-
Title: Compact Operators
-
Series: Functional Analysis
-
YouTube-Title: Functional Analysis 18 | Compact Operators
-
Bright video: https://youtu.be/tiR9-u6K5EE
-
Dark video: https://youtu.be/7nV3a1eClWA
-
Ad-free video: Watch Vimeo video
-
Dark-PDF: Download PDF version of the dark video
-
Print-PDF: Download printable PDF version
-
Thumbnail (bright): Download PNG
-
Thumbnail (dark): Download PNG
-
Subtitle on GitHub: fa18_sub_eng.srt
-
Timestamps
00:00 Introduction
02:39 Definition
03:13 Example
-
Subtitle in English
1 00:00:00,349 –> 00:00:02,210 Hello and welcome back to
2 00:00:02,220 –> 00:00:03,490 functional analysis
3 00:00:03,670 –> 00:00:05,289 and as always many, many
4 00:00:05,300 –> 00:00:07,070 thanks to all the nice people
5 00:00:07,079 –> 00:00:08,449 that support me on Steady
6 00:00:08,460 –> 00:00:09,250 or PayPal.
7 00:00:10,020 –> 00:00:11,390 We’ve already reached part
8 00:00:11,399 –> 00:00:12,829 18 in our course
9 00:00:12,840 –> 00:00:14,090 and today we want to talk
10 00:00:14,100 –> 00:00:16,010 about compact operators.
11 00:00:16,719 –> 00:00:18,440 As the name suggests, this
12 00:00:18,450 –> 00:00:19,579 has something to do with
13 00:00:19,590 –> 00:00:21,389 the compact sets we’ve already
14 00:00:21,399 –> 00:00:21,899 studied.
15 00:00:22,780 –> 00:00:24,149 Therefore, recall that the
16 00:00:24,159 –> 00:00:25,989 idea of compactness has
17 00:00:26,000 –> 00:00:27,950 been to extend the notion
18 00:00:27,959 –> 00:00:29,469 of finiteness a little bit.
19 00:00:30,229 –> 00:00:31,709 A similar thing we will now
20 00:00:31,719 –> 00:00:33,220 do for operators
21 00:00:33,849 –> 00:00:34,310 Here
22 00:00:34,319 –> 00:00:36,080 the analogue of finiteness would
23 00:00:36,090 –> 00:00:37,799 be linear operators between
24 00:00:37,810 –> 00:00:39,139 finite dimensional vector
25 00:00:39,150 –> 00:00:39,799 spaces.
26 00:00:40,340 –> 00:00:41,880 So let’s look at an operator
27 00:00:41,889 –> 00:00:43,340 from F^n to
28 00:00:43,349 –> 00:00:45,319 F^m. On both sides,
29 00:00:45,330 –> 00:00:46,889 we can just choose the normal
30 00:00:46,900 –> 00:00:48,459 Euclidean norm, the standard
31 00:00:48,470 –> 00:00:48,880 norm.
32 00:00:49,319 –> 00:00:50,599 However, everything I say
33 00:00:50,610 –> 00:00:51,729 now does not depend on the
34 00:00:51,740 –> 00:00:52,220 norm.
35 00:00:52,229 –> 00:00:53,529 It holds no matter which
36 00:00:53,540 –> 00:00:54,529 norm you choose here.
37 00:00:55,389 –> 00:00:56,409 The first step is not so
38 00:00:56,419 –> 00:00:57,970 hard to show, it tells us
39 00:00:57,979 –> 00:00:59,150 that the linear operator
40 00:00:59,159 –> 00:01:00,590 between finite dimensional
41 00:01:00,599 –> 00:01:02,479 normed spaces is always
42 00:01:02,490 –> 00:01:03,340 continuous
43 00:01:03,729 –> 00:01:04,888 and of course, you already
44 00:01:04,900 –> 00:01:06,330 know continuous means the
45 00:01:06,339 –> 00:01:07,889 same as bounded for linear
46 00:01:07,900 –> 00:01:08,669 operators.
