WEBVTT 00:00:00.000 --> 00:00:01.996 align:middle line:90% [SQUEAKING] 00:00:01.996 --> 00:00:03.992 align:middle line:90% [RUSTLING] 00:00:03.992 --> 00:00:13.208 align:middle line:90% [CLICKING] 00:00:13.208 --> 00:00:15.250 align:middle line:84% CASEY RODRIGUEZ: So let's continue our discussion 00:00:15.250 --> 00:00:16.870 align:middle line:90% of sequences of functions. 00:00:16.870 --> 00:00:29.060 align:middle line:90% 00:00:29.060 --> 00:00:31.810 align:middle line:84% So we had two different notions of convergence 00:00:31.810 --> 00:00:34.430 align:middle line:90% of sequences of functions. 00:00:34.430 --> 00:00:37.750 align:middle line:84% So the first notion was pointwise convergence. 00:00:37.750 --> 00:00:41.590 align:middle line:90% 00:00:41.590 --> 00:00:43.180 align:middle line:84% So we have a sequence of functions, 00:00:43.180 --> 00:00:49.330 align:middle line:84% f n from some set S to R, another function, fixed 00:00:49.330 --> 00:00:55.260 align:middle line:84% function f from S to R. And then we 00:00:55.260 --> 00:01:07.920 align:middle line:84% say f n converges to f pointwise if for every x in S, 00:01:07.920 --> 00:01:11.070 align:middle line:84% the sequence of real numbers converges to f of x. 00:01:11.070 --> 00:01:19.050 align:middle line:84% So if for all x and S, the limit as n 00:01:19.050 --> 00:01:24.450 align:middle line:84% goes to infinity of f n of x equals f of x. 00:01:24.450 --> 00:01:28.740 align:middle line:84% So I take an x out of S, stick it into f n. 00:01:28.740 --> 00:01:30.630 align:middle line:84% So now I get a sequence of real numbers. 00:01:30.630 --> 00:01:35.400 align:middle line:84% And I should get f of x as n goes to infinity. 00:01:35.400 --> 00:01:38.640 align:middle line:84% And then we introduced the stronger notion 00:01:38.640 --> 00:01:45.150 align:middle line:84% of convergence of functions, which was the following-- 00:01:45.150 --> 00:01:53.760 align:middle line:84% so we have a sequence going from S to R, function from S to R, 00:01:53.760 --> 00:02:08.979 align:middle line:84% then we said f n converges to f uniformly on S if somehow, 00:02:08.979 --> 00:02:13.450 align:middle line:84% across the entire set, f n gets close to f. 00:02:13.450 --> 00:02:19.150 align:middle line:84% So here, this statement says if for each fixed x, eventually, 00:02:19.150 --> 00:02:22.720 align:middle line:84% f n of x-- so f n evaluated at that point-- 00:02:22.720 --> 00:02:25.900 align:middle line:84% is close to f of x, f evaluated at that point. 00:02:25.900 --> 00:02:30.130 align:middle line:84% But uniform convergence says f n is close to f 00:02:30.130 --> 00:02:31.660 align:middle line:90% across the entire set. 00:02:31.660 --> 00:02:37.690 align:middle line:84% So if for all epsilon positive, there exists natural number 00:02:37.690 --> 00:02:48.200 align:middle line:84% M so that for all n bigger than or equal to M, for all x in S, 00:02:48.200 --> 00:02:55.980 align:middle line:84% f n of x minus f of x is less than epsilon. 00:02:55.980 --> 00:03:01.430 align:middle line:84% Now, if you write out what this means in terms of epsilon M-- 00:03:01.430 --> 00:03:04.230 align:middle line:84% remember, this is a limit, so this is a limit of sequences, 00:03:04.230 --> 00:03:08.150 align:middle line:84% so this means something in terms of epsilons and M's-- 00:03:08.150 --> 00:03:11.570 align:middle line:84% this would say for all x in S, for all epsilon positive, 00:03:11.570 --> 00:03:15.010 align:middle line:90% there exists an M, so on. 00:03:15.010 --> 00:03:19.750 align:middle line:84% So the x appears at the front of this definition, 00:03:19.750 --> 00:03:21.700 align:middle line:84% while here for uniform convergence, 00:03:21.700 --> 00:03:23.590 align:middle line:90% it appears at the end. 00:03:23.590 --> 00:03:33.830 align:middle line:84% That is not just a kind of meaningless difference 00:03:33.830 --> 00:03:37.070 align:middle line:84% in the way you write the definition, 00:03:37.070 --> 00:03:43.100 align:middle line:84% meaning this is a stronger statement than if the x is 00:03:43.100 --> 00:03:43.950 align:middle line:90% appearing here. 00:03:43.950 --> 00:03:46.050 align:middle line:90% So what am I going on about? 00:03:46.050 --> 00:03:49.130 align:middle line:84% So first off, we proved last time 00:03:49.130 --> 00:03:55.430 align:middle line:84% that if I have a sequence of functions, 00:03:55.430 --> 00:04:05.060 align:middle line:84% again from some subset S to R, converging to f uniformly, then 00:04:05.060 --> 00:04:10.430 align:middle line:84% this implies f n convergence to f pointwise. 00:04:10.430 --> 00:04:15.590 align:middle line:90% 00:04:15.590 --> 00:04:19.630 align:middle line:84% But now, what I'm going to prove is 00:04:19.630 --> 00:04:21.399 align:middle line:84% that in fact, uniform convergence 00:04:21.399 --> 00:04:24.340 align:middle line:84% is something stronger than pointwise convergence. 00:04:24.340 --> 00:04:27.550 align:middle line:84% In other words, this is a one-way street. 00:04:27.550 --> 00:04:31.690 align:middle line:84% Pointwise convergence does not imply uniform convergence. 00:04:31.690 --> 00:04:39.590 align:middle line:90% 00:04:39.590 --> 00:04:46.890 align:middle line:84% And we were just going to look at a very specific example, 00:04:46.890 --> 00:04:55.260 align:middle line:84% which I was going to state as a theorem, which 00:04:55.260 --> 00:04:58.430 align:middle line:90% is the following-- 00:04:58.430 --> 00:05:01.880 align:middle line:90% let f n of x be x to the n. 00:05:01.880 --> 00:05:04.955 align:middle line:84% And now we're looking on the unit interval, 0, 1. 00:05:04.955 --> 00:05:07.720 align:middle line:90% 00:05:07.720 --> 00:05:22.308 align:middle line:84% And let f the function that is 0 if x is in 0, 1, 1 00:05:22.308 --> 00:05:23.680 align:middle line:90% if x equals 1. 00:05:23.680 --> 00:05:30.080 align:middle line:90% 00:05:30.080 --> 00:05:33.920 align:middle line:84% So first off, let me from last time-- 00:05:33.920 --> 00:05:40.850 align:middle line:84% or you can even just check by looking at the form of f n-- 00:05:40.850 --> 00:05:45.800 align:middle line:84% that f n converges to f pointwise. 00:05:45.800 --> 00:05:48.590 align:middle line:90% 00:05:48.590 --> 00:05:55.400 align:middle line:84% If I take x in this interval here, so not equal to 1, 00:05:55.400 --> 00:05:58.100 align:middle line:90% then x is strictly less than 1. 00:05:58.100 --> 00:06:00.050 align:middle line:84% And if I raise it to a high enough power over 00:06:00.050 --> 00:06:02.690 align:middle line:84% and over again, that's converging to 0. 00:06:02.690 --> 00:06:05.270 align:middle line:90% So it converges to f of x, 0. 00:06:05.270 --> 00:06:09.950 align:middle line:84% Now at 1, I just get 1, and that clearly converges to 1. 00:06:09.950 --> 00:06:12.170 align:middle line:90% So f n converges to f pointwise. 00:06:12.170 --> 00:06:16.940 align:middle line:90% 00:06:16.940 --> 00:06:27.640 align:middle line:84% And so the claim is that for all b between 0 and 1, 00:06:27.640 --> 00:06:29.270 align:middle line:90% f n converges to f. 00:06:29.270 --> 00:06:30.525 align:middle line:90% And I guess I could include 0. 00:06:30.525 --> 00:06:32.150 align:middle line:84% That would just be looking at one point 00:06:32.150 --> 00:06:38.000 align:middle line:84% and not very interesting, but f n converges to f uniformly. 00:06:38.000 --> 00:06:40.800 align:middle line:90% 00:06:40.800 --> 00:06:47.120 align:middle line:84% And the second is f n does not converge to f uniformly 00:06:47.120 --> 00:06:50.995 align:middle line:90% on the whole interval, however. 00:06:50.995 --> 00:06:54.670 align:middle line:90% 00:06:54.670 --> 00:06:59.830 align:middle line:84% So maybe it's best to again draw this picture 00:06:59.830 --> 00:07:01.900 align:middle line:84% that you're supposed to think about when it 00:07:01.900 --> 00:07:03.940 align:middle line:90% comes to uniform convergence. 00:07:03.940 --> 00:07:09.080 align:middle line:84% So we have the limiting function, f. 00:07:09.080 --> 00:07:20.190 align:middle line:84% And then we draw a little epsilon-sized collar 00:07:20.190 --> 00:07:23.610 align:middle line:90% around the function f. 00:07:23.610 --> 00:07:25.920 align:middle line:90% And then uniform convergence-- 00:07:25.920 --> 00:07:28.500 align:middle line:90% so this is f-- 00:07:28.500 --> 00:07:33.450 align:middle line:84% uniform convergence says that as long as I go far enough out, 00:07:33.450 --> 00:07:39.910 align:middle line:84% the graph of f n should be within this little epsilon 00:07:39.910 --> 00:07:44.020 align:middle line:84% collar, so that f n is getting close to f uniformly 00:07:44.020 --> 00:07:45.160 align:middle line:90% across the entire set. 00:07:45.160 --> 00:07:53.840 align:middle line:90% 00:07:53.840 --> 00:07:57.560 align:middle line:84% So number 2 will give us a chance 00:07:57.560 --> 00:08:00.980 align:middle line:84% to negate this definition of uniform convergence, which like 00:08:00.980 --> 00:08:03.133 align:middle line:90% I said, you should always-- 00:08:03.133 --> 00:08:04.800 align:middle line:84% so we're actually doing two things here. 00:08:04.800 --> 00:08:08.210 align:middle line:84% We're giving an example of uniform convergence, 00:08:08.210 --> 00:08:13.010 align:middle line:84% and also a sequence of functions which does not converge 00:08:13.010 --> 00:08:16.740 align:middle line:90% uniformly to this function. 00:08:16.740 --> 00:08:19.710 align:middle line:84% So we're doing both an example and non-example, 00:08:19.710 --> 00:08:22.440 align:middle line:84% which is the best thing to do for a new definition. 00:08:22.440 --> 00:08:27.890 align:middle line:84% So for the proof of 1, let's prove uniform convergence. 00:08:27.890 --> 00:08:32.054 align:middle line:90% 00:08:32.054 --> 00:08:40.020 align:middle line:90% So let b be in 0, 1. 00:08:40.020 --> 00:08:43.090 align:middle line:90% 00:08:43.090 --> 00:08:48.320 align:middle line:84% Then the limit as n goes to infinity of b to the n 00:08:48.320 --> 00:08:51.170 align:middle line:90% is 0, which implies-- 00:08:51.170 --> 00:08:53.502 align:middle line:84% so now I'm getting ahead of myself. 00:08:53.502 --> 00:08:56.920 align:middle line:90% Don't write that yet. 00:08:56.920 --> 00:09:00.485 align:middle line:84% So now we want to prove uniform convergence of f n to f on-- 00:09:00.485 --> 00:09:03.850 align:middle line:90% 00:09:03.850 --> 00:09:08.710 align:middle line:84% I didn't finish this statement, f n to f uniformly on b. 00:09:08.710 --> 00:09:11.620 align:middle line:90% 00:09:11.620 --> 00:09:16.270 align:middle line:84% So I have uniform convergence on any smaller interval 00:09:16.270 --> 00:09:17.560 align:middle line:90% other than 0, 1. 00:09:17.560 --> 00:09:21.950 align:middle line:90% 00:09:21.950 --> 00:09:24.155 align:middle line:84% Sorry about that if that looked a little weird. 00:09:24.155 --> 00:09:27.290 align:middle line:90% 00:09:27.290 --> 00:09:32.250 align:middle line:84% So then b n converges to 0 as n goes to infinity. 00:09:32.250 --> 00:09:35.390 align:middle line:84% And now we want to prove that f n converges to f uniformly 00:09:35.390 --> 00:09:38.190 align:middle line:90% on this interval 0, b. 00:09:38.190 --> 00:09:40.500 align:middle line:90% So let epsilon be positive. 