• Title: Example

  • Series: Unbounded Operators

  • YouTube-Title: Unbounded Operators - Part 5 - Example

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    1 00:00:00,449 –> 00:00:02,109 Hello and welcome back

    2 00:00:02,119 –> 00:00:04,000 to unbounded operators,

    3 00:00:04,019 –> 00:00:05,260 the video series where we

    4 00:00:05,269 –> 00:00:06,880 talk a lot about some nice

    5 00:00:06,889 –> 00:00:09,179 topics of functional analysis.

    6 00:00:09,510 –> 00:00:12,630 And indeed, in today’s part 5 we will continue with

    7 00:00:12,640 –> 00:00:14,539 the closable operators by

    8 00:00:14,550 –> 00:00:16,068 looking at an example.

    9 00:00:16,629 –> 00:00:17,879 But you might already know

    10 00:00:17,889 –> 00:00:19,190 before we can do that,

    11 00:00:19,200 –> 00:00:20,639 I first want to thank all

    12 00:00:20,649 –> 00:00:21,889 the nice people who support

    13 00:00:21,899 –> 00:00:23,059 the channel on Steady here

    14 00:00:23,069 –> 00:00:24,870 on YouTube or on Patreon.

    15 00:00:25,610 –> 00:00:26,950 And please don’t forget on

    16 00:00:26,959 –> 00:00:27,950 these platforms

    17 00:00:27,959 –> 00:00:29,309 you find PDF versions,

    18 00:00:29,319 –> 00:00:32,939 quizzes and early access to all new videos.

    19 00:00:33,580 –> 00:00:34,000 OK.

    20 00:00:34,009 –> 00:00:35,310 Then let’s start this video

    21 00:00:35,319 –> 00:00:36,970 here by considering a

    22 00:00:36,979 –> 00:00:38,950 closable operator T

    23 00:00:39,560 –> 00:00:40,779 and there you already know

    24 00:00:40,790 –> 00:00:42,290 for the definition, we need

    25 00:00:42,299 –> 00:00:44,790 two normed spaces X and Y.

    26 00:00:45,169 –> 00:00:46,220 And at this point, I can

    27 00:00:46,229 –> 00:00:47,849 already tell you in order

    28 00:00:47,860 –> 00:00:49,580 to get nice theorems

    29 00:00:49,590 –> 00:00:51,380 out, we have to assume that

    30 00:00:51,389 –> 00:00:52,729 X and Y are Banach

    31 00:00:52,750 –> 00:00:53,549 spaces.

    32 00:00:53,659 –> 00:00:54,909 So complete normed

    33 00:00:54,919 –> 00:00:55,740 spaces.

    34 00:00:56,500 –> 00:00:58,439 In fact, also in this example

    35 00:00:58,450 –> 00:01:00,209 today, we will take Banach

    36 00:01:00,240 –> 00:01:00,959 spaces.

    37 00:01:01,770 –> 00:01:03,180 But let’s first write down

    38 00:01:03,189 –> 00:01:04,459 the general sequence

    39 00:01:04,470 –> 00:01:06,080 characterization for a

    40 00:01:06,089 –> 00:01:07,419 closable operator.

    41 00:01:08,110 –> 00:01:09,680 There, we just have to look

    42 00:01:09,690 –> 00:01:11,129 at sequences that

    43 00:01:11,139 –> 00:01:12,769 converge to the origin

    44 00:01:13,400 –> 00:01:14,819 and they have to have the

    45 00:01:14,830 –> 00:01:16,139 property that the

    46 00:01:16,150 –> 00:01:17,449 images also

    47 00:01:17,459 –> 00:01:19,180 converge and

    48 00:01:19,190 –> 00:01:21,059 then the conclusion is that

    49 00:01:21,069 –> 00:01:22,860 only the zero vector in

    50 00:01:22,870 –> 00:01:24,739 Y is possible for this

    51 00:01:24,750 –> 00:01:25,260 limit.

    52 00:01:26,000 –> 00:01:27,279 Therefore, please remember

    53 00:01:27,290 –> 00:01:28,699 here, if we have this

    54 00:01:28,709 –> 00:01:30,580 implication, we know that

    55 00:01:30,589 –> 00:01:32,430 the operator T is closable.

    56 00:01:33,410 –> 00:01:34,540 And this is exactly what

    57 00:01:34,550 –> 00:01:36,099 we now apply to the

    58 00:01:36,110 –> 00:01:37,650 following example.

