
Title: Uniform Convergence

Series: Real Analysis

Chapter: Continuous Functions

YouTubeTitle: Real Analysis 25  Uniform Convergence

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Quiz: Test your knowledge

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Subtitle on GitHub: ra25_sub_eng.srt

Timestamps
00:00 Intro
00:44 Uniform convergence
02:52 Measuring distance between functions
04:30 Uniform convergence using supremum norm
05:06 Example
06:28 Uniform convergence is stronger than pointwise convergence
07:30 Credits

Subtitle in English
1 00:00:00,389 –> 00:00:02,170 Hello and welcome to
2 00:00:02,180 –> 00:00:03,500 Real Analysis part
3 00:00:03,509 –> 00:00:04,530 25.
4 00:00:05,059 –> 00:00:06,409 And first, I want to thank
5 00:00:06,420 –> 00:00:07,809 all the nice supporters on
6 00:00:07,820 –> 00:00:09,170 Steady or paypal.
7 00:00:09,939 –> 00:00:10,800 In this video.
8 00:00:10,810 –> 00:00:12,180 Today, we will talk about
9 00:00:12,189 –> 00:00:13,930 the uniform convergence of
10 00:00:13,939 –> 00:00:15,359 a sequence of functions.
11 00:00:16,100 –> 00:00:17,379 In order to understand this,
12 00:00:17,389 –> 00:00:19,229 please first recall that
13 00:00:19,239 –> 00:00:21,079 we call a sequence of functions
14 00:00:21,090 –> 00:00:22,850 pointwisely convergent to
15 00:00:22,860 –> 00:00:24,829 another function F
16 00:00:24,840 –> 00:00:26,670 if for all points X tilde,
17 00:00:26,709 –> 00:00:28,600 we have the convergence of
18 00:00:28,610 –> 00:00:30,100 the values at this point.
19 00:00:30,750 –> 00:00:32,200 This means that in this
20 00:00:32,209 –> 00:00:34,150 formulation with the quantifiers,
21 00:00:34,200 –> 00:00:35,599 you are allowed to choose
22 00:00:35,610 –> 00:00:37,450 different capital NS depending
23 00:00:37,459 –> 00:00:38,479 on X tilde.
24 00:00:39,029 –> 00:00:40,759 And indeed, this is the only
25 00:00:40,770 –> 00:00:42,180 thing we will change to get
26 00:00:42,189 –> 00:00:43,639 the uniform convergence.
27 00:00:44,319 –> 00:00:46,000 So we call a sequence of
28 00:00:46,009 –> 00:00:47,830 functions uniformly
29 00:00:47,840 –> 00:00:49,000 convergent to F
30 00:00:49,689 –> 00:00:51,240 if for all epsilon
31 00:00:51,250 –> 00:00:53,150 greater than zero, there
32 00:00:53,159 –> 00:00:55,110 exists a capital N such that
33 00:00:55,119 –> 00:00:56,819 for all indices N
34 00:00:56,830 –> 00:00:58,430 greater or equal than N
35 00:00:58,909 –> 00:01:00,869 and for all points x tilde
36 00:01:00,880 –> 00:01:02,869 in I, we have that
37 00:01:02,880 –> 00:01:03,990 the distance between the
38 00:01:04,000 –> 00:01:05,510 values is less than
39 00:01:05,519 –> 00:01:06,230 epsilon.
40 00:01:06,919 –> 00:01:08,699 So you see this quantifier
41 00:01:08,709 –> 00:01:10,699 from before is pushed to
42 00:01:10,709 –> 00:01:12,459 the right, of course.
43 00:01:12,470 –> 00:01:13,959 Therefore, this looks similar.
44 00:01:13,970 –> 00:01:15,559 But now this capital
45 00:01:15,569 –> 00:01:17,400 N works uniformly
46 00:01:17,410 –> 00:01:19,150 for all points X tilde.
47 00:01:19,669 –> 00:01:20,379 Of course.
48 00:01:20,389 –> 00:01:22,300 Therefore, this looks similarly.
49 00:01:22,309 –> 00:01:23,940 But now this capital N
50 00:01:23,949 –> 00:01:25,839 here works uniformly
51 00:01:25,849 –> 00:01:27,709 for all points X tilde in
52 00:01:27,720 –> 00:01:29,650 I that’s different from
53 00:01:29,660 –> 00:01:31,519 before because here you
54 00:01:31,529 –> 00:01:32,860 could choose capital N
55 00:01:32,870 –> 00:01:34,510 depending on X tilde.