00:09:40.500 --> 00:09:44.570 align:middle line:84% We now have to find an M so that f n is close to 0 00:09:44.570 --> 00:09:48.380 align:middle line:84% because f is 0 on such an interval. 00:09:48.380 --> 00:09:52.364 align:middle line:90% 00:09:52.364 --> 00:09:55.880 align:middle line:84% Now, since b to the n converges to 0, 00:09:55.880 --> 00:10:02.020 align:middle line:84% there exists natural number M such 00:10:02.020 --> 00:10:06.880 align:middle line:84% that for all n bigger than or equal to M, b to the n 00:10:06.880 --> 00:10:07.765 align:middle line:90% is less than epsilon. 00:10:07.765 --> 00:10:10.670 align:middle line:90% 00:10:10.670 --> 00:10:16.270 align:middle line:84% Then for all n bigger than or equal to M, we get-- 00:10:16.270 --> 00:10:17.485 align:middle line:90% and also for all x-- 00:10:17.485 --> 00:10:20.790 align:middle line:90% 00:10:20.790 --> 00:10:29.220 align:middle line:84% 0, b, we get that f n of x minus f of x-- 00:10:29.220 --> 00:10:36.700 align:middle line:84% b is less than 1, so when I stick it into f, I just get 0. 00:10:36.700 --> 00:10:40.360 align:middle line:84% This is equal to x to the n minus 0 00:10:40.360 --> 00:10:42.400 align:middle line:84% in absolute value is just x to the n 00:10:42.400 --> 00:10:47.410 align:middle line:84% because we're looking at non-negative x. 00:10:47.410 --> 00:10:50.350 align:middle line:84% And now x is in this interval 0, b, 00:10:50.350 --> 00:10:52.210 align:middle line:90% so it's less than or equal to b. 00:10:52.210 --> 00:10:54.400 align:middle line:84% So x to the n is going to be less than 00:10:54.400 --> 00:10:58.150 align:middle line:84% or equal to b to the n, and this is less than epsilon. 00:10:58.150 --> 00:11:01.310 align:middle line:90% 00:11:01.310 --> 00:11:09.510 align:middle line:84% So just as when we looked at uniform continuity 00:11:09.510 --> 00:11:13.500 align:middle line:84% of a function, it was a statement like, for all epsilon 00:11:13.500 --> 00:11:18.202 align:middle line:84% there exists a delta which just depends on, basically, 00:11:18.202 --> 00:11:19.410 align:middle line:90% the epsilon and the function. 00:11:19.410 --> 00:11:22.200 align:middle line:90% 00:11:22.200 --> 00:11:27.540 align:middle line:84% Now for uniform convergence, it looks kind of like pointwise 00:11:27.540 --> 00:11:29.790 align:middle line:84% convergence, except now for every epsilon 00:11:29.790 --> 00:11:34.830 align:middle line:84% you can find an M which does not depend on the point x. 00:11:34.830 --> 00:11:37.020 align:middle line:84% So for every epsilon, you can find an M 00:11:37.020 --> 00:11:42.300 align:middle line:84% depending only on epsilon, and maybe the function little f. 00:11:42.300 --> 00:11:44.590 align:middle line:90% But that doesn't depend on x. 00:11:44.590 --> 00:11:49.060 align:middle line:84% This n here that I chose, M, depended only 00:11:49.060 --> 00:11:55.550 align:middle line:84% on b, not the point x, which I have to stick into this here. 00:11:55.550 --> 00:12:00.930 align:middle line:90% 00:12:00.930 --> 00:12:11.440 align:middle line:90% So now, let's prove number 2. 00:12:11.440 --> 00:12:17.380 align:middle line:84% So first off, let's negate the definition so that we 00:12:17.380 --> 00:12:18.760 align:middle line:90% know what we're talking about. 00:12:18.760 --> 00:12:31.080 align:middle line:84% So f n does not converge to f uniformly on a set S 00:12:31.080 --> 00:12:33.960 align:middle line:84% if there exists a bad-- so every "for all" 00:12:33.960 --> 00:12:36.750 align:middle line:90% becomes a "there exists. 00:12:36.750 --> 00:12:42.240 align:middle line:84% So if there exists some bad epsilon 0 positive so that 00:12:42.240 --> 00:12:48.820 align:middle line:84% for all M, a natural number, there exists n bigger than 00:12:48.820 --> 00:12:54.620 align:middle line:84% or equal to M, and there exists an x in S 00:12:54.620 --> 00:13:04.100 align:middle line:84% so that f n of x minus f of x is bigger than or equal to epsilon 00:13:04.100 --> 00:13:04.600 align:middle line:90% 0. 00:13:04.600 --> 00:13:10.150 align:middle line:90% 00:13:10.150 --> 00:13:13.480 align:middle line:90% So this is the negation. 00:13:13.480 --> 00:13:17.590 align:middle line:84% But why should we not be surprised that f n does not 00:13:17.590 --> 00:13:21.590 align:middle line:84% converge to f uniformly on 0, 1, if you believe this picture? 00:13:21.590 --> 00:13:26.275 align:middle line:84% So let's look at what's going on here. 00:13:26.275 --> 00:13:29.940 align:middle line:90% 00:13:29.940 --> 00:13:37.445 align:middle line:84% So let me draw, now, the graph of f. 00:13:37.445 --> 00:13:41.340 align:middle line:90% 00:13:41.340 --> 00:13:46.470 align:middle line:84% And let's say I take epsilon to be 1/4, say. 00:13:46.470 --> 00:14:04.690 align:middle line:90% 00:14:04.690 --> 00:14:08.890 align:middle line:84% Now this is what my little epsilon neighborhood of f 00:14:08.890 --> 00:14:18.250 align:middle line:84% looks like or epsilon tube, or for epsilon equals 1/4, say. 00:14:18.250 --> 00:14:23.080 align:middle line:84% It's a tube around 0 up to x equals 1, 00:14:23.080 --> 00:14:27.220 align:middle line:84% and then it's a little area around 1. 00:14:27.220 --> 00:14:31.150 align:middle line:84% And now if I were to have uniform convergence, then as 00:14:31.150 --> 00:14:37.690 align:middle line:84% long as n is very large, f n has to be within this area 00:14:37.690 --> 00:14:41.180 align:middle line:90% that I have here. 00:14:41.180 --> 00:14:43.300 align:middle line:90% So in fact, let me shade it in. 00:14:43.300 --> 00:14:48.120 align:middle line:90% 00:14:48.120 --> 00:14:52.050 align:middle line:84% So for f n large, it has to be within this region 00:14:52.050 --> 00:14:55.310 align:middle line:90% that I'm coloring in. 00:14:55.310 --> 00:14:59.470 align:middle line:84% But now, what do we know about x to the n? 00:14:59.470 --> 00:15:02.560 align:middle line:84% Well, it starts at 0, and ends at 1, 00:15:02.560 --> 00:15:08.190 align:middle line:84% and looks something like that, which means here, 00:15:08.190 --> 00:15:16.040 align:middle line:84% always, it leaves this epsilon tube around f, this epsilon 00:15:16.040 --> 00:15:16.895 align:middle line:90% collar, I guess. 00:15:16.895 --> 00:15:22.015 align:middle line:90% 00:15:22.015 --> 00:15:24.600 align:middle line:84% f n leaves the shaded area, which 00:15:24.600 --> 00:15:28.410 align:middle line:90% is where it's supposed to stay. 00:15:28.410 --> 00:15:34.420 align:middle line:84% So I hope this intuitive explanation is clear, 00:15:34.420 --> 00:15:39.905 align:middle line:84% and why it shouldn't be too big of a surprise that f 00:15:39.905 --> 00:15:42.350 align:middle line:84% n does not converge to f uniformly on 0, 1. 00:15:42.350 --> 00:15:44.330 align:middle line:84% We'll see another reason in a little bit 00:15:44.330 --> 00:15:47.150 align:middle line:84% why it's impossible for f n to converge 00:15:47.150 --> 00:15:49.710 align:middle line:84% to f uniformly when we talk about the interchange 00:15:49.710 --> 00:15:50.210 align:middle line:90% of limits. 00:15:50.210 --> 00:15:53.150 align:middle line:84% But just using the definition, we 00:15:53.150 --> 00:15:57.500 align:middle line:84% can prove that f n does not converge to f uniformly. 00:15:57.500 --> 00:16:03.770 align:middle line:84% So the negation is that there exists a bad epsilon 0 so 00:16:03.770 --> 00:16:05.510 align:middle line:90% that we have all of this. 00:16:05.510 --> 00:16:10.750 align:middle line:84% So the point is to choose epsilon 00:16:10.750 --> 00:16:17.040 align:middle line:84% so that this never intersects with this epsilon 00:16:17.040 --> 00:16:19.800 align:middle line:84% neighborhood of the function down here, 00:16:19.800 --> 00:16:21.900 align:middle line:84% never intersects with this epsilon 00:16:21.900 --> 00:16:26.100 align:middle line:84% neighborhood of just the point x equals 1, y equals 1. 00:16:26.100 --> 00:16:34.270 align:middle line:84% So let's choose epsilon 0 to be 1/4. 00:16:34.270 --> 00:16:42.180 align:middle line:90% 00:16:42.180 --> 00:16:44.850 align:middle line:84% So now we have to prove for all n, there exists an n. 00:16:44.850 --> 00:16:48.505 align:middle line:90% So let M be a natural number. 00:16:48.505 --> 00:16:53.050 align:middle line:90% 00:16:53.050 --> 00:17:07.619 align:middle line:84% Choose n to be M. And choose x to be, let's say, 1 over 4 00:17:07.619 --> 00:17:14.230 align:middle line:90% to the 1 over M. 00:17:14.230 --> 00:17:15.849 align:middle line:84% Now, this is a number less than 1. 00:17:15.849 --> 00:17:20.140 align:middle line:84% So its n-th root is less than 1 and also positive. 00:17:20.140 --> 00:17:22.700 align:middle line:90% 00:17:22.700 --> 00:17:34.170 align:middle line:84% Then this f of x equals 0, and f sub M 00:17:34.170 --> 00:17:38.280 align:middle line:84% of this x, which is just 1 over 4 to the 1 over M, 00:17:38.280 --> 00:17:41.850 align:middle line:84% now raised to the M-th power, 1 over 4, 00:17:41.850 --> 00:17:51.510 align:middle line:84% which implies that f sub M of x minus f of x is equal 1/4, 00:17:51.510 --> 00:17:55.430 align:middle line:90% which equals epsilon 0. 00:17:55.430 --> 00:17:57.950 align:middle line:84% Or I guess if you like, you can write bigger 00:17:57.950 --> 00:17:59.360 align:middle line:90% than or equal to epsilon 0. 00:17:59.360 --> 00:18:07.860 align:middle line:90% 00:18:07.860 --> 00:18:15.370 align:middle line:84% So basically, if you choose any-- so 00:18:15.370 --> 00:18:17.770 align:middle line:84% there was nothing special about epsilon 0 being 1/4. 00:18:17.770 --> 00:18:21.640 align:middle line:84% If you choose anything less than 1, that would do 00:18:21.640 --> 00:18:22.780 align:middle line:90% You can check that. 00:18:22.780 --> 00:18:26.080 align:middle line:84% If I chose 1/2 here, I could then choose this point x 00:18:26.080 --> 00:18:32.050 align:middle line:84% where f M is far away from f of x to be 1 over 1/2, 00:18:32.050 --> 00:18:35.440 align:middle line:84% 1/2 to the 1 over M, just as long 00:18:35.440 --> 00:18:38.620 align:middle line:84% as I don't choose epsilon 0 to be equal to 1. 00:18:38.620 --> 00:18:55.600 align:middle line:90% 00:18:55.600 --> 00:18:59.050 align:middle line:90% So we have that example. 00:18:59.050 --> 00:19:03.840 align:middle line:84% And we had another sequence of functions 00:19:03.840 --> 00:19:08.700 align:middle line:84% we had looked at last time, which were these functions that 00:19:08.700 --> 00:19:11.308 align:middle line:90% look like tents. 00:19:11.308 --> 00:19:13.350 align:middle line:84% I don't know why I'm making this axis bigger when 00:19:13.350 --> 00:19:14.475 align:middle line:90% it should be the other one. 00:19:14.475 --> 00:19:20.830 align:middle line:90% 00:19:20.830 --> 00:19:27.530 align:middle line:84% So there's 1, 1 over n, 1 over 2n, then way up here 2n, 00:19:27.530 --> 00:19:30.135 align:middle line:90% so at this point. 00:19:30.135 --> 00:19:34.000 align:middle line:90% 00:19:34.000 --> 00:19:35.380 align:middle line:90% And then it's piecewise linear. 