Quasinilpotent , non-compact operatorsBanach spaces admitting no proper quasi-affinityExtending compact operatorsNon strictly-singular operators and complemented subspacesWeakly compact operators between Banach spacesCompact non-nuclear operators$C(X)$-compact operators and families of compact operatorsSpace of compact operators defined on separable Hilbert spaceCompact restrictions of the inclusion of $J:L_infty(0,1)to L_1(0,1)$Do sufficiently large Banach spaces admit non-compact operators with not too large range?Kernel of compact operators
Multi tool use
Quasinilpotent , non-compact operators
Banach spaces admitting no proper quasi-affinityExtending compact operatorsNon strictly-singular operators and complemented subspacesWeakly compact operators between Banach spacesCompact non-nuclear operators$C(X)$-compact operators and families of compact operatorsSpace of compact operators defined on separable Hilbert spaceCompact restrictions of the inclusion of $J:L_infty(0,1)to L_1(0,1)$Do sufficiently large Banach spaces admit non-compact operators with not too large range?Kernel of compact operators
5
$begingroup$
If $X$ is a separable Banach space and $(epsilon_n)downarrow 0$, can we find a quasinilpotent, non-compact operator on $X$ such that $||T^n||^1/n<epsilon_n$ for all $n$? I suspect the answer is positive, but cannot come up with an example.
fa.functional-analysis banach-spaces operator-theory hilbert-spaces
asked yesterday
MarkusMarkus
37018
$endgroup$
add a comment |
5
$begingroup$
If $X$ is a separable Banach space and $(epsilon_n)downarrow 0$, can we find a quasinilpotent, non-compact operator on $X$ such that $||T^n||^1/n<epsilon_n$ for all $n$? I suspect the answer is positive, but cannot come up with an example.
fa.functional-analysis banach-spaces operator-theory hilbert-spaces
asked yesterday
MarkusMarkus
37018
$endgroup$
2
$begingroup$
What is wrong with a nilpotent shift on $L^p[0,1]$? For example $(Tf)(x):=f(x+frac12)$ if $x<frac12$ and $(Tf)(x)=0$ elswhere?
$endgroup$
– András Bátkai
yesterday
2
$begingroup$
@AndrásBátkai: What is probably a bit more interesting is to find such an operator which is, in addition, not power-compact, and thus in particular not nilpotent. (An example would be the resolvent of a nil-potent vector-valued shift semigroup.)
$endgroup$
– Jochen Glueck
yesterday
$begingroup$
@AndrásBátkai: Oh, I think we misunderstood the question. I guess, Markus asks whether such an operator exists for every separable space $X$ and every sequence $(epsilon_n)$.
$endgroup$
– Jochen Glueck
yesterday
add a comment |
5
5
5
$begingroup$
If $X$ is a separable Banach space and $(epsilon_n)downarrow 0$, can we find a quasinilpotent, non-compact operator on $X$ such that $||T^n||^1/n<epsilon_n$ for all $n$? I suspect the answer is positive, but cannot come up with an example.
fa.functional-analysis banach-spaces operator-theory hilbert-spaces
asked yesterday
MarkusMarkus
37018
$endgroup$
If $X$ is a separable Banach space and $(epsilon_n)downarrow 0$, can we find a quasinilpotent, non-compact operator on $X$ such that $||T^n||^1/n<epsilon_n$ for all $n$? I suspect the answer is positive, but cannot come up with an example.
fa.functional-analysis banach-spaces operator-theory hilbert-spaces
fa.functional-analysis banach-spaces operator-theory hilbert-spaces
asked yesterday
MarkusMarkus
37018
asked yesterday
MarkusMarkus
37018
asked yesterday
MarkusMarkus
37018
asked yesterday
MarkusMarkus
37018
asked yesterday
MarkusMarkus
37018
37018
2
$begingroup$
What is wrong with a nilpotent shift on $L^p[0,1]$? For example $(Tf)(x):=f(x+frac12)$ if $x<frac12$ and $(Tf)(x)=0$ elswhere?