56 00:01:35,080 –> 00:01:36,180 So there could be different
57 00:01:36,190 –> 00:01:37,389 capital Ns here.
58 00:01:37,739 –> 00:01:39,110 However, for the uniform
59 00:01:39,120 –> 00:01:40,870 convergence, this N here
60 00:01:40,879 –> 00:01:42,760 has to work for all x tilde
61 00:01:42,769 –> 00:01:43,809 at the same time.
62 00:01:44,319 –> 00:01:46,290 And exactly this is the reason
63 00:01:46,300 –> 00:01:47,709 we call it the uniform
64 00:01:47,720 –> 00:01:48,489 convergence.
65 00:01:48,900 –> 00:01:50,389 And please keep in mind this
66 00:01:50,400 –> 00:01:52,220 new order for the quantifier
67 00:01:52,230 –> 00:01:53,419 changes a lot.
68 00:01:54,000 –> 00:01:55,489 Let’s visualize that in a
69 00:01:55,500 –> 00:01:56,199 picture.
70 00:01:56,709 –> 00:01:58,069 So here you see the graph
71 00:01:58,080 –> 00:01:59,709 of the function F which is
72 00:01:59,720 –> 00:02:00,830 our limit function.
73 00:02:01,400 –> 00:02:03,139 Now if you fix one point
74 00:02:03,150 –> 00:02:04,919 X tilde on the X axis,
75 00:02:04,980 –> 00:02:06,830 this inequality here tells
76 00:02:06,839 –> 00:02:08,830 us that the, value f_n(x tilde)
77 00:02:08,839 –> 00:02:10,809 lies somewhere in this
78 00:02:10,820 –> 00:02:11,339 region.
79 00:02:11,979 –> 00:02:13,839 So we have our epsilon here
80 00:02:13,850 –> 00:02:14,600 and here.
81 00:02:15,199 –> 00:02:17,139 However, we also know this
82 00:02:17,149 –> 00:02:18,869 inequality here should hold
83 00:02:18,880 –> 00:02:20,289 for all X tilde.
84 00:02:21,000 –> 00:02:22,660 Hence we have to move this
85 00:02:22,669 –> 00:02:24,610 region here through all the
86 00:02:24,619 –> 00:02:25,339 points.
87 00:02:25,929 –> 00:02:27,580 In other words, what we get
88 00:02:27,589 –> 00:02:29,130 is such an epsilon tube
89 00:02:29,139 –> 00:02:30,729 around the graph of the
90 00:02:30,740 –> 00:02:31,639 function F.
91 00:02:32,350 –> 00:02:33,899 So the graph of the function
92 00:02:33,910 –> 00:02:35,690 FN needs to lie
93 00:02:35,699 –> 00:02:37,419 inside this epsilon tube.
94 00:02:38,039 –> 00:02:39,949 And of course, also the graphs
95 00:02:39,960 –> 00:02:41,690 of the functions in the sequence
96 00:02:41,699 –> 00:02:43,240 that come after this capital
97 00:02:43,270 –> 00:02:43,649 N.
98 00:02:43,759 –> 00:02:45,429 This is exactly what the
99 00:02:45,440 –> 00:02:47,259 uniform convergence tells
100 00:02:47,270 –> 00:02:47,669 us.
101 00:02:48,210 –> 00:02:49,789 So you immediately see this
102 00:02:49,800 –> 00:02:51,360 is more than we had for the
103 00:02:51,369 –> 00:02:52,649 pointwise convergence.
104 00:02:53,419 –> 00:02:55,179 And also this picture tells
105 00:02:55,190 –> 00:02:56,800 us that it should be possible
106 00:02:56,809 –> 00:02:58,509 to measure a distance between
107 00:02:58,520 –> 00:02:59,360 two functions.
108 00:03:00,009 –> 00:03:01,350 By having this distance.
109 00:03:01,360 –> 00:03:02,869 You can see that the uniform
110 00:03:02,880 –> 00:03:04,830 convergence is indeed very
111 00:03:04,839 –> 00:03:06,669 natural for this,
112 00:03:06,679 –> 00:03:07,990 we first should look at the
113 00:03:08,000 –> 00:03:09,839 picture again, let’s
114 00:03:09,850 –> 00:03:11,699 take two functions and their
115 00:03:11,710 –> 00:03:12,309 graphs.
116 00:03:12,800 –> 00:03:14,669 Now the question is how can
117 00:03:14,679 –> 00:03:16,320 we measure a suitable distance
118 00:03:16,330 –> 00:03:17,610 between both functions?