    59 00:01:38,139 –> 00:01:39,319 And in order to make it very

    60 00:01:39,330 –> 00:01:41,089 concrete, let’s choose a

    61 00:01:41,099 –> 00:01:42,919 very nice Banach space for

    62 00:01:42,930 –> 00:01:44,900 X. Namely

    63 00:01:44,910 –> 00:01:46,739 X should be given as lower

    64 00:01:46,750 –> 00:01:47,940 case l^2.

    65 00:01:48,680 –> 00:01:50,099 So you know this is a space

    66 00:01:50,110 –> 00:01:51,629 of sequences with

    67 00:01:51,639 –> 00:01:53,400 index set N and

    68 00:01:53,410 –> 00:01:54,809 entries in C.

    69 00:01:55,650 –> 00:01:56,989 And please recall from the

    70 00:01:57,000 –> 00:01:58,699 functional analysis series

    71 00:01:58,760 –> 00:02:00,230 that this is a well defined

    72 00:02:00,239 –> 00:02:01,150 Banach space.

    73 00:02:01,830 –> 00:02:03,569 Moreover, it’s even a Hilbert

    74 00:02:03,580 –> 00:02:04,940 space, but this is not so

    75 00:02:04,949 –> 00:02:06,129 important right now.

    76 00:02:06,849 –> 00:02:07,349 OK.

    77 00:02:07,370 –> 00:02:09,089 In addition, let’s also fix

    78 00:02:09,100 –> 00:02:10,660 the canonical unit vectors

    79 00:02:10,669 –> 00:02:13,559 denoted by e_1, e_2, e_3 and so on.

    80 00:02:14,470 –> 00:02:15,889 Now, if you don’t know them,

    81 00:02:15,899 –> 00:02:17,369 then maybe very quickly,

    82 00:02:17,380 –> 00:02:18,550 e_2 is given as the

    83 00:02:18,559 –> 00:02:19,479 sequence

    84 00:02:19,850 –> 00:02:22,660 0100

    85 00:02:22,669 –> 00:02:23,410 and so on.

    86 00:02:23,850 –> 00:02:25,309 In other words, everything

    87 00:02:25,320 –> 00:02:26,929 is 0 in the sequence

    88 00:02:26,940 –> 00:02:28,649 except the second position,

    89 00:02:28,660 –> 00:02:29,610 which is a 1.

    90 00:02:30,389 –> 00:02:31,770 So you see this is exactly

    91 00:02:31,779 –> 00:02:33,009 the same as you would have

    92 00:02:33,020 –> 00:02:34,110 it in a finite dimensional

    93 00:02:34,119 –> 00:02:35,860 vector space, but now in

    94 00:02:35,869 –> 00:02:37,279 infinite dimensions.

    95 00:02:37,990 –> 00:02:38,429 OK.

    96 00:02:38,440 –> 00:02:39,699 And now I want the operator

    97 00:02:39,710 –> 00:02:41,490 T going from X to

    98 00:02:41,500 –> 00:02:42,940 C and

    99 00:02:42,949 –> 00:02:44,919 obviously C is also a normed

    100 00:02:44,929 –> 00:02:46,339 space, even a Banach

    101 00:02:46,520 –> 00:02:47,039 space.

    102 00:02:47,839 –> 00:02:49,259 Then in the next step, I

    103 00:02:49,270 –> 00:02:50,589 want to define the domain

    104 00:02:50,600 –> 00:02:52,410 of T and this

    105 00:02:52,419 –> 00:02:54,179 should be given as the span

    106 00:02:54,190 –> 00:02:55,520 of the canonical unit

    107 00:02:55,529 –> 00:02:56,220 vectors.

    108 00:02:57,139 –> 00:02:58,630 This means all the linear

    109 00:02:58,639 –> 00:03:00,220 combinations of the canonical

    110 00:03:00,229 –> 00:03:01,919 unit vectors lie in the

    111 00:03:01,929 –> 00:03:02,960 domain of T.

    112 00:03:03,639 –> 00:03:04,979 And here please note this

    113 00:03:04,990 –> 00:03:06,660 is not equal to the whole

    114 00:03:06,669 –> 00:03:08,300 space X, but it’s a

    115 00:03:08,309 –> 00:03:09,720 dense subset in it.

    116 00:03:10,520 –> 00:03:11,000 OK.

    117 00:03:11,009 –> 00:03:12,500 And now this domain makes

    118 00:03:12,509 –> 00:03:14,139 it very easy to write

    119 00:03:14,149 –> 00:03:15,949 down a nice definition for

    120 00:03:15,960 –> 00:03:16,960 an operator T.