$endgroup$
– András Bátkai
yesterday
2
$begingroup$
@AndrásBátkai: What is probably a bit more interesting is to find such an operator which is, in addition, not power-compact, and thus in particular not nilpotent. (An example would be the resolvent of a nil-potent vector-valued shift semigroup.)
$endgroup$
– Jochen Glueck
yesterday
$begingroup$
@AndrásBátkai: Oh, I think we misunderstood the question. I guess, Markus asks whether such an operator exists for every separable space $X$ and every sequence $(epsilon_n)$.
$endgroup$
– Jochen Glueck
yesterday
add a comment |
2
$begingroup$
What is wrong with a nilpotent shift on $L^p[0,1]$? For example $(Tf)(x):=f(x+frac12)$ if $x<frac12$ and $(Tf)(x)=0$ elswhere?
$endgroup$
– András Bátkai
yesterday
2
$begingroup$
@AndrásBátkai: What is probably a bit more interesting is to find such an operator which is, in addition, not power-compact, and thus in particular not nilpotent. (An example would be the resolvent of a nil-potent vector-valued shift semigroup.)
$endgroup$
– Jochen Glueck
yesterday
$begingroup$
@AndrásBátkai: Oh, I think we misunderstood the question. I guess, Markus asks whether such an operator exists for every separable space $X$ and every sequence $(epsilon_n)$.
$endgroup$
– Jochen Glueck
yesterday
2
2
$begingroup$
What is wrong with a nilpotent shift on $L^p[0,1]$? For example $(Tf)(x):=f(x+frac12)$ if $x<frac12$ and $(Tf)(x)=0$ elswhere?
$endgroup$
– András Bátkai
yesterday
$begingroup$
What is wrong with a nilpotent shift on $L^p[0,1]$? For example $(Tf)(x):=f(x+frac12)$ if $x<frac12$ and $(Tf)(x)=0$ elswhere?
$endgroup$
– András Bátkai
yesterday
2
2
$begingroup$
@AndrásBátkai: What is probably a bit more interesting is to find such an operator which is, in addition, not power-compact, and thus in particular not nilpotent. (An example would be the resolvent of a nil-potent vector-valued shift semigroup.)
$endgroup$
– Jochen Glueck
yesterday
$begingroup$
@AndrásBátkai: What is probably a bit more interesting is to find such an operator which is, in addition, not power-compact, and thus in particular not nilpotent. (An example would be the resolvent of a nil-potent vector-valued shift semigroup.)
$endgroup$
– Jochen Glueck
yesterday
$begingroup$
@AndrásBátkai: Oh, I think we misunderstood the question. I guess, Markus asks whether such an operator exists for every separable space $X$ and every sequence $(epsilon_n)$.
$endgroup$
– Jochen Glueck
yesterday
$begingroup$
@AndrásBátkai: Oh, I think we misunderstood the question. I guess, Markus asks whether such an operator exists for every separable space $X$ and every sequence $(epsilon_n)$.
$endgroup$
– Jochen Glueck
yesterday
add a comment |
1 Answer
1
active
oldest
votes
7
$begingroup$
On the Argyros-Haydon space every operator is a compact perturbation of a scalar multiple of the identity, and hence every quasinilpotent operator is compact.
answered yesterday
Bill JohnsonBill Johnson
24.4k371117
$endgroup$
3
$begingroup$
Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"?
$endgroup$
– Fedor Petrov
yesterday
$begingroup$
Probably not, Fedor. Maybe there is a space with unconditional basis on which every operator is diagonal plus compact. Gowers built a space with unconditional basis on which every operator is diagonal plus strictly singular.
$endgroup$
– Bill Johnson
yesterday
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326074%2fquasinilpotent-non-compact-operators%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
$begingroup$
On the Argyros-Haydon space every operator is a compact perturbation of a scalar multiple of the identity, and hence every quasinilpotent operator is compact.
answered yesterday
Bill JohnsonBill Johnson
24.4k371117
$endgroup$
3
$begingroup$
Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"?