119 00:03:17,669 –> 00:03:19,490 Or in other words, when would
120 00:03:19,500 –> 00:03:20,860 you say that the functions
121 00:03:20,869 –> 00:03:22,369 are close to each other?
122 00:03:22,800 –> 00:03:24,250 Of course, what you already
123 00:03:24,259 –> 00:03:25,830 know is that for each point
124 00:03:25,839 –> 00:03:27,559 X tilde, we can measure the
125 00:03:27,570 –> 00:03:29,020 distance of the values.
126 00:03:29,710 –> 00:03:31,130 So what you have to calculate
127 00:03:31,139 –> 00:03:32,839 is the absolute value as
128 00:03:32,850 –> 00:03:33,529 before.
129 00:03:34,020 –> 00:03:35,220 So if we call the point,
130 00:03:35,229 –> 00:03:37,220 just X, we have F of X
131 00:03:37,229 –> 00:03:39,039 minus G of X and then
132 00:03:39,050 –> 00:03:40,080 the absolute value.
133 00:03:40,509 –> 00:03:41,970 And of course, as before,
134 00:03:41,979 –> 00:03:43,399 we could do that for all
135 00:03:43,410 –> 00:03:45,309 the points X form the domain
136 00:03:45,320 –> 00:03:47,169 I there, you see,
137 00:03:47,179 –> 00:03:48,960 we find small distances
138 00:03:48,970 –> 00:03:50,869 but also larger ones.
139 00:03:51,399 –> 00:03:53,050 And indeed the largest one
140 00:03:53,059 –> 00:03:54,710 we would find if we take
141 00:03:54,720 –> 00:03:56,559 the maximum of these numbers
142 00:03:56,570 –> 00:03:56,929 here.
143 00:03:57,460 –> 00:03:58,830 However, because we don’t
144 00:03:58,839 –> 00:04:00,410 know if the maximum really
145 00:04:00,419 –> 00:04:02,080 exists, we should take the
146 00:04:02,089 –> 00:04:02,809 supremum.
147 00:04:03,449 –> 00:04:05,320 Hence this number here tells
148 00:04:05,330 –> 00:04:07,259 us in fact how close
149 00:04:07,270 –> 00:04:08,460 the two graphs are.
150 00:04:08,720 –> 00:04:10,350 So we have a distance measure
151 00:04:10,360 –> 00:04:11,550 for two functions
152 00:04:12,240 –> 00:04:13,899 and usually we just call
153 00:04:13,910 –> 00:04:15,419 it the supremum norm.
154 00:04:15,970 –> 00:04:17,238 The common short notation
155 00:04:17,250 –> 00:04:18,899 for this would be two bars
156 00:04:18,910 –> 00:04:20,700 left and right, and the
157 00:04:20,709 –> 00:04:22,100 infinity symbol as an
158 00:04:22,108 –> 00:04:22,739 index.
159 00:04:23,290 –> 00:04:24,910 And then we have F minus
160 00:04:24,920 –> 00:04:25,980 G inside.
161 00:04:26,640 –> 00:04:28,320 So this is the supremum norm
162 00:04:28,329 –> 00:04:29,899 of F minus G.
163 00:04:30,640 –> 00:04:32,220 Now, as promised with this,
164 00:04:32,230 –> 00:04:34,000 we can rewrite the uniform
165 00:04:34,010 –> 00:04:35,220 convergence definition.
166 00:04:36,019 –> 00:04:37,339 It simply means that the
167 00:04:37,350 –> 00:04:39,100 supremum norm gives us a
168 00:04:39,109 –> 00:04:40,920 sequence that goes to zero.
169 00:04:41,529 –> 00:04:43,230 So FN minus the limit
170 00:04:43,239 –> 00:04:44,760 function F measured in the
171 00:04:44,769 –> 00:04:46,209 supremum norm goes to
172 00:04:46,220 –> 00:04:47,799 zero when N goes to
173 00:04:47,809 –> 00:04:48,559 infinity.
174 00:04:49,019 –> 00:04:50,679 So we started with a sequence
175 00:04:50,690 –> 00:04:51,619 of functions.
176 00:04:51,630 –> 00:04:53,279 But what we got here is an
177 00:04:53,290 –> 00:04:54,760 ordinary sequence of
178 00:04:54,769 –> 00:04:55,470 numbers.
179 00:04:56,179 –> 00:04:57,839 Therefore, this is also the
180 00:04:57,850 –> 00:04:59,500 ordinary convergence for
181 00:04:59,510 –> 00:05:01,160 a sequence of real numbers.
