1 00:00:00,360 –> 00:00:02,140 Hello and welcome back to

2 00:00:02,150 –> 00:00:03,789 functional analysis.

3 00:00:04,139 –> 00:00:05,570 And first, as always, I want

4 00:00:05,579 –> 00:00:06,920 to thank all the nice people

5 00:00:06,929 –> 00:00:07,949 that support this channel

6 00:00:07,960 –> 00:00:09,319 on Steady or paypal.

7 00:00:09,739 –> 00:00:11,569 Today’s part 31 is

8 00:00:11,579 –> 00:00:12,840 still about spectral

9 00:00:12,850 –> 00:00:13,619 theory.

10 00:00:14,130 –> 00:00:15,369 In particular, we will talk

11 00:00:15,380 –> 00:00:17,000 about the so called spectral

12 00:00:17,010 –> 00:00:17,700 radius.

13 00:00:18,420 –> 00:00:18,950 For this.

14 00:00:18,959 –> 00:00:20,610 First, please recall that

15 00:00:20,620 –> 00:00:22,059 we work in a complex Banach

16 00:00:22,100 –> 00:00:23,969 space X and

17 00:00:23,979 –> 00:00:25,840 with a bounded linear operator

18 00:00:25,850 –> 00:00:27,530 T there, we have

19 00:00:27,540 –> 00:00:28,809 already discussed that the

20 00:00:28,819 –> 00:00:30,100 spectrum of the operator

21 00:00:30,110 –> 00:00:31,690 T is a subset of the

22 00:00:31,700 –> 00:00:32,750 complex numbers.

23 00:00:33,200 –> 00:00:35,009 Indeed, we have shown it’s

24 00:00:35,020 –> 00:00:35,950 a close set.

25 00:00:36,599 –> 00:00:38,060 And as usual, we denote the

26 00:00:38,069 –> 00:00:40,060 spectrum with sigma of T.

27 00:00:40,810 –> 00:00:42,650 Now the spectral radius just

28 00:00:42,659 –> 00:00:44,500 measures how far the spectrum

29 00:00:44,509 –> 00:00:45,909 is away from zero.

30 00:00:46,659 –> 00:00:48,000 And this is what we denote

31 00:00:48,009 –> 00:00:49,479 by R of T.

32 00:00:49,950 –> 00:00:51,680 So by definition simply the

33 00:00:51,689 –> 00:00:53,099 suprema of the absolute

34 00:00:53,110 –> 00:00:53,849 values.

35 00:00:54,279 –> 00:00:55,659 So you see we can immediately

36 00:00:55,669 –> 00:00:57,360 define this number here.

37 00:00:57,880 –> 00:00:59,069 And in this video, we will

38 00:00:59,080 –> 00:01:00,319 show that this is indeed

39 00:01:00,330 –> 00:01:02,319 always a finite number

40 00:01:02,819 –> 00:01:03,720 more concretely.

41 00:01:03,729 –> 00:01:05,169 The spectrum is not the empty

42 00:01:05,180 –> 00:01:06,610 set and also a

43 00:01:06,620 –> 00:01:07,559 bounded set.

44 00:01:08,269 –> 00:01:08,629 OK.

45 00:01:08,639 –> 00:01:10,370 Now let’s formulate it and

46 00:01:10,379 –> 00:01:11,449 then prove it.

47 00:01:11,889 –> 00:01:13,050 So we have a theorem with

48 00:01:13,059 –> 00:01:14,849 the same assumptions as before

49 00:01:15,209 –> 00:01:16,680 most importantly, we have

50 00:01:16,690 –> 00:01:18,169 a bounded linear operator

51 00:01:18,180 –> 00:01:19,370 from X to X.

52 00:01:20,080 –> 00:01:21,599 Then our first statement

53 00:01:21,610 –> 00:01:23,540 is that the spectrum of T

54 00:01:23,550 –> 00:01:25,089 as a subset of the complex

55 00:01:25,099 –> 00:01:26,919 numbers is a compact set.

56 00:01:27,739 –> 00:01:29,199 Please recall in the last

57 00:01:29,209 –> 00:01:30,839 video, we have already shown

58 00:01:30,849 –> 00:01:32,160 that it is a closed set.

59 00:01:32,839 –> 00:01:33,879 Therefore, the new thing

60 00:01:33,889 –> 00:01:35,489 here is that the set is

61 00:01:35,500 –> 00:01:36,690 also bounded.

