
Title: Spectrum of Multiplication Operator

Series: Functional Analysis

YouTubeTitle: Functional Analysis 29  Spectrum of Multiplication Operator

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Subtitle in English
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 socalled
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 eigen vector
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.