47 00:01:09,330 –> 00:01:10,809 In other words, the image
48 00:01:10,819 –> 00:01:12,690 of the unit ball under T
49 00:01:12,699 –> 00:01:13,889 is a bounded set.
50 00:01:14,379 –> 00:01:15,839 Now, this is a bounded set
51 00:01:15,849 –> 00:01:17,269 in our finite dimensional
52 00:01:17,279 –> 00:01:18,769 normed space F^m
53 00:01:19,199 –> 00:01:20,610 and there you know this is
54 00:01:20,620 –> 00:01:22,089 one half of the things you
55 00:01:22,099 –> 00:01:23,430 need for being compact.
56 00:01:24,019 –> 00:01:25,510 However, the second ingredient
57 00:01:25,519 –> 00:01:26,989 we just get when we form
58 00:01:27,000 –> 00:01:28,410 the closure of this set.
59 00:01:28,819 –> 00:01:30,389 This means that we just can
60 00:01:30,400 –> 00:01:32,260 use a bar over the set
61 00:01:32,989 –> 00:01:34,449 and then we get a set that
62 00:01:34,459 –> 00:01:36,330 is always compact in
63 00:01:36,339 –> 00:01:36,919 F^m.
64 00:01:37,629 –> 00:01:38,989 With this essentially, you
65 00:01:39,000 –> 00:01:40,510 now know what a compact
66 00:01:40,519 –> 00:01:41,290 operators is.
67 00:01:42,250 –> 00:01:43,470 When we look at the image
68 00:01:43,480 –> 00:01:45,139 of the unit ball and form
69 00:01:45,150 –> 00:01:45,819 the closure
70 00:01:45,830 –> 00:01:47,309 and this set is compact,
71 00:01:47,319 –> 00:01:49,279 we speak of a compact operator.
72 00:01:49,830 –> 00:01:51,349 Of course, this always holds
73 00:01:51,360 –> 00:01:52,910 in this case, but not in
74 00:01:52,919 –> 00:01:53,500 general.
75 00:01:54,099 –> 00:01:55,510 In order to see that let’s
76 00:01:55,519 –> 00:01:56,730 consider the identity
77 00:01:56,739 –> 00:01:57,580 operator
78 00:01:57,589 –> 00:01:58,330 in l^p.
79 00:01:59,110 –> 00:02:00,610 Of course, identity operator
80 00:02:00,620 –> 00:02:02,309 just means we take a sequence
81 00:02:02,319 –> 00:02:03,980 x and send it to itself.
82 00:02:04,449 –> 00:02:05,779 Therefore, calculating the
83 00:02:05,790 –> 00:02:07,099 image of the unit ball is
84 00:02:07,110 –> 00:02:08,660 not so hard. It stays the
85 00:02:08,669 –> 00:02:09,380 unit ball.
86 00:02:10,119 –> 00:02:11,160 Now we are interested in
87 00:02:11,169 –> 00:02:12,259 the closure of the set.
88 00:02:12,270 –> 00:02:13,779 So we use the bars again.
89 00:02:14,490 –> 00:02:15,580 So what you have on the right-
90 00:02:15,589 –> 00:02:17,009 hand side is just a closed
91 00:02:17,029 –> 00:02:18,429 unit ball in l^p
92 00:02:19,089 –> 00:02:20,169 and there we know from the
93 00:02:20,179 –> 00:02:21,820 last video it’s closed and
94 00:02:21,830 –> 00:02:23,789 bounded, but not compact.
95 00:02:24,509 –> 00:02:25,570 Hence, we have something
96 00:02:25,580 –> 00:02:25,809 here,
97 00:02:25,820 –> 00:02:27,350 we would not call a compact
98 00:02:27,360 –> 00:02:29,059 operator, because the image
99 00:02:29,070 –> 00:02:30,509 is just too large in this
100 00:02:30,520 –> 00:02:30,929 case.
