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Title: Bounded Operators
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Series: Functional Analysis
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YouTube-Title: Functional Analysis 13 | Bounded Operators
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Bright video: https://youtu.be/442PxdU35q4
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Quiz: Test your knowledge
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Subtitle on GitHub: fa13_sub_eng.srt
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Timestamps
00:00 Introduction
01:40 Definition - bounded operator
04:35 Proposition - continuous equivalent to bounded
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Subtitle in English
1 00:00:00,370 –> 00:00:02,190 Hello and welcome back to
2 00:00:02,200 –> 00:00:03,400 functional analysis
3 00:00:03,480 –> 00:00:05,070 and as always, many thanks
4 00:00:05,079 –> 00:00:06,480 to all the nice people that
5 00:00:06,489 –> 00:00:08,148 support this channel on Steady
6 00:00:08,159 –> 00:00:08,949 or PayPal.
7 00:00:09,310 –> 00:00:11,210 Today in part 13, we
8 00:00:11,220 –> 00:00:12,670 talk about operators
9 00:00:12,680 –> 00:00:14,550 between normed spaces.
10 00:00:15,319 –> 00:00:16,309 So the picture should look
11 00:00:16,318 –> 00:00:16,930 like this.
12 00:00:16,940 –> 00:00:18,469 We have one normed space on
13 00:00:18,479 –> 00:00:19,879 the left-hand side and one
14 00:00:19,889 –> 00:00:20,899 on the right-hand side.
15 00:00:21,469 –> 00:00:23,250 Now an operator T is just
16 00:00:23,260 –> 00:00:25,010 a map that conserves some
17 00:00:25,020 –> 00:00:26,450 structures of our spaces.
18 00:00:27,200 –> 00:00:29,190 We don’t call T a function,
19 00:00:29,200 –> 00:00:30,700 but an operator. Simply
20 00:00:30,709 –> 00:00:32,549 because often we have a
21 00:00:32,560 –> 00:00:34,330 space of functions as the domain
22 00:00:34,340 –> 00:00:35,290 or the codomain.
23 00:00:36,130 –> 00:00:37,479 So every time you see the
24 00:00:37,490 –> 00:00:39,229 notion operator, you know,
25 00:00:39,240 –> 00:00:40,770 it’s just another name for
26 00:00:40,779 –> 00:00:41,639 a special map.
27 00:00:42,340 –> 00:00:43,939 Now the first property that
28 00:00:43,950 –> 00:00:45,560 T should conserve is the
29 00:00:45,569 –> 00:00:47,130 algebraic structure. The
30 00:00:47,139 –> 00:00:48,790 linear structure given in
31 00:00:48,799 –> 00:00:49,639 the vector space.
32 00:00:50,409 –> 00:00:52,090 In other words, the map should
33 00:00:52,099 –> 00:00:53,330 be a linear map.
34 00:00:54,169 –> 00:00:54,470 OK.
35 00:00:54,479 –> 00:00:56,119 There we have one half of
36 00:00:56,130 –> 00:00:57,490 our functional analytical
37 00:00:57,500 –> 00:00:59,220 world and the other half
38 00:00:59,229 –> 00:01:00,549 should be the topological
39 00:01:00,560 –> 00:01:01,069 structure.
40 00:01:01,709 –> 00:01:02,970 This is what we already know,
41 00:01:02,979 –> 00:01:04,638 because in a metric space,
42 00:01:04,650 –> 00:01:06,169 we have a notion of open
43 00:01:06,180 –> 00:01:06,650 sets
44 00:01:07,339 –> 00:01:09,089 and the property that conserves
45 00:01:09,099 –> 00:01:10,830 these open sets is what we
46 00:01:10,839 –> 00:01:12,080 learned in the last video
47 00:01:12,209 –> 00:01:13,959 and is called continuity.
48 00:01:14,870 –> 00:01:16,120 Of course, this seems a little
49 00:01:16,129 –> 00:01:16,940 bit abstract,
50 00:01:16,949 –> 00:01:18,300 but we will now see that
51 00:01:18,309 –> 00:01:20,150 for normed spaces as we have
52 00:01:20,160 –> 00:01:21,739 it here, we can define
53 00:01:21,750 –> 00:01:23,459 another property we call
54 00:01:23,470 –> 00:01:24,099 bounded
55 00:01:24,849 –> 00:01:26,510 and we will see that together
56 00:01:26,519 –> 00:01:27,870 with the linearity, this
57 00:01:27,879 –> 00:01:29,489 is indeed equivalent to the
58 00:01:29,500 –> 00:01:31,389 continuity. In this
59 00:01:31,400 –> 00:01:33,029 sense one often speaks of
60 00:01:33,040 –> 00:01:34,870 linear bounded operators
61 00:01:34,879 –> 00:01:36,709 between two normed spaces.