    121 00:03:17,570 –> 00:03:19,550 Namely we just have to say

    122 00:03:19,559 –> 00:03:21,539 what D does to a canonical

    123 00:03:21,550 –> 00:03:22,320 unit vector.

    124 00:03:23,089 –> 00:03:24,350 And for example, we could

    125 00:03:24,360 –> 00:03:25,639 just say take the

    126 00:03:25,649 –> 00:03:27,509 corresponding index of the

    127 00:03:27,520 –> 00:03:28,320 unit vector.

    128 00:03:29,119 –> 00:03:30,759 Therefore T would

    129 00:03:30,770 –> 00:03:33,300 send e_2 to the number 2

    130 00:03:33,830 –> 00:03:34,940 and now, since we want a

    131 00:03:34,949 –> 00:03:36,729 linear map, the extension

    132 00:03:36,740 –> 00:03:38,440 to the whole span here is

    133 00:03:38,449 –> 00:03:39,570 uniquely given.

    134 00:03:40,020 –> 00:03:41,649 But for the sake of completeness,

    135 00:03:41,660 –> 00:03:43,000 let’s write it down.

    136 00:03:43,729 –> 00:03:45,589 So let’s say we have a finite

    137 00:03:45,600 –> 00:03:48,719 linear combination lambda_j times e_j.

    138 00:03:49,190 –> 00:03:50,490 So please note this here

    139 00:03:50,500 –> 00:03:52,070 is always a finite sum,

    140 00:03:52,080 –> 00:03:53,110 because this is what the

    141 00:03:53,119 –> 00:03:54,089 span means.

    142 00:03:54,100 –> 00:03:56,080 It means the set of the finite

    143 00:03:56,089 –> 00:03:57,990 linear combinations of these

    144 00:03:58,000 –> 00:03:58,710 vectors.

    145 00:03:59,270 –> 00:04:00,279 And therefore the mapping

    146 00:04:00,289 –> 00:04:01,559 here is totally clear, we

    147 00:04:01,570 –> 00:04:04,500 just add up lambda_j times j

    148 00:04:05,289 –> 00:04:06,880 This gives us a complex number

    149 00:04:06,889 –> 00:04:08,979 which is the output of T.

    150 00:04:09,619 –> 00:04:10,029 OK.

    151 00:04:10,039 –> 00:04:11,160 And the first question from

    152 00:04:11,169 –> 00:04:12,960 my side would be is

    153 00:04:12,970 –> 00:04:14,529 this an unbounded or a

    154 00:04:14,539 –> 00:04:15,660 bounded operator?

    155 00:04:16,250 –> 00:04:17,750 To answer that we can just

    156 00:04:17,760 –> 00:04:19,559 calculate the operator norm

    157 00:04:19,570 –> 00:04:20,343 of T.

    158 00:04:20,759 –> 00:04:22,609 By definition, this is the

    159 00:04:22,619 –> 00:04:24,410 supremum over all

    160 00:04:24,420 –> 00:04:26,760 inputs X with norm 1.

    162 00:04:27,459 –> 00:04:29,220 And then we look at the norm

    163 00:04:29,230 –> 00:04:30,480 of the images.

    164 00:04:31,000 –> 00:04:32,779 Which in this case is the

    165 00:04:32,790 –> 00:04:33,890 norm of C.

    166 00:04:34,700 –> 00:04:36,040 But of course, the norm in

    167 00:04:36,049 –> 00:04:37,820 C is just the absolute

    168 00:04:37,829 –> 00:04:39,709 value of complex numbers.

    169 00:04:40,230 –> 00:04:41,649 And now let’s say instead

    170 00:04:41,660 –> 00:04:43,459 of every vector x, we just

    171 00:04:43,470 –> 00:04:45,089 put in the canonical unit

    172 00:04:45,100 –> 00:04:46,980 vectors. Then the

    173 00:04:46,989 –> 00:04:48,880 whole supremum cannot get

    174 00:04:48,890 –> 00:04:49,440 bigger.

    175 00:04:49,760 –> 00:04:51,450 Then everything is much simpler.

    176 00:04:51,459 –> 00:04:52,850 We just have to calculate

    177 00:04:52,859 –> 00:04:54,279 the absolute value of

    178 00:04:54,290 –> 00:04:55,200 Te_j.

    179 00:04:55,700 –> 00:04:56,869 But by the definition of

    180 00:04:56,880 –> 00:04:58,600 the operator T, we already

    181 00:04:58,609 –> 00:05:00,770 know this is just the number j.

