1 00:00:00,319 –> 00:00:02,099 Hello and welcome back to

2 00:00:02,109 –> 00:00:03,819 functional analysis.

3 00:00:04,420 –> 00:00:05,699 And first, I want to thank

4 00:00:05,710 –> 00:00:06,860 all the nice people that

5 00:00:06,869 –> 00:00:08,420 support this channel on Steady

6 00:00:08,430 –> 00:00:09,220 or paypal.

7 00:00:09,880 –> 00:00:11,220 This is the second part in

8 00:00:11,229 –> 00:00:12,609 our spectral theory.

9 00:00:12,619 –> 00:00:13,729 And we will talk about

10 00:00:13,739 –> 00:00:14,869 examples.

11 00:00:15,369 –> 00:00:16,850 The overall assumption here

12 00:00:16,860 –> 00:00:18,170 is that we have a complex

13 00:00:18,180 –> 00:00:19,719 Banach space X and a

14 00:00:19,729 –> 00:00:21,239 bounded linear operated

15 00:00:21,250 –> 00:00:21,729 T.

16 00:00:22,290 –> 00:00:23,770 For this, we define the set

17 00:00:23,780 –> 00:00:25,489 sigma T, the so-called

18 00:00:25,500 –> 00:00:26,729 spectrum of T.

19 00:00:27,329 –> 00:00:29,049 Now you already know lambda

20 00:00:29,059 –> 00:00:30,850 is an element of Sigma T

21 00:00:30,860 –> 00:00:32,348 if and only if

22 00:00:32,950 –> 00:00:34,369 T minus lambda

23 00:00:34,380 –> 00:00:35,759 identity is not

24 00:00:35,770 –> 00:00:37,369 invertible as a bounded

25 00:00:37,380 –> 00:00:38,130 operator.

26 00:00:38,569 –> 00:00:39,810 The first new thing I can

27 00:00:39,819 –> 00:00:41,029 tell you today is that

28 00:00:41,040 –> 00:00:42,830 usually we will omit the

29 00:00:42,840 –> 00:00:44,689 identity in our notation.

30 00:00:45,110 –> 00:00:46,529 One can ignore the identity

31 00:00:46,540 –> 00:00:48,259 here because everyone knows

32 00:00:48,270 –> 00:00:49,549 what it should actually mean

33 00:00:49,709 –> 00:00:51,700 and we save space every time

34 00:00:51,709 –> 00:00:53,180 we use this expression.

35 00:00:53,720 –> 00:00:55,139 And of course, when we talk

36 00:00:55,150 –> 00:00:56,590 about the spectrum, this

37 00:00:56,599 –> 00:00:58,540 operator here occurs a lot.

38 00:00:58,869 –> 00:00:59,330 OK.

39 00:00:59,340 –> 00:01:00,639 Then let’s go immediately

40 00:01:00,650 –> 00:01:02,229 to our first example.

41 00:01:02,779 –> 00:01:03,840 A good start would be to

42 00:01:03,849 –> 00:01:05,769 look at a simple finite dimensional

43 00:01:05,779 –> 00:01:06,449 example.

44 00:01:07,110 –> 00:01:08,400 Therefore, our Banach space

45 00:01:08,410 –> 00:01:10,150 X is then given by

46 00:01:10,160 –> 00:01:12,019 CN and

47 00:01:12,029 –> 00:01:13,720 our linear operator T should

48 00:01:13,730 –> 00:01:15,160 act on any vector

49 00:01:15,169 –> 00:01:16,589 X like the

50 00:01:16,599 –> 00:01:18,199 diagonal matrix with

51 00:01:18,209 –> 00:01:19,809 entries lambda one, lambda

52 00:01:19,819 –> 00:01:21,139 two and so on, on the

53 00:01:21,150 –> 00:01:21,819 diagonal.

54 00:01:22,629 –> 00:01:23,940 In other words, the result

55 00:01:23,949 –> 00:01:25,660 of TX is simply the

56 00:01:25,669 –> 00:01:27,389 vector lambda one X

57 00:01:27,400 –> 00:01:29,000 one until lambda

58 00:01:29,010 –> 00:01:30,059 NXN.

59 00:01:30,680 –> 00:01:32,139 And here you see merely that

60 00:01:32,150 –> 00:01:33,959 the set lambda one to lambda

61 00:01:33,970 –> 00:01:35,529 N is exactly our

62 00:01:35,540 –> 00:01:36,580 spectrum of T.

