Mixing
Let $T: C[0,1] to C^1[0,1]$ defined by $T(f) = f'$, all with the norm of the max.
$begingroup$
Let $T: C[0,1] to C^1[0,1]$ defined by $T(f) = f'$, all with the norm of the max. First of all, I think that $C^1[0,1]$ is not a banach space with the max norm. So I do not have the result that T has closed graph iff T is limited. I just have that T limited implies graph of T closed.
However, I think that T is not limited so I cannot apply that.
Is there another theorem that can help me?
I tried doing it "by hands": let $(f_n(x),T(f_n(x)) in graph(T)$. Suppose this is a convergent sequence with limit $(X,Y)$. If I say that X,Y belong to $C[0,1]$ and $C^1[0,1]$ respectively then I am done.
For sure, X belongs to C[0,1] because it is complete.
However, I do not know what to do with Y. Any hints?
functional-analysis
asked Nov 30 '18 at 6:57
qcc101qcc101
627213
$endgroup$
add a comment |
$begingroup$
Let $T: C[0,1] to C^1[0,1]$ defined by $T(f) = f'$, all with the norm of the max. First of all, I think that $C^1[0,1]$ is not a banach space with the max norm. So I do not have the result that T has closed graph iff T is limited. I just have that T limited implies graph of T closed.
However, I think that T is not limited so I cannot apply that.
Is there another theorem that can help me?
I tried doing it "by hands": let $(f_n(x),T(f_n(x)) in graph(T)$. Suppose this is a convergent sequence with limit $(X,Y)$. If I say that X,Y belong to $C[0,1]$ and $C^1[0,1]$ respectively then I am done.
For sure, X belongs to C[0,1] because it is complete.
However, I do not know what to do with Y. Any hints?
functional-analysis
asked Nov 30 '18 at 6:57
qcc101qcc101
627213
$endgroup$
$begingroup$
My gues is that you meant to write that $T$ goes from $C^1[0,1]$ into $C[0,1]$.
$endgroup$
– José Carlos Santos
Nov 30 '18 at 7:23
$begingroup$
I agree with @JoséCarlosSantos
$endgroup$
– DisintegratingByParts
Nov 30 '18 at 10:39
add a comment |
$begingroup$
Let $T: C[0,1] to C^1[0,1]$ defined by $T(f) = f'$, all with the norm of the max. First of all, I think that $C^1[0,1]$ is not a banach space with the max norm. So I do not have the result that T has closed graph iff T is limited. I just have that T limited implies graph of T closed.
However, I think that T is not limited so I cannot apply that.
Is there another theorem that can help me?
I tried doing it "by hands": let $(f_n(x),T(f_n(x)) in graph(T)$. Suppose this is a convergent sequence with limit $(X,Y)$. If I say that X,Y belong to $C[0,1]$ and $C^1[0,1]$ respectively then I am done.
For sure, X belongs to C[0,1] because it is complete.
However, I do not know what to do with Y. Any hints?
functional-analysis
asked Nov 30 '18 at 6:57
qcc101qcc101
627213
$endgroup$
Let $T: C[0,1] to C^1[0,1]$ defined by $T(f) = f'$, all with the norm of the max. First of all, I think that $C^1[0,1]$ is not a banach space with the max norm. So I do not have the result that T has closed graph iff T is limited. I just have that T limited implies graph of T closed.
However, I think that T is not limited so I cannot apply that.
Is there another theorem that can help me?
I tried doing it "by hands": let $(f_n(x),T(f_n(x)) in graph(T)$. Suppose this is a convergent sequence with limit $(X,Y)$. If I say that X,Y belong to $C[0,1]$ and $C^1[0,1]$ respectively then I am done.
For sure, X belongs to C[0,1] because it is complete.
However, I do not know what to do with Y. Any hints?
functional-analysis
functional-analysis
asked Nov 30 '18 at 6:57
qcc101qcc101
627213
asked Nov 30 '18 at 6:57
qcc101qcc101
627213
asked Nov 30 '18 at 6:57
qcc101qcc101
627213
asked Nov 30 '18 at 6:57
qcc101qcc101
627213
asked Nov 30 '18 at 6:57
qcc101qcc101
627213
627213
$begingroup$
My gues is that you meant to write that $T$ goes from $C^1[0,1]$ into $C[0,1]$.
$endgroup$
– José Carlos Santos
Nov 30 '18 at 7:23
$begingroup$
I agree with @JoséCarlosSantos
$endgroup$
– DisintegratingByParts
Nov 30 '18 at 10:39
add a comment |
$begingroup$
My gues is that you meant to write that $T$ goes from $C^1[0,1]$ into $C[0,1]$.