00:19:35.380 --> 00:19:38.730 align:middle line:84% So then it's just a straight line down to here. 00:19:38.730 --> 00:19:41.790 align:middle line:84% I'm not going to write down exactly the equation 00:19:41.790 --> 00:19:51.450 align:middle line:90% for each piece, and then 0 to 1. 00:19:51.450 --> 00:19:54.480 align:middle line:90% So this is f n. 00:19:54.480 --> 00:20:01.500 align:middle line:84% And last time, we proved that f n converges 00:20:01.500 --> 00:20:05.415 align:middle line:90% to the function 0 pointwise. 00:20:05.415 --> 00:20:10.480 align:middle line:90% 00:20:10.480 --> 00:20:12.900 align:middle line:84% But it does not converge uniformly to 0. 00:20:12.900 --> 00:20:15.550 align:middle line:90% Again, what's the point? 00:20:15.550 --> 00:20:21.930 align:middle line:84% The point here is that if I were to draw a little epsilon 00:20:21.930 --> 00:20:29.630 align:middle line:84% collar around the function 0, it would look something like this. 00:20:29.630 --> 00:20:33.460 align:middle line:84% And f n would have to be within this little strip 00:20:33.460 --> 00:20:35.290 align:middle line:90% for all n sufficiently large. 00:20:35.290 --> 00:20:38.200 align:middle line:84% But f n is getting taller and taller, 00:20:38.200 --> 00:20:42.670 align:middle line:84% so it always leaves any strip that I put around the function 00:20:42.670 --> 00:20:44.230 align:middle line:90% 0. 00:20:44.230 --> 00:20:47.110 align:middle line:84% So f n does not convert to 0 uniformly. 00:20:47.110 --> 00:20:48.910 align:middle line:84% We can make a proof out of that, though. 00:20:48.910 --> 00:20:56.290 align:middle line:84% So f n does not converge to 0 uniformly on-- 00:20:56.290 --> 00:21:02.000 align:middle line:84% so I should always tell you where I'm talking about. 00:21:02.000 --> 00:21:05.410 align:middle line:90% And this should-- on 0, 1. 00:21:05.410 --> 00:21:08.750 align:middle line:90% 00:21:08.750 --> 00:21:09.650 align:middle line:90% So why not? 00:21:09.650 --> 00:21:11.810 align:middle line:84% Well, we can take any epsilon 0, really. 00:21:11.810 --> 00:21:18.530 align:middle line:84% So let's choose epsilon 0 to be 1. 00:21:18.530 --> 00:21:21.800 align:middle line:90% 00:21:21.800 --> 00:21:23.347 align:middle line:90% Let M be a natural number. 00:21:23.347 --> 00:21:29.100 align:middle line:90% 00:21:29.100 --> 00:21:33.450 align:middle line:84% So we should find an n and a x so that this inequality 00:21:33.450 --> 00:21:34.660 align:middle line:90% is satisfied. 00:21:34.660 --> 00:21:39.720 align:middle line:84% So let's choose n to be M, x to be 1 over 2M, 00:21:39.720 --> 00:21:40.845 align:middle line:90% at this point where I peak. 00:21:40.845 --> 00:21:43.970 align:middle line:90% 00:21:43.970 --> 00:21:50.820 align:middle line:90% Then f of M of x minus f of x-- 00:21:50.820 --> 00:21:53.850 align:middle line:90% f here is just 0. 00:21:53.850 --> 00:21:57.810 align:middle line:84% The limiting function is 0, so let me not even put f, 00:21:57.810 --> 00:22:00.370 align:middle line:90% let me put this 0. 00:22:00.370 --> 00:22:07.665 align:middle line:84% This equals f M of 1 over 2M, which equals 2M. 00:22:07.665 --> 00:22:09.040 align:middle line:84% And this is certainly bigger than 00:22:09.040 --> 00:22:12.745 align:middle line:84% or equal to 1, which is epsilon 0. 00:22:12.745 --> 00:22:21.770 align:middle line:90% 00:22:21.770 --> 00:22:31.310 align:middle line:84% Now in a little bit, I'll give you a whole host of examples-- 00:22:31.310 --> 00:22:33.830 align:middle line:84% it seems like I've only given you maybe one example up 00:22:33.830 --> 00:22:34.910 align:middle line:90% to this point-- 00:22:34.910 --> 00:22:38.330 align:middle line:84% of functions which converge uniformly to something. 00:22:38.330 --> 00:22:44.870 align:middle line:84% But in a minute, I'll give you a very useful test 00:22:44.870 --> 00:22:48.800 align:middle line:84% to decide when a series involving functions 00:22:48.800 --> 00:22:50.120 align:middle line:90% converges uniformly. 00:22:50.120 --> 00:22:56.880 align:middle line:90% 00:22:56.880 --> 00:23:01.860 align:middle line:84% But before we get to that, let's revisit these three questions 00:23:01.860 --> 00:23:04.800 align:middle line:84% that we asked in the last lecture 00:23:04.800 --> 00:23:07.440 align:middle line:84% in the context of power series, but now 00:23:07.440 --> 00:23:11.790 align:middle line:84% in this more general setting of convergence of functions. 00:23:11.790 --> 00:23:17.475 align:middle line:84% So now what I'm talking about-- and this is really-- 00:23:17.475 --> 00:23:23.960 align:middle line:90% 00:23:23.960 --> 00:23:29.610 align:middle line:84% although what I'm about to say may sound a bit alarming, 00:23:29.610 --> 00:23:32.490 align:middle line:84% that we're now in essentially the last week of the class 00:23:32.490 --> 00:23:40.760 align:middle line:84% and we're really getting to the first real part of analysis. 00:23:40.760 --> 00:23:46.290 align:middle line:84% So I was talking with a professor from Duke one time, 00:23:46.290 --> 00:23:50.997 align:middle line:84% and he made this funny observation that somehow 00:23:50.997 --> 00:23:52.830 align:middle line:84% in math-- and I think this is not just math, 00:23:52.830 --> 00:23:58.390 align:middle line:84% but this is a lot of science-based classes-- 00:23:58.390 --> 00:24:02.530 align:middle line:84% are taught, if you were to translate it 00:24:02.530 --> 00:24:06.790 align:middle line:84% into studying a book, it's like you spend a year studying 00:24:06.790 --> 00:24:11.110 align:middle line:84% introductions to a book, and then you 00:24:11.110 --> 00:24:13.630 align:middle line:84% spend a year studying middle parts of books, 00:24:13.630 --> 00:24:16.810 align:middle line:84% and then you spend another year studying 00:24:16.810 --> 00:24:18.550 align:middle line:90% the last parts of books. 00:24:18.550 --> 00:24:21.550 align:middle line:84% So now here, we're starting to get 00:24:21.550 --> 00:24:25.547 align:middle line:84% into the middle part of the book of analysis, not the textbook. 00:24:25.547 --> 00:24:27.130 align:middle line:84% We're at the end of that, but at least 00:24:27.130 --> 00:24:29.162 align:middle line:84% in the grand scheme of things, which 00:24:29.162 --> 00:24:30.370 align:middle line:90% is the interchange of limits. 00:24:30.370 --> 00:24:38.650 align:middle line:84% You have two limiting processes that you want to interchange. 00:24:38.650 --> 00:24:44.650 align:middle line:84% In analysis, first God created the limit, and then man asked, 00:24:44.650 --> 00:24:46.180 align:middle line:90% can we interchange limits? 00:24:46.180 --> 00:24:48.870 align:middle line:90% 00:24:48.870 --> 00:24:50.650 align:middle line:90% And so what do I mean by that? 00:24:50.650 --> 00:24:56.830 align:middle line:84% This is not always a thing you can do. 00:24:56.830 --> 00:25:01.510 align:middle line:84% And this is at the very heart of analysis 00:25:01.510 --> 00:25:03.190 align:middle line:84% is when can we interchange limits. 00:25:03.190 --> 00:25:05.990 align:middle line:84% So let me give you the simplest example. 00:25:05.990 --> 00:25:15.340 align:middle line:84% Let's say we have the following limit-- 00:25:15.340 --> 00:25:18.070 align:middle line:84% so I take the limit as k goes to infinity of the limit 00:25:18.070 --> 00:25:24.550 align:middle line:84% as n goes to infinity of n over k, n over k plus 1. 00:25:24.550 --> 00:25:28.270 align:middle line:84% So that's just a sequence depending on n and k. 00:25:28.270 --> 00:25:31.900 align:middle line:84% Now, as n goes to infinity, what do I get? 00:25:31.900 --> 00:25:35.290 align:middle line:90% This is just equal to-- 00:25:35.290 --> 00:25:39.700 align:middle line:84% I can multiply the bottom by k to get rid of this one. 00:25:39.700 --> 00:25:41.830 align:middle line:84% So then I just get n over n plus k. 00:25:41.830 --> 00:25:44.150 align:middle line:90% And remember, k is fixed. 00:25:44.150 --> 00:25:46.160 align:middle line:84% And then I take the limit as n goes to infinity. 00:25:46.160 --> 00:25:46.990 align:middle line:90% So I get 1. 00:25:46.990 --> 00:25:50.880 align:middle line:90% 00:25:50.880 --> 00:25:52.470 align:middle line:84% So I'm taking the limit as k goes 00:25:52.470 --> 00:25:54.690 align:middle line:90% to infinity of this expression. 00:25:54.690 --> 00:25:57.850 align:middle line:84% Now for each k, when I take the limit as n goes to infinity, 00:25:57.850 --> 00:26:00.000 align:middle line:90% I get 1. 00:26:00.000 --> 00:26:01.680 align:middle line:90% So I get 1 there. 00:26:01.680 --> 00:26:04.920 align:middle line:84% Now, what happens if I interchange the limit, 00:26:04.920 --> 00:26:09.810 align:middle line:84% and now look at the limit as n goes to infinity of the limit 00:26:09.810 --> 00:26:18.990 align:middle line:84% as k goes to infinity of n over k, n over k plus 1? 00:26:18.990 --> 00:26:23.880 align:middle line:84% Well, now as k goes to infinity for each n-- 00:26:23.880 --> 00:26:25.440 align:middle line:84% so remember, I'm taking the limit 00:26:25.440 --> 00:26:31.230 align:middle line:84% as n goes to infinity of this expression, which is formerly 00:26:31.230 --> 00:26:35.480 align:middle line:84% this expression with the limits interchanged-- 00:26:35.480 --> 00:26:40.860 align:middle line:84% this is equal to 0 over 0 plus 1 equals 0. 00:26:40.860 --> 00:26:44.560 align:middle line:84% And these two do not equal each other. 00:26:44.560 --> 00:26:47.565 align:middle line:84% So it's not always the case that you can interchange limits. 00:26:47.565 --> 00:26:51.430 align:middle line:90% 00:26:51.430 --> 00:26:54.460 align:middle line:90% That's the simple fact of life. 00:26:54.460 --> 00:27:00.780 align:middle line:84% And to be able to make certain statements, 00:27:00.780 --> 00:27:04.800 align:middle line:84% do certain computations, you need 00:27:04.800 --> 00:27:06.270 align:middle line:84% to be able to interchange limits. 00:27:06.270 --> 00:27:11.160 align:middle line:90% And what kind of limits? 00:27:11.160 --> 00:27:15.570 align:middle line:84% Maybe taking an infinite sum and integrating, 00:27:15.570 --> 00:27:19.320 align:middle line:84% or like we stated in the beginning, 00:27:19.320 --> 00:27:22.110 align:middle line:84% power series are, in a sense, a certain limit. 00:27:22.110 --> 00:27:24.660 align:middle line:84% They're a limit of partial sums of-- 00:27:24.660 --> 00:27:27.730 align:middle line:90% they're a limit of polynomials. 00:27:27.730 --> 00:27:30.430 align:middle line:84% And then let's say differentiation-- 00:27:30.430 --> 00:27:31.160 align:middle line:90% that's a limit. 00:27:31.160 --> 00:27:36.200 align:middle line:84% So a natural question, like we said last time is, 00:27:36.200 --> 00:27:39.650 align:middle line:84% is the derivative of this infinite sum the infinite sum 00:27:39.650 --> 00:27:40.400 align:middle line:90% of the derivative? 00:27:40.400 --> 00:27:43.640 align:middle line:84% Those are two limiting processes which we're 00:27:43.640 --> 00:27:46.520 align:middle line:90% asking can we interchange. 