101 00:02:31,679 –> 00:02:33,229 So you should always remember
102 00:02:33,240 –> 00:02:34,910 compact operators is what
103 00:02:34,919 –> 00:02:36,589 we get when we extend the
104 00:02:36,600 –> 00:02:38,059 finite dimensional operators
105 00:02:38,070 –> 00:02:39,070 here a little bit.
106 00:02:39,949 –> 00:02:41,559 OK, then let’s write down
107 00:02:41,570 –> 00:02:42,740 the formal definition.
108 00:02:43,210 –> 00:02:45,139 So we need two normed spaces
109 00:02:45,149 –> 00:02:46,789 and often there will be Banach
110 00:02:46,800 –> 00:02:48,449 spaces. Then a
111 00:02:48,460 –> 00:02:49,910 bounded linear operator
112 00:02:49,919 –> 00:02:51,500 T from X to
113 00:02:51,509 –> 00:02:52,720 Y is called
114 00:02:52,729 –> 00:02:53,559 compact,
115 00:02:54,050 –> 00:02:55,720 if we have the thing we discussed
116 00:02:55,729 –> 00:02:57,199 before, that this set
117 00:02:57,210 –> 00:02:58,880 here is compact
118 00:02:58,889 –> 00:03:00,699 in Y. Therefore, in the
119 00:03:00,710 –> 00:03:02,509 case that Y is a finite
120 00:03:02,520 –> 00:03:04,250 dimensional normed space, this
121 00:03:04,259 –> 00:03:05,559 here is nothing special.
122 00:03:06,250 –> 00:03:07,320 However, in the infinite
123 00:03:07,330 –> 00:03:08,860 dimensional case, it really
124 00:03:08,869 –> 00:03:09,309 is.
125 00:03:09,949 –> 00:03:11,309 Hence, I would suggest that
126 00:03:11,320 –> 00:03:12,710 we now look at a common
127 00:03:12,720 –> 00:03:14,559 example. Let’s
128 00:03:14,570 –> 00:03:16,139 look at an integral operator
129 00:03:16,149 –> 00:03:17,750 defined for the continuous
130 00:03:17,759 –> 00:03:18,389 functions.
131 00:03:18,960 –> 00:03:20,179 So it should take a function
132 00:03:20,190 –> 00:03:22,149 defined from 0 to 1 and
133 00:03:22,160 –> 00:03:23,850 send that to another function
134 00:03:23,860 –> 00:03:24,710 in the same space
135 00:03:25,360 –> 00:03:26,820 and as often the space of
136 00:03:26,830 –> 00:03:28,139 continuous functions should
137 00:03:28,149 –> 00:03:29,740 carry the supremum norm.
138 00:03:30,509 –> 00:03:32,270 So what we can do is apply
139 00:03:32,279 –> 00:03:33,960 the operator T to a function
140 00:03:33,970 –> 00:03:35,880 f and then we get out a new
141 00:03:35,889 –> 00:03:36,520 function.
142 00:03:36,979 –> 00:03:38,169 Hence, we can look what the
143 00:03:38,179 –> 00:03:39,410 function does at a given
144 00:03:39,419 –> 00:03:41,410 point s, where s comes from
145 00:03:41,419 –> 00:03:42,330 the unit interval.
146 00:03:43,029 –> 00:03:44,210 Now the number that comes
147 00:03:44,220 –> 00:03:45,449 out here should be given
148 00:03:45,460 –> 00:03:47,220 by an integral from 0 to
149 00:03:47,229 –> 00:03:49,190 1 where we have the function
150 00:03:49,199 –> 00:03:50,690 f involved, but
151 00:03:50,699 –> 00:03:52,570 also a fixed function
152 00:03:52,580 –> 00:03:53,050 k
153 00:03:53,550 –> 00:03:54,570 and this function should
154 00:03:54,580 –> 00:03:56,500 get the variable s and also
155 00:03:56,509 –> 00:03:57,550 the integration variable
156 00:03:57,559 –> 00:03:57,949 t.
