Mixing
Compact operator $L:ell^2toell^2$ with $Vert LVert=1$ such that $Vert L(x)Vert<Vert xVert$ for all $x$
$begingroup$
Let $ell^2$ denote the space of square summable sequences of complex numbers. Let $L:ell^2toell^2$ be a linear operator with $Vert LVert=1$ such that for all $xinell^2setminus{0}$, $Vert L(x)Vert_2<Vert xVert_2$. Give an example of a compact operator with these properties or show that no such compact operator exists.
I thought of some bounded operators with the above properties and none of them were compact, so I feel like no such compact operator exists. Since $Vert LVert=sup_{Vert xVert_2=1}Vert L(x)Vert_2=1$, there exists a sequence $x_ninell^2$ such that $Vert x_nVert_2=1$ for all $n$ and $Vert L(x_n)Vert_2to 1$. My idea was to show that no subsequence of $L(x_n)$ converges (as that tended to be the case in the examples I thought of) which would show that the image of the closed unit ball is not contained in any compact set and hence, $L$ is not compact. But I'm not sure whether that is actually true, and if it is true, I'm not sure how to prove it.
functional-analysis operator-theory compact-operators
asked Jan 16 at 23:30
AnonymousAnonymous
29713
$endgroup$
add a comment |
$begingroup$
Let $ell^2$ denote the space of square summable sequences of complex numbers. Let $L:ell^2toell^2$ be a linear operator with $Vert LVert=1$ such that for all $xinell^2setminus{0}$, $Vert L(x)Vert_2<Vert xVert_2$. Give an example of a compact operator with these properties or show that no such compact operator exists.
I thought of some bounded operators with the above properties and none of them were compact, so I feel like no such compact operator exists. Since $Vert LVert=sup_{Vert xVert_2=1}Vert L(x)Vert_2=1$, there exists a sequence $x_ninell^2$ such that $Vert x_nVert_2=1$ for all $n$ and $Vert L(x_n)Vert_2to 1$. My idea was to show that no subsequence of $L(x_n)$ converges (as that tended to be the case in the examples I thought of) which would show that the image of the closed unit ball is not contained in any compact set and hence, $L$ is not compact. But I'm not sure whether that is actually true, and if it is true, I'm not sure how to prove it.
functional-analysis operator-theory compact-operators
asked Jan 16 at 23:30
AnonymousAnonymous
29713
$endgroup$
3
$begingroup$
The eigenvalues of compact operators can only accumulate at 0. Guess no.
$endgroup$
– JRen
Jan 16 at 23:40
$begingroup$
@T.Bongers I don't get the part where you assume that $x_n$ can pass to a subsequence that converges. Why is that so?
$endgroup$
– BigbearZzz
Jan 16 at 23:46
$begingroup$
@JRen Oh, that's right, I totally forgot that property. For any compact operator, that operator only has finitely many eigenvalues outside a ball centered at the origin of the complex plane. Thus, the sequence $x_n$ can't have the property that $Vert L(x_n)Vert_2to 1$ if $L$ were compact. Thanks!
$endgroup$
– Anonymous
Jan 16 at 23:48
$begingroup$
@JRen On second thought, looking at the eigenvalues doesn't appear to work. The mapping $L(x_1,x_2,...)=(0,0,frac{1}{2}x_2,frac{2}{3}x_3,frac{3}{4}x_4,...)$ is a (not compact) bounded linear operator with the above properties which has no nonzero eigenvalues.
$endgroup$
– Anonymous
Jan 17 at 0:29
$begingroup$
@Anonymous I'm not generating to all the bounded linear operator.
$endgroup$
– JRen
Jan 17 at 17:13
add a comment |
$begingroup$
Let $ell^2$ denote the space of square summable sequences of complex numbers. Let $L:ell^2toell^2$ be a linear operator with $Vert LVert=1$ such that for all $xinell^2setminus{0}$, $Vert L(x)Vert_2<Vert xVert_2$. Give an example of a compact operator with these properties or show that no such compact operator exists.
I thought of some bounded operators with the above properties and none of them were compact, so I feel like no such compact operator exists. Since $Vert LVert=sup_{Vert xVert_2=1}Vert L(x)Vert_2=1$, there exists a sequence $x_ninell^2$ such that $Vert x_nVert_2=1$ for all $n$ and $Vert L(x_n)Vert_2to 1$. My idea was to show that no subsequence of $L(x_n)$ converges (as that tended to be the case in the examples I thought of) which would show that the image of the closed unit ball is not contained in any compact set and hence, $L$ is not compact. But I'm not sure whether that is actually true, and if it is true, I'm not sure how to prove it.
functional-analysis operator-theory compact-operators
asked Jan 16 at 23:30
AnonymousAnonymous
29713
$endgroup$
Let $ell^2$ denote the space of square summable sequences of complex numbers. Let $L:ell^2toell^2$ be a linear operator with $Vert LVert=1$ such that for all $xinell^2setminus{0}$, $Vert L(x)Vert_2<Vert xVert_2$. Give an example of a compact operator with these properties or show that no such compact operator exists.
I thought of some bounded operators with the above properties and none of them were compact, so I feel like no such compact operator exists. Since $Vert LVert=sup_{Vert xVert_2=1}Vert L(x)Vert_2=1$, there exists a sequence $x_ninell^2$ such that $Vert x_nVert_2=1$ for all $n$ and $Vert L(x_n)Vert_2to 1$. My idea was to show that no subsequence of $L(x_n)$ converges (as that tended to be the case in the examples I thought of) which would show that the image of the closed unit ball is not contained in any compact set and hence, $L$ is not compact. But I'm not sure whether that is actually true, and if it is true, I'm not sure how to prove it.
functional-analysis operator-theory compact-operators
functional-analysis operator-theory compact-operators
asked Jan 16 at 23:30
AnonymousAnonymous
29713
asked Jan 16 at 23:30
AnonymousAnonymous
29713
edited Jan 17 at 0:31
Anonymous
asked Jan 16 at 23:30
AnonymousAnonymous
29713
asked Jan 16 at 23:30
AnonymousAnonymous
29713
asked Jan 16 at 23:30
AnonymousAnonymous
29713
29713
3
$begingroup$
The eigenvalues of compact operators can only accumulate at 0. Guess no.
$endgroup$
– JRen
Jan 16 at 23:40
$begingroup$
@T.Bongers I don't get the part where you assume that $x_n$ can pass to a subsequence that converges. Why is that so?
$endgroup$
– BigbearZzz
Jan 16 at 23:46
$begingroup$
@JRen Oh, that's right, I totally forgot that property. For any compact operator, that operator only has finitely many eigenvalues outside a ball centered at the origin of the complex plane. Thus, the sequence $x_n$ can't have the property that $Vert L(x_n)Vert_2to 1$ if $L$ were compact. Thanks!
$endgroup$
– Anonymous
Jan 16 at 23:48
$begingroup$
@JRen On second thought, looking at the eigenvalues doesn't appear to work. The mapping $L(x_1,x_2,...)=(0,0,frac{1}{2}x_2,frac{2}{3}x_3,frac{3}{4}x_4,...)$ is a (not compact) bounded linear operator with the above properties which has no nonzero eigenvalues.
$endgroup$
– Anonymous
Jan 17 at 0:29
$begingroup$
@Anonymous I'm not generating to all the bounded linear operator.
$endgroup$
– JRen
Jan 17 at 17:13
add a comment |
3
$begingroup$
The eigenvalues of compact operators can only accumulate at 0. Guess no.
$endgroup$
– JRen
Jan 16 at 23:40
$begingroup$
@T.Bongers I don't get the part where you assume that $x_n$ can pass to a subsequence that converges. Why is that so?
$endgroup$
– BigbearZzz
Jan 16 at 23:46
$begingroup$
@JRen Oh, that's right, I totally forgot that property. For any compact operator, that operator only has finitely many eigenvalues outside a ball centered at the origin of the complex plane. Thus, the sequence $x_n$ can't have the property that $Vert L(x_n)Vert_2to 1$ if $L$ were compact. Thanks!
$endgroup$
– Anonymous
Jan 16 at 23:48
$begingroup$
@JRen On second thought, looking at the eigenvalues doesn't appear to work. The mapping $L(x_1,x_2,...)=(0,0,frac{1}{2}x_2,frac{2}{3}x_3,frac{3}{4}x_4,...)$ is a (not compact) bounded linear operator with the above properties which has no nonzero eigenvalues.
$endgroup$
– Anonymous
Jan 17 at 0:29
$begingroup$
@Anonymous I'm not generating to all the bounded linear operator.
$endgroup$
– JRen
Jan 17 at 17:13
3
3
$begingroup$
The eigenvalues of compact operators can only accumulate at 0. Guess no.
$endgroup$
– JRen
Jan 16 at 23:40
$begingroup$
The eigenvalues of compact operators can only accumulate at 0. Guess no.
$endgroup$
– JRen
Jan 16 at 23:40
$begingroup$
@T.Bongers I don't get the part where you assume that $x_n$ can pass to a subsequence that converges. Why is that so?
$endgroup$
– BigbearZzz
Jan 16 at 23:46
$begingroup$
@T.Bongers I don't get the part where you assume that $x_n$ can pass to a subsequence that converges. Why is that so?
$endgroup$
– BigbearZzz
Jan 16 at 23:46
$begingroup$
@JRen Oh, that's right, I totally forgot that property. For any compact operator, that operator only has finitely many eigenvalues outside a ball centered at the origin of the complex plane. Thus, the sequence $x_n$ can't have the property that $Vert L(x_n)Vert_2to 1$ if $L$ were compact. Thanks!
$endgroup$
– Anonymous
Jan 16 at 23:48
$begingroup$
@JRen Oh, that's right, I totally forgot that property. For any compact operator, that operator only has finitely many eigenvalues outside a ball centered at the origin of the complex plane. Thus, the sequence $x_n$ can't have the property that $Vert L(x_n)Vert_2to 1$ if $L$ were compact. Thanks!
$endgroup$
– Anonymous
Jan 16 at 23:48
$begingroup$
@JRen On second thought, looking at the eigenvalues doesn't appear to work. The mapping $L(x_1,x_2,...)=(0,0,frac{1}{2}x_2,frac{2}{3}x_3,frac{3}{4}x_4,...)$ is a (not compact) bounded linear operator with the above properties which has no nonzero eigenvalues.
$endgroup$
– Anonymous
Jan 17 at 0:29
$begingroup$
@JRen On second thought, looking at the eigenvalues doesn't appear to work. The mapping $L(x_1,x_2,...)=(0,0,frac{1}{2}x_2,frac{2}{3}x_3,frac{3}{4}x_4,...)$ is a (not compact) bounded linear operator with the above properties which has no nonzero eigenvalues.
$endgroup$
– Anonymous
Jan 17 at 0:29
$begingroup$
@Anonymous I'm not generating to all the bounded linear operator.
$endgroup$
– JRen
Jan 17 at 17:13
$begingroup$
@Anonymous I'm not generating to all the bounded linear operator.
$endgroup$
– JRen
Jan 17 at 17:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $T$ be a compact operator with $|T|=1$. Then there is a sequence $(x_n)$ in $H$ with $|x_n|=1$ with $|Tx_n|to1$ as $ntoinfty$. As $T$ is compact, there is a subsequence $(x_{n_k})$ of $(x_n)$ such that $Tx_{n_k}to y$ for some $yin H$. But as the image closed unit ball $B$ under $T$ is compact (see Douglas, Banach Algebra Techniques in Operator Theory, Corollary 5.4), it follows that $yin T(B)$, i.e., there is some $xin H$ with $|x|leq 1$ and $y=Tx$. But then $|x|=1$, and $1=|y|=|Tx|$.
answered Jan 16 at 23:48
AweyganAweygan
14.3k21441
$endgroup$
2
$begingroup$
math.stackexchange.com/questions/404516/…
$endgroup$
– Tsemo Aristide
Jan 16 at 23:53
1
$begingroup$
The image of the closed unit ball under a compact operator is relatively compact. Why is it compact?
$endgroup$
– Kavi Rama Murthy
Jan 16 at 23:56
1
$begingroup$
@KaviRamaMurthy It's actually compact in the Hilbert space setting. See the book I referenced.
$endgroup$
– Aweygan
Jan 16 at 23:58
1
$begingroup$
Thanks guys for educating me on the image of the closed unit ball under a compact operator under reflexivity.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 0:36
1
$begingroup$
@Anonymous I would be very interested to see if the approach hinted at by JRen does work. It's fairly easy if you assume the operator is normal, but in general I can't see it.
$endgroup$
– Aweygan
Jan 17 at 0:58
|
show 3 more comments
$begingroup$
The same argument without touching sequences: As $T$ is weakly continuous and the unit ball of $ell^2$ is weakly compact its impage $T(B)$ is weakly compact hence weakly closed and thus closed in $ell^2$. At the same time it is relatively compact and thus compact. Hence the continuous map $ymapsto |y|$ realizes its supremum on $T(B)$.
answered Jan 17 at 13:41
JochenJochen
6,8131023
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $T$ be a compact operator with $|T|=1$. Then there is a sequence $(x_n)$ in $H$ with $|x_n|=1$ with $|Tx_n|to1$ as $ntoinfty$. As $T$ is compact, there is a subsequence $(x_{n_k})$ of $(x_n)$ such that $Tx_{n_k}to y$ for some $yin H$. But as the image closed unit ball $B$ under $T$ is compact (see Douglas, Banach Algebra Techniques in Operator Theory, Corollary 5.4), it follows that $yin T(B)$, i.e., there is some $xin H$ with $|x|leq 1$ and $y=Tx$. But then $|x|=1$, and $1=|y|=|Tx|$.
answered Jan 16 at 23:48
AweyganAweygan
14.3k21441
$endgroup$
2
$begingroup$
math.stackexchange.com/questions/404516/…
$endgroup$
– Tsemo Aristide
Jan 16 at 23:53
1
$begingroup$
The image of the closed unit ball under a compact operator is relatively compact. Why is it compact?
$endgroup$
– Kavi Rama Murthy
Jan 16 at 23:56
1
$begingroup$
@KaviRamaMurthy It's actually compact in the Hilbert space setting. See the book I referenced.
$endgroup$
– Aweygan
Jan 16 at 23:58
1
$begingroup$
Thanks guys for educating me on the image of the closed unit ball under a compact operator under reflexivity.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 0:36
1
$begingroup$
@Anonymous I would be very interested to see if the approach hinted at by JRen does work. It's fairly easy if you assume the operator is normal, but in general I can't see it.
$endgroup$
– Aweygan
Jan 17 at 0:58
|
show 3 more comments
$begingroup$
Let $T$ be a compact operator with $|T|=1$. Then there is a sequence $(x_n)$ in $H$ with $|x_n|=1$ with $|Tx_n|to1$ as $ntoinfty$. As $T$ is compact, there is a subsequence $(x_{n_k})$ of $(x_n)$ such that $Tx_{n_k}to y$ for some $yin H$. But as the image closed unit ball $B$ under $T$ is compact (see Douglas, Banach Algebra Techniques in Operator Theory, Corollary 5.4), it follows that $yin T(B)$, i.e., there is some $xin H$ with $|x|leq 1$ and $y=Tx$. But then $|x|=1$, and $1=|y|=|Tx|$.
answered Jan 16 at 23:48
AweyganAweygan
14.3k21441
$endgroup$
2
$begingroup$
math.stackexchange.com/questions/404516/…
$endgroup$
– Tsemo Aristide
Jan 16 at 23:53
1
$begingroup$
The image of the closed unit ball under a compact operator is relatively compact. Why is it compact?
$endgroup$
– Kavi Rama Murthy
Jan 16 at 23:56
1
$begingroup$
@KaviRamaMurthy It's actually compact in the Hilbert space setting. See the book I referenced.
$endgroup$
– Aweygan
Jan 16 at 23:58
1
$begingroup$
Thanks guys for educating me on the image of the closed unit ball under a compact operator under reflexivity.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 0:36
1
$begingroup$
@Anonymous I would be very interested to see if the approach hinted at by JRen does work. It's fairly easy if you assume the operator is normal, but in general I can't see it.
$endgroup$
– Aweygan
Jan 17 at 0:58
|
show 3 more comments
$begingroup$
Let $T$ be a compact operator with $|T|=1$. Then there is a sequence $(x_n)$ in $H$ with $|x_n|=1$ with $|Tx_n|to1$ as $ntoinfty$. As $T$ is compact, there is a subsequence $(x_{n_k})$ of $(x_n)$ such that $Tx_{n_k}to y$ for some $yin H$. But as the image closed unit ball $B$ under $T$ is compact (see Douglas, Banach Algebra Techniques in Operator Theory, Corollary 5.4), it follows that $yin T(B)$, i.e., there is some $xin H$ with $|x|leq 1$ and $y=Tx$. But then $|x|=1$, and $1=|y|=|Tx|$.
answered Jan 16 at 23:48
AweyganAweygan
14.3k21441
$endgroup$
Let $T$ be a compact operator with $|T|=1$. Then there is a sequence $(x_n)$ in $H$ with $|x_n|=1$ with $|Tx_n|to1$ as $ntoinfty$. As $T$ is compact, there is a subsequence $(x_{n_k})$ of $(x_n)$ such that $Tx_{n_k}to y$ for some $yin H$. But as the image closed unit ball $B$ under $T$ is compact (see Douglas, Banach Algebra Techniques in Operator Theory, Corollary 5.4), it follows that $yin T(B)$, i.e., there is some $xin H$ with $|x|leq 1$ and $y=Tx$. But then $|x|=1$, and $1=|y|=|Tx|$.
answered Jan 16 at 23:48
AweyganAweygan
14.3k21441
edited Jan 16 at 23:56
answered Jan 16 at 23:48
AweyganAweygan
14.3k21441
answered Jan 16 at 23:48
AweyganAweygan
14.3k21441
answered Jan 16 at 23:48
AweyganAweygan
14.3k21441
14.3k21441
2
$begingroup$
math.stackexchange.com/questions/404516/…
$endgroup$
– Tsemo Aristide
Jan 16 at 23:53
1
$begingroup$
The image of the closed unit ball under a compact operator is relatively compact. Why is it compact?
$endgroup$
– Kavi Rama Murthy
Jan 16 at 23:56
1
$begingroup$
@KaviRamaMurthy It's actually compact in the Hilbert space setting. See the book I referenced.
$endgroup$
– Aweygan
Jan 16 at 23:58
1
$begingroup$
Thanks guys for educating me on the image of the closed unit ball under a compact operator under reflexivity.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 0:36
1
$begingroup$
@Anonymous I would be very interested to see if the approach hinted at by JRen does work. It's fairly easy if you assume the operator is normal, but in general I can't see it.
$endgroup$
– Aweygan
Jan 17 at 0:58
|
show 3 more comments
2
$begingroup$
math.stackexchange.com/questions/404516/…
$endgroup$
– Tsemo Aristide
Jan 16 at 23:53
1
$begingroup$
The image of the closed unit ball under a compact operator is relatively compact. Why is it compact?
$endgroup$
– Kavi Rama Murthy
Jan 16 at 23:56
1
$begingroup$
@KaviRamaMurthy It's actually compact in the Hilbert space setting. See the book I referenced.
$endgroup$
– Aweygan
Jan 16 at 23:58
1
$begingroup$
Thanks guys for educating me on the image of the closed unit ball under a compact operator under reflexivity.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 0:36
1
$begingroup$
@Anonymous I would be very interested to see if the approach hinted at by JRen does work. It's fairly easy if you assume the operator is normal, but in general I can't see it.
$endgroup$
– Aweygan
Jan 17 at 0:58
2
2
$begingroup$
math.stackexchange.com/questions/404516/…
$endgroup$
– Tsemo Aristide
Jan 16 at 23:53
$begingroup$
math.stackexchange.com/questions/404516/…
$endgroup$
– Tsemo Aristide
Jan 16 at 23:53
1
1
$begingroup$
The image of the closed unit ball under a compact operator is relatively compact. Why is it compact?
$endgroup$
– Kavi Rama Murthy
Jan 16 at 23:56
$begingroup$
The image of the closed unit ball under a compact operator is relatively compact. Why is it compact?
$endgroup$
– Kavi Rama Murthy
Jan 16 at 23:56
1
1
$begingroup$
@KaviRamaMurthy It's actually compact in the Hilbert space setting. See the book I referenced.
$endgroup$
– Aweygan
Jan 16 at 23:58
$begingroup$
@KaviRamaMurthy It's actually compact in the Hilbert space setting. See the book I referenced.
$endgroup$
– Aweygan
Jan 16 at 23:58
1
1
$begingroup$
Thanks guys for educating me on the image of the closed unit ball under a compact operator under reflexivity.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 0:36
$begingroup$
Thanks guys for educating me on the image of the closed unit ball under a compact operator under reflexivity.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 0:36
1
1
$begingroup$
@Anonymous I would be very interested to see if the approach hinted at by JRen does work. It's fairly easy if you assume the operator is normal, but in general I can't see it.
$endgroup$
– Aweygan
Jan 17 at 0:58
$begingroup$
@Anonymous I would be very interested to see if the approach hinted at by JRen does work. It's fairly easy if you assume the operator is normal, but in general I can't see it.
$endgroup$
– Aweygan
Jan 17 at 0:58
|
show 3 more comments
$begingroup$
The same argument without touching sequences: As $T$ is weakly continuous and the unit ball of $ell^2$ is weakly compact its impage $T(B)$ is weakly compact hence weakly closed and thus closed in $ell^2$. At the same time it is relatively compact and thus compact. Hence the continuous map $ymapsto |y|$ realizes its supremum on $T(B)$.
answered Jan 17 at 13:41
JochenJochen
6,8131023
$endgroup$
add a comment |
$begingroup$
The same argument without touching sequences: As $T$ is weakly continuous and the unit ball of $ell^2$ is weakly compact its impage $T(B)$ is weakly compact hence weakly closed and thus closed in $ell^2$. At the same time it is relatively compact and thus compact. Hence the continuous map $ymapsto |y|$ realizes its supremum on $T(B)$.
answered Jan 17 at 13:41
JochenJochen
6,8131023
$endgroup$
add a comment |
$begingroup$
The same argument without touching sequences: As $T$ is weakly continuous and the unit ball of $ell^2$ is weakly compact its impage $T(B)$ is weakly compact hence weakly closed and thus closed in $ell^2$. At the same time it is relatively compact and thus compact. Hence the continuous map $ymapsto |y|$ realizes its supremum on $T(B)$.
answered Jan 17 at 13:41
JochenJochen
6,8131023
$endgroup$
The same argument without touching sequences: As $T$ is weakly continuous and the unit ball of $ell^2$ is weakly compact its impage $T(B)$ is weakly compact hence weakly closed and thus closed in $ell^2$. At the same time it is relatively compact and thus compact. Hence the continuous map $ymapsto |y|$ realizes its supremum on $T(B)$.
answered Jan 17 at 13:41
JochenJochen
6,8131023
answered Jan 17 at 13:41
JochenJochen
6,8131023
answered Jan 17 at 13:41
JochenJochen
6,8131023
answered Jan 17 at 13:41
JochenJochen
6,8131023
6,8131023
add a comment |
add a comment |
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3
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The eigenvalues of compact operators can only accumulate at 0. Guess no.
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– JRen
Jan 16 at 23:40
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@T.Bongers I don't get the part where you assume that $x_n$ can pass to a subsequence that converges. Why is that so?
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– BigbearZzz
Jan 16 at 23:46
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@JRen Oh, that's right, I totally forgot that property. For any compact operator, that operator only has finitely many eigenvalues outside a ball centered at the origin of the complex plane. Thus, the sequence $x_n$ can't have the property that $Vert L(x_n)Vert_2to 1$ if $L$ were compact. Thanks!
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– Anonymous
Jan 16 at 23:48
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@JRen On second thought, looking at the eigenvalues doesn't appear to work. The mapping $L(x_1,x_2,...)=(0,0,frac{1}{2}x_2,frac{2}{3}x_3,frac{3}{4}x_4,...)$ is a (not compact) bounded linear operator with the above properties which has no nonzero eigenvalues.
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– Anonymous
Jan 17 at 0:29
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@Anonymous I'm not generating to all the bounded linear operator.
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– JRen
Jan 17 at 17:13