# Information about Functional Analysis - Part 18

• 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

• 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 finite 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 analogon of finite 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 FN to

28 00:00:43,349 –> 00:00:45,319 FM 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 norm 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 norm space FM.

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 FM.

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.

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 LP.

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 LP.

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 norm 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 finer

120 00:03:02,520 –> 00:03:04,250 dimensional norm 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 supreme 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 TK 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 crater zero,

181 00:04:37,549 –> 00:04:39,260 there exists a delta crater

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 two.

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 one, 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 as one and two

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 as one as two 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 as one as two 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 supreme

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 supreme norm.

260 00:06:41,739 –> 00:06:43,239 And because the supreme 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 one S two 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

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 as one as two 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 equi continuous.

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 equi continuous

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 supreme 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 supreme norm.

322 00:08:21,239 –> 00:08:22,769 This is the supreme 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 supreme 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 supreme 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 TK 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 ATSA Ascoli.

354 00:09:12,280 –> 00:09:13,830 This said 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 equi continuous 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 TK

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.

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