62 00:01:37,080 –> 00:01:38,529 Then our next statement here

63 00:01:38,540 –> 00:01:40,209 is that the spectrum is never

64 00:01:40,220 –> 00:01:40,809 empty.

65 00:01:41,459 –> 00:01:43,080 However, this is not completely

66 00:01:43,089 –> 00:01:44,790 true because there is one

67 00:01:44,800 –> 00:01:46,199 case where this can happen,

68 00:01:46,639 –> 00:01:47,760 it happens when you choose

69 00:01:47,769 –> 00:01:49,339 the smallest possible vector

70 00:01:49,349 –> 00:01:49,879 space.

71 00:01:50,400 –> 00:01:51,449 So when the whole vector

72 00:01:51,459 –> 00:01:53,410 space is only the zero vector,

73 00:01:53,419 –> 00:01:54,980 then there’s only one linear

74 00:01:54,989 –> 00:01:56,800 map T and is

75 00:01:56,809 –> 00:01:58,160 always invertible.

76 00:01:58,169 –> 00:01:59,559 Therefore, the spectrum needs

77 00:01:59,569 –> 00:02:00,440 to be empty.

78 00:02:00,910 –> 00:02:02,360 However, when we exclude

79 00:02:02,370 –> 00:02:03,800 this in fairness boring

80 00:02:03,809 –> 00:02:05,419 case, then the

81 00:02:05,430 –> 00:02:07,000 implication is correct, the

82 00:02:07,010 –> 00:02:08,809 spectrum of T is non

83 00:02:08,880 –> 00:02:09,449 empty.

84 00:02:09,809 –> 00:02:10,130 OK.

85 00:02:10,139 –> 00:02:12,070 Then the last part C is about

86 00:02:12,080 –> 00:02:13,440 the spectral radius.

87 00:02:13,600 –> 00:02:15,070 As I already told you the

88 00:02:15,080 –> 00:02:16,710 spectral radius are of

89 00:02:16,720 –> 00:02:17,970 T is given by the

90 00:02:17,979 –> 00:02:19,750 supreme where we go

91 00:02:19,759 –> 00:02:21,309 through all the points lambda

92 00:02:21,320 –> 00:02:22,250 in the spectrum.

93 00:02:22,259 –> 00:02:23,410 And we look at the absolute

94 00:02:23,419 –> 00:02:23,910 value.

95 00:02:24,320 –> 00:02:25,809 Now, one important result

96 00:02:25,820 –> 00:02:27,529 here is that instead of calculating

97 00:02:27,539 –> 00:02:29,270 a supreme, we can look at

98 00:02:29,279 –> 00:02:31,210 a limit and there we take

99 00:02:31,220 –> 00:02:32,509 the operator norm of the

100 00:02:32,520 –> 00:02:33,649 powers of T.

101 00:02:34,429 –> 00:02:35,720 Then outside of the norm

102 00:02:35,729 –> 00:02:37,360 we take the k-th root again,

103 00:02:37,869 –> 00:02:39,339 one additional result is

104 00:02:39,350 –> 00:02:41,259 that this sequence is monotonically

105 00:02:41,270 –> 00:02:42,059 decreasing.

106 00:02:42,070 –> 00:02:43,699 So the limit is equal to

107 00:02:43,710 –> 00:02:44,199 the infimum.

108 00:02:45,119 –> 00:02:46,639 However, most importantly,

109 00:02:46,649 –> 00:02:47,990 we get the result that the

110 00:02:48,000 –> 00:02:49,470 whole spectral radius is

111 00:02:49,479 –> 00:02:51,130 bounded by the operator norm

112 00:02:51,139 –> 00:02:51,649 of T.

113 00:02:52,100 –> 00:02:53,309 Therefore, please always

114 00:02:53,320 –> 00:02:55,020 remember the spectral radius

115 00:02:55,029 –> 00:02:56,570 can never be bigger than

116 00:02:56,580 –> 00:02:57,669 the operator norm.

117 00:02:58,029 –> 00:02:59,350 And this is not something

118 00:02:59,360 –> 00:03:00,979 we can immediately prove

119 00:03:01,080 –> 00:03:01,850 for this.

120 00:03:01,860 –> 00:03:03,199 Let’s use the properties

121 00:03:03,210 –> 00:03:04,710 we used in the last video.