157 00:03:58,720 –> 00:03:59,949 So we have a function with
158 00:03:59,960 –> 00:04:01,539 two variables and for us
159 00:04:01,550 –> 00:04:03,270 it should be also a continuous
160 00:04:03,279 –> 00:04:03,710 function.
161 00:04:04,520 –> 00:04:06,449 So we have it from C defined
162 00:04:06,460 –> 00:04:08,350 on the unit interval times
163 00:04:08,360 –> 00:04:09,330 the unit interval.
164 00:04:10,100 –> 00:04:11,360 Now, since the function k
165 00:04:11,369 –> 00:04:12,759 goes into the definition
166 00:04:12,770 –> 00:04:14,710 of T, I put it into the
167 00:04:14,720 –> 00:04:15,509 index here.
168 00:04:16,160 –> 00:04:17,690 OK, then let’s check if
169 00:04:17,700 –> 00:04:19,059 T_k is indeed a
170 00:04:19,070 –> 00:04:20,339 compact operator.
171 00:04:21,170 –> 00:04:22,429 An important fact we will
172 00:04:22,440 –> 00:04:24,230 need here is that the function
173 00:04:24,239 –> 00:04:25,609 k is defined on a
174 00:04:25,619 –> 00:04:26,709 compact set.
175 00:04:26,720 –> 00:04:28,709 So it’s not just continuous,
176 00:04:28,720 –> 00:04:30,660 it’s uniformly continuous.
177 00:04:31,420 –> 00:04:32,730 To refresh your memory,
178 00:04:32,739 –> 00:04:34,059 let’s write down what this
179 00:04:34,070 –> 00:04:34,929 exactly means.
180 00:04:35,720 –> 00:04:37,540 For all epsilon greater zero,
181 00:04:37,549 –> 00:04:39,260 there exists a delta greater
182 00:04:39,269 –> 00:04:41,140 zero such that for
183 00:04:41,149 –> 00:04:42,859 all points we put in and
184 00:04:42,869 –> 00:04:44,440 now we need two variables
185 00:04:45,010 –> 00:04:46,339 and for them, it should hold
186 00:04:46,350 –> 00:04:47,779 if the distance is less than
187 00:04:47,790 –> 00:04:49,420 delta, the distance of the
188 00:04:49,429 –> 00:04:50,820 images should be less than
189 00:04:50,829 –> 00:04:51,589 epsilon.
190 00:04:52,059 –> 00:04:53,100 On the left-hand side, we
191 00:04:53,109 –> 00:04:54,170 measure the distance with
192 00:04:54,179 –> 00:04:55,720 the Euclidean norm in R^2
193 00:04:55,929 –> 00:04:57,100 and on the right-hand side,
194 00:04:57,109 –> 00:04:58,399 we measure with the Euclidean
195 00:04:58,410 –> 00:05:00,239 norm in R^1, which is the
196 00:05:00,250 –> 00:05:00,940 absolute value.
197 00:05:01,640 –> 00:05:03,600 Why we need the uniform continuity
198 00:05:03,609 –> 00:05:05,299 here, we see in a moment.
199 00:05:05,869 –> 00:05:06,859 The first step we have to
200 00:05:06,869 –> 00:05:08,540 do when we see such an integral
201 00:05:08,549 –> 00:05:10,399 operator is to check if
202 00:05:10,410 –> 00:05:12,220 this integral defines
203 00:05:12,230 –> 00:05:14,140 indeed a continuous function.
204 00:05:15,029 –> 00:05:16,709 Simply because otherwise
205 00:05:16,720 –> 00:05:18,029 the operator wouldn’t be
206 00:05:18,040 –> 00:05:18,850 well defined.
207 00:05:19,489 –> 00:05:21,309 It really should map continuous
208 00:05:21,320 –> 00:05:22,760 functions to continuous
209 00:05:22,769 –> 00:05:23,440 functions.
210 00:05:23,850 –> 00:05:25,140 Checking the continuity
211 00:05:25,149 –> 00:05:26,549 then means we look at the
212 00:05:26,559 –> 00:05:28,029 difference of the images
213 00:05:28,040 –> 00:05:29,459 when we put in different
214 00:05:29,470 –> 00:05:29,989 points.