63 00:01:37,269 –> 00:01:38,370 Now, please remember that

64 00:01:38,379 –> 00:01:39,769 we learned in the last video

65 00:01:39,779 –> 00:01:41,059 that in the finite dimensional

66 00:01:41,069 –> 00:01:42,529 case, all the points in the

67 00:01:42,540 –> 00:01:44,440 spectrum are eigenvalues.

68 00:01:44,900 –> 00:01:46,489 So the spectrum is equal

69 00:01:46,500 –> 00:01:48,330 to the so called point spectrum.

70 00:01:48,860 –> 00:01:50,279 Indeed, in this simple case

71 00:01:50,290 –> 00:01:51,389 here, we can immediately

72 00:01:51,400 –> 00:01:53,209 write down the eigenvectors.

73 00:01:53,589 –> 00:01:54,730 They are just given by the

74 00:01:54,739 –> 00:01:56,559 standard basis in CN.

75 00:01:57,220 –> 00:01:57,610 OK.

76 00:01:57,620 –> 00:01:58,680 I think that’s enough, we

77 00:01:58,690 –> 00:02:00,470 can say in finite dimensions,

78 00:02:00,970 –> 00:02:02,180 the interesting things happen

79 00:02:02,190 –> 00:02:03,709 of course, in an infinite

80 00:02:03,720 –> 00:02:05,569 dimensional case for

81 00:02:05,580 –> 00:02:07,150 this, let’s choose the LP

82 00:02:07,160 –> 00:02:07,779 space.

83 00:02:08,589 –> 00:02:09,940 We’ve already proven that

84 00:02:09,949 –> 00:02:11,339 this is a Banach space for

85 00:02:11,350 –> 00:02:12,750 P between one and

86 00:02:12,759 –> 00:02:13,490 infinity.

87 00:02:14,220 –> 00:02:15,800 So here you should see this

88 00:02:15,809 –> 00:02:17,580 is a straightforward generalization

89 00:02:17,589 –> 00:02:18,699 of our CN.

90 00:02:19,220 –> 00:02:20,600 The only difference is that

91 00:02:20,610 –> 00:02:22,529 a chosen vector X here does

92 00:02:22,539 –> 00:02:23,500 not have an end.

93 00:02:24,259 –> 00:02:25,320 Otherwise I want to have

94 00:02:25,330 –> 00:02:26,910 the operator T to do the

95 00:02:26,919 –> 00:02:28,110 same as before.

96 00:02:28,770 –> 00:02:29,899 Therefore, we could write

97 00:02:29,910 –> 00:02:31,880 this as a one sided infinite

98 00:02:31,889 –> 00:02:32,820 matrix.

99 00:02:33,059 –> 00:02:34,800 So you see this is the overall

100 00:02:34,809 –> 00:02:36,619 idea to find the generalization

101 00:02:36,630 –> 00:02:37,889 of the finite dimensional

102 00:02:37,899 –> 00:02:39,039 example from above.

103 00:02:39,580 –> 00:02:40,830 But of course, we really

104 00:02:40,839 –> 00:02:42,750 should state the formal definition

105 00:02:42,759 –> 00:02:44,020 of the operator T.

106 00:02:44,539 –> 00:02:46,179 So we take complex numbers

107 00:02:46,190 –> 00:02:47,880 lambda one lambda two and

108 00:02:47,889 –> 00:02:49,830 so on with the proper, that

109 00:02:49,839 –> 00:02:51,809 they form a bounded set.

110 00:02:52,619 –> 00:02:53,960 And this can be stated that

111 00:02:53,970 –> 00:02:55,500 the supreme of the absolute

112 00:02:55,509 –> 00:02:57,279 values is finite.

113 00:02:57,800 –> 00:02:59,119 With this, we then define

114 00:02:59,130 –> 00:03:00,960 T as a bounded linear

115 00:03:00,970 –> 00:03:02,770 operator from LP

116 00:03:02,809 –> 00:03:03,889 into LP

117 00:03:04,429 –> 00:03:06,320 simply by setting the J component

118 00:03:06,330 –> 00:03:08,080 of TX as lambda

119 00:03:08,089 –> 00:03:09,869 J times XJ.

120 00:03:10,440 –> 00:03:11,929 Of course, this fits in with

121 00:03:11,940 –> 00:03:13,410 our definition from above.

122 00:03:13,440 –> 00:03:15,119 But now you see we need this

123 00:03:15,130 –> 00:03:16,619 condition to get a bounded

124 00:03:16,630 –> 00:03:17,279 operator.