122 00:03:05,119 –> 00:03:06,669 For example, the Neumann

123 00:03:06,679 –> 00:03:08,350 series is very helpful here.

124 00:03:08,889 –> 00:03:10,369 Therefore, let’s take a complex

125 00:03:10,380 –> 00:03:12,029 number lambda which is in

126 00:03:12,039 –> 00:03:13,259 the absolute value greater

127 00:03:13,270 –> 00:03:14,410 than the operator norm.

128 00:03:14,669 –> 00:03:15,949 And now we want to show that

129 00:03:15,960 –> 00:03:17,389 this number can’t be in the

130 00:03:17,399 –> 00:03:18,509 spectrum of T.

131 00:03:18,789 –> 00:03:20,479 And in fact, this is immediately

132 00:03:20,490 –> 00:03:22,250 given by the Neumann series.

133 00:03:22,479 –> 00:03:24,169 It tells us that the identity

134 00:03:24,179 –> 00:03:25,860 operator minus an operator

135 00:03:25,869 –> 00:03:27,250 S is always

136 00:03:27,259 –> 00:03:28,979 invertible if the

137 00:03:28,990 –> 00:03:30,770 norm of S is less

138 00:03:30,779 –> 00:03:31,550 than one.

139 00:03:32,089 –> 00:03:33,820 And in this case, the inverse

140 00:03:33,830 –> 00:03:35,600 that exists is given by this

141 00:03:35,610 –> 00:03:36,750 infinite series.

142 00:03:37,289 –> 00:03:38,429 And now, in our case, of

143 00:03:38,440 –> 00:03:40,309 course, instead of S we take

144 00:03:40,320 –> 00:03:42,059 T divided by Lambda

145 00:03:42,529 –> 00:03:44,220 by assumption this operator

146 00:03:44,229 –> 00:03:45,869 here has operator norm

147 00:03:45,880 –> 00:03:47,080 less than one.

148 00:03:47,419 –> 00:03:48,529 And now you should see it’s

149 00:03:48,539 –> 00:03:50,330 a very short way from this

150 00:03:50,339 –> 00:03:51,789 to T minus Lambda,

151 00:03:52,309 –> 00:03:53,679 we just have to multiply

152 00:03:53,690 –> 00:03:55,440 this with minus one

153 00:03:55,449 –> 00:03:56,509 over Lambda.

154 00:03:57,139 –> 00:03:58,270 In the end, we get a very

155 00:03:58,279 –> 00:04:00,199 nice expression for T minus

156 00:04:00,210 –> 00:04:01,470 Lambda inverse.

157 00:04:01,850 –> 00:04:02,949 Of course, the important

158 00:04:02,960 –> 00:04:04,419 result is this inverse

159 00:04:04,429 –> 00:04:06,389 exists as a bounded linear

160 00:04:06,399 –> 00:04:07,149 operator.

161 00:04:07,679 –> 00:04:09,059 Moreover, I would say you

162 00:04:09,070 –> 00:04:10,339 should also remember this

163 00:04:10,350 –> 00:04:12,149 formula here that holds for

164 00:04:12,160 –> 00:04:13,660 Lambda that are large enough.

165 00:04:14,130 –> 00:04:14,500 OK.

166 00:04:14,509 –> 00:04:16,059 Now we conclude this lambda

167 00:04:16,119 –> 00:04:17,928 here is not in the spectrum.

168 00:04:17,940 –> 00:04:19,579 Therefore, this supreme here

169 00:04:19,589 –> 00:04:21,298 is indeed less or equal than

170 00:04:21,309 –> 00:04:22,829 the operator norm of T.

171 00:04:22,869 –> 00:04:24,760 In addition, this implies

172 00:04:24,769 –> 00:04:26,540 that the spectrum of T is

173 00:04:26,549 –> 00:04:27,250 bounded.

174 00:04:28,010 –> 00:04:29,579 And I already told you, we

175 00:04:29,589 –> 00:04:30,660 know from the last video

176 00:04:30,670 –> 00:04:32,059 that the set is closed.

177 00:04:32,070 –> 00:04:33,450 Therefore, we have proven

178 00:04:33,459 –> 00:04:35,190 a additionally, you

179 00:04:35,200 –> 00:04:37,079 see that one part of C is

180 00:04:37,089 –> 00:04:38,649 also already proven.