00:27:46.520 --> 00:27:50.060 align:middle line:84% So the three questions, again, that I 00:27:50.060 --> 00:27:52.130 align:middle line:84% asked in terms of power series, I'm 00:27:52.130 --> 00:27:58.430 align:middle line:84% going to now frame again in this general setting. 00:27:58.430 --> 00:28:14.090 align:middle line:84% So if f n, S to R, f from S to R, and f n converges to f-- 00:28:14.090 --> 00:28:16.790 align:middle line:84% maybe pointwise or uniformly, so let's 00:28:16.790 --> 00:28:21.840 align:middle line:84% leave this open ended for now because these are the only two 00:28:21.840 --> 00:28:23.700 align:middle line:90% notions we have of convergence-- 00:28:23.700 --> 00:28:30.620 align:middle line:90% 00:28:30.620 --> 00:28:31.970 align:middle line:90% and let me make-- 00:28:31.970 --> 00:28:39.010 align:middle line:90% 00:28:39.010 --> 00:28:42.310 align:middle line:84% so suppose f n is from S to R, f is S to R, 00:28:42.310 --> 00:28:45.980 align:middle line:84% and f n converges to f pointwise or uniformly, 00:28:45.980 --> 00:28:48.250 align:middle line:84% and this sentence is not written very well, 00:28:48.250 --> 00:29:02.910 align:middle line:84% and f n is continuous for all n, then is f continuous? 00:29:02.910 --> 00:29:07.020 align:middle line:84% So suppose we have a sequence of continuous functions converging 00:29:07.020 --> 00:29:12.120 align:middle line:84% to another function f, either pointwise or uniformly. 00:29:12.120 --> 00:29:13.632 align:middle line:84% Is the limiting function continuous? 00:29:13.632 --> 00:29:15.090 align:middle line:84% And I'll explain it in a minute why 00:29:15.090 --> 00:29:17.040 align:middle line:84% this is a limiting, kind of asking 00:29:17.040 --> 00:29:19.080 align:middle line:90% can we interchange two limits. 00:29:19.080 --> 00:29:30.330 align:middle line:84% Second question is suppose f n from a, b to R 00:29:30.330 --> 00:29:46.860 align:middle line:84% is differentiable for all n, f from a, b to R, 00:29:46.860 --> 00:29:51.600 align:middle line:84% and we have that f n converges to f-- 00:29:51.600 --> 00:29:53.670 align:middle line:84% either pointwise or uniform, we're 00:29:53.670 --> 00:29:56.950 align:middle line:84% leaving this open ended for now-- 00:29:56.950 --> 00:30:00.460 align:middle line:84% and the derivatives also converge, say to some function 00:30:00.460 --> 00:30:14.010 align:middle line:90% g, then is f differentiable? 00:30:14.010 --> 00:30:19.910 align:middle line:84% And is the limit of the derivatives 00:30:19.910 --> 00:30:22.040 align:middle line:84% equal to the derivative of the limit? 00:30:22.040 --> 00:30:30.870 align:middle line:90% 00:30:30.870 --> 00:30:36.300 align:middle line:84% And then the last question-- so this is the third main limiting 00:30:36.300 --> 00:30:39.150 align:middle line:84% process we've seen in this class, which is integration-- 00:30:39.150 --> 00:30:44.540 align:middle line:90% 00:30:44.540 --> 00:30:53.110 align:middle line:84% suppose f n is a sequence of continuous functions, 00:30:53.110 --> 00:31:03.900 align:middle line:84% f is a continuous function, and f n converges 00:31:03.900 --> 00:31:07.350 align:middle line:84% to f-- again, maybe either pointwise or uniformly, 00:31:07.350 --> 00:31:09.240 align:middle line:84% we're leaving this open ended for now. 00:31:09.240 --> 00:31:12.950 align:middle line:90% 00:31:12.950 --> 00:31:19.730 align:middle line:84% Does limit of the integrals equal 00:31:19.730 --> 00:31:21.110 align:middle line:90% the integral of the limit? 00:31:21.110 --> 00:31:33.190 align:middle line:90% 00:31:33.190 --> 00:31:35.990 align:middle line:90% Now again, I want you to-- 00:31:35.990 --> 00:31:40.090 align:middle line:84% it's really, I guess, more clear here, 00:31:40.090 --> 00:31:43.040 align:middle line:84% the fact that we're asking about interchanging two limits. 00:31:43.040 --> 00:31:46.180 align:middle line:84% So here, we have this integration 00:31:46.180 --> 00:31:49.600 align:middle line:84% is just a symbol for taking this limiting process, 00:31:49.600 --> 00:31:52.210 align:middle line:84% where you take a sequence of partitions of a, b 00:31:52.210 --> 00:31:55.810 align:middle line:84% with norm converging to 0, then the integral from a to b of f 00:31:55.810 --> 00:32:01.120 align:middle line:84% n-- this is defined to be this limit of Riemann sums. 00:32:01.120 --> 00:32:03.490 align:middle line:84% So this is a limiting process here, 00:32:03.490 --> 00:32:06.950 align:middle line:84% although I'm writing it with this simple notation. 00:32:06.950 --> 00:32:09.130 align:middle line:84% So I'm asking it, can I take this limit 00:32:09.130 --> 00:32:10.840 align:middle line:90% as n goes to infinity? 00:32:10.840 --> 00:32:13.510 align:middle line:84% And can I interchange it with this limiting process? 00:32:13.510 --> 00:32:19.800 align:middle line:84% Now, for continuity, it's maybe a little more hidden 00:32:19.800 --> 00:32:21.540 align:middle line:84% on what's the interchange of limits 00:32:21.540 --> 00:32:23.070 align:middle line:90% that you're really looking at. 00:32:23.070 --> 00:32:26.940 align:middle line:90% So it looks more like this. 00:32:26.940 --> 00:32:44.164 align:middle line:84% So remember, for 1, we're asking suppose x is in set S, 00:32:44.164 --> 00:32:59.510 align:middle line:84% x n is a sequence converging to x, then basically, 00:32:59.510 --> 00:33:00.280 align:middle line:90% can we do this? 00:33:00.280 --> 00:33:05.910 align:middle line:90% 00:33:05.910 --> 00:33:10.800 align:middle line:84% Now I would like to show that f of x is equal to the limit 00:33:10.800 --> 00:33:14.070 align:middle line:84% as n goes to infinity of f of x sub n. 00:33:14.070 --> 00:33:22.000 align:middle line:84% So let me actually make this an m, and use two different-- 00:33:22.000 --> 00:33:22.960 align:middle line:90% let's use a k. 00:33:22.960 --> 00:33:26.620 align:middle line:90% 00:33:26.620 --> 00:33:31.390 align:middle line:84% So if I compute limit as k goes to infinity of f of x sub k, 00:33:31.390 --> 00:33:33.190 align:middle line:84% would like to show that equals f of x 00:33:33.190 --> 00:33:36.220 align:middle line:84% if I'm trying to show that the limit is continuous, 00:33:36.220 --> 00:33:40.120 align:middle line:84% assuming the functions that are converging to f are continuous. 00:33:40.120 --> 00:33:42.110 align:middle line:84% If I look at the limit as k goes to infinity, 00:33:42.110 --> 00:33:44.110 align:middle line:84% then I'm tempted to do the following-- that this 00:33:44.110 --> 00:33:51.700 align:middle line:84% is equal to the limit as k goes to infinity of the limit as n 00:33:51.700 --> 00:33:55.330 align:middle line:84% goes to infinity of f n of x, assuming 00:33:55.330 --> 00:34:00.520 align:middle line:84% either pointwise convergence or uniform convergence, x k. 00:34:00.520 --> 00:34:03.050 align:middle line:90% 00:34:03.050 --> 00:34:08.980 align:middle line:84% And now, if I'm just being a little bit careless, 00:34:08.980 --> 00:34:13.440 align:middle line:84% I interchange limits and write this as the limit 00:34:13.440 --> 00:34:18.230 align:middle line:84% as n goes to infinity of the limit as k goes to infinity. 00:34:18.230 --> 00:34:22.635 align:middle line:84% So here is where I'm asking, can I interchange limits? 00:34:22.635 --> 00:34:25.330 align:middle line:90% 00:34:25.330 --> 00:34:27.010 align:middle line:84% Again, this is not a proof, this is 00:34:27.010 --> 00:34:29.980 align:middle line:84% discussion on what is the interchange of limits 00:34:29.980 --> 00:34:32.139 align:middle line:84% that I'm looking at for question number 1. 00:34:32.139 --> 00:34:34.540 align:middle line:90% Question number 3 is more clear. 00:34:34.540 --> 00:34:37.179 align:middle line:84% And then for this, once I've done this, 00:34:37.179 --> 00:34:40.030 align:middle line:84% differentiability will be a little bit clearer 00:34:40.030 --> 00:34:46.420 align:middle line:84% on what is the interchange of limits I'm looking at. 00:34:46.420 --> 00:34:49.772 align:middle line:84% So let's say I was just doing whatever I like 00:34:49.772 --> 00:34:50.730 align:middle line:90% and going through this. 00:34:50.730 --> 00:34:53.190 align:middle line:84% Then I would interchange these limits. 00:34:53.190 --> 00:34:57.080 align:middle line:90% 00:34:57.080 --> 00:35:00.680 align:middle line:84% And since each of these f n's are continuous, 00:35:00.680 --> 00:35:04.310 align:middle line:84% then the limit as k goes to infinity of f n of x sub k, 00:35:04.310 --> 00:35:09.970 align:middle line:84% and x sub k converges to x, this just gives me f n of x. 00:35:09.970 --> 00:35:14.680 align:middle line:84% And so since f n, again, converges to f in some sense, 00:35:14.680 --> 00:35:16.730 align:middle line:90% this should give me f of x. 00:35:16.730 --> 00:35:21.790 align:middle line:84% So what I'm asking is, was all this OK? 00:35:21.790 --> 00:35:24.550 align:middle line:84% Because at some point, at this point in particular, 00:35:24.550 --> 00:35:28.098 align:middle line:90% I had to interchange a limit. 00:35:28.098 --> 00:35:30.140 align:middle line:84% And we've just seen that we can't always do that. 00:35:30.140 --> 00:35:34.090 align:middle line:84% We can't always interchange limits and get the same thing. 00:35:34.090 --> 00:35:36.162 align:middle line:90% Here, we got 1. 00:35:36.162 --> 00:35:37.870 align:middle line:84% When we interchanged the limit, we got 0. 00:35:37.870 --> 00:35:41.500 align:middle line:84% So it's not always the case that I can interchange limits. 00:35:41.500 --> 00:35:46.180 align:middle line:84% This equality between this limit and the interchanged limit 00:35:46.180 --> 00:35:47.740 align:middle line:90% is the big question mark. 00:35:47.740 --> 00:35:52.000 align:middle line:84% So that's the whole basis for question number 1. 00:35:52.000 --> 00:35:55.350 align:middle line:90% 00:35:55.350 --> 00:36:02.530 align:middle line:84% So I hope that discussion was clear enough. 00:36:02.530 --> 00:36:05.140 align:middle line:84% Now, the answer to these three questions 00:36:05.140 --> 00:36:11.770 align:middle line:84% is, in fact, yes, but only for uniform convergence. 00:36:11.770 --> 00:36:15.890 align:middle line:90% 00:36:15.890 --> 00:36:41.080 align:middle line:84% If the convergence-- and this is the mode of convergence 00:36:41.080 --> 00:36:43.690 align:middle line:84% we must have so that the answer to all of these three 00:36:43.690 --> 00:36:49.710 align:middle line:84% questions, which we'll state and prove as theorems, is correct. 00:36:49.710 --> 00:36:55.680 align:middle line:84% Now the natural question is, what 00:36:55.680 --> 00:36:57.750 align:middle line:90% if we have a weaker hypothesis? 00:36:57.750 --> 00:37:01.560 align:middle line:84% Namely, what if we only assume pointwise convergence? 00:37:01.560 --> 00:37:05.610 align:middle line:84% Is the answer to any of these questions yes? 00:37:05.610 --> 00:37:08.520 align:middle line:90% And, well, no. 00:37:08.520 --> 00:37:10.800 align:middle line:84% So the answer to all three of these questions 00:37:10.800 --> 00:37:13.530 align:middle line:84% is no if we only assume pointwise convergence. 