215 00:05:30,660 –> 00:05:32,059 So this should be small when
216 00:05:32,070 –> 00:05:33,579 the points s_1 and s_2
217 00:05:33,589 –> 00:05:34,630 are close together.
218 00:05:35,399 –> 00:05:37,100 Therefore, we first calculate
219 00:05:37,109 –> 00:05:38,290 and then we can look what
220 00:05:38,299 –> 00:05:38,859 happens.
221 00:05:39,750 –> 00:05:40,730 Now, the first thing you
222 00:05:40,739 –> 00:05:41,769 should see here is that we
223 00:05:41,779 –> 00:05:43,220 can put that into one
224 00:05:43,230 –> 00:05:43,809 integral.
225 00:05:44,649 –> 00:05:46,440 So use some parentheses here
226 00:05:46,450 –> 00:05:48,119 and delete this integral
227 00:05:48,130 –> 00:05:48,399 here.
228 00:05:49,299 –> 00:05:50,640 Then of course, we pull in
229 00:05:50,649 –> 00:05:52,239 the absolute value. Then we
230 00:05:52,250 –> 00:05:53,279 get an inequality
231 00:05:54,130 –> 00:05:55,589 and with this, we have everything
232 00:05:55,600 –> 00:05:57,369 we need, because this one
233 00:05:57,380 –> 00:05:59,000 here is less than a supremum
234 00:05:59,059 –> 00:05:59,989 norm of f
235 00:06:00,140 –> 00:06:01,899 and this one by the uniform
236 00:06:01,910 –> 00:06:03,859 continuity of k can be as
237 00:06:03,869 –> 00:06:05,149 small as we want
238 00:06:05,890 –> 00:06:07,390 and exactly this is what
239 00:06:07,399 –> 00:06:08,670 we should formally write
240 00:06:08,679 –> 00:06:10,230 in front of the whole calculation.
241 00:06:10,899 –> 00:06:12,559 So for a given epsilon greater
242 00:06:12,570 –> 00:06:13,700 than zero, we choose the
243 00:06:13,709 –> 00:06:15,309 delta in such a way that
244 00:06:15,320 –> 00:06:16,739 this whole thing holds.
245 00:06:17,390 –> 00:06:18,440 Therefore, we can choose
246 00:06:18,450 –> 00:06:20,000 s_1, s_2 from the unit
247 00:06:20,010 –> 00:06:21,799 interval such that the distance
248 00:06:21,809 –> 00:06:22,950 is less than delta.
249 00:06:23,510 –> 00:06:25,489 For this s_1, s_2 here
250 00:06:25,500 –> 00:06:27,420 and the same t on both sides,
251 00:06:27,429 –> 00:06:28,910 we can apply what we know,
252 00:06:28,980 –> 00:06:30,779 that this is less than epsilon.
253 00:06:31,579 –> 00:06:33,500 By using also that this one
254 00:06:33,510 –> 00:06:34,570 is less than the supremum
255 00:06:34,720 –> 00:06:35,160 norm,
256 00:06:35,170 –> 00:06:36,260 we are finished with the
257 00:06:36,269 –> 00:06:37,820 whole integral. It’s
258 00:06:37,829 –> 00:06:39,690 simply less than epsilon
259 00:06:39,700 –> 00:06:41,179 times the supremum norm
260 00:06:41,739 –> 00:06:43,239 and because the supremum norm
261 00:06:43,250 –> 00:06:44,910 of f is just a constant in
262 00:06:44,920 –> 00:06:46,440 the whole calculation, we
263 00:06:46,450 –> 00:06:48,369 now know that this function
264 00:06:48,380 –> 00:06:49,839 is indeed continuous.
265 00:06:50,339 –> 00:06:51,700 So we can note, our
266 00:06:51,709 –> 00:06:53,109 operator as written as
267 00:06:53,119 –> 00:06:54,950 here, is well defined.