181 00:04:39,000 –> 00:04:40,450 However, now I want to skip

182 00:04:40,459 –> 00:04:41,929 the other parts of C and

183 00:04:41,940 –> 00:04:43,329 concentrate on B

184 00:04:43,739 –> 00:04:45,170 because there we can also

185 00:04:45,179 –> 00:04:46,600 use immediately some facts

186 00:04:46,609 –> 00:04:47,679 from the last video.

187 00:04:48,239 –> 00:04:49,440 So here we can do a proof

188 00:04:49,450 –> 00:04:50,470 by contraposition.

189 00:04:50,480 –> 00:04:51,559 So let’s assume that the

190 00:04:51,570 –> 00:04:53,480 spectrum of T is the empty

191 00:04:53,489 –> 00:04:53,869 set.

192 00:04:54,359 –> 00:04:55,869 No spectrum at all means

193 00:04:55,880 –> 00:04:57,130 that the resolvent set is

194 00:04:57,140 –> 00:04:58,470 the whole complex plane.

195 00:04:58,920 –> 00:05:00,839 So row of T is C

196 00:05:01,480 –> 00:05:03,000 and now you can recall the

197 00:05:03,010 –> 00:05:04,899 fact that the resolvent map

198 00:05:04,910 –> 00:05:06,179 is an analytic map

199 00:05:06,869 –> 00:05:08,470 more concretely the map that

200 00:05:08,480 –> 00:05:10,109 takes lambda and sends it

201 00:05:10,119 –> 00:05:11,290 to T minus lambda

202 00:05:11,309 –> 00:05:12,890 inverse can be

203 00:05:12,899 –> 00:05:14,640 locally expressed by a Taylor

204 00:05:14,649 –> 00:05:15,329 series.

205 00:05:15,619 –> 00:05:16,709 This is what we have shown

206 00:05:16,720 –> 00:05:17,570 last time.

207 00:05:17,579 –> 00:05:19,250 And now we can use it.

208 00:05:19,750 –> 00:05:20,070 OK.

209 00:05:20,079 –> 00:05:21,739 Since you know a lot of functional

210 00:05:21,750 –> 00:05:23,519 analysis at this point, it

211 00:05:23,529 –> 00:05:24,829 shouldn’t be a problem for

212 00:05:24,839 –> 00:05:25,059 you.

213 00:05:25,070 –> 00:05:26,579 When we now use the dual

214 00:05:26,589 –> 00:05:28,109 space of BX,

215 00:05:28,179 –> 00:05:30,089 this space of bounded linear

216 00:05:30,100 –> 00:05:32,049 operators is a Banach space.

217 00:05:32,059 –> 00:05:33,250 Therefore, it has a well

218 00:05:33,260 –> 00:05:34,649 defined dual space

219 00:05:35,149 –> 00:05:36,700 which I always denote with

220 00:05:36,709 –> 00:05:37,299 the prime.

221 00:05:37,940 –> 00:05:39,390 Therefore, this L is now

222 00:05:39,399 –> 00:05:41,100 a linear functional defined

223 00:05:41,109 –> 00:05:42,290 on the operators.

224 00:05:42,640 –> 00:05:44,019 And of course, now I want

225 00:05:44,029 –> 00:05:45,980 to apply the L to this

226 00:05:45,989 –> 00:05:47,010 operator here.

227 00:05:47,489 –> 00:05:48,660 Now you see we get a nice

228 00:05:48,670 –> 00:05:50,230 map fl from the resolvent

229 00:05:50,239 –> 00:05:51,519 set into C

230 00:05:51,950 –> 00:05:53,450 because the linear function

231 00:05:53,459 –> 00:05:55,279 in L maps into the complex

232 00:05:55,290 –> 00:05:55,880 numbers.

233 00:05:56,049 –> 00:05:57,600 Also by assumption we know

234 00:05:57,609 –> 00:05:59,000 the resolvent set is given

235 00:05:59,010 –> 00:06:00,600 by the whole complex plane.

236 00:06:00,940 –> 00:06:02,100 So what we have here is a

237 00:06:02,109 –> 00:06:03,600 very nice analytic

238 00:06:03,609 –> 00:06:05,000 function, an

239 00:06:05,010 –> 00:06:06,399 alternative name for this

240 00:06:06,410 –> 00:06:08,279 is the notion holomorphic.