00:37:13.530 --> 00:37:36.170 align:middle line:90% 00:37:36.170 --> 00:37:38.930 align:middle line:84% So let's go through an example showing 00:37:38.930 --> 00:37:40.580 align:middle line:84% that each of these three questions 00:37:40.580 --> 00:37:45.700 align:middle line:84% is no if we only assume pointwise convergence. 00:37:45.700 --> 00:37:48.990 align:middle line:84% So let's look at an example showing 1 00:37:48.990 --> 00:37:52.863 align:middle line:84% is no if we only assume pointwise convergence. 00:37:52.863 --> 00:37:54.780 align:middle line:84% And we basically already have it on the board. 00:37:54.780 --> 00:37:58.970 align:middle line:90% 00:37:58.970 --> 00:38:12.260 align:middle line:84% Take f n of x to be x to the n on 0, 1, x is in-- 00:38:12.260 --> 00:38:18.290 align:middle line:90% 00:38:18.290 --> 00:38:25.650 align:middle line:84% then each of these functions, and for all n, 00:38:25.650 --> 00:38:31.650 align:middle line:84% f n is a continuous function on 0, 1, 00:38:31.650 --> 00:38:36.915 align:middle line:90% f n converges to f pointwise. 00:38:36.915 --> 00:38:40.670 align:middle line:90% 00:38:40.670 --> 00:38:48.010 align:middle line:84% But f itself is not continuous function. 00:38:48.010 --> 00:38:50.590 align:middle line:84% So this provides an example of a sequence of functions 00:38:50.590 --> 00:38:53.470 align:middle line:84% which converges pointwise to a function which 00:38:53.470 --> 00:38:55.330 align:middle line:90% is not continuous. 00:38:55.330 --> 00:38:57.190 align:middle line:84% Again, what I was saying there in the answer 00:38:57.190 --> 00:39:00.100 align:middle line:84% is that if I have a sequence of functions which 00:39:00.100 --> 00:39:02.200 align:middle line:84% converges uniformly to a function f, 00:39:02.200 --> 00:39:04.780 align:middle line:84% then that function is continuous. 00:39:04.780 --> 00:39:07.870 align:middle line:84% Here, if we only assume pointwise convergence, 00:39:07.870 --> 00:39:10.780 align:middle line:84% we may not get a continuous function in the end. 00:39:10.780 --> 00:39:12.610 align:middle line:84% And this is what this example shows-- 00:39:12.610 --> 00:39:15.160 align:middle line:90% f n of x equals x to the n. 00:39:15.160 --> 00:39:16.930 align:middle line:84% These are all continuous functions. 00:39:16.930 --> 00:39:18.790 align:middle line:84% They converge to a function which 00:39:18.790 --> 00:39:20.800 align:middle line:90% is not continuous pointwise. 00:39:20.800 --> 00:39:27.690 align:middle line:90% 00:39:27.690 --> 00:39:29.560 align:middle line:84% I'll say something else in a minute. 00:39:29.560 --> 00:39:33.000 align:middle line:90% So let's look at example 2 now. 00:39:33.000 --> 00:39:35.730 align:middle line:84% Basically, we take the previous example and kind 00:39:35.730 --> 00:39:38.610 align:middle line:84% of integrate it to get an example of a sequence 00:39:38.610 --> 00:39:42.660 align:middle line:84% of functions that converges, and its derivative converges 00:39:42.660 --> 00:39:45.270 align:middle line:90% pointwise. 00:39:45.270 --> 00:40:00.170 align:middle line:84% 2, take f n of x to be x to the n over n on 0, 1. 00:40:00.170 --> 00:40:05.480 align:middle line:90% Then a couple of things-- 00:40:05.480 --> 00:40:06.935 align:middle line:90% f n converges to 0. 00:40:06.935 --> 00:40:10.900 align:middle line:90% 00:40:10.900 --> 00:40:15.670 align:middle line:84% f n prime converges to the function 00:40:15.670 --> 00:40:18.640 align:middle line:84% f, which is the same function from here. 00:40:18.640 --> 00:40:28.200 align:middle line:90% 00:40:28.200 --> 00:40:29.620 align:middle line:90% So let me actually make this g. 00:40:29.620 --> 00:40:35.380 align:middle line:90% 00:40:35.380 --> 00:40:42.530 align:middle line:84% And these are pointwise, so I have 00:40:42.530 --> 00:40:45.530 align:middle line:84% a sequence of differentiable functions 00:40:45.530 --> 00:40:47.870 align:middle line:84% that converge pointwise to something. 00:40:47.870 --> 00:40:49.580 align:middle line:84% And the derivatives also converge 00:40:49.580 --> 00:40:50.960 align:middle line:90% pointwise to something. 00:40:50.960 --> 00:40:57.220 align:middle line:90% 00:40:57.220 --> 00:41:06.110 align:middle line:84% Let's call f equals 0, but g is not equal to f prime. 00:41:06.110 --> 00:41:08.600 align:middle line:84% The derivative of f, the constant function 0, 00:41:08.600 --> 00:41:09.550 align:middle line:90% which is 0. 00:41:09.550 --> 00:41:11.920 align:middle line:84% g is equal to this function, which 00:41:11.920 --> 00:41:16.030 align:middle line:84% is 0 if x is in 0, 1 and 1 if x is equal to 1. 00:41:16.030 --> 00:41:20.410 align:middle line:84% So here, we see that the derivative of the limit 00:41:20.410 --> 00:41:22.240 align:middle line:84% is not the limit of the derivatives, 00:41:22.240 --> 00:41:26.660 align:middle line:84% if we only assume pointwise convergence of the functions. 00:41:26.660 --> 00:41:29.970 align:middle line:90% 00:41:29.970 --> 00:41:33.540 align:middle line:84% Now, the last example showing that 3 does not 00:41:33.540 --> 00:41:36.030 align:middle line:84% hold if we only assume pointwise convergence 00:41:36.030 --> 00:41:37.240 align:middle line:90% is also on the board. 00:41:37.240 --> 00:41:40.230 align:middle line:84% It's the whole point of this one up there. 00:41:40.230 --> 00:41:46.400 align:middle line:90% So let's go on the next board. 00:41:46.400 --> 00:42:12.240 align:middle line:90% 00:42:12.240 --> 00:42:23.960 align:middle line:84% So here, f n from 0 to 1 to R is this tent function, 1 over n, 00:42:23.960 --> 00:42:26.760 align:middle line:90% 1 over 2n, 2n. 00:42:26.760 --> 00:42:31.110 align:middle line:90% 00:42:31.110 --> 00:42:33.100 align:middle line:90% It's just piecewise linear. 00:42:33.100 --> 00:42:40.620 align:middle line:84% So it's 0 from 1 over n to 1, and it goes up to 1 00:42:40.620 --> 00:42:44.910 align:middle line:84% over 2n, 2n, and then back down to the origin. 00:42:44.910 --> 00:42:48.940 align:middle line:90% 00:42:48.940 --> 00:42:54.880 align:middle line:84% And so we know that f n converges to 0 pointwise. 00:42:54.880 --> 00:42:57.560 align:middle line:90% 00:42:57.560 --> 00:42:59.810 align:middle line:84% Now, the integral of 0 is just 0. 00:42:59.810 --> 00:43:03.770 align:middle line:84% Let's look at the integral of f n from 0 to 1. 00:43:03.770 --> 00:43:08.710 align:middle line:90% 00:43:08.710 --> 00:43:12.250 align:middle line:90% The integral of 0, 1 of f n-- 00:43:12.250 --> 00:43:14.080 align:middle line:84% now if we were all together, this 00:43:14.080 --> 00:43:22.420 align:middle line:84% would be the point where I stop and ask if anybody can remember 00:43:22.420 --> 00:43:25.090 align:middle line:84% the area of a triangle, even though we're 00:43:25.090 --> 00:43:27.460 align:middle line:90% in this advanced analysis class. 00:43:27.460 --> 00:43:29.020 align:middle line:90% But I don't get to ask you that. 00:43:29.020 --> 00:43:32.230 align:middle line:84% I get to just ask myself that and I know the answer 00:43:32.230 --> 00:43:34.180 align:middle line:84% because I prepared and only because of that. 00:43:34.180 --> 00:43:40.850 align:middle line:84% This is 1/2 base times the height. 00:43:40.850 --> 00:43:42.640 align:middle line:84% So remember, the integral is an area. 00:43:42.640 --> 00:43:46.120 align:middle line:90% 00:43:46.120 --> 00:43:49.300 align:middle line:84% And I could write down the formula 00:43:49.300 --> 00:43:53.453 align:middle line:84% of what this function is and actually integrate it 00:43:53.453 --> 00:43:55.370 align:middle line:84% out using the fundamental theorem of calculus, 00:43:55.370 --> 00:43:56.810 align:middle line:90% but just go with me on this-- 00:43:56.810 --> 00:44:00.380 align:middle line:84% that the integral of f n is just the area of this triangle that 00:44:00.380 --> 00:44:04.730 align:middle line:84% has base starting at 0 and going to 1 over n, 00:44:04.730 --> 00:44:07.670 align:middle line:90% and it peaks at 2n. 00:44:07.670 --> 00:44:13.460 align:middle line:90% So the base is 1 over n. 00:44:13.460 --> 00:44:16.850 align:middle line:90% The height is 2n. 00:44:16.850 --> 00:44:20.720 align:middle line:90% So this equals 1 for all n. 00:44:20.720 --> 00:44:22.370 align:middle line:84% See, this is why I was making it peak. 00:44:22.370 --> 00:44:26.150 align:middle line:90% 00:44:26.150 --> 00:44:34.310 align:middle line:84% So the integral from 0 to 1, which is just 1 for all n, 00:44:34.310 --> 00:44:38.510 align:middle line:84% does not converge to the integral of the limit. 00:44:38.510 --> 00:44:45.030 align:middle line:90% 00:44:45.030 --> 00:44:48.810 align:middle line:84% So in this case, the limit of the integrals 00:44:48.810 --> 00:44:51.660 align:middle line:84% is not the integral of the limit. 00:44:51.660 --> 00:44:53.370 align:middle line:90% And what's the reason? 00:44:53.370 --> 00:44:56.430 align:middle line:84% Because we only have pointwise convergence. 00:44:56.430 --> 00:44:58.890 align:middle line:84% Like I said a minute ago, the answer is yes to all three 00:44:58.890 --> 00:45:01.650 align:middle line:84% of these questions if the convergence is uniform. 00:45:01.650 --> 00:45:03.150 align:middle line:84% And what these three examples are 00:45:03.150 --> 00:45:04.567 align:middle line:84% supposed to show you is the answer 00:45:04.567 --> 00:45:08.910 align:middle line:84% is no if I only assume pointwise convergence, the weaker 00:45:08.910 --> 00:45:10.260 align:middle line:90% notion of convergence. 00:45:10.260 --> 00:45:19.100 align:middle line:90% 00:45:19.100 --> 00:45:23.405 align:middle line:84% So let's now prove some theorems. 00:45:23.405 --> 00:45:29.300 align:middle line:90% 00:45:29.300 --> 00:45:32.180 align:middle line:84% So this first theorem addresses question 1. 00:45:32.180 --> 00:45:37.500 align:middle line:90% 00:45:37.500 --> 00:45:47.790 align:middle line:84% Suppose f n from S to R, f from S to R, f n is continuous, 00:45:47.790 --> 00:45:56.450 align:middle line:84% meaning it's continuous at every point in S for all n, 00:45:56.450 --> 00:46:14.850 align:middle line:84% and f n convergence to f uniformly on S. Then 00:46:14.850 --> 00:46:19.320 align:middle line:84% the conclusion is that f is continuous. 00:46:19.320 --> 00:46:29.000 align:middle line:90% 00:46:29.000 --> 00:46:34.250 align:middle line:84% So the proof-- we've done several proofs like this 00:46:34.250 --> 00:46:37.790 align:middle line:84% before but for some reason, at least I 00:46:37.790 --> 00:46:40.940 align:middle line:84% see in textbooks for this proof, they always refer to it 00:46:40.940 --> 00:46:43.178 align:middle line:90% as an epsilon over 3 argument. 00:46:43.178 --> 00:46:44.720 align:middle line:84% And then it's the last time they call 00:46:44.720 --> 00:46:48.410 align:middle line:90% it an epsilon over 3 argument. 00:46:48.410 --> 00:46:55.600 align:middle line:84% So we have to show f is continuous at every point in S. 00:46:55.600 --> 00:47:01.710 align:middle line:84% So let c be a point in S. Let epsilon be positive. 