00:00 Introduction

01:40 Definition - bounded operator

04:35 Proposition - continuous equivalent to bounded

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 non 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 norm 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 a domain

22 00:00:34,340 –> 00:00:35,290 or the co domain.

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 non 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 the

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 norm 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 norm 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 norm 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 And why?

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 Norman 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 Norman 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 xs.

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 lengths 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 supreme 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 supreme 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 bounded 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 norm 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 when 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 norm 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 bounded this 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 convergence

222 00:05:51,089 –> 00:05:52,660 sequences but only with

223 00:05:52,670 –> 00:05:54,010 limit 0.0.

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 star 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 one 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 star in wet

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 XN

258 00:06:47,890 –> 00:06:49,600 with length X and

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 TXN 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 applies 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 we call 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 star

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 supreme

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 supreme

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 bounded.

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 TXN

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 nom.

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 XN 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 norm

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 be 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.