 0 equals u of 1 01:17:33.670 --> 01:17:34.427 align:middle line:90% equals 0. 01:17:34.427 --> 01:17:40.670 align:middle line:90% 01:17:40.670 --> 01:17:51.470 align:middle line:84% All right, so I'll just recall for you 01:17:51.470 --> 01:17:57.110 align:middle line:84% the plan was to define u to be A to the 1/2 pi plus A 01:17:57.110 --> 01:18:03.830 align:middle line:84% to the 1/2 mv, A to the 1/2 inverse, A to the 1/2 01:18:03.830 --> 01:18:06.750 align:middle line:90% applied to f. 01:18:06.750 --> 01:18:10.290 align:middle line:84% Now we just need to say why this thing exists, right. 01:18:10.290 --> 01:18:12.720 align:middle line:84% Why is this operator in the middle invertible? 01:18:12.720 --> 01:18:15.470 align:middle line:90% 01:18:15.470 --> 01:18:18.290 align:middle line:90% And then we'll get what we need. 01:18:18.290 --> 01:18:24.485 align:middle line:84% So proof by the Fredholm alternative, right. 01:18:24.485 --> 01:18:30.230 align:middle line:90% 01:18:30.230 --> 01:18:35.640 align:middle line:84% So let me not skip ahead by previous theorem, 01:18:35.640 --> 01:18:40.920 align:middle line:84% this operator mv is sandwiched between A to the 1/2 01:18:40.920 --> 01:18:47.150 align:middle line:84% is a self-adjoint compact operator. 01:18:47.150 --> 01:18:50.391 align:middle line:90% 01:18:50.391 --> 01:19:05.770 align:middle line:84% And therefore, the Fredholm alternative, the inverse 01:19:05.770 --> 01:19:20.545 align:middle line:84% exists if and only if the null space of the operator 01:19:20.545 --> 01:19:21.045 align:middle line:90% is trivial. 01:19:21.045 --> 01:19:32.260 align:middle line:90% 01:19:32.260 --> 01:19:37.600 align:middle line:84% Now suppose we have something in the null space, 01:19:37.600 --> 01:19:47.850 align:middle line:84% and we'll show that it has to be 0, 01:19:47.850 --> 01:19:52.770 align:middle line:84% then if I pair this with sub g is an element in L2, 01:19:52.770 --> 01:19:55.320 align:middle line:84% if I pair it with g, so I get that 0 01:19:55.320 --> 01:20:05.450 align:middle line:84% is equal to I plus A to the 1/2, applied to g, inner product g. 01:20:05.450 --> 01:20:08.040 align:middle line:90% 01:20:08.040 --> 01:20:10.350 align:middle line:84% Now carrying this g through, and it'll 01:20:10.350 --> 01:20:11.800 align:middle line:90% get inner product with itself. 01:20:11.800 --> 01:20:19.770 align:middle line:84% So I get g squared plus A to the 1/2 multiplication 01:20:19.770 --> 01:20:23.940 align:middle line:84% by V of A to the 1/2 applied to g, inner product g. 01:20:23.940 --> 01:20:26.670 align:middle line:84% Now since A to the 1/2 is self-adjoint, 01:20:26.670 --> 01:20:30.330 align:middle line:84% I can move it over to the second entry. 01:20:30.330 --> 01:20:37.630 align:middle line:90% 01:20:37.630 --> 01:20:39.280 align:middle line:90% And let me just write this out. 01:20:39.280 --> 01:20:42.130 align:middle line:84% This is equal to g L2 plus the inner world 01:20:42.130 --> 01:20:48.340 align:middle line:84% from 0 to 1 of V times A to the 1/2 g, times-- 01:20:48.340 --> 01:20:52.240 align:middle line:84% so this is all multiplication, pointwise multiplication-- 01:20:52.240 --> 01:20:58.010 align:middle line:84% times A to the 1/2 g complex conjugate, 01:20:58.010 --> 01:21:08.725 align:middle line:84% dx, which this gives me A to the 1/2 g on squared. 01:21:08.725 --> 01:21:11.450 align:middle line:90% 01:21:11.450 --> 01:21:13.130 align:middle line:90% Now V is non-negative. 01:21:13.130 --> 01:21:14.810 align:middle line:84% So this quantity here is non-negative. 01:21:14.810 --> 01:21:20.360 align:middle line:84% So this is bigger than or equal to g squared, an L2 norm of g. 01:21:20.360 --> 01:21:22.310 align:middle line:90% And we started out with 0. 01:21:22.310 --> 01:21:26.600 align:middle line:84% So I get that g is 0, and therefore 01:21:26.600 --> 01:21:31.190 align:middle line:84% the null space of I plus this self-adjoint compact operator 01:21:31.190 --> 01:21:34.310 align:middle line:90% is trivial. 01:21:34.310 --> 01:21:37.220 align:middle line:84% And therefore I plus this compact self-adjoint operator 01:21:37.220 --> 01:21:38.810 align:middle line:90% is invertible-- 01:21:38.810 --> 01:21:40.204 align:middle line:90% the Fredholm alternative. 01:21:40.204 --> 01:21:44.850 align:middle line:90% 01:21:44.850 --> 01:21:52.100 align:middle line:90% So this inverse exists. 01:21:52.100 --> 01:22:00.040 align:middle line:90% And I define u to be what? 01:22:00.040 --> 01:22:03.510 align:middle line:90% A to the 1/2. 01:22:03.510 --> 01:22:12.990 align:middle line:84% Let's say this way I'll define V to be I plus A to the 1/2 01:22:12.990 --> 01:22:16.860 align:middle line:84% multiplication by V, A to the 1/2 01:22:16.860 --> 01:22:25.575 align:middle line:84% inverse applied to A to the 1/2 f, and u to be A to the 1/2 01:22:25.575 --> 01:22:28.550 align:middle line:90% of V. 01:22:28.550 --> 01:22:29.705 align:middle line:90% So then what do I get? 01:22:29.705 --> 01:22:32.810 align:middle line:90% 01:22:32.810 --> 01:22:46.050 align:middle line:84% Then u plus A applied to multiplication by V of-- 01:22:46.050 --> 01:22:52.530 align:middle line:84% so I'll say A Vu, this is equal to, by definition, 01:22:52.530 --> 01:22:59.490 align:middle line:84% A to the 1/2 V plus A to the 1/2, 01:22:59.490 --> 01:23:20.050 align:middle line:84% simply because A is equal to A to the 1/2 squared applied 01:23:20.050 --> 01:23:21.630 align:middle line:90% to V. 01:23:21.630 --> 01:23:34.110 align:middle line:84% So I get A to the 1/2, I plus A to the 1/2 of V, A 01:23:34.110 --> 01:23:39.490 align:middle line:84% to the 1/2 applied to V. And now V is given by this thing. 01:23:39.490 --> 01:23:42.330 align:middle line:84% So I just get, when it hits that, I just get back A 01:23:42.330 --> 01:23:47.450 align:middle line:84% to the 1/2 f multiplied by A to the 1/2 to get Af. 01:23:47.450 --> 01:23:53.000 align:middle line:90%