00:47:01.710 --> 00:47:05.370 align:middle line:84% We have to find delta so that for all x 00:47:05.370 --> 00:47:07.890 align:middle line:84% minus c less than delta in absolute value, 00:47:07.890 --> 00:47:11.835 align:middle line:84% f of x minus f of c is less than epsilon. 00:47:11.835 --> 00:47:13.710 align:middle line:84% And what we're going to do is we're basically 00:47:13.710 --> 00:47:18.490 align:middle line:84% going to replace f by some f m for m large enough. 00:47:18.490 --> 00:47:20.580 align:middle line:84% And the fact that we have uniform convergence 00:47:20.580 --> 00:47:22.600 align:middle line:90% is what allows us to do that. 00:47:22.600 --> 00:47:25.320 align:middle line:90% So let epsilon be positive. 00:47:25.320 --> 00:47:29.720 align:middle line:90% 00:47:29.720 --> 00:47:35.570 align:middle line:84% Since the f n's converge to f uniformly, 00:47:35.570 --> 00:47:38.840 align:middle line:84% there exists a natural number M so 00:47:38.840 --> 00:47:48.244 align:middle line:84% that for all n bigger than or equal to M, for all y in S, 00:47:48.244 --> 00:47:56.110 align:middle line:84% f M of y minus f of y is less than epsilon over 3. 00:47:56.110 --> 00:48:09.440 align:middle line:90% 00:48:09.440 --> 00:48:13.490 align:middle line:90% Now, this f sub capital M-- 00:48:13.490 --> 00:48:16.880 align:middle line:84% I'm just going to fix that so it should be n. 00:48:16.880 --> 00:48:19.870 align:middle line:84% But now, this is for all n bigger than or equal to capital 00:48:19.870 --> 00:48:24.370 align:middle line:84% M. I really just need one, so let's look at the f sub capital 00:48:24.370 --> 00:48:25.750 align:middle line:90% M. That's a continuous function. 00:48:25.750 --> 00:48:29.210 align:middle line:90% 00:48:29.210 --> 00:48:38.490 align:middle line:84% Since f sub capital M is continuous, 00:48:38.490 --> 00:48:46.080 align:middle line:84% there exists a delta positive so that for all x minus, if x 00:48:46.080 --> 00:48:58.110 align:middle line:84% is within distance delta to c, then 00:48:58.110 --> 00:49:06.415 align:middle line:84% I get that f sub M of x minus f sub M of c is less than epsilon 00:49:06.415 --> 00:49:06.915 align:middle line:90% over 3. 00:49:06.915 --> 00:49:11.690 align:middle line:90% 00:49:11.690 --> 00:49:18.200 align:middle line:84% So then this delta here-- so I stopped 00:49:18.200 --> 00:49:19.700 align:middle line:84% doing something which I was doing 00:49:19.700 --> 00:49:22.490 align:middle line:84% in all previous lectures, when I would say there exists delta 0. 00:49:22.490 --> 00:49:24.170 align:middle line:90% Choose delta to be this delta 0. 00:49:24.170 --> 00:49:25.430 align:middle line:90% There exists M0. 00:49:25.430 --> 00:49:27.110 align:middle line:90% Choose M to be this M0. 00:49:27.110 --> 00:49:32.350 align:middle line:84% Now I've dropped that and stopped 00:49:32.350 --> 00:49:35.540 align:middle line:84% doing that because it should be clear from the context 00:49:35.540 --> 00:49:37.270 align:middle line:84% now what I'm choosing delta to be. 00:49:37.270 --> 00:49:43.690 align:middle line:84% So then for all x minus c less than delta-- 00:49:43.690 --> 00:49:45.990 align:middle line:84% so I'm saying that I'm now choosing this delta that 00:49:45.990 --> 00:49:51.120 align:middle line:84% came from here, which came from f sub capital M. Capital M came 00:49:51.120 --> 00:49:55.890 align:middle line:90% from this, what I needed here. 00:49:55.890 --> 00:49:58.410 align:middle line:90% 00:49:58.410 --> 00:50:02.780 align:middle line:84% So now, if I look f of x minus f of c, which I want to show 00:50:02.780 --> 00:50:07.950 align:middle line:84% is less than epsilon, now if I add and subtract 00:50:07.950 --> 00:50:12.120 align:middle line:84% f sub M of x and f sub M of c and use the triangle 00:50:12.120 --> 00:50:14.910 align:middle line:84% inequality, this is less than or equal to f 00:50:14.910 --> 00:50:25.740 align:middle line:84% of x minus f sub M of x plus f sub M of x minus f sub M of c 00:50:25.740 --> 00:50:34.540 align:middle line:90% plus f sub M of c minus f of c. 00:50:34.540 --> 00:50:39.430 align:middle line:84% So I just added and subtracted f sub M of x and f of M of c, 00:50:39.430 --> 00:50:43.510 align:middle line:84% and then used the triangle inequality. 00:50:43.510 --> 00:50:49.230 align:middle line:84% Now, by this estimate here, since I'm 00:50:49.230 --> 00:50:53.250 align:middle line:84% looking at a particular n basically equal to M, 00:50:53.250 --> 00:50:56.340 align:middle line:84% I have this is less than epsilon over 3 00:50:56.340 --> 00:50:59.770 align:middle line:90% no matter what y is in this set. 00:50:59.770 --> 00:51:02.880 align:middle line:84% So then certainly for x, I will have this is less than epsilon 00:51:02.880 --> 00:51:05.420 align:middle line:90% over 3. 00:51:05.420 --> 00:51:08.240 align:middle line:84% How I chose delta, remember, was to guarantee 00:51:08.240 --> 00:51:11.000 align:middle line:84% that this would be less than epsilon over 3 00:51:11.000 --> 00:51:13.235 align:middle line:84% as long as x minus c is less than delta. 00:51:13.235 --> 00:51:15.602 align:middle line:90% 00:51:15.602 --> 00:51:17.310 align:middle line:84% And then of course, this one is less than 00:51:17.310 --> 00:51:21.840 align:middle line:84% epsilon over 3 again because of this uniform closeness of f M 00:51:21.840 --> 00:51:22.740 align:middle line:90% to f. 00:51:22.740 --> 00:51:26.360 align:middle line:90% 00:51:26.360 --> 00:51:29.180 align:middle line:84% And therefore, for all absolute value 00:51:29.180 --> 00:51:31.250 align:middle line:84% of x minus c less than delta, I have 00:51:31.250 --> 00:51:36.690 align:middle line:84% f of x minus f of c is less than epsilon. 00:51:36.690 --> 00:51:39.200 align:middle line:90% And that finishes the proof. 00:51:39.200 --> 00:51:42.350 align:middle line:84% So if we have uniform convergence, then-- 00:51:42.350 --> 00:51:45.410 align:middle line:90% 00:51:45.410 --> 00:51:47.930 align:middle line:84% a good theorem should be able to be 00:51:47.930 --> 00:51:54.020 align:middle line:84% stated in one sentence, or at least some simple and easy way 00:51:54.020 --> 00:51:55.160 align:middle line:90% to remember it. 00:51:55.160 --> 00:51:58.550 align:middle line:84% What this says is the uniform limit of continuous functions 00:51:58.550 --> 00:51:59.570 align:middle line:90% is continuous. 00:51:59.570 --> 00:52:04.180 align:middle line:90% 00:52:04.180 --> 00:52:07.420 align:middle line:84% So the uniform limit of continuous functions 00:52:07.420 --> 00:52:08.500 align:middle line:90% is continuous. 00:52:08.500 --> 00:52:11.550 align:middle line:90% 00:52:11.550 --> 00:52:16.560 align:middle line:84% Next, we'll show, in a sense, the uniform limit 00:52:16.560 --> 00:52:21.360 align:middle line:84% of differentiable functions is continuous. 00:52:21.360 --> 00:52:24.300 align:middle line:90% 00:52:24.300 --> 00:52:32.870 align:middle line:84% We're going to do the simplest statement of that, 00:52:32.870 --> 00:52:35.870 align:middle line:84% although one can make stronger statements. 00:52:35.870 --> 00:52:38.550 align:middle line:90% 00:52:38.550 --> 00:52:44.340 align:middle line:84% But in practice, this one suffices, really, at least 00:52:44.340 --> 00:52:46.155 align:middle line:90% where it pops up later in life. 00:52:46.155 --> 00:52:50.050 align:middle line:90% 00:52:50.050 --> 00:52:56.355 align:middle line:84% But before I get to that one, let's do 3, which is-- 00:52:56.355 --> 00:52:59.270 align:middle line:90% 00:52:59.270 --> 00:53:01.520 align:middle line:84% let's see, what would be the short way of saying that? 00:53:01.520 --> 00:53:06.620 align:middle line:90% 00:53:06.620 --> 00:53:09.130 align:middle line:84% The integral of the uniform limit 00:53:09.130 --> 00:53:12.340 align:middle line:84% is the limit of the integrals, something like that. 00:53:12.340 --> 00:53:14.500 align:middle line:84% That would be the short and sweet way 00:53:14.500 --> 00:53:18.970 align:middle line:84% of stating the following theorem, which is-- 00:53:18.970 --> 00:53:20.440 align:middle line:90% so this is an answer to 3. 00:53:20.440 --> 00:53:24.480 align:middle line:84% So we skipped over 2 for a minute. 00:53:24.480 --> 00:53:29.800 align:middle line:84% Suppose f n is a sequence of continuous functions on a, b, 00:53:29.800 --> 00:53:33.100 align:middle line:84% because we're going to be talking about integrals 00:53:33.100 --> 00:53:34.780 align:middle line:84% and we've only spoken about integrals 00:53:34.780 --> 00:53:37.620 align:middle line:90% for continuous functions-- 00:53:37.620 --> 00:53:48.000 align:middle line:84% suppose f n is a continuous function on a, b, f from a, b 00:53:48.000 --> 00:53:58.840 align:middle line:84% to R, and f n converges to F uniformly. 00:53:58.840 --> 00:54:05.740 align:middle line:84% Note that by what we just proven the previous theorem, 00:54:05.740 --> 00:54:10.390 align:middle line:84% this automatically guarantees that f 00:54:10.390 --> 00:54:11.600 align:middle line:90% is a continuous function. 00:54:11.600 --> 00:54:14.170 align:middle line:84% So we can ask about the relationship 00:54:14.170 --> 00:54:16.810 align:middle line:84% between the integral of f and the limit 00:54:16.810 --> 00:54:17.920 align:middle line:90% of the integrals of f n. 00:54:17.920 --> 00:54:21.330 align:middle line:90% 00:54:21.330 --> 00:54:26.250 align:middle line:84% Then the limit as n goes to infinity, 00:54:26.250 --> 00:54:32.040 align:middle line:84% integral from a to b of f n equals the integral 00:54:32.040 --> 00:54:34.120 align:middle line:90% from a, b to f. 00:54:34.120 --> 00:54:37.080 align:middle line:84% So the integral of the uniform limit 00:54:37.080 --> 00:54:39.360 align:middle line:90% is the limit of the integrals. 00:54:39.360 --> 00:54:49.720 align:middle line:90% 00:54:49.720 --> 00:54:51.580 align:middle line:84% So this is just a sequence of numbers, 00:54:51.580 --> 00:54:54.200 align:middle line:84% and we want to show it converges to this number here. 00:54:54.200 --> 00:54:56.620 align:middle line:84% So let's do this the old-fashioned way. 00:54:56.620 --> 00:55:00.250 align:middle line:90% 00:55:00.250 --> 00:55:01.630 align:middle line:90% Let epsilon be positive. 00:55:01.630 --> 00:55:04.460 align:middle line:90% 00:55:04.460 --> 00:55:11.210 align:middle line:84% Since f n converges to f uniformly, 00:55:11.210 --> 00:55:21.750 align:middle line:84% there exists a natural number M such that 00:55:21.750 --> 00:55:25.610 align:middle line:84% for all n bigger than or equal to M, for all x in a, 00:55:25.610 --> 00:55:33.200 align:middle line:84% b, f n of x minus f of x is less than epsilon. 00:55:33.200 --> 00:55:37.488 align:middle line:90% 00:55:37.488 --> 00:55:39.030 align:middle line:84% Now, basically what we're going to do 00:55:39.030 --> 00:55:41.070 align:middle line:90% is integrate this inequality. 00:55:41.070 --> 00:55:43.860 align:middle line:84% And remember, unlike differentiation, integration 00:55:43.860 --> 00:55:45.270 align:middle line:90% respects inequalities. 