125 00:03:17,860 –> 00:03:18,320 OK.

126 00:03:18,330 –> 00:03:19,240 At this point, you should

127 00:03:19,250 –> 00:03:20,830 see what we did in the finite

128 00:03:20,839 –> 00:03:22,600 dimension example, still

129 00:03:22,610 –> 00:03:24,369 works in this infinite dimensional

130 00:03:24,380 –> 00:03:25,039 example.

131 00:03:25,639 –> 00:03:27,350 In particular, we find simple

132 00:03:27,460 –> 00:03:29,100 EIG vectors such that all

133 00:03:29,110 –> 00:03:30,429 these values are

134 00:03:30,460 –> 00:03:31,289 eigenvalues.

135 00:03:31,690 –> 00:03:33,089 So let’s immediately write

136 00:03:33,100 –> 00:03:33,880 that down.

137 00:03:34,350 –> 00:03:36,039 For example, E one, the

138 00:03:36,050 –> 00:03:36,740 sequence

139 00:03:36,750 –> 00:03:38,669 100 and so

140 00:03:38,679 –> 00:03:40,639 on is an eigenvector

141 00:03:40,649 –> 00:03:42,139 corresponding to the eigen

142 00:03:42,360 –> 00:03:43,039 value lambda.

143 00:03:43,050 –> 00:03:44,889 One of course, in the

144 00:03:44,899 –> 00:03:46,529 same way we have a two

145 00:03:46,539 –> 00:03:47,929 with one at the second

146 00:03:47,940 –> 00:03:49,690 position as an eyeing vector

147 00:03:49,699 –> 00:03:50,949 with corresponding eigenvalue

148 00:03:50,960 –> 00:03:52,130 will you lambda two?

149 00:03:52,899 –> 00:03:54,610 Hence, you see we continue

150 00:03:54,619 –> 00:03:55,899 this for all our

151 00:03:55,910 –> 00:03:56,660 lambdas.

152 00:03:57,419 –> 00:03:58,470 Therefore, we have shown

153 00:03:58,479 –> 00:04:00,429 now that all these lambdas

154 00:04:00,440 –> 00:04:02,130 lie in the spectrum of tea.

155 00:04:02,990 –> 00:04:04,559 In other words, at least

156 00:04:04,570 –> 00:04:05,910 we have a subset relation

157 00:04:05,919 –> 00:04:06,199 here.

158 00:04:06,869 –> 00:04:08,240 On the other hand, we immediately

159 00:04:08,250 –> 00:04:10,110 see any other complex

160 00:04:10,119 –> 00:04:12,050 number can’t be an eigenvalue

161 00:04:12,059 –> 00:04:12,610 of T.

162 00:04:13,309 –> 00:04:15,089 Therefore, this set is indeed

163 00:04:15,100 –> 00:04:16,589 the point spectrum of T.

164 00:04:17,209 –> 00:04:18,690 However, you already know

165 00:04:18,700 –> 00:04:20,358 this is in general not the

166 00:04:20,369 –> 00:04:21,670 whole spectrum of T.

167 00:04:22,200 –> 00:04:23,570 Indeed the important thing

168 00:04:23,579 –> 00:04:24,690 that could happen is that

169 00:04:24,700 –> 00:04:26,329 this set is actually

170 00:04:26,339 –> 00:04:28,320 infinite and then

171 00:04:28,329 –> 00:04:30,049 it has an accumulation point

172 00:04:30,630 –> 00:04:32,160 which is something that cannot

173 00:04:32,170 –> 00:04:33,829 happen in the finite dimensional

174 00:04:33,839 –> 00:04:34,290 case.

175 00:04:34,950 –> 00:04:35,329 OK.

176 00:04:35,339 –> 00:04:36,799 Then in the next step, let’s

177 00:04:36,809 –> 00:04:38,320 consider such an accumulation

178 00:04:38,329 –> 00:04:38,799 point.

179 00:04:39,369 –> 00:04:41,089 So let you be a complex

180 00:04:41,100 –> 00:04:42,519 number with the property

181 00:04:43,100 –> 00:04:44,519 that it is not in the set

182 00:04:44,529 –> 00:04:46,410 itself but

183 00:04:46,420 –> 00:04:47,929 in the closure of the set,

184 00:04:48,760 –> 00:04:49,059 OK.