62 00:01:37,610 –> 00:01:38,099 OK.
63 00:01:38,110 –> 00:01:39,260 Then let’s start with the
64 00:01:39,269 –> 00:01:40,660 definition of the notion
65 00:01:40,669 –> 00:01:42,610 bounded. As before we
66 00:01:42,620 –> 00:01:44,339 have two normed spaces and
67 00:01:44,349 –> 00:01:45,519 a linear map T.
68 00:01:46,379 –> 00:01:47,779 Of course, linear always
69 00:01:47,790 –> 00:01:49,349 means it conserves the vector
70 00:01:49,360 –> 00:01:50,669 addition and the scalar
71 00:01:50,680 –> 00:01:51,559 multiplication.
72 00:01:51,989 –> 00:01:53,330 This means that we have these
73 00:01:53,339 –> 00:01:54,169 two formulas
74 00:01:54,180 –> 00:01:55,580 and maybe you also recognize
75 00:01:55,589 –> 00:01:57,180 another common notation here.
76 00:01:57,529 –> 00:01:58,870 For linear maps, we will
77 00:01:58,879 –> 00:02:00,500 omit parentheses, if
78 00:02:00,510 –> 00:02:01,139 possible.
79 00:02:01,720 –> 00:02:03,180 Now, what we want is the
80 00:02:03,190 –> 00:02:05,080 length of this linear map.
81 00:02:05,089 –> 00:02:06,639 So we want a norm for the
82 00:02:06,650 –> 00:02:07,300 operator.
83 00:02:07,949 –> 00:02:09,300 Hence this is then often
84 00:02:09,309 –> 00:02:10,960 called just the operator
85 00:02:10,970 –> 00:02:11,740 norm of T
86 00:02:12,330 –> 00:02:13,809 and if we want to emphasize
87 00:02:13,820 –> 00:02:15,169 the spaces the map acts
88 00:02:15,179 –> 00:02:17,089 on, we have to put them in
89 00:02:17,100 –> 00:02:17,850 the index.
90 00:02:18,440 –> 00:02:19,490 However, most of the time
91 00:02:19,500 –> 00:02:21,240 we will omit them there, because
92 00:02:21,250 –> 00:02:22,800 the corresponding normed spaces
93 00:02:22,809 –> 00:02:24,250 are known from the beginning.
94 00:02:24,899 –> 00:02:25,330 OK,
95 00:02:25,339 –> 00:02:26,490 but we still don’t know how
96 00:02:26,500 –> 00:02:27,520 to measure this length.
97 00:02:27,550 –> 00:02:29,240 So let’s start with an idea.
98 00:02:29,970 –> 00:02:31,130 Now, imagine we have our
99 00:02:31,139 –> 00:02:32,429 normed space X on the left-
100 00:02:32,440 –> 00:02:34,380 hand side and Y on the right-
101 00:02:34,389 –> 00:02:36,139 hand side. Let’s pick a
102 00:02:36,149 –> 00:02:36,940 vector x.
103 00:02:36,949 –> 00:02:38,330 and we know we can measure
104 00:02:38,339 –> 00:02:39,509 the length of x
105 00:02:39,949 –> 00:02:41,880 and now we know the map T
106 00:02:41,889 –> 00:02:43,570 acting on this vector will
107 00:02:43,580 –> 00:02:44,639 give us a new vector
108 00:02:44,649 –> 00:02:45,190 in Y.
109 00:02:45,990 –> 00:02:47,130 So here we have the vector
110 00:02:47,139 –> 00:02:48,850 Tx with length
111 00:02:48,860 –> 00:02:50,160 norm of Tx.
112 00:02:50,750 –> 00:02:51,929 Of course, here we have to
113 00:02:51,940 –> 00:02:53,139 measure the length with the
114 00:02:53,149 –> 00:02:54,070 norm in Y
115 00:02:54,210 –> 00:02:55,449 and here we have to measure
116 00:02:55,460 –> 00:02:56,649 it with the Norm in X.