00:55:45.270 --> 00:55:48.410 align:middle line:90% 00:55:48.410 --> 00:55:50.500 align:middle line:84% And for all n bigger than or equal to M, 00:55:50.500 --> 00:55:57.920 align:middle line:84% meaning I'm choosing my M as this guy for proving 00:55:57.920 --> 00:56:04.085 align:middle line:84% this limit, if I look at the integral from a, b of f 00:56:04.085 --> 00:56:07.910 align:middle line:84% n minus the integral of a, b to f, 00:56:07.910 --> 00:56:11.240 align:middle line:84% this is, by linearity of the integral, 00:56:11.240 --> 00:56:14.990 align:middle line:84% the integral of f n minus f, absolute value. 00:56:14.990 --> 00:56:18.530 align:middle line:84% By the triangle inequality for integrals, which we proved, 00:56:18.530 --> 00:56:21.470 align:middle line:84% this is less than or equal to the integral from a, 00:56:21.470 --> 00:56:25.490 align:middle line:90% b of f n minus f. 00:56:25.490 --> 00:56:29.600 align:middle line:84% And what do we know for all n bigger than or equal to M? 00:56:29.600 --> 00:56:35.420 align:middle line:84% This function here, f n of x minus f of x in absolute value, 00:56:35.420 --> 00:56:38.240 align:middle line:90% is bounded by epsilon. 00:56:38.240 --> 00:56:40.040 align:middle line:84% And therefore, the integral of this side 00:56:40.040 --> 00:56:42.830 align:middle line:84% is going to be less than the integral of the right side. 00:56:42.830 --> 00:56:45.150 align:middle line:90% So I did this wrong. 00:56:45.150 --> 00:56:48.330 align:middle line:90% Let's put a b minus a over this. 00:56:48.330 --> 00:56:55.730 align:middle line:84% So this is less than integral a, b epsilon over b minus a. 00:56:55.730 --> 00:56:57.750 align:middle line:84% And so then I just pick up this number 00:56:57.750 --> 00:57:06.370 align:middle line:84% times the length of the interval, which equals epsilon. 00:57:06.370 --> 00:57:08.520 align:middle line:84% So for all n bigger than or equal to M, 00:57:08.520 --> 00:57:10.500 align:middle line:84% the absolute value of this integral 00:57:10.500 --> 00:57:14.445 align:middle line:84% of f sub n minus the integral of f is less than epsilon. 00:57:14.445 --> 00:57:17.510 align:middle line:90% 00:57:17.510 --> 00:57:20.530 align:middle line:84% So the integral of the uniform limit 00:57:20.530 --> 00:57:24.020 align:middle line:90% is the limit of the integrals. 00:57:24.020 --> 00:57:31.300 align:middle line:84% So now we'll use this to do the last interchange of limits 00:57:31.300 --> 00:57:35.655 align:middle line:84% theorem that I had in mind, so number 2. 00:57:35.655 --> 00:57:38.520 align:middle line:90% 00:57:38.520 --> 00:57:41.670 align:middle line:84% And again, this is kind of the simplest statement, and maybe 00:57:41.670 --> 00:57:44.040 align:middle line:84% easiest to prove, that one can make. 00:57:44.040 --> 00:57:46.380 align:middle line:84% One can make stronger statements and prove them, 00:57:46.380 --> 00:57:49.150 align:middle line:84% but in most cases, this suffices. 00:57:49.150 --> 00:58:09.560 align:middle line:84% So suppose f in a, b to R. So this is continuously 00:58:09.560 --> 00:58:16.290 align:middle line:84% differentiable for all n, f and g-- 00:58:16.290 --> 00:58:20.880 align:middle line:84% these are two functions from a, b to R-- 00:58:20.880 --> 00:58:33.310 align:middle line:84% and f n converges to f uniformly on a, b, and the derivatives 00:58:33.310 --> 00:58:35.815 align:middle line:84% converge uniformly to this function g. 00:58:35.815 --> 00:58:45.530 align:middle line:90% 00:58:45.530 --> 00:58:49.390 align:middle line:84% And in fact, I don't even need this. 00:58:49.390 --> 00:58:50.710 align:middle line:90% We just say pointwise. 00:58:50.710 --> 00:58:53.760 align:middle line:90% 00:58:53.760 --> 00:58:58.770 align:middle line:84% So I only need uniform convergence of the derivatives 00:58:58.770 --> 00:59:02.820 align:middle line:84% if I'm assuming everything is continuously differentiable, 00:59:02.820 --> 00:59:04.950 align:middle line:84% if I'm assuming the sequences are continuously 00:59:04.950 --> 00:59:06.660 align:middle line:90% differentiable. 00:59:06.660 --> 00:59:16.320 align:middle line:84% Then f is differentiable, in fact, 00:59:16.320 --> 00:59:19.230 align:middle line:84% continuously differentiable, meaning 00:59:19.230 --> 00:59:21.030 align:middle line:90% the derivative is continuous. 00:59:21.030 --> 00:59:26.140 align:middle line:90% 00:59:26.140 --> 00:59:33.670 align:middle line:84% And the derivative of f is equal to g, 00:59:33.670 --> 00:59:40.630 align:middle line:84% meaning that the uniform limit of the derivatives 00:59:40.630 --> 00:59:44.590 align:middle line:84% converges to the derivative of f. 00:59:44.590 --> 00:59:51.120 align:middle line:84% So again, what this says is the uniform limit of-- 00:59:51.120 --> 00:59:52.628 align:middle line:84% or at least in essence and spirit, 00:59:52.628 --> 00:59:55.170 align:middle line:84% it's not exactly what it says-- but it says the uniform limit 00:59:55.170 --> 01:00:02.060 align:middle line:84% of continuously differentiable functions is differentiable, 01:00:02.060 --> 01:00:07.640 align:middle line:84% and that the derivative of the uniform limit is the limit of-- 01:00:07.640 --> 01:00:10.620 align:middle line:90% 01:00:10.620 --> 01:00:12.270 align:middle line:84% the derivative of the uniform limit 01:00:12.270 --> 01:00:14.400 align:middle line:84% is the uniform limit of the derivatives. 01:00:14.400 --> 01:00:19.850 align:middle line:90% 01:00:19.850 --> 01:00:22.185 align:middle line:84% So to prove this, we use the fundamental theorem 01:00:22.185 --> 01:00:22.685 align:middle line:90% of calculus. 01:00:22.685 --> 01:00:25.690 align:middle line:90% 01:00:25.690 --> 01:00:32.970 align:middle line:90% So let x be a point in a, b. 01:00:32.970 --> 01:00:42.390 align:middle line:84% Then by the fundamental theorem of calculus, 01:00:42.390 --> 01:00:45.615 align:middle line:84% if I take f n of x minus f n of a, 01:00:45.615 --> 01:00:51.265 align:middle line:84% this is equal to the integral from a to x of f n prime. 01:00:51.265 --> 01:00:53.640 align:middle line:84% The integral of the derivative gives me back the function 01:00:53.640 --> 01:00:55.140 align:middle line:90% evaluated at the endpoints. 01:00:55.140 --> 01:00:59.780 align:middle line:90% 01:00:59.780 --> 01:01:02.600 align:middle line:84% And therefore, since I pointwise convergence, 01:01:02.600 --> 01:01:07.280 align:middle line:84% these two numbers converge pointwise to f of x and f 01:01:07.280 --> 01:01:12.450 align:middle line:90% of a respectively. 01:01:12.450 --> 01:01:15.920 align:middle line:84% So f of x minus f of a equals the limit 01:01:15.920 --> 01:01:20.720 align:middle line:84% as n goes to infinity of f n of x minus f n of a. 01:01:20.720 --> 01:01:23.648 align:middle line:90% 01:01:23.648 --> 01:01:31.590 align:middle line:84% And this is equal to, just by the previous expression, limit 01:01:31.590 --> 01:01:33.870 align:middle line:84% as n goes to infinity of the integral from a 01:01:33.870 --> 01:01:38.030 align:middle line:90% to x of f n prime. 01:01:38.030 --> 01:01:41.340 align:middle line:90% 01:01:41.340 --> 01:01:44.610 align:middle line:84% Now, these f n primes are converging to g, this function 01:01:44.610 --> 01:01:46.260 align:middle line:90% g, uniformly. 01:01:46.260 --> 01:01:48.315 align:middle line:84% And therefore, the integrals converge. 01:01:48.315 --> 01:01:49.440 align:middle line:90% that's what we just proved. 01:01:49.440 --> 01:01:52.170 align:middle line:90% 01:01:52.170 --> 01:01:53.990 align:middle line:84% So this equals the integral of the limit. 01:01:53.990 --> 01:01:56.710 align:middle line:90% 01:01:56.710 --> 01:02:01.290 align:middle line:84% And therefore-- so we started off with f of x minus f of a, 01:02:01.290 --> 01:02:04.560 align:middle line:84% and we showed it's equal to the integral from a to x of g. 01:02:04.560 --> 01:02:14.610 align:middle line:84% So f of x equals f of a plus the integral from a to x of g. 01:02:14.610 --> 01:02:17.250 align:middle line:84% But no, again by the fundamental theorem of calculus, 01:02:17.250 --> 01:02:26.490 align:middle line:84% this implies that f is differentiable. 01:02:26.490 --> 01:02:31.580 align:middle line:84% And the derivative of f of x, the derivative of f, 01:02:31.580 --> 01:02:34.070 align:middle line:84% is equal to the derivative-- so remember, 01:02:34.070 --> 01:02:37.820 align:middle line:84% f of a, that's just a constant-- equals the derivative 01:02:37.820 --> 01:02:39.710 align:middle line:90% of this function, which is g. 01:02:39.710 --> 01:02:51.580 align:middle line:90% 01:02:51.580 --> 01:02:55.030 align:middle line:84% I think we have just enough time to prove 01:02:55.030 --> 01:03:01.770 align:middle line:90% one more pretty good theorem. 01:03:01.770 --> 01:03:07.050 align:middle line:90% 01:03:07.050 --> 01:03:08.550 align:middle line:90% Pretty good-- it's very good. 01:03:08.550 --> 01:03:14.000 align:middle line:90% 01:03:14.000 --> 01:03:21.500 align:middle line:84% And then we'll use it in our next lecture 01:03:21.500 --> 01:03:25.130 align:middle line:84% to discuss, or at least conclude, 01:03:25.130 --> 01:03:27.545 align:middle line:84% some answers that we originally asked about power series. 01:03:27.545 --> 01:03:32.090 align:middle line:90% 01:03:32.090 --> 01:03:37.090 align:middle line:84% So this theorem is due to the godfather, Weierstrass. 01:03:37.090 --> 01:03:40.372 align:middle line:84% I don't know, did I refer to him as godfather or Riemann 01:03:40.372 --> 01:03:41.080 align:middle line:90% as the godfather? 01:03:41.080 --> 01:03:42.970 align:middle line:84% Anyway, I think it was Weierstrass 01:03:42.970 --> 01:03:45.670 align:middle line:84% that I refer to as the godfather. 01:03:45.670 --> 01:03:47.110 align:middle line:84% He proved the following theorem-- 01:03:47.110 --> 01:03:50.320 align:middle line:90% 01:03:50.320 --> 01:03:53.950 align:middle line:84% when can you guarantee uniform convergence 01:03:53.950 --> 01:03:55.480 align:middle line:84% for at least a series of functions, 01:03:55.480 --> 01:04:00.340 align:middle line:84% which are limits of partial sums, so limits of functions? 01:04:00.340 --> 01:04:04.540 align:middle line:84% And he proved this basically so he could come up 01:04:04.540 --> 01:04:09.190 align:middle line:84% with a whole host of examples of continuous functions 01:04:09.190 --> 01:04:12.825 align:middle line:84% which are nowhere differentiable. 01:04:12.825 --> 01:04:14.450 align:middle line:84% We kind of gave a proof of this theorem 01:04:14.450 --> 01:04:17.763 align:middle line:84% when we looked at that example, when 01:04:17.763 --> 01:04:19.430 align:middle line:84% we were talking about differentiability, 01:04:19.430 --> 01:04:24.320 align:middle line:84% but we didn't state it as a theorem there. 01:04:24.320 --> 01:04:34.340 align:middle line:84% So let's take a sequence of functions f j from S to R, 01:04:34.340 --> 01:04:50.170 align:middle line:84% and suppose there exists a sequence M j such that two 01:04:50.170 --> 01:04:51.040 align:middle line:90% things hold. 01:04:51.040 --> 01:04:53.080 align:middle line:84% So these are a sequence of positive numbers. 