241 00:06:08,670 –> 00:06:10,260 Therefore, now at this point,

242 00:06:10,269 –> 00:06:11,609 you can use everything you

243 00:06:11,619 –> 00:06:13,450 know from complex analysis.

244 00:06:14,000 –> 00:06:15,480 Indeed, what we get is that

245 00:06:15,489 –> 00:06:17,059 FL is a bounded

246 00:06:17,070 –> 00:06:18,320 entire function

247 00:06:18,790 –> 00:06:19,130 here.

248 00:06:19,140 –> 00:06:20,700 The term entire is not so

249 00:06:20,709 –> 00:06:21,640 complicated.

250 00:06:22,100 –> 00:06:23,570 It simply means analytic

251 00:06:23,579 –> 00:06:25,160 together with the domain

252 00:06:25,170 –> 00:06:25,679 C.

253 00:06:25,890 –> 00:06:27,480 So the entire complex

254 00:06:27,489 –> 00:06:28,959 plane is the domain

255 00:06:29,359 –> 00:06:30,480 and that the function is

256 00:06:30,489 –> 00:06:31,959 bounded we can easily show.

257 00:06:31,970 –> 00:06:33,859 Now of course, there we

258 00:06:33,869 –> 00:06:35,119 just have to look what happens

259 00:06:35,130 –> 00:06:36,790 for large lambda values.

260 00:06:37,329 –> 00:06:38,500 And here for us, it turns

261 00:06:38,510 –> 00:06:40,239 out that two times the operator

262 00:06:40,250 –> 00:06:41,920 norm of T is already

263 00:06:41,929 –> 00:06:43,820 large enough because for

264 00:06:43,829 –> 00:06:45,290 these LAMBDAS, we can simply

265 00:06:45,299 –> 00:06:46,970 use our equation star from

266 00:06:46,980 –> 00:06:47,500 above.

267 00:06:47,820 –> 00:06:49,369 More concretely, let’s take

268 00:06:49,380 –> 00:06:51,179 fl of lambda in the

269 00:06:51,190 –> 00:06:52,079 absolute value.

270 00:06:52,109 –> 00:06:53,880 Then by definition, this

271 00:06:53,890 –> 00:06:55,079 is less or equal than the

272 00:06:55,089 –> 00:06:56,510 operator norm of L

273 00:06:56,809 –> 00:06:58,570 times the operator norm of

274 00:06:58,579 –> 00:07:00,410 T minus lambda inverse.

275 00:07:00,839 –> 00:07:02,309 And exactly for this norm,

276 00:07:02,320 –> 00:07:03,829 we can use our Neumann series

277 00:07:03,839 –> 00:07:04,260 here.

278 00:07:04,750 –> 00:07:06,269 This means that this is less

279 00:07:06,279 –> 00:07:07,929 or equal where we have one

280 00:07:07,940 –> 00:07:09,290 over the absolute value of

281 00:07:09,299 –> 00:07:10,829 Lambda in front and then

282 00:07:10,839 –> 00:07:12,160 comes the infinite sum of

283 00:07:12,170 –> 00:07:14,089 the operator norm of T divided

284 00:07:14,100 –> 00:07:15,799 by lambda to the power K.

285 00:07:16,470 –> 00:07:17,869 And now, by assumption, we

286 00:07:17,880 –> 00:07:19,630 know this operator here

287 00:07:19,640 –> 00:07:21,380 has an operator norm which

288 00:07:21,390 –> 00:07:22,910 is less or equal than one

289 00:07:22,920 –> 00:07:23,410 half.

290 00:07:24,000 –> 00:07:25,790 So what remains here is just

291 00:07:25,799 –> 00:07:27,250 a geometric series, you can

292 00:07:27,260 –> 00:07:29,190 calculate more concretely.

293 00:07:29,200 –> 00:07:31,040 The value is less or equal

294 00:07:31,049 –> 00:07:31,609 than two.

295 00:07:32,119 –> 00:07:34,100 And because we can also estimate

296 00:07:34,109 –> 00:07:35,809 this one here, we have our

297 00:07:35,820 –> 00:07:37,660 bound, the absolute value

298 00:07:37,670 –> 00:07:39,329 of our function F Lambda

299 00:07:39,339 –> 00:07:41,290 is simply less or equal

300 00:07:41,299 –> 00:07:43,029 than the operator norm of

301 00:07:43,040 –> 00:07:44,829 L divided by the operator

302 00:07:44,839 –> 00:07:45,609 norm of T.