    182 00:05:01,450 –> 00:05:02,859 And at this point, we already

    183 00:05:02,869 –> 00:05:04,470 see the supremum over

    184 00:05:04,480 –> 00:05:06,269 j is not finite.

    185 00:05:06,940 –> 00:05:08,480 Therefore, the operator norm

    186 00:05:08,489 –> 00:05:10,279 of T can also not be

    187 00:05:10,290 –> 00:05:10,959 finite.

    188 00:05:11,660 –> 00:05:12,869 So the first conclusion here

    189 00:05:12,880 –> 00:05:14,660 is we have an unbounded

    190 00:05:14,670 –> 00:05:15,359 operator.

    191 00:05:16,089 –> 00:05:17,579 Therefore my next question

    192 00:05:17,589 –> 00:05:19,149 would be, is it at least a

    193 00:05:19,160 –> 00:05:20,660 closable operator?

    194 00:05:21,109 –> 00:05:22,500 And in order to answer this

    195 00:05:22,510 –> 00:05:24,380 question, we will use our

    196 00:05:24,390 –> 00:05:26,160 sequence property for closable

    197 00:05:26,170 –> 00:05:27,820 operators from before.

    198 00:05:28,369 –> 00:05:29,859 And we can combine that with

    199 00:05:29,869 –> 00:05:31,179 the fact that we have an

    200 00:05:31,190 –> 00:05:32,660 operator which is not

    201 00:05:32,670 –> 00:05:33,980 continuous at the

    202 00:05:33,989 –> 00:05:34,690 origin.

    203 00:05:35,089 –> 00:05:36,640 So please never forget for

    204 00:05:36,649 –> 00:05:38,450 linear maps, boundedness

    205 00:05:38,459 –> 00:05:39,959 and continuity are

    206 00:05:39,970 –> 00:05:40,720 connected.

    207 00:05:41,470 –> 00:05:41,920 OK.

    208 00:05:41,929 –> 00:05:43,250 Then let’s use this fact

    209 00:05:43,260 –> 00:05:44,799 here for a sequence that

    210 00:05:44,809 –> 00:05:46,079 converges to zero.

    211 00:05:46,739 –> 00:05:48,309 So what we can do is to choose

    212 00:05:48,320 –> 00:05:50,040 a sequence inside the

    213 00:05:50,049 –> 00:05:52,019 domain of T that converges

    214 00:05:52,029 –> 00:05:53,019 to the origin.

    215 00:05:53,309 –> 00:05:54,989 And now we know that we have

    216 00:05:55,000 –> 00:05:56,679 at least one sequence with

    217 00:05:56,690 –> 00:05:58,040 the property, that the

    218 00:05:58,049 –> 00:05:59,839 images don’t converge to

    219 00:05:59,850 –> 00:06:00,359 zero.

    220 00:06:01,049 –> 00:06:02,980 The operator T is not continuous

    221 00:06:02,989 –> 00:06:03,690 at zero.

    222 00:06:03,769 –> 00:06:05,070 So this is possible

    223 00:06:05,760 –> 00:06:07,350 however, still it could

    224 00:06:07,359 –> 00:06:09,170 happen that the images here

    225 00:06:09,179 –> 00:06:11,070 accumulate at zero.

    226 00:06:11,510 –> 00:06:13,190 But we also know that for

    227 00:06:13,200 –> 00:06:14,790 infinitely many sequence

    228 00:06:14,799 –> 00:06:16,500 members that does not happen.

    229 00:06:17,000 –> 00:06:18,459 So it’s possible to choose

    230 00:06:18,470 –> 00:06:20,369 a safety distant epsilon

    231 00:06:20,380 –> 00:06:21,269 from zero

    232 00:06:22,160 –> 00:06:23,739 and a subsequence

    233 00:06:23,750 –> 00:06:25,100 x_nk

    234 00:06:25,480 –> 00:06:27,239 such that the images

    235 00:06:27,250 –> 00:06:29,019 stay away from zero

    236 00:06:29,029 –> 00:06:30,769 with distance epsilon.

    237 00:06:31,290 –> 00:06:32,709 So the absolute value is

    238 00:06:32,720 –> 00:06:34,660 greater or equal than our

    239 00:06:34,670 –> 00:06:35,260 epsilon.

    240 00:06:35,899 –> 00:06:37,220 Therefore, now we can

    241 00:06:37,230 –> 00:06:39,029 conclude it’s allowed to

    242 00:06:39,040 –> 00:06:40,720 divide by this number.