117 00:02:57,250 –> 00:02:58,850 What we now can put in relation
118 00:02:58,860 –> 00:03:00,610 is; how much did the length
119 00:03:00,619 –> 00:03:02,369 change from left to right?
120 00:03:03,119 –> 00:03:04,520 In other words, what is the
121 00:03:04,529 –> 00:03:06,199 quotient of the length of
122 00:03:06,210 –> 00:03:07,960 Tx divided by the
123 00:03:07,970 –> 00:03:09,070 length of x?
124 00:03:10,229 –> 00:03:11,399 Of course, this is now a
125 00:03:11,410 –> 00:03:13,139 number we could use as a
126 00:03:13,149 –> 00:03:15,070 definition for the norm
127 00:03:15,080 –> 00:03:15,649 of T.
128 00:03:16,369 –> 00:03:17,509 However, of course, this
129 00:03:17,520 –> 00:03:19,089 can only be meaningful if
130 00:03:19,100 –> 00:03:21,009 we look at all possible inputs
131 00:03:21,020 –> 00:03:22,110 on the left-hand side.
132 00:03:22,600 –> 00:03:24,339 So we’re looking at all possible
133 00:03:24,350 –> 00:03:26,089 ratios that can come out
134 00:03:26,100 –> 00:03:27,509 by going through all the
135 00:03:27,520 –> 00:03:27,699 x’s.
136 00:03:28,199 –> 00:03:29,729 Obviously, the only exception
137 00:03:29,740 –> 00:03:31,050 should be the zero vector,
138 00:03:31,080 –> 00:03:32,639 because this one is the only
139 00:03:32,649 –> 00:03:33,929 one with length of zero.
140 00:03:34,669 –> 00:03:35,970 Now, you might already see
141 00:03:35,979 –> 00:03:37,529 that the norm of T should
142 00:03:37,539 –> 00:03:39,169 be the biggest ratio we can
143 00:03:39,179 –> 00:03:39,970 get out here.
144 00:03:39,979 –> 00:03:41,800 So the maximum of the set.
145 00:03:42,619 –> 00:03:44,520 However, we can have an infinite
146 00:03:44,529 –> 00:03:45,649 dimension in X.
147 00:03:45,660 –> 00:03:47,309 So we have infinitely many
148 00:03:47,320 –> 00:03:48,820 directions we can look at.
149 00:03:49,539 –> 00:03:50,570 Therefore, it can happen
150 00:03:50,580 –> 00:03:52,149 that the maximum does not
151 00:03:52,160 –> 00:03:52,990 exist.
152 00:03:53,000 –> 00:03:54,740 So what we need here is the
153 00:03:54,750 –> 00:03:56,059 supremum of this set.
154 00:03:56,899 –> 00:03:58,330 We are in the real numbers.
155 00:03:58,339 –> 00:04:00,070 So we know the supremum always
156 00:04:00,080 –> 00:04:01,470 exists. In the worst case
157 00:04:01,479 –> 00:04:03,229 it would be the symbol infinity
158 00:04:03,729 –> 00:04:05,110 and there we have our notion.
159 00:04:05,160 –> 00:04:06,860 If this norm of T is
160 00:04:06,869 –> 00:04:08,619 finite, we call T
161 00:04:08,630 –> 00:04:09,330 bounded
162 00:04:10,020 –> 00:04:11,759 and I already mentioned that
163 00:04:11,770 –> 00:04:13,679 non bounded linear operators
164 00:04:13,690 –> 00:04:15,559 can only happen, if X is
165 00:04:15,570 –> 00:04:16,720 of infinite dimension
166 00:04:17,570 –> 00:04:19,019 and please note that this
167 00:04:19,029 –> 00:04:20,950 notion of boundedness for linear
168 00:04:20,959 –> 00:04:22,850 operators is different than
169 00:04:22,859 –> 00:04:24,829 the notion bounded for normal
170 00:04:24,839 –> 00:04:26,549 functions in R for example.
171 00:04:27,179 –> 00:04:28,799 So please don’t get confused
172 00:04:28,809 –> 00:04:29,200 there.
173 00:04:29,989 –> 00:04:31,049 Now, with the definition,
174 00:04:31,059 –> 00:04:32,380 out of the way, let’s go
175 00:04:32,390 –> 00:04:33,940 to the proposition that connects
176 00:04:33,950 –> 00:04:35,630 this to the continuity.
