Continuous embeddings of $C(a,b)$ into itself with $L^p$ norms.
$begingroup$
Let $1leq p<qleqinfty$
Consider the spaces $X=(C(a,b), |cdot|_{L^q(a,b)})$ and $Y=(C(a,b), |cdot|_{L^p(a,b)})$. I want to show $X$ is continuously embedded in $Y$, but $Y$ is not continuously embedded in $X$.
Proof attempt:
Note an operator between normed spaces is continuous if and only if bounded, so it suffices to show that $exists C>0$ s.t.
$|x|_Yleq C|x|_X$
We want the inclusion map $iota(f) mapsto f$, to be bounded for $fin C(a,b)$ where $|f|_{L^q}$.
Note for a finite measure space $E$, (Lebesgue measure), there is a nesting of $L^p(E)$ spaces. We have the following relationship derived from Holder's inequality:
for $1leq p<qleq infty$
$|f|_{L^p}leq mu(E)^{1/p-1/q}|f|_{L^q}$
So we have $|f|_Y leq C|f|_X$ where $C=mu(E)^{1/p-1/q}$. So the inclusion map is continuous, and so $X$ is continuously embedded into $Y$.
However, $Y$ is not continuously embedded into $X$.
we can show the inverse inclusion map $iota^{-1}(f)mapsto f$ in general is not bounded by taking $fin L^p(a,b)setminus L^q(a,b)$, and noting the norm in $L^p$ is finite but the norm in $L^q$ is infinite, showing the inverse/backwards inclusion map is not continuous. I could use the explicit counterexample: $x^{-1/q}{chi(a,b)}$ to show this.
real-analysis functional-analysis operator-theory lp-spaces
edited Dec 15 '18 at 7:49
Yiorgos S. Smyrlis
63.4k1385163
asked Sep 21 '17 at 2:22
Kernel_DirichletKernel_Dirichlet
1,147416
$endgroup$
add a comment |
$begingroup$
Let $1leq p<qleqinfty$
Consider the spaces $X=(C(a,b), |cdot|_{L^q(a,b)})$ and $Y=(C(a,b), |cdot|_{L^p(a,b)})$. I want to show $X$ is continuously embedded in $Y$, but $Y$ is not continuously embedded in $X$.
Proof attempt:
Note an operator between normed spaces is continuous if and only if bounded, so it suffices to show that $exists C>0$ s.t.
$|x|_Yleq C|x|_X$
We want the inclusion map $iota(f) mapsto f$, to be bounded for $fin C(a,b)$ where $|f|_{L^q}$.
Note for a finite measure space $E$, (Lebesgue measure), there is a nesting of $L^p(E)$ spaces. We have the following relationship derived from Holder's inequality:
for $1leq p<qleq infty$
$|f|_{L^p}leq mu(E)^{1/p-1/q}|f|_{L^q}$
So we have $|f|_Y leq C|f|_X$ where $C=mu(E)^{1/p-1/q}$. So the inclusion map is continuous, and so $X$ is continuously embedded into $Y$.
However, $Y$ is not continuously embedded into $X$.
we can show the inverse inclusion map $iota^{-1}(f)mapsto f$ in general is not bounded by taking $fin L^p(a,b)setminus L^q(a,b)$, and noting the norm in $L^p$ is finite but the norm in $L^q$ is infinite, showing the inverse/backwards inclusion map is not continuous. I could use the explicit counterexample: $x^{-1/q}{chi(a,b)}$ to show this.
real-analysis functional-analysis operator-theory lp-spaces
edited Dec 15 '18 at 7:49
Yiorgos S. Smyrlis
63.4k1385163
asked Sep 21 '17 at 2:22
Kernel_DirichletKernel_Dirichlet
1,147416
$endgroup$
add a comment |
$begingroup$
Let $1leq p<qleqinfty$
Consider the spaces $X=(C(a,b), |cdot|_{L^q(a,b)})$ and $Y=(C(a,b), |cdot|_{L^p(a,b)})$. I want to show $X$ is continuously embedded in $Y$, but $Y$ is not continuously embedded in $X$.
Proof attempt:
Note an operator between normed spaces is continuous if and only if bounded, so it suffices to show that $exists C>0$ s.t.
$|x|_Yleq C|x|_X$
We want the inclusion map $iota(f) mapsto f$, to be bounded for $fin C(a,b)$ where $|f|_{L^q}$.
Note for a finite measure space $E$, (Lebesgue measure), there is a nesting of $L^p(E)$ spaces. We have the following relationship derived from Holder's inequality:
for $1leq p<qleq infty$
$|f|_{L^p}leq mu(E)^{1/p-1/q}|f|_{L^q}$
So we have $|f|_Y leq C|f|_X$ where $C=mu(E)^{1/p-1/q}$. So the inclusion map is continuous, and so $X$ is continuously embedded into $Y$.
However, $Y$ is not continuously embedded into $X$.
we can show the inverse inclusion map $iota^{-1}(f)mapsto f$ in general is not bounded by taking $fin L^p(a,b)setminus L^q(a,b)$, and noting the norm in $L^p$ is finite but the norm in $L^q$ is infinite, showing the inverse/backwards inclusion map is not continuous. I could use the explicit counterexample: $x^{-1/q}{chi(a,b)}$ to show this.
real-analysis functional-analysis operator-theory lp-spaces
edited Dec 15 '18 at 7:49
Yiorgos S. Smyrlis
63.4k1385163
asked Sep 21 '17 at 2:22
Kernel_DirichletKernel_Dirichlet
1,147416
$endgroup$
Let $1leq p<qleqinfty$
Consider the spaces $X=(C(a,b), |cdot|_{L^q(a,b)})$ and $Y=(C(a,b), |cdot|_{L^p(a,b)})$. I want to show $X$ is continuously embedded in $Y$, but $Y$ is not continuously embedded in $X$.