182 00:05:01,549 –> 00:05:01,920 OK.
183 00:05:01,929 –> 00:05:03,200 Now, after all these
184 00:05:03,209 –> 00:05:04,829 definitions, let’s look at
185 00:05:04,839 –> 00:05:05,920 an example.
186 00:05:06,450 –> 00:05:07,929 So maybe we should take one,
187 00:05:07,940 –> 00:05:09,440 we already discussed in the
188 00:05:09,450 –> 00:05:11,339 last video there we had
189 00:05:11,350 –> 00:05:12,779 some simple functions that
190 00:05:12,790 –> 00:05:14,570 got steeper and steeper here
191 00:05:14,579 –> 00:05:16,149 until the pointwise limit
192 00:05:16,160 –> 00:05:17,440 was this function.
193 00:05:18,130 –> 00:05:19,429 We want to calculate a little
194 00:05:19,440 –> 00:05:19,750 bit.
195 00:05:19,760 –> 00:05:20,959 Therefore, let’s say this
196 00:05:20,970 –> 00:05:22,429 value here is minus
197 00:05:22,440 –> 00:05:24,190 one and the one above is
198 00:05:24,200 –> 00:05:24,589 one.
199 00:05:25,299 –> 00:05:26,760 What we will see now is that
200 00:05:26,769 –> 00:05:28,619 we have the pointwise convergence
201 00:05:28,630 –> 00:05:30,230 but not the uniform
202 00:05:30,239 –> 00:05:31,000 convergence.
203 00:05:31,750 –> 00:05:32,790 Therefore, what we should
204 00:05:32,799 –> 00:05:34,540 look at is the distance between
205 00:05:34,549 –> 00:05:36,369 FN and F in the
206 00:05:36,380 –> 00:05:37,269 supremum norm.
207 00:05:37,799 –> 00:05:39,290 Indeed, this is what we can
208 00:05:39,299 –> 00:05:40,750 immediately calculate when
209 00:05:40,760 –> 00:05:41,970 we look at the jump.
210 00:05:42,649 –> 00:05:44,230 So here we have the jump
211 00:05:44,239 –> 00:05:45,869 of our limit function F.
212 00:05:46,559 –> 00:05:48,190 Now it does not matter which
213 00:05:48,200 –> 00:05:49,260 of the functions from our
214 00:05:49,269 –> 00:05:50,609 sequence we choose
215 00:05:50,929 –> 00:05:52,459 because all of them are
216 00:05:52,470 –> 00:05:53,709 connected from bottom to
217 00:05:53,720 –> 00:05:54,190 top.
218 00:05:54,890 –> 00:05:56,410 Hence, you can always just
219 00:05:56,420 –> 00:05:57,880 take the middle point here
220 00:05:57,959 –> 00:05:59,190 and measure the distance
221 00:05:59,200 –> 00:06:00,190 to the limit function,
222 00:06:01,000 –> 00:06:02,649 which is by our given values
223 00:06:02,660 –> 00:06:04,260 here, always one.
224 00:06:04,690 –> 00:06:06,130 Hence, we don’t need to calculate
225 00:06:06,140 –> 00:06:06,820 anything.
226 00:06:06,829 –> 00:06:08,640 We immediately know the supremum
227 00:06:08,649 –> 00:06:10,559 norm is always greater or
228 00:06:10,570 –> 00:06:11,779 equal than one.
229 00:06:12,549 –> 00:06:13,850 And we don’t need anything
230 00:06:13,859 –> 00:06:15,549 more because we immediately
231 00:06:15,559 –> 00:06:17,279 know we can’t have this
232 00:06:17,290 –> 00:06:18,679 converted to zero
233 00:06:18,890 –> 00:06:20,649 because this distance does
234 00:06:20,660 –> 00:06:22,290 not get smaller than one.
235 00:06:22,980 –> 00:06:24,519 Now, this simple example,
236 00:06:24,529 –> 00:06:26,220 without much calculation
237 00:06:26,260 –> 00:06:28,119 directly gives us the following
238 00:06:28,130 –> 00:06:28,760 result.
239 00:06:29,470 –> 00:06:31,220 Having the pointwise convergence
240 00:06:31,230 –> 00:06:32,890 does not tell us that we
241 00:06:32,899 –> 00:06:34,329 also have the uniform
242 00:06:34,339 –> 00:06:35,089 convergence.