177 00:04:36,359 –> 00:04:36,600 Here
178 00:04:36,609 –> 00:04:37,989 again, we have a linear map
179 00:04:38,000 –> 00:04:39,809 between two normed spaces.
180 00:04:40,399 –> 00:04:41,929 So the same thing as before
181 00:04:41,940 –> 00:04:43,540 and then we know the following
182 00:04:43,549 –> 00:04:45,290 three things are equivalent.
183 00:04:46,149 –> 00:04:47,799 The first is that T as a
184 00:04:47,809 –> 00:04:49,380 map between metric spaces
185 00:04:49,390 –> 00:04:50,579 is continuous.
186 00:04:51,470 –> 00:04:53,309 Now (b) is similar, but
187 00:04:53,320 –> 00:04:54,630 here T has only to be
188 00:04:54,640 –> 00:04:56,559 continuous at one point,
189 00:04:56,570 –> 00:04:57,890 where we choose the origin
190 00:04:57,899 –> 00:04:58,690 at this point
191 00:04:59,339 –> 00:05:00,899 and the last one is that
192 00:05:00,910 –> 00:05:02,570 T as a linear operator
193 00:05:02,579 –> 00:05:04,130 between normed spaces
194 00:05:04,209 –> 00:05:05,799 is a bounded operator.
195 00:05:06,630 –> 00:05:08,279 So this is the fact I already
196 00:05:08,290 –> 00:05:09,279 told you at the beginning.
197 00:05:09,290 –> 00:05:10,799 Continuity and
198 00:05:10,809 –> 00:05:12,750 boundedness exactly in this
199 00:05:12,760 –> 00:05:14,730 sense, are equivalent terms
200 00:05:14,739 –> 00:05:16,279 for linear operators
201 00:05:17,070 –> 00:05:18,540 and of course, this is so
202 00:05:18,549 –> 00:05:19,809 important that we should
203 00:05:19,820 –> 00:05:20,970 write down a proof.
204 00:05:22,170 –> 00:05:23,339 The first implication we
205 00:05:23,350 –> 00:05:25,010 should show is (a) to
206 00:05:25,019 –> 00:05:26,950 (b) which is obviously
207 00:05:26,959 –> 00:05:28,649 immediately fulfilled, because
208 00:05:28,660 –> 00:05:30,230 being continuous at all
209 00:05:30,239 –> 00:05:31,760 points implies being
210 00:05:31,769 –> 00:05:33,350 continuous at zero.
211 00:05:34,070 –> 00:05:35,410 So let’s go to the next one
212 00:05:35,420 –> 00:05:36,859 which would be the implication
213 00:05:36,869 –> 00:05:38,380 from (b) to (c).
214 00:05:39,190 –> 00:05:40,480 Here we really have to do
215 00:05:40,489 –> 00:05:41,190 something.
216 00:05:41,200 –> 00:05:42,540 So let’s start by writing
217 00:05:42,549 –> 00:05:43,859 down what it means to be
218 00:05:43,869 –> 00:05:45,339 continuous at zero.
219 00:05:46,000 –> 00:05:47,410 As often we want to use the
220 00:05:47,420 –> 00:05:49,089 characterization with sequences.
221 00:05:49,100 –> 00:05:51,079 So here we consider convergent
222 00:05:51,089 –> 00:05:52,660 sequences, but only with
223 00:05:52,670 –> 00:05:54,010 limit point 0.
224 00:05:54,480 –> 00:05:56,200 The continuity then implies
225 00:05:56,209 –> 00:05:57,709 that the images also
226 00:05:57,720 –> 00:05:58,399 converge
227 00:05:58,410 –> 00:05:59,480 and because we have a linear
228 00:05:59,489 –> 00:06:01,459 map, the limit point is also
229 00:06:01,470 –> 00:06:02,000 zero.
230 00:06:02,579 –> 00:06:03,989 However, here it might be
231 00:06:04,000 –> 00:06:05,119 easier to work with an epsilon-
232 00:06:05,149 –> 00:06:06,920 delta characterization for
233 00:06:06,929 –> 00:06:07,929 the continuity.
234 00:06:08,000 –> 00:06:09,220 So let me write down the
235 00:06:09,230 –> 00:06:10,369 claim we need here.