268 00:06:55,769 –> 00:06:57,390 However, our calculation
269 00:06:57,399 –> 00:06:58,989 here shows us even more.
270 00:06:59,869 –> 00:07:01,730 If we define the set A
271 00:07:01,779 –> 00:07:03,299 as the image of the unit
272 00:07:03,309 –> 00:07:05,279 ball, then we see
273 00:07:05,290 –> 00:07:07,049 by this whole estimate here,
274 00:07:07,170 –> 00:07:08,450 that the set A is
275 00:07:08,459 –> 00:07:09,609 uniformly equi
276 00:07:09,809 –> 00:07:10,470 continuous.
277 00:07:11,059 –> 00:07:12,299 If you don’t know what this
278 00:07:12,309 –> 00:07:13,899 means anymore, let’s write
279 00:07:13,910 –> 00:07:14,609 it down again.
280 00:07:15,369 –> 00:07:16,910 It just means that for all
281 00:07:16,920 –> 00:07:18,279 epsilon greater zero, there
282 00:07:18,290 –> 00:07:19,869 exists a delta such that
283 00:07:19,880 –> 00:07:21,869 for all s_1, s_3 and all
284 00:07:21,880 –> 00:07:23,600 g in A, we have the
285 00:07:23,609 –> 00:07:24,959 uniform continuity
286 00:07:24,970 –> 00:07:25,910 implication here.
287 00:07:26,570 –> 00:07:27,709 From the definition of the
288 00:07:27,720 –> 00:07:29,220 last video, I only had to
289 00:07:29,230 –> 00:07:30,410 change some names.
290 00:07:30,940 –> 00:07:32,260 I used the name g for the
291 00:07:32,269 –> 00:07:33,540 function here, because we
292 00:07:33,549 –> 00:07:35,279 already had a function f
293 00:07:35,940 –> 00:07:37,269 and of course, we needed
294 00:07:37,279 –> 00:07:38,859 the names s_1, s_2 for
295 00:07:38,869 –> 00:07:39,660 the variables.
296 00:07:40,329 –> 00:07:41,299 However, what you should
297 00:07:41,309 –> 00:07:42,799 see is that this one is the
298 00:07:42,809 –> 00:07:44,279 definition of a being
299 00:07:44,290 –> 00:07:46,279 uniformly equicontinuous
300 00:07:46,350 –> 00:07:48,119 and it’s the same thing as
301 00:07:48,130 –> 00:07:49,119 we have written here.
302 00:07:50,109 –> 00:07:51,079 In other words, with the
303 00:07:51,089 –> 00:07:52,559 calculation above, we have
304 00:07:52,570 –> 00:07:54,279 proven that A is indeed
305 00:07:54,290 –> 00:07:56,220 uniformly equicontinuous
306 00:07:56,970 –> 00:07:58,029 at this point. You might
307 00:07:58,040 –> 00:07:59,609 already guess that we want
308 00:07:59,619 –> 00:08:01,350 to apply the Arzelà–Ascoli
309 00:08:01,369 –> 00:08:02,079 theorem here.
310 00:08:03,029 –> 00:08:04,329 Therefore, another step we
311 00:08:04,339 –> 00:08:06,019 have to do is showing that
312 00:08:06,029 –> 00:08:07,519 the whole set is bounded.
313 00:08:07,529 –> 00:08:08,720 Or in other words that the
314 00:08:08,730 –> 00:08:09,869 operator is bounded.
315 00:08:10,570 –> 00:08:12,109 Hence, let’s calculate the
316 00:08:12,119 –> 00:08:13,070 operator norm.
317 00:08:13,619 –> 00:08:14,989 So we have the supremum of
318 00:08:15,000 –> 00:08:16,839 all the norms of the images,
319 00:08:16,850 –> 00:08:18,600 where f has norm one
320 00:08:19,040 –> 00:08:20,230 by the definition of the
321 00:08:20,239 –> 00:08:21,230 supremum norm.