243 00:06:35,549 –> 00:06:36,890 However, it’s also a nice
244 00:06:36,899 –> 00:06:38,579 result and not hard to show
245 00:06:38,649 –> 00:06:40,619 that we have the other implication.
246 00:06:41,420 –> 00:06:43,160 Hence, the uniform convergence
247 00:06:43,170 –> 00:06:44,839 is indeed stronger than the
248 00:06:44,850 –> 00:06:46,200 pointwise convergence.
249 00:06:46,730 –> 00:06:48,100 And for us, the uniform
250 00:06:48,109 –> 00:06:49,500 convergence will be very
251 00:06:49,510 –> 00:06:51,250 important later because it
252 00:06:51,260 –> 00:06:53,029 conserves nice properties
253 00:06:53,040 –> 00:06:53,920 of functions.
254 00:06:54,429 –> 00:06:55,829 For example, it conserves
255 00:06:55,839 –> 00:06:57,480 continuity we will
256 00:06:57,489 –> 00:06:58,480 define later.
257 00:06:59,309 –> 00:07:00,579 That’s something that just
258 00:07:00,589 –> 00:07:02,170 a pointwise convergence can’t
259 00:07:02,179 –> 00:07:02,570 do.
260 00:07:03,010 –> 00:07:04,489 However, one property you
261 00:07:04,500 –> 00:07:06,420 already know is the bounded
262 00:07:06,429 –> 00:07:07,309 of functions.
263 00:07:07,750 –> 00:07:09,339 And the uniform convergence
264 00:07:09,350 –> 00:07:10,839 also conserves this
265 00:07:10,850 –> 00:07:11,589 property.
266 00:07:11,910 –> 00:07:13,480 This means if you have a
267 00:07:13,489 –> 00:07:15,380 sequence of bounded functions
268 00:07:15,489 –> 00:07:16,609 and they converge
269 00:07:16,619 –> 00:07:18,299 uniformly, then the
270 00:07:18,309 –> 00:07:20,170 limit function is also
271 00:07:20,179 –> 00:07:20,750 bounded.
272 00:07:21,429 –> 00:07:21,890 OK.
273 00:07:21,899 –> 00:07:23,309 All details about these
274 00:07:23,320 –> 00:07:25,000 properties we will discuss
275 00:07:25,010 –> 00:07:26,290 in the next videos.
276 00:07:26,769 –> 00:07:27,959 Therefore, I hope I see you
277 00:07:27,970 –> 00:07:29,489 there and have a nice day.
278 00:07:29,630 –> 00:07:30,359 Bye.

Quiz Content
Q1: Let $(f_1, f_2, f_3, \ldots)$ be a sequence of functions $f_n: I \rightarrow \mathbb{R}$. What is the correct definition for the uniform convergence to a function $f: I \rightarrow \mathbb{R}$?
A1: $ \displaystyle \forall x \in I ~~ \forall \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall n \leq N ~ : ~ f_n(x)f(x)<\varepsilon $
A2: $ \displaystyle \forall \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall n \geq N ~~ \forall x \in I ~ : ~ f_n(x)f(x)<\varepsilon $
A3: $ \displaystyle \forall x \in I ~~ \forall \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall n \geq N ~ : ~ f_n(x)f(x)<\varepsilon $
A4: $ \displaystyle \forall \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall x \in I ~~ \forall n \leq N ~ : ~ f_n(x)f(x)<\varepsilon $
Q2: Let $f$, $g$ be two functions $f,g: [0,1] \rightarrow \mathbb{R}$ given by $f(x) = x1 $ and $g(x) = x^2$. What is the distance between both functions $ f  g _{\infty}$?
A1: $ f  g _{\infty} = 1 $
A2: $ f  g _{\infty} = 2 $
A3: $ f  g _{\infty} = 1 $
A4: $ f  g _{\infty} = \sqrt{2} $
A5: $ f  g _{\infty} = 0 $
Q3: Let $(f_1, f_2, f_3, \ldots)$ be a sequence of functions $f_n: [0,1] \rightarrow \mathbb{R}$. Which implication is correct?
A1: If it is pointwisely convergent, then also uniformly convergent.
A2: If it is uniformly convergent, then also pointwisely convergent.
Q4: Let $(f_1, f_2, f_3, \ldots)$ be a sequence of functions $f_n: [0,1] \rightarrow \mathbb{R}$ given by $f_n(x) = \frac{x}{n^2}  2  \frac{1}{n} $. Is the sequence uniformly convergent?
A1: No, it is also not pointwisely convergent.
A2: No, but it is pointwisely convergent.
A3: Yes!