236 00:06:10,760 –> 00:06:12,209 The formulation before, which
237 00:06:12,220 –> 00:06:14,119 we call * now, implies
238 00:06:14,130 –> 00:06:15,149 there is a delta
239 00:06:15,730 –> 00:06:17,279 such that the norm of
240 00:06:17,290 –> 00:06:19,140 Tx is always less
241 00:06:19,149 –> 00:06:20,890 than 1, for
242 00:06:20,899 –> 00:06:22,839 all x with length less
243 00:06:22,850 –> 00:06:23,640 than delta.
244 00:06:24,820 –> 00:06:26,279 If you know continuity, you
245 00:06:26,290 –> 00:06:27,790 already know that. You already
246 00:06:27,799 –> 00:06:28,660 believe that,
247 00:06:28,799 –> 00:06:30,220 but for the sake of completeness,
248 00:06:30,230 –> 00:06:31,549 let’s write down the proof.
249 00:06:32,260 –> 00:06:33,570 Let’s do a proof by contra
250 00:06:33,579 –> 00:06:34,079 position.
251 00:06:34,089 –> 00:06:35,790 So let’s call the whole right-
252 00:06:35,799 –> 00:06:36,799 hand side here
253 00:06:37,149 –> 00:06:38,869 just * in red.
254 00:06:40,100 –> 00:06:41,540 The negation of the red star
255 00:06:41,549 –> 00:06:42,799 then implies
256 00:06:43,839 –> 00:06:45,779 that for all n, we
257 00:06:45,790 –> 00:06:46,980 find an x_n
258 00:06:47,890 –> 00:06:49,600 with length x_n
259 00:06:49,720 –> 00:06:51,470 less than one over n.
260 00:06:52,369 –> 00:06:53,829 The one over n corresponds
261 00:06:53,839 –> 00:06:54,820 to the delta here.
262 00:06:54,850 –> 00:06:56,399 So we say there is no such
263 00:06:56,410 –> 00:06:56,950 delta.
264 00:06:57,089 –> 00:06:58,660 So we can do that for all
265 00:06:58,670 –> 00:07:00,640 n here. Which also
266 00:07:00,649 –> 00:07:02,230 means that the norm of
267 00:07:02,970 –> 00:07:04,809 Tx_n is greater or equal than
268 00:07:04,820 –> 00:07:05,230 one
269 00:07:06,010 –> 00:07:07,519 and there you see, we found
270 00:07:07,529 –> 00:07:08,950 a sequence that converges
271 00:07:08,959 –> 00:07:09,630 to zero,
272 00:07:09,640 –> 00:07:11,470 but the images don’t converge
273 00:07:11,480 –> 00:07:12,109 to zero.
274 00:07:12,690 –> 00:07:14,220 So this implies then
275 00:07:14,279 –> 00:07:15,790 not green star
276 00:07:16,500 –> 00:07:17,760 and by contraposition,
277 00:07:17,769 –> 00:07:19,320 this proves the claim we
278 00:07:19,329 –> 00:07:20,000 want to use
279 00:07:20,010 –> 00:07:20,440 now.
280 00:07:21,779 –> 00:07:23,049 Recall that we want to
281 00:07:23,059 –> 00:07:24,970 calculate the quotient of
282 00:07:24,980 –> 00:07:26,660 the norm of Tx
283 00:07:26,700 –> 00:07:28,329 divided by the norm of
284 00:07:28,339 –> 00:07:28,890 x.
285 00:07:29,649 –> 00:07:30,869 However, at the moment, we
286 00:07:30,880 –> 00:07:32,279 can only say something about
287 00:07:32,290 –> 00:07:34,269 the vector x that have length
288 00:07:34,279 –> 00:07:35,559 less than delta.
289 00:07:36,089 –> 00:07:37,320 Of course, this is something
290 00:07:37,329 –> 00:07:38,829 we can use here, because we
291 00:07:38,839 –> 00:07:40,760 could multiply with delta
292 00:07:40,769 –> 00:07:42,390 half, which is less than
293 00:07:42,399 –> 00:07:44,119 a delta, times
294 00:07:44,130 –> 00:07:45,809 one over the norm of x.
295 00:07:46,390 –> 00:07:47,730 So why do we do that?
296 00:07:47,799 –> 00:07:49,209 Simply because with that
297 00:07:49,220 –> 00:07:50,880 factor, we can scale the
298 00:07:50,890 –> 00:07:52,230 length of the vector x.