322 00:08:21,239 –> 00:08:22,769 This is the supremum over
323 00:08:22,779 –> 00:08:24,670 all s, where we calculate
324 00:08:24,679 –> 00:08:25,690 the absolute value of the
325 00:08:25,700 –> 00:08:26,299 integral.
326 00:08:26,760 –> 00:08:28,630 As often, we can just pull
327 00:08:28,640 –> 00:08:29,929 in the absolute value into
328 00:08:29,940 –> 00:08:31,910 the integral and get an inequality
329 00:08:31,920 –> 00:08:32,308 out
330 00:08:32,820 –> 00:08:34,308 and with this, we are almost
331 00:08:34,320 –> 00:08:35,840 finished, because the last
332 00:08:35,849 –> 00:08:37,590 part here is less or
333 00:08:37,599 –> 00:08:38,849 equal than the supremum norm
334 00:08:38,859 –> 00:08:39,229 of f
335 00:08:39,969 –> 00:08:41,260 and this is by assumption
336 00:08:41,270 –> 00:08:43,028 just one. This
337 00:08:43,039 –> 00:08:44,239 means that we don’t need
338 00:08:44,249 –> 00:08:45,708 the outer supremum anymore
339 00:08:45,778 –> 00:08:47,179 and can just write down
340 00:08:48,000 –> 00:08:49,789 everything is less or equal
341 00:08:49,799 –> 00:08:50,979 than this integral.
342 00:08:51,380 –> 00:08:52,940 However, this is also less
343 00:08:52,950 –> 00:08:54,580 or equal than the supremum
344 00:08:54,659 –> 00:08:56,419 norm of our function k.
345 00:08:56,859 –> 00:08:58,320 As often, it’s not important
346 00:08:58,330 –> 00:09:00,080 what the number here is exactly.
347 00:09:00,140 –> 00:09:01,599 It’s only important that
348 00:09:01,609 –> 00:09:03,500 it is constant, because then
349 00:09:03,510 –> 00:09:05,280 we know that T_k is a
350 00:09:05,289 –> 00:09:06,500 bounded operator
351 00:09:06,950 –> 00:09:08,780 and now finally comes our
352 00:09:08,789 –> 00:09:10,609 conclusion, we can apply
353 00:09:11,090 –> 00:09:11,590 Arzelà–Ascoli.
354 00:09:12,280 –> 00:09:13,830 This set, the image of the
355 00:09:13,840 –> 00:09:15,690 unit ball is uniformly
356 00:09:15,700 –> 00:09:17,630 equicontinuous and bounded.
357 00:09:18,340 –> 00:09:19,940 Therefore, both things also
358 00:09:19,950 –> 00:09:21,109 hold for the closure of the
359 00:09:21,119 –> 00:09:21,630 set
360 00:09:21,820 –> 00:09:23,669 and by Arzelà–Ascoli, we now
361 00:09:23,679 –> 00:09:25,020 have a compact set
362 00:09:25,530 –> 00:09:27,070 and by our definition, we
363 00:09:27,080 –> 00:09:28,909 also know T_k,
364 00:09:29,010 –> 00:09:30,750 the integral operator is
365 00:09:30,760 –> 00:09:32,020 a compact operator.
366 00:09:32,799 –> 00:09:34,270 So you see this was a long
367 00:09:34,280 –> 00:09:36,030 example, but it is a standard
368 00:09:36,039 –> 00:09:37,489 example for a compact
369 00:09:37,500 –> 00:09:38,150 operator.
370 00:09:38,599 –> 00:09:40,070 Of course, we will talk about
371 00:09:40,080 –> 00:09:41,669 compact operators in later
372 00:09:41,679 –> 00:09:42,869 videos even more.
373 00:09:43,580 –> 00:09:45,330 Therefore, as always, I hope
374 00:09:45,340 –> 00:09:46,150 I see you there.
375 00:09:46,380 –> 00:09:46,950 Bye.
-
Quiz Content (n/a)
-
Last update: 2024-10