01:04:53.080 --> 01:05:02.520 align:middle line:90% 01:05:02.520 --> 01:05:04.350 align:middle line:84% I don't know why I just erased that, but-- 01:05:04.350 --> 01:05:08.070 align:middle line:84% so we have a sequence of functions from S to R 01:05:08.070 --> 01:05:12.660 align:middle line:84% and a sequence of positive numbers so that these numbers 01:05:12.660 --> 01:05:13.845 align:middle line:90% dominate the f j's. 01:05:13.845 --> 01:05:24.640 align:middle line:90% 01:05:24.640 --> 01:05:30.540 align:middle line:84% And 2, they're summable, so sum from j 01:05:30.540 --> 01:05:41.230 align:middle line:84% equals 1 to infinity, which is convergent. 01:05:41.230 --> 01:05:52.160 align:middle line:90% 01:05:52.160 --> 01:05:56.370 align:middle line:84% Then the conclusion-- and since I've been using numbers 01:05:56.370 --> 01:05:58.590 align:middle line:84% for conclusions and letters for hypotheses, 01:05:58.590 --> 01:05:59.730 align:middle line:90% let's go back to that-- 01:05:59.730 --> 01:06:02.380 align:middle line:90% then we have two conclusions. 01:06:02.380 --> 01:06:04.890 align:middle line:90% The first one is pretty obvious. 01:06:04.890 --> 01:06:17.560 align:middle line:84% For all x in S, this series where I just 01:06:17.560 --> 01:06:28.370 align:middle line:84% take x and stick it into f sub j, this converges absolutely. 01:06:28.370 --> 01:06:34.950 align:middle line:84% And the second is if I define, now, the function 01:06:34.950 --> 01:06:43.510 align:middle line:84% to be the sum of these series, then the partial sums converge 01:06:43.510 --> 01:06:44.590 align:middle line:90% to f uniformly-- 01:06:44.590 --> 01:06:54.380 align:middle line:90% 01:06:54.380 --> 01:07:10.970 align:middle line:84% converges to f uniformly as n goes to infinity on S. 01:07:10.970 --> 01:07:14.570 align:middle line:84% So if I have a sequence of functions, 01:07:14.570 --> 01:07:17.660 align:middle line:84% each one bounded by some number of M sub j, positive number-- 01:07:17.660 --> 01:07:21.080 align:middle line:84% or not negative number, at least-- 01:07:21.080 --> 01:07:26.690 align:middle line:84% and the series involving the M sub j's is convergent, then 01:07:26.690 --> 01:07:30.200 align:middle line:84% the series of the f sub j's converges uniformly. 01:07:30.200 --> 01:07:31.560 align:middle line:90% That's the conclusion. 01:07:31.560 --> 01:07:34.520 align:middle line:90% 01:07:34.520 --> 01:07:36.890 align:middle line:84% So in our example of that continuous, nowhere 01:07:36.890 --> 01:07:40.430 align:middle line:84% differentiable functions, each of these f sub j's 01:07:40.430 --> 01:07:51.140 align:middle line:84% were cosine of 160j x over 4 to the j. 01:07:51.140 --> 01:07:54.864 align:middle line:90% 01:07:54.864 --> 01:08:00.190 align:middle line:84% So before we prove this theorem, we 01:08:00.190 --> 01:08:02.020 align:middle line:84% can combine all of what we've done 01:08:02.020 --> 01:08:04.490 align:middle line:90% so far to state the following-- 01:08:04.490 --> 01:08:08.900 align:middle line:90% 01:08:08.900 --> 01:08:14.972 align:middle line:84% the function f of x equals sum from j equals 1 to infinity, 01:08:14.972 --> 01:08:16.430 align:middle line:84% and I could make it something else, 01:08:16.430 --> 01:08:29.240 align:middle line:84% but let's say sine of 40k over 2 to the k x is continuous on, 01:08:29.240 --> 01:08:31.460 align:middle line:90% let's say, 0, 1 for now. 01:08:31.460 --> 01:08:34.090 align:middle line:90% 01:08:34.090 --> 01:08:42.279 align:middle line:90% I could say on R. 01:08:42.279 --> 01:08:43.779 align:middle line:90% Why is that? 01:08:43.779 --> 01:08:46.524 align:middle line:90% Each of these f sub-- 01:08:46.524 --> 01:08:48.399 align:middle line:84% I'm really butchering this trying to go fast. 01:08:48.399 --> 01:08:54.689 align:middle line:90% 01:08:54.689 --> 01:08:58.010 align:middle line:90% This is 2 to the j, not 2. 01:08:58.010 --> 01:09:02.029 align:middle line:84% Each of these functions is bounded by 1 over 2 to the j, 01:09:02.029 --> 01:09:05.600 align:middle line:84% and 1 over 2 to the j is summable. 01:09:05.600 --> 01:09:10.279 align:middle line:84% So then by this theorem, this convergence, 01:09:10.279 --> 01:09:13.520 align:middle line:84% so this function here, is the uniform limit 01:09:13.520 --> 01:09:16.430 align:middle line:90% of the partial sums. 01:09:16.430 --> 01:09:21.140 align:middle line:84% The partial sums are just a finite sum involving sine, 01:09:21.140 --> 01:09:23.220 align:middle line:90% so they're continuous functions. 01:09:23.220 --> 01:09:26.450 align:middle line:84% So f is the uniform limit of continuous functions, 01:09:26.450 --> 01:09:30.000 align:middle line:90% and is, therefore, continuous. 01:09:30.000 --> 01:09:32.220 align:middle line:84% So we could have used this theorem 01:09:32.220 --> 01:09:35.729 align:middle line:84% to prove that function we looked at back in differentiability 01:09:35.729 --> 01:09:38.279 align:middle line:84% was continuous, but we didn't have 01:09:38.279 --> 01:09:40.770 align:middle line:84% uniform convergence and all that back then. 01:09:40.770 --> 01:09:42.060 align:middle line:90% So we just proved it by hand. 01:09:42.060 --> 01:09:49.120 align:middle line:84% But using this, this theorem gives you 01:09:49.120 --> 01:09:54.730 align:middle line:84% a big class of sequences of functions which converge 01:09:54.730 --> 01:09:56.065 align:middle line:90% uniformly to some function. 01:09:56.065 --> 01:10:01.630 align:middle line:90% 01:10:01.630 --> 01:10:05.440 align:middle line:90% So let's prove this quickly. 01:10:05.440 --> 01:10:24.220 align:middle line:84% So one follows from the assumption a, b, 01:10:24.220 --> 01:10:25.300 align:middle line:90% and the comparison test. 01:10:25.300 --> 01:10:28.650 align:middle line:90% 01:10:28.650 --> 01:10:34.430 align:middle line:84% If I take an x in S f sub j of x is bounded by M sub j, 01:10:34.430 --> 01:10:39.570 align:middle line:84% and M sub j is summable, this series converges. 01:10:39.570 --> 01:10:43.730 align:middle line:84% So by the comparison test, the sum, 01:10:43.730 --> 01:10:48.160 align:middle line:84% with absolute values here for each fixed x, converges. 01:10:48.160 --> 01:10:49.645 align:middle line:90% And that's absolute convergence. 01:10:49.645 --> 01:10:55.400 align:middle line:90% 01:10:55.400 --> 01:10:58.430 align:middle line:84% Now, let's show that the partial sums converge 01:10:58.430 --> 01:11:00.170 align:middle line:90% to the limit uniformly. 01:11:00.170 --> 01:11:03.910 align:middle line:90% 01:11:03.910 --> 01:11:07.570 align:middle line:90% So let epsilon be positive. 01:11:07.570 --> 01:11:11.380 align:middle line:90% 01:11:11.380 --> 01:11:23.330 align:middle line:84% Since this sum converges, there exists an M, 01:11:23.330 --> 01:11:25.610 align:middle line:84% a natural number so this-- remember, 01:11:25.610 --> 01:11:27.395 align:middle line:84% we have this Cauchy criterion for-- 01:11:27.395 --> 01:11:31.278 align:middle line:90% 01:11:31.278 --> 01:11:32.820 align:middle line:84% well, I guess we don't have to have-- 01:11:32.820 --> 01:11:43.340 align:middle line:84% so that sum from n equals M plus 1 to infinity of M 01:11:43.340 --> 01:11:48.230 align:middle line:90% sub j, so j, which is equal to-- 01:11:48.230 --> 01:12:00.270 align:middle line:90% 01:12:00.270 --> 01:12:03.960 align:middle line:90% I chose a poor letter. 01:12:03.960 --> 01:12:15.370 align:middle line:84% So this M here should not-- let's change this to an N. 01:12:15.370 --> 01:12:17.030 align:middle line:90% So this is less than epsilon. 01:12:17.030 --> 01:12:18.790 align:middle line:84% So the point is that the tail is small. 01:12:18.790 --> 01:12:22.180 align:middle line:84% The sum from j equals n plus 1 to infinity of M sub j 01:12:22.180 --> 01:12:23.095 align:middle line:90% is less than epsilon. 01:12:23.095 --> 01:12:31.400 align:middle line:90% 01:12:31.400 --> 01:12:35.020 align:middle line:84% And for all n bigger than or equal to this capital N, 01:12:35.020 --> 01:12:39.520 align:middle line:84% for all x in S, if I look at the limit, which 01:12:39.520 --> 01:12:44.500 align:middle line:84% is f of x, this is sum of the series minus sum from j 01:12:44.500 --> 01:12:51.080 align:middle line:84% equals 1 to n, this is equal to-- 01:12:51.080 --> 01:12:53.010 align:middle line:84% now, f of x is equal to the whole sum, 01:12:53.010 --> 01:12:54.470 align:middle line:84% so minus this first part, this is 01:12:54.470 --> 01:12:57.290 align:middle line:84% equal to sum from j equals n plus 1 01:12:57.290 --> 01:13:02.370 align:middle line:90% to infinity of f of j of x. 01:13:02.370 --> 01:13:04.680 align:middle line:84% And by the triangle inequality, which 01:13:04.680 --> 01:13:06.300 align:middle line:84% does hold for a convergent series, 01:13:06.300 --> 01:13:10.320 align:middle line:84% this is less than or equal to sum from j 01:13:10.320 --> 01:13:29.890 align:middle line:84% equals n plus 1 to infinity of f sub j of x. 01:13:29.890 --> 01:13:37.430 align:middle line:84% And by assumption a, this is less than 01:13:37.430 --> 01:13:43.280 align:middle line:84% or equal to sum from j equals n plus 1 to infinity of M sub j. 01:13:43.280 --> 01:13:46.100 align:middle line:84% And since n is bigger than or equal to capital N, 01:13:46.100 --> 01:13:49.970 align:middle line:84% I have this is less than or equal 01:13:49.970 --> 01:13:55.460 align:middle line:84% to sum from j equals capital N plus 1 to infinity of M sub j. 01:13:55.460 --> 01:13:59.120 align:middle line:84% And this thing, we chose capital N 01:13:59.120 --> 01:14:01.748 align:middle line:84% so that that's less than epsilon. 01:14:01.748 --> 01:14:02.540 align:middle line:90% And that's the end. 01:14:02.540 --> 01:14:07.440 align:middle line:84% So again, this n was chosen depending only on the series. 01:14:07.440 --> 01:14:11.240 align:middle line:84% It didn't depend on any x at any-- 01:14:11.240 --> 01:14:14.928 align:middle line:84% it didn't depend on the point x in S. 01:14:14.928 --> 01:14:16.720 align:middle line:84% So I had to go through that kind of quickly 01:14:16.720 --> 01:14:18.610 align:middle line:84% because I'm up against the time crunch. 01:14:18.610 --> 01:14:23.470 align:middle line:84% But the point is, for this theorem, 01:14:23.470 --> 01:14:28.240 align:middle line:84% is that if I have a sequence of functions that are bounded 01:14:28.240 --> 01:14:32.365 align:middle line:90% by these numbers M sub j-- 01:14:32.365 --> 01:14:39.220 align:middle line:84% I was going so fast I didn't even label it correctly-- 01:14:39.220 --> 01:14:42.790 align:middle line:84% this is called the Weierstrass M test, 01:14:42.790 --> 01:14:46.520 align:middle line:90% M because you see M here-- 01:14:46.520 --> 01:14:48.470 align:middle line:84% if I have a sequence of functions bounded 01:14:48.470 --> 01:14:52.010 align:middle line:84% by some numbers M sub j, And those M sub j's are summable, 01:14:52.010 --> 01:14:57.530 align:middle line:84% then what it says is that the series of the functions 01:14:57.530 --> 01:14:58.790 align:middle line:90% converges uniformly. 01:14:58.790 --> 01:15:01.680 align:middle line:90% 01:15:01.680 --> 01:15:04.010 align:middle line:90% So we'll stop there. 01:15:04.010 --> 01:15:10.000 align:middle line:90%