$endgroup$
– José Carlos Santos
Nov 30 '18 at 7:23
$begingroup$
I agree with @JoséCarlosSantos
$endgroup$
– DisintegratingByParts
Nov 30 '18 at 10:39
$begingroup$
My gues is that you meant to write that $T$ goes from $C^1[0,1]$ into $C[0,1]$.
$endgroup$
– José Carlos Santos
Nov 30 '18 at 7:23
$begingroup$
My gues is that you meant to write that $T$ goes from $C^1[0,1]$ into $C[0,1]$.
$endgroup$
– José Carlos Santos
Nov 30 '18 at 7:23
$begingroup$
I agree with @JoséCarlosSantos
$endgroup$
– DisintegratingByParts
Nov 30 '18 at 10:39
$begingroup$
I agree with @JoséCarlosSantos
$endgroup$
– DisintegratingByParts
Nov 30 '18 at 10:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assuming that you meant to write $Tcolon C^1[0,1]longrightarrow C[0,1]$ (otherwsise, $T$ doesn't make sense), then $T$ is not bouned. For each $ninmathbb N$, let$$begin{array}{rccc}f_ncolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&dfrac{x^{n+1}}{n+1}.end{array}$$Then $lim_{ntoinfty}f_n=0$, but $(forall ninmathbb{N}):bigllVert T(f_n)bigrrVert=1$, and therefore we don't have $lim_{ntoinfty}T(f_n)=0$.
answered Nov 30 '18 at 7:28
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
add a comment |
$begingroup$
I think that $T: C[0,1] to C^1[0,1]$ is not correct. Correct is $T: C^1[0,1] to C[0,1]$. Let $D(T):=C^1[0,1]$.
Now let $(f_n)$ be a sequence in $D(T)$ such that $f_n to f in C[0,1]$ (norm convergent) and $Tf_n to g in C[0,1]$ (norm convergent).
Then the sequences $(f_n)$ and $(Tf_n)$ are uniformly convergent on $[0,1]$.
A theorem in Analysis now says: $f in D(T)$ and $g=f'$.
This shows that $T$ is closed.
Observe that $T$ is not bounded !
answered Nov 30 '18 at 7:31
FredFred
48.6k11849
$endgroup$
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: "69"
};
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
});
}
});
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%2fmath.stackexchange.com%2fquestions%2f3019760%2flet-t-c0-1-to-c10-1-defined-by-tf-f-all-with-the-norm-of-the-ma%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming that you meant to write $Tcolon C^1[0,1]longrightarrow C[0,1]$ (otherwsise, $T$ doesn't make sense), then $T$ is not bouned. For each $ninmathbb N$, let$$begin{array}{rccc}f_ncolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&dfrac{x^{n+1}}{n+1}.end{array}$$Then $lim_{ntoinfty}f_n=0$, but $(forall ninmathbb{N}):bigllVert T(f_n)bigrrVert=1$, and therefore we don't have $lim_{ntoinfty}T(f_n)=0$.
answered Nov 30 '18 at 7:28
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
add a comment |
$begingroup$
Assuming that you meant to write $Tcolon C^1[0,1]longrightarrow C[0,1]$ (otherwsise, $T$ doesn't make sense), then $T$ is not bouned. For each $ninmathbb N$, let$$begin{array}{rccc}f_ncolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&dfrac{x^{n+1}}{n+1}.end{array}$$Then $lim_{ntoinfty}f_n=0$, but $(forall ninmathbb{N}):bigllVert T(f_n)bigrrVert=1$, and therefore we don't have $lim_{ntoinfty}T(f_n)=0$.
answered Nov 30 '18 at 7:28
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
add a comment |
$begingroup$
Assuming that you meant to write $Tcolon C^1[0,1]longrightarrow C[0,1]$ (otherwsise, $T$ doesn't make sense), then $T$ is not bouned. For each $ninmathbb N$, let$$begin{array}{rccc}f_ncolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&dfrac{x^{n+1}}{n+1}.end{array}$$Then $lim_{ntoinfty}f_n=0$, but $(forall ninmathbb{N}):bigllVert T(f_n)bigrrVert=1$, and therefore we don't have $lim_{ntoinfty}T(f_n)=0$.