185 00:04:49,070 –> 00:04:50,380 Maybe an example is very

186 00:04:50,390 –> 00:04:51,100 helpful here.

187 00:04:51,109 –> 00:04:52,529 So it could happen that Lambda

188 00:04:52,540 –> 00:04:54,399 J is given as one

189 00:04:54,410 –> 00:04:56,130 over J, then the

190 00:04:56,140 –> 00:04:57,579 only possible accumulation

191 00:04:57,589 –> 00:04:59,489 point would be mu as

192 00:04:59,500 –> 00:05:00,049 zero.

193 00:05:00,470 –> 00:05:01,970 Hence, in this example, we

194 00:05:01,980 –> 00:05:03,410 would have our eigenvalues

195 00:05:03,420 –> 00:05:04,600 as one over J.

196 00:05:04,609 –> 00:05:06,480 So they tend to zero but

197 00:05:06,489 –> 00:05:08,220 zero itself is not an eye

198 00:05:08,260 –> 00:05:08,850 value.

199 00:05:09,459 –> 00:05:10,880 In fact, this is the general

200 00:05:10,890 –> 00:05:12,570 result T minus

201 00:05:12,579 –> 00:05:14,109 mu is always

202 00:05:14,119 –> 00:05:14,920 injective.

203 00:05:15,470 –> 00:05:16,820 Now, at this point, you might

204 00:05:16,829 –> 00:05:18,329 already guess in the next

205 00:05:18,339 –> 00:05:19,529 step, we want to show that

206 00:05:19,540 –> 00:05:21,209 this operator is not

207 00:05:21,220 –> 00:05:21,929 subjective.

208 00:05:22,609 –> 00:05:23,910 Now, with this result, we

209 00:05:23,920 –> 00:05:25,910 know that Mu is in the spectrum

210 00:05:25,920 –> 00:05:27,640 of T as well for

211 00:05:27,649 –> 00:05:29,100 showing this, let’s use a

212 00:05:29,109 –> 00:05:30,559 proof by contradiction.

213 00:05:31,010 –> 00:05:32,739 Now assuming that the operator

214 00:05:32,750 –> 00:05:34,709 is surjective, then we know

215 00:05:34,720 –> 00:05:36,230 it’s also by bijective.

216 00:05:36,359 –> 00:05:38,109 Please remember the injectivity

217 00:05:38,119 –> 00:05:39,600 was not a problem at

218 00:05:39,609 –> 00:05:40,029 all.

219 00:05:40,630 –> 00:05:41,010 OK?

220 00:05:41,019 –> 00:05:42,859 And now we can use the famous

221 00:05:42,869 –> 00:05:44,670 bounded inverse theorem to

222 00:05:44,679 –> 00:05:46,269 conclude that the inverse

223 00:05:46,279 –> 00:05:48,029 of our operator is also

224 00:05:48,040 –> 00:05:48,709 bounded.

225 00:05:49,329 –> 00:05:50,730 Knowing this, let’s try

226 00:05:50,739 –> 00:05:52,540 calculating the corresponding

227 00:05:52,549 –> 00:05:53,489 operator norm.

228 00:05:54,100 –> 00:05:55,329 At least we know this is

229 00:05:55,339 –> 00:05:56,769 greater or equal

230 00:05:57,200 –> 00:05:58,149 than the norm.

231 00:05:58,160 –> 00:06:00,089 When we put in one of our

232 00:06:00,100 –> 00:06:00,899 e_js,

233 00:06:01,899 –> 00:06:03,529 please recall, this is just

234 00:06:03,540 –> 00:06:04,970 a sequence of zeros with

235 00:06:04,980 –> 00:06:06,309 the exception that at the

236 00:06:06,320 –> 00:06:08,160 J position there is a one

237 00:06:08,690 –> 00:06:10,160 for this reason, the acting

238 00:06:10,170 –> 00:06:11,750 of the inverse operator is

239 00:06:11,760 –> 00:06:13,000 easy to calculate.

240 00:06:13,459 –> 00:06:14,880 We just gave the inverse

241 00:06:14,890 –> 00:06:16,619 of the number lambda J

242 00:06:16,630 –> 00:06:17,600 minus mu.

243 00:06:18,390 –> 00:06:19,829 Also very simple is then

244 00:06:19,839 –> 00:06:21,630 calculating the P norm we

245 00:06:21,640 –> 00:06:23,029 just get the absolute value

246 00:06:23,040 –> 00:06:23,869 of this number.