299 00:07:52,420 –> 00:07:53,769 Since you know the properties
300 00:07:53,779 –> 00:07:55,070 of the norm, you know, we
301 00:07:55,079 –> 00:07:56,549 can push that inside the
302 00:07:56,559 –> 00:07:57,089 norm
303 00:07:57,100 –> 00:07:58,410 and you also know that T
304 00:07:58,420 –> 00:07:59,100 is linear.
305 00:07:59,820 –> 00:08:00,970 So what we get in the
306 00:08:00,980 –> 00:08:02,829 denominator is a vector
307 00:08:02,839 –> 00:08:04,410 that has exactly length
308 00:08:04,420 –> 00:08:05,450 delta half
309 00:08:06,029 –> 00:08:07,059 and that’s something that
310 00:08:07,070 –> 00:08:09,019 reminds us of our red *
311 00:08:09,029 –> 00:08:09,730 property.
312 00:08:10,470 –> 00:08:11,970 The vector has length less
313 00:08:11,980 –> 00:08:12,690 than delta.
314 00:08:12,760 –> 00:08:14,250 So the corresponding image
315 00:08:14,260 –> 00:08:15,890 has length less than one.
316 00:08:16,649 –> 00:08:18,350 Hence, the numerator is now
317 00:08:18,359 –> 00:08:20,049 less than one, which means
318 00:08:20,059 –> 00:08:21,640 the whole thing is less than
319 00:08:21,649 –> 00:08:22,829 two over delta.
320 00:08:23,600 –> 00:08:25,140 So the only thing that remains
321 00:08:25,149 –> 00:08:26,980 is applying the supremum
322 00:08:26,989 –> 00:08:28,820 on both sides. Now,
323 00:08:28,829 –> 00:08:29,929 since we exclude the zero
324 00:08:29,940 –> 00:08:31,179 vector, we know this all
325 00:08:31,190 –> 00:08:33,058 works. Also the supremum
326 00:08:33,070 –> 00:08:34,929 has to be less than two over
327 00:08:34,940 –> 00:08:35,419 delta.
328 00:08:35,929 –> 00:08:36,929 The important thing is of
329 00:08:36,940 –> 00:08:38,200 course, this is not
330 00:08:38,210 –> 00:08:38,940 infinity.
331 00:08:39,669 –> 00:08:41,429 Well, this was (b) to (c).
332 00:08:41,438 –> 00:08:43,169 Continuity at zero
333 00:08:43,179 –> 00:08:44,578 implies boundedness
334 00:08:45,590 –> 00:08:47,119 and now the last part is
335 00:08:47,130 –> 00:08:48,570 a bounded operator is
336 00:08:48,580 –> 00:08:50,010 continuous everywhere.
337 00:08:50,739 –> 00:08:52,469 So let’s consider any point
338 00:08:52,479 –> 00:08:54,469 x tilde in X and any
339 00:08:54,479 –> 00:08:56,239 sequence that is convergent
340 00:08:56,250 –> 00:08:57,159 to this point
341 00:08:57,799 –> 00:08:58,799 and then we want to look
342 00:08:58,809 –> 00:09:00,219 what happens to the images.
343 00:09:00,229 –> 00:09:01,320 So we look at
344 00:09:01,340 –> 00:09:02,409 Tx_n
345 00:09:02,419 –> 00:09:04,260 minus Tx tilde
346 00:09:04,270 –> 00:09:06,099 inside the norm of Y.
347 00:09:06,869 –> 00:09:08,330 Then the linearity tells
348 00:09:08,340 –> 00:09:10,080 us that we can apply T to
349 00:09:10,090 –> 00:09:11,700 the difference vector and
350 00:09:11,710 –> 00:09:13,039 then calculate the norm
351 00:09:13,700 –> 00:09:15,299 and now we can use what we
352 00:09:15,309 –> 00:09:16,840 know. We have to find an
353 00:09:16,849 –> 00:09:18,280 operator norm of T,
354 00:09:18,799 –> 00:09:20,239 which is by definition the
355 00:09:20,250 –> 00:09:22,039 largest possible scaling
356 00:09:22,049 –> 00:09:23,400 for the length of the image.