Proof attempt:
Note an operator between normed spaces is continuous if and only if bounded, so it suffices to show that $exists C>0$ s.t.
$|x|_Yleq C|x|_X$
We want the inclusion map $iota(f) mapsto f$, to be bounded for $fin C(a,b)$ where $|f|_{L^q}$.
Note for a finite measure space $E$, (Lebesgue measure), there is a nesting of $L^p(E)$ spaces. We have the following relationship derived from Holder's inequality:
for $1leq p<qleq infty$
$|f|_{L^p}leq mu(E)^{1/p-1/q}|f|_{L^q}$
So we have $|f|_Y leq C|f|_X$ where $C=mu(E)^{1/p-1/q}$. So the inclusion map is continuous, and so $X$ is continuously embedded into $Y$.
However, $Y$ is not continuously embedded into $X$.
we can show the inverse inclusion map $iota^{-1}(f)mapsto f$ in general is not bounded by taking $fin L^p(a,b)setminus L^q(a,b)$, and noting the norm in $L^p$ is finite but the norm in $L^q$ is infinite, showing the inverse/backwards inclusion map is not continuous. I could use the explicit counterexample: $x^{-1/q}{chi(a,b)}$ to show this.
real-analysis functional-analysis operator-theory lp-spaces
real-analysis functional-analysis operator-theory lp-spaces
edited Dec 15 '18 at 7:49
Yiorgos S. Smyrlis
63.4k1385163
asked Sep 21 '17 at 2:22
Kernel_DirichletKernel_Dirichlet
1,147416
edited Dec 15 '18 at 7:49
Yiorgos S. Smyrlis
63.4k1385163
asked Sep 21 '17 at 2:22
Kernel_DirichletKernel_Dirichlet
1,147416
edited Dec 15 '18 at 7:49
Yiorgos S. Smyrlis
63.4k1385163
edited Dec 15 '18 at 7:49
Yiorgos S. Smyrlis
63.4k1385163
edited Dec 15 '18 at 7:49
Yiorgos S. Smyrlis
63.4k1385163
63.4k1385163
asked Sep 21 '17 at 2:22
Kernel_DirichletKernel_Dirichlet
1,147416
asked Sep 21 '17 at 2:22
Kernel_DirichletKernel_Dirichlet
1,147416
asked Sep 21 '17 at 2:22
Kernel_DirichletKernel_Dirichlet
1,147416
1,147416
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider $(a,b)=(0,1)$, $rin (p,q)$ and
$$
f_varepsilon(x)=(x+varepsilon)^{-frac{1}{r}}.
$$
Then
$$
int_0^1 |,f_varepsilon|^p<int_0^1 x^{-p/r},dx=frac{1}{1-p/r}=frac{r}{r-p}
$$
and hence
$$
|,f_varepsilon|_p<left(frac{r}{r-p}right)^{1/p}.
$$
Meanwhile
$$
lim_{varepsilonto 0}int_0^1 |,f_varepsilon|^q=lim_{varepsilonto 0}int_0^1 (x+varepsilon)^{-frac{q}{r}},dx=lim_{varepsilonto 0}frac{1}{q/r-1}left(varepsilon^{-q/r}-(1+varepsilon)^{-q/r}right)=infty.
$$
answered Sep 21 '17 at 9:27
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
$endgroup$
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%2f2438323%2fcontinuous-embeddings-of-ca-b-into-itself-with-lp-norms%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
$begingroup$
Consider $(a,b)=(0,1)$, $rin (p,q)$ and
$$
f_varepsilon(x)=(x+varepsilon)^{-frac{1}{r}}.
$$
Then
$$
int_0^1 |,f_varepsilon|^p<int_0^1 x^{-p/r},dx=frac{1}{1-p/r}=frac{r}{r-p}
$$
and hence
$$
|,f_varepsilon|_p<left(frac{r}{r-p}right)^{1/p}.
$$
Meanwhile
$$
lim_{varepsilonto 0}int_0^1 |,f_varepsilon|^q=lim_{varepsilonto 0}int_0^1 (x+varepsilon)^{-frac{q}{r}},dx=lim_{varepsilonto 0}frac{1}{q/r-1}left(varepsilon^{-q/r}-(1+varepsilon)^{-q/r}right)=infty.
$$
answered Sep 21 '17 at 9:27
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
$endgroup$
add a comment |
$begingroup$
Consider $(a,b)=(0,1)$, $rin (p,q)$ and
$$
f_varepsilon(x)=(x+varepsilon)^{-frac{1}{r}}.
$$
Then
$$
int_0^1 |,f_varepsilon|^p<int_0^1 x^{-p/r},dx=frac{1}{1-p/r}=frac{r}{r-p}
$$
and hence
$$
|,f_varepsilon|_p<left(frac{r}{r-p}right)^{1/p}.
$$
Meanwhile
$$
lim_{varepsilonto 0}int_0^1 |,f_varepsilon|^q=lim_{varepsilonto 0}int_0^1 (x+varepsilon)^{-frac{q}{r}},dx=lim_{varepsilonto 0}frac