    243 00:06:41,209 –> 00:06:42,829 Hence, we can define a new

    244 00:06:42,839 –> 00:06:44,679 sequence and let’s call it

    245 00:06:44,690 –> 00:06:45,640 z_k.

    246 00:06:46,079 –> 00:06:48,000 So we take x_nk and then

    247 00:06:48,010 –> 00:06:49,950 we divide this by the number

    248 00:06:49,959 –> 00:06:50,869 Tx_nk.

    249 00:06:51,540 –> 00:06:53,000 So this one is definitely

    250 00:06:53,010 –> 00:06:54,850 a well-defined element in

    251 00:06:54,859 –> 00:06:55,959 the domain of T.

    252 00:06:56,609 –> 00:06:58,549 And moreover, if we increase

    253 00:06:58,559 –> 00:07:00,269 k, we also see that

    254 00:07:00,279 –> 00:07:02,250 this sequence tends to zero.

    255 00:07:02,839 –> 00:07:04,450 So in this regard, it works

    256 00:07:04,459 –> 00:07:06,070 the same as our original

    257 00:07:06,079 –> 00:07:07,309 sequence x_n.

    258 00:07:08,140 –> 00:07:10,049 However, for the images,

    259 00:07:10,059 –> 00:07:11,850 something different happens,

    260 00:07:12,510 –> 00:07:13,970 namely, we immediately

    261 00:07:13,980 –> 00:07:16,269 see that we divide 1 by 1.

    262 00:07:16,279 –> 00:07:17,769 So the outcome is always

    263 00:07:17,779 –> 00:07:19,750

    1. No matter which

    264 00:07:19,760 –> 00:07:21,549 k from N we choose.

    265 00:07:22,220 –> 00:07:23,730 And with that, we now have

    266 00:07:23,739 –> 00:07:25,579 our counter example for the

    267 00:07:25,589 –> 00:07:27,570 characterization for a closable

    268 00:07:27,709 –> 00:07:28,940 operator with

    269 00:07:28,950 –> 00:07:29,720 sequences.

    270 00:07:30,500 –> 00:07:31,649 In fact, we can immediately

    271 00:07:31,660 –> 00:07:32,619 compare it here.

    272 00:07:32,670 –> 00:07:34,519 Now we have a sequence in

    273 00:07:34,529 –> 00:07:36,130 the domain of T which

    274 00:07:36,140 –> 00:07:37,579 converges to zero

    275 00:07:37,839 –> 00:07:39,470 and the images also

    276 00:07:39,480 –> 00:07:40,859 converge to a y,

    277 00:07:41,429 –> 00:07:43,220 in particular, the y in

    278 00:07:43,230 –> 00:07:45,620 our case is given by 1

    279 00:07:46,019 –> 00:07:47,660 and that’s already the contradiction

    280 00:07:47,670 –> 00:07:49,630 because y has to be equal

    281 00:07:49,640 –> 00:07:51,510 to zero if the operator

    282 00:07:51,519 –> 00:07:52,399 T is closable.

    283 00:07:53,209 –> 00:07:54,700 And with that, we can make

    284 00:07:54,709 –> 00:07:56,700 our conclusion T is not

    285 00:07:56,709 –> 00:07:58,040 a closable operator.

    286 00:07:58,730 –> 00:08:00,649 And this also includes that

    287 00:08:00,660 –> 00:08:02,269 T is not a closed operator

    288 00:08:02,279 –> 00:08:02,910 as well.

    289 00:08:03,459 –> 00:08:04,989 Therefore, this is important

    290 00:08:05,000 –> 00:08:06,510 to remember in infinite

    291 00:08:06,519 –> 00:08:08,290 dimensions, we can easily

    292 00:08:08,299 –> 00:08:09,579 define unbounded

    293 00:08:09,589 –> 00:08:11,290 operators that are not even

    294 00:08:11,299 –> 00:08:11,690 closable.

    295 00:08:12,589 –> 00:08:13,910 But I can already tell you

    296 00:08:13,920 –> 00:08:15,410 in this course, we will also

    297 00:08:15,420 –> 00:08:16,989 see a lot of closed

    298 00:08:17,000 –> 00:08:17,709 operators.

    299 00:08:17,720 –> 00:08:19,549 So a lot of positive examples.

    300 00:08:20,369 –> 00:08:20,790 OK.

    301 00:08:20,799 –> 00:08:21,790 Then I really hope we meet

    302 00:08:21,799 –> 00:08:23,190 again and have a nice day.

    303 00:08:23,200 –> 00:08:23,910 Bye bye.

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