357 00:09:24,030 –> 00:09:25,210 In other words, we know this
358 00:09:25,219 –> 00:09:26,760 is less or equal than the
359 00:09:26,770 –> 00:09:28,520 operator norm times the
360 00:09:28,530 –> 00:09:29,679 length of the input
361 00:09:30,130 –> 00:09:31,809 and that’s x_n minus
362 00:09:31,890 –> 00:09:32,599 x tilde.
363 00:09:33,460 –> 00:09:34,719 Since we already know this
364 00:09:34,729 –> 00:09:36,090 is convergent, we know this
365 00:09:36,099 –> 00:09:37,979 goes to zero when n goes
366 00:09:37,989 –> 00:09:38,969 to infinity.
367 00:09:38,979 –> 00:09:40,320 So the whole right-hand side
368 00:09:40,330 –> 00:09:41,270 goes to zero.
369 00:09:42,090 –> 00:09:43,469 Hence also the left-hand
370 00:09:43,479 –> 00:09:45,150 side which tells us that
371 00:09:45,159 –> 00:09:46,789 also the images converge
372 00:09:47,500 –> 00:09:48,679 and that’s by definition,
373 00:09:48,690 –> 00:09:49,679 the continuity.
374 00:09:49,690 –> 00:09:51,359 So our proof is finished.
375 00:09:52,049 –> 00:09:52,390 OK.
376 00:09:52,400 –> 00:09:54,349 So this was our first important
377 00:09:54,359 –> 00:09:55,830 result for linear
378 00:09:55,840 –> 00:09:57,340 operators between normed
379 00:09:57,349 –> 00:09:57,989 spaces.
380 00:09:58,729 –> 00:10:00,099 What you can do for yourself
381 00:10:00,109 –> 00:10:01,489 now is showing that this
382 00:10:01,500 –> 00:10:03,309 operator norm we defined
383 00:10:03,320 –> 00:10:04,969 is indeed a norm in the
384 00:10:04,979 –> 00:10:06,809 usual sense. With
385 00:10:06,820 –> 00:10:07,119 that
386 00:10:07,130 –> 00:10:08,109 I think it’s good enough
387 00:10:08,119 –> 00:10:08,750 for today.
388 00:10:08,840 –> 00:10:10,109 Thanks for listening.
389 00:10:10,159 –> 00:10:11,500 Thanks for supporting me
390 00:10:11,510 –> 00:10:12,989 and see you next time.
391 00:10:13,080 –> 00:10:13,710 Bye.
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Quiz Content
Q1: Let $(X, |\cdot |_X)$, $(Y, |\cdot |_Y)$ be two finite-dimensional normed vector spaces and $T: X \rightarrow Y$ a linear operator. What is not possible for the operator norm of $T$?
A1: $| T | = 0$
A2: $| T | = \infty$
A3: $| T | = 1$
A4: $| T | = \sup \Big{ \frac{ | Tx |_Y }{ | x |_X } ~ \Big| ~ x \in X \setminus { 0 } \Big} $
A5: $| T | = \sup \Big{ | Tx |_Y ~ \Big| ~ x \in X \text{ with } | x | = 1 \Big} $
Q2: Let $\ell^2(\mathbb{N}, \mathbb{C})$ be the Hilbert space with the associated norm $| \cdot |{\ell^2}$ and let $T: \ell^2(\mathbb{N}, \mathbb{C}) \rightarrow \ell^2(\mathbb{N}, \mathbb{C})$ be a linear operator given by $T( (x_n){n \in \mathbb{N}} ) = (2 x_n )_{n \in \mathbb{N}} $. What is not correct?
A1: $| T | = 2 $
A2: $T$ is continuous
A3: $T$ is bounded
A4: $| T | < \infty $
A5: $| T | = \sqrt{2} $
Q3: Consider the space of eventually zero sequences $c_{00}$ given by $${ (x_n){n \in \mathbb{N}} \mid x_n \neq 0 \text{ only for finitely many indices}}$$ with norm $| (x_n){n \in \mathbb{N}} | = \sup_{n \in \mathbb{N} } | x_n |$. There we can define the linear operator $T : c_{00} \rightarrow c_{00}$ with $T( (x_n){n \in \mathbb{N}} ) = (n x_n ){n \in \mathbb{N}} $. What is correct?
A1: $| T | = 2 $
A2: $T$ is continuous
A3: $T$ is bounded
A4: $| T | = \infty $
A5: $| T | = \sqrt{2} $
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Last update: 2024-10