answered Nov 30 '18 at 7:28
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
Assuming that you meant to write $Tcolon C^1[0,1]longrightarrow C[0,1]$ (otherwsise, $T$ doesn't make sense), then $T$ is not bouned. For each $ninmathbb N$, let$$begin{array}{rccc}f_ncolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&dfrac{x^{n+1}}{n+1}.end{array}$$Then $lim_{ntoinfty}f_n=0$, but $(forall ninmathbb{N}):bigllVert T(f_n)bigrrVert=1$, and therefore we don't have $lim_{ntoinfty}T(f_n)=0$.
answered Nov 30 '18 at 7:28
José Carlos SantosJosé Carlos Santos
169k23132237
edited Jan 26 at 10:05
answered Nov 30 '18 at 7:28
José Carlos SantosJosé Carlos Santos
169k23132237
answered Nov 30 '18 at 7:28
José Carlos SantosJosé Carlos Santos
169k23132237
answered Nov 30 '18 at 7:28
José Carlos SantosJosé Carlos Santos
169k23132237
169k23132237
add a comment |
add a comment |
$begingroup$
I think that $T: C[0,1] to C^1[0,1]$ is not correct. Correct is $T: C^1[0,1] to C[0,1]$. Let $D(T):=C^1[0,1]$.
Now let $(f_n)$ be a sequence in $D(T)$ such that $f_n to f in C[0,1]$ (norm convergent) and $Tf_n to g in C[0,1]$ (norm convergent).
Then the sequences $(f_n)$ and $(Tf_n)$ are uniformly convergent on $[0,1]$.
A theorem in Analysis now says: $f in D(T)$ and $g=f'$.
This shows that $T$ is closed.
Observe that $T$ is not bounded !
answered Nov 30 '18 at 7:31
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
I think that $T: C[0,1] to C^1[0,1]$ is not correct. Correct is $T: C^1[0,1] to C[0,1]$. Let $D(T):=C^1[0,1]$.
Now let $(f_n)$ be a sequence in $D(T)$ such that $f_n to f in C[0,1]$ (norm convergent) and $Tf_n to g in C[0,1]$ (norm convergent).
Then the sequences $(f_n)$ and $(Tf_n)$ are uniformly convergent on $[0,1]$.
A theorem in Analysis now says: $f in D(T)$ and $g=f'$.
This shows that $T$ is closed.
Observe that $T$ is not bounded !
answered Nov 30 '18 at 7:31
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
I think that $T: C[0,1] to C^1[0,1]$ is not correct. Correct is $T: C^1[0,1] to C[0,1]$. Let $D(T):=C^1[0,1]$.
Now let $(f_n)$ be a sequence in $D(T)$ such that $f_n to f in C[0,1]$ (norm convergent) and $Tf_n to g in C[0,1]$ (norm convergent).
Then the sequences $(f_n)$ and $(Tf_n)$ are uniformly convergent on $[0,1]$.
A theorem in Analysis now says: $f in D(T)$ and $g=f'$.
This shows that $T$ is closed.
Observe that $T$ is not bounded !
answered Nov 30 '18 at 7:31
FredFred
48.6k11849
$endgroup$
I think that $T: C[0,1] to C^1[0,1]$ is not correct. Correct is $T: C^1[0,1] to C[0,1]$. Let $D(T):=C^1[0,1]$.
Now let $(f_n)$ be a sequence in $D(T)$ such that $f_n to f in C[0,1]$ (norm convergent) and $Tf_n to g in C[0,1]$ (norm convergent).
Then the sequences $(f_n)$ and $(Tf_n)$ are uniformly convergent on $[0,1]$.
A theorem in Analysis now says: $f in D(T)$ and $g=f'$.
This shows that $T$ is closed.
Observe that $T$ is not bounded !
answered Nov 30 '18 at 7:31
FredFred
48.6k11849
answered Nov 30 '18 at 7:31
FredFred
48.6k11849
answered Nov 30 '18 at 7:31
FredFred
48.6k11849
answered Nov 30 '18 at 7:31
FredFred
48.6k11849
48.6k11849
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- 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.
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%2fmath.stackexchange.com%2fquestions%2f3019760%2flet-t-c0-1-to-c10-1-defined-by-tf-f-all-with-the-norm-of-the-ma%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
$begingroup$
My gues is that you meant to write that $T$ goes from $C^1[0,1]$ into $C[0,1]$.
$endgroup$
– José Carlos Santos
Nov 30 '18 at 7:23
$begingroup$
I agree with @JoséCarlosSantos
$endgroup$
– DisintegratingByParts
Nov 30 '18 at 10:39