247 00:06:24,649 –> 00:06:25,170 OK.

248 00:06:25,190 –> 00:06:26,660 And at this point, you should

249 00:06:26,670 –> 00:06:27,760 see a problem here.

250 00:06:28,390 –> 00:06:30,179 This estimate we have shown

251 00:06:30,190 –> 00:06:32,149 holds for all natural numbers

252 00:06:32,160 –> 00:06:32,619 J.

253 00:06:33,359 –> 00:06:35,040 And that’s a problem because

254 00:06:35,049 –> 00:06:36,619 we know with the lambdas,

255 00:06:36,850 –> 00:06:38,739 we can get as close as we

256 00:06:38,750 –> 00:06:39,549 want to mu.

257 00:06:40,279 –> 00:06:41,880 In other words, the reciprocal

258 00:06:41,890 –> 00:06:43,299 here will explode

259 00:06:44,019 –> 00:06:45,450 or more formally, you can

260 00:06:45,459 –> 00:06:46,989 choose a subsequence of the

261 00:06:47,000 –> 00:06:48,170 natural numbers here.

262 00:06:48,190 –> 00:06:49,660 And then this sequence goes

263 00:06:49,670 –> 00:06:50,480 to infinity.

264 00:06:51,299 –> 00:06:52,720 So no matter how we put it,

265 00:06:52,730 –> 00:06:54,600 this number here can’t be

266 00:06:54,609 –> 00:06:55,230 finite.

267 00:06:55,899 –> 00:06:57,320 And that’s the contradiction

268 00:06:57,329 –> 00:06:58,600 because we know it should

269 00:06:58,609 –> 00:06:59,239 be bounded.

270 00:07:00,100 –> 00:07:01,359 Therefore, our conclusion

271 00:07:01,369 –> 00:07:02,989 is it’s not subjective.

272 00:07:03,829 –> 00:07:04,209 OK.

273 00:07:04,220 –> 00:07:05,850 Let’s summarize what we have

274 00:07:05,859 –> 00:07:06,410 found.

275 00:07:06,980 –> 00:07:08,700 The spectrum of T contains

276 00:07:08,709 –> 00:07:10,049 our lambda values

277 00:07:10,760 –> 00:07:12,630 and also the accumulation

278 00:07:12,640 –> 00:07:13,160 points.

279 00:07:13,200 –> 00:07:15,049 You now it’s not

280 00:07:15,059 –> 00:07:16,480 hard to show that these are

281 00:07:16,489 –> 00:07:18,309 indeed the only possibilities

282 00:07:18,320 –> 00:07:19,170 for the spectrum.

283 00:07:19,730 –> 00:07:21,130 So this is what the spectrum

284 00:07:21,140 –> 00:07:22,809 of the operator T looks

285 00:07:22,820 –> 00:07:23,200 like.

286 00:07:23,850 –> 00:07:25,019 The first part is the point

287 00:07:25,029 –> 00:07:26,570 spectrum of T, the eigen

288 00:07:26,579 –> 00:07:27,260 values.

289 00:07:27,589 –> 00:07:28,869 And the second part is the

290 00:07:28,880 –> 00:07:30,510 continuous spectrum together

291 00:07:30,519 –> 00:07:32,029 with the residual spectrum

292 00:07:32,660 –> 00:07:34,230 in fact, later, we will be

293 00:07:34,239 –> 00:07:35,929 able to show that for P

294 00:07:35,940 –> 00:07:37,760 less than infinity, the

295 00:07:37,769 –> 00:07:39,510 residual spectrum is indeed

296 00:07:39,519 –> 00:07:40,109 empty.

297 00:07:40,799 –> 00:07:42,459 Therefore, in our case, this

298 00:07:42,470 –> 00:07:43,899 is actually the continuous

299 00:07:43,910 –> 00:07:44,540 spectrum.

300 00:07:45,239 –> 00:07:45,720 OK.

301 00:07:45,730 –> 00:07:46,700 I think that’s good enough

302 00:07:46,709 –> 00:07:48,179 for this example today,

303 00:07:48,910 –> 00:07:50,290 let’s use the next videos

304 00:07:50,299 –> 00:07:52,140 to talk about general results

305 00:07:52,149 –> 00:07:53,329 about the spectrum.

306 00:07:53,809 –> 00:07:55,079 Therefore, I hope I see you

307 00:07:55,089 –> 00:07:56,609 there and have a nice day.

308 00:07:56,739 –> 00:07:57,440 Bye.