303 00:07:46,059 –> 00:07:46,440 OK.

304 00:07:46,450 –> 00:07:48,209 So this tells us that FL

305 00:07:48,220 –> 00:07:50,179 is indeed a bounded entire

306 00:07:50,190 –> 00:07:50,799 function.

307 00:07:51,369 –> 00:07:53,160 And now from complex analysis,

308 00:07:53,170 –> 00:07:55,079 we can apply Liouville’s

309 00:07:55,089 –> 00:07:56,929 theorem, it tells

310 00:07:56,940 –> 00:07:58,279 us that a bounded entire

311 00:07:58,290 –> 00:07:59,880 function needs to be a

312 00:07:59,890 –> 00:08:01,049 constant function.

313 00:08:01,440 –> 00:08:02,720 And please note here this

314 00:08:02,730 –> 00:08:04,309 fact holds no matter which

315 00:08:04,320 –> 00:08:06,160 linear functional L we choose.

316 00:08:06,869 –> 00:08:07,170 OK.

317 00:08:07,179 –> 00:08:08,230 Since we have a constant

318 00:08:08,239 –> 00:08:09,630 function, we can simply

319 00:08:09,640 –> 00:08:11,200 calculate this constant,

320 00:08:11,420 –> 00:08:12,880 we can simply put in one

321 00:08:12,890 –> 00:08:14,410 point into the function.

322 00:08:14,820 –> 00:08:16,029 And of course, the simplest

323 00:08:16,040 –> 00:08:17,410 one would be zero

324 00:08:17,829 –> 00:08:19,329 and there we just have L

325 00:08:19,339 –> 00:08:20,660 of T inverse.

326 00:08:21,149 –> 00:08:22,779 And now this value is the

327 00:08:22,790 –> 00:08:24,480 only value the function can

328 00:08:24,489 –> 00:08:25,459 have.

329 00:08:25,470 –> 00:08:27,269 Therefore, we have this equality

330 00:08:27,279 –> 00:08:28,890 for all complex numbers

331 00:08:28,899 –> 00:08:29,369 lambda.

332 00:08:30,040 –> 00:08:30,359 OK.

333 00:08:30,369 –> 00:08:31,540 Now this is the definition

334 00:08:31,549 –> 00:08:32,489 of the function.

335 00:08:32,530 –> 00:08:34,010 And now I want to put in

336 00:08:34,020 –> 00:08:35,830 what we already know of this

337 00:08:35,840 –> 00:08:36,539 inverse.

338 00:08:36,849 –> 00:08:38,340 As a reminder we often call

339 00:08:38,349 –> 00:08:40,058 this inverse simply the

340 00:08:40,070 –> 00:08:41,940 resolve from the

341 00:08:41,950 –> 00:08:43,739 last video we know this is

342 00:08:43,750 –> 00:08:44,580 analytic.

343 00:08:44,590 –> 00:08:45,780 Therefore, we can put in

344 00:08:45,789 –> 00:08:47,539 the Taylor series we calculated

345 00:08:47,549 –> 00:08:47,900 there.

346 00:08:48,460 –> 00:08:49,750 And it simply looked like

347 00:08:49,760 –> 00:08:51,460 this not so

348 00:08:51,469 –> 00:08:52,859 complicated because here

349 00:08:52,869 –> 00:08:54,150 you see the coefficients

350 00:08:54,219 –> 00:08:55,460 and there we have lambda

351 00:08:55,469 –> 00:08:56,719 minus lambda zero.

352 00:08:57,150 –> 00:08:58,289 And to make it simpler, we

353 00:08:58,299 –> 00:09:00,020 can just put lambda 0 to

354 00:09:00,030 –> 00:09:00,609 0.

355 00:09:01,000 –> 00:09:02,330 So we can simply erase it

356 00:09:02,340 –> 00:09:04,260 here and there.

357 00:09:05,380 –> 00:09:05,739 OK.

358 00:09:05,750 –> 00:09:07,400 Now this is much nicer and

359 00:09:07,409 –> 00:09:09,159 now we can pull in L

360 00:09:09,710 –> 00:09:11,000 and of course, then we get

361 00:09:11,010 –> 00:09:12,460 our Taylor series for

362 00:09:12,469 –> 00:09:14,109 fl so you see

363 00:09:14,119 –> 00:09:16,080 coefficients here and lambda

364 00:09:16,090 –> 00:09:17,239 to the power K here.

365 00:09:17,750 –> 00:09:19,299 OK, maybe still a little

366 00:09:19,309 –> 00:09:20,280 bit complicated.

367 00:09:20,289 –> 00:09:22,039 But remember we know the

368 00:09:22,049 –> 00:09:23,289 function is constant.

369 00:09:24,020 –> 00:09:25,710 Therefore all the coefficients

370 00:09:25,719 –> 00:09:27,229 after the term K is equal

371 00:09:27,239 –> 00:09:28,770 to zero will vanish.

372 00:09:29,299 –> 00:09:30,530 Therefore, for example, the

373 00:09:30,539 –> 00:09:31,750 term that corresponds to

374 00:09:31,760 –> 00:09:33,549 K is equal to one is

375 00:09:33,559 –> 00:09:34,030 zero.

376 00:09:34,650 –> 00:09:36,630 So this means L of T to the

377 00:09:36,640 –> 00:09:38,460 power minus two is zero.

378 00:09:38,960 –> 00:09:40,510 And again, as a reminder,

379 00:09:40,520 –> 00:09:42,070 this fact holds now for

380 00:09:42,080 –> 00:09:43,570 all linear functionals

381 00:09:44,239 –> 00:09:45,609 and now every time you see

382 00:09:45,619 –> 00:09:46,869 something like this, you

383 00:09:46,880 –> 00:09:47,989 should immediately think

384 00:09:48,000 –> 00:09:49,739 of the Hahn-Banach theorem.

385 00:09:49,969 –> 00:09:50,940 And in the case, you don’t

386 00:09:50,950 –> 00:09:51,619 remember it.

387 00:09:51,630 –> 00:09:53,260 Look at part 25.

388 00:09:53,270 –> 00:09:55,190 Again, it tells us

389 00:09:55,200 –> 00:09:56,900 that if we put in the same

390 00:09:56,909 –> 00:09:58,820 point into all functionals

391 00:09:58,979 –> 00:10:00,299 and the result is always

392 00:10:00,309 –> 00:10:02,109 zero, then the point has

393 00:10:02,119 –> 00:10:03,340 to be zero as well.

394 00:10:03,659 –> 00:10:05,140 So for us, this means this

395 00:10:05,150 –> 00:10:06,469 operator is the zero

396 00:10:06,479 –> 00:10:08,419 operator however, it’s an

397 00:10:08,429 –> 00:10:09,179 inverse.

398 00:10:09,190 –> 00:10:10,299 Therefore, this shouldn’t

399 00:10:10,309 –> 00:10:11,159 happen at all.

400 00:10:11,880 –> 00:10:13,289 The only case where this

401 00:10:13,299 –> 00:10:14,349 strange thing can happen

402 00:10:14,359 –> 00:10:16,190 is if we start with the trivial

403 00:10:16,200 –> 00:10:17,820 vector space and

404 00:10:17,830 –> 00:10:19,489 usually we don’t do that.

405 00:10:20,219 –> 00:10:21,580 Therefore, the result is

406 00:10:21,590 –> 00:10:23,080 that in all other interesting

407 00:10:23,090 –> 00:10:24,729 cases, especially for

408 00:10:24,739 –> 00:10:25,890 infinite dimensional vector

409 00:10:25,900 –> 00:10:27,619 spaces, the spectrum of an

410 00:10:27,630 –> 00:10:29,460 operator is never empty.

411 00:10:30,210 –> 00:10:30,570 OK.

412 00:10:30,580 –> 00:10:32,179 So this was a lot but I hope

413 00:10:32,190 –> 00:10:33,190 you could follow here.

414 00:10:33,700 –> 00:10:34,900 In that case, I see you in

415 00:10:34,909 –> 00:10:36,179 the next video when we go

416 00:10:36,190 –> 00:10:37,549 deeper into the field of

417 00:10:37,559 –> 00:10:38,789 spectral theory.

418 00:10:39,229 –> 00:10:40,150 Have a nice day.

419 00:10:40,159 –> 00:10:40,820 Bye.