$endgroup$
– Fedor Petrov
yesterday
$begingroup$
Probably not, Fedor. Maybe there is a space with unconditional basis on which every operator is diagonal plus compact. Gowers built a space with unconditional basis on which every operator is diagonal plus strictly singular.
$endgroup$
– Bill Johnson
yesterday
add a comment |
7
$begingroup$
On the Argyros-Haydon space every operator is a compact perturbation of a scalar multiple of the identity, and hence every quasinilpotent operator is compact.
answered yesterday
Bill JohnsonBill Johnson
24.4k371117
$endgroup$
3
$begingroup$
Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"?
$endgroup$
– Fedor Petrov
yesterday
$begingroup$
Probably not, Fedor. Maybe there is a space with unconditional basis on which every operator is diagonal plus compact. Gowers built a space with unconditional basis on which every operator is diagonal plus strictly singular.
$endgroup$
– Bill Johnson
yesterday
add a comment |
7
7
7
$begingroup$
On the Argyros-Haydon space every operator is a compact perturbation of a scalar multiple of the identity, and hence every quasinilpotent operator is compact.
answered yesterday
Bill JohnsonBill Johnson
24.4k371117
$endgroup$
On the Argyros-Haydon space every operator is a compact perturbation of a scalar multiple of the identity, and hence every quasinilpotent operator is compact.
answered yesterday
Bill JohnsonBill Johnson
24.4k371117
answered yesterday
Bill JohnsonBill Johnson
24.4k371117
answered yesterday
Bill JohnsonBill Johnson
24.4k371117
answered yesterday
Bill JohnsonBill Johnson
24.4k371117
24.4k371117
3
$begingroup$
Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"?
$endgroup$
– Fedor Petrov
yesterday
$begingroup$
Probably not, Fedor. Maybe there is a space with unconditional basis on which every operator is diagonal plus compact. Gowers built a space with unconditional basis on which every operator is diagonal plus strictly singular.
$endgroup$
– Bill Johnson
yesterday
add a comment |
3
$begingroup$
Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"?
$endgroup$
– Fedor Petrov
yesterday
$begingroup$
Probably not, Fedor. Maybe there is a space with unconditional basis on which every operator is diagonal plus compact. Gowers built a space with unconditional basis on which every operator is diagonal plus strictly singular.
$endgroup$
– Bill Johnson
yesterday
3
3
$begingroup$
Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"?
$endgroup$
– Fedor Petrov
yesterday
$begingroup$
Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"?
$endgroup$
– Fedor Petrov
yesterday
$begingroup$
Probably not, Fedor. Maybe there is a space with unconditional basis on which every operator is diagonal plus compact. Gowers built a space with unconditional basis on which every operator is diagonal plus strictly singular.
$endgroup$
– Bill Johnson
yesterday
$begingroup$
Probably not, Fedor. Maybe there is a space with unconditional basis on which every operator is diagonal plus compact. Gowers built a space with unconditional basis on which every operator is diagonal plus strictly singular.
$endgroup$
– Bill Johnson
yesterday
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326074%2fquasinilpotent-non-compact-operators%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
4u7Oz8WSx,WMPZ6cSF,Z94xAYJEc lPWmO
2
$begingroup$
What is wrong with a nilpotent shift on $L^p[0,1]$? For example $(Tf)(x):=f(x+frac12)$ if $x<frac12$ and $(Tf)(x)=0$ elswhere?
$endgroup$
– András Bátkai
yesterday
2
$begingroup$
@AndrásBátkai: What is probably a bit more interesting is to find such an operator which is, in addition, not power-compact, and thus in particular not nilpotent. (An example would be the resolvent of a nil-potent vector-valued shift semigroup.)
$endgroup$
– Jochen Glueck
yesterday
$begingroup$
@AndrásBátkai: Oh, I think we misunderstood the question. I guess, Markus asks whether such an operator exists for every separable space $X$ and every sequence $(epsilon_n)$.
$endgroup$
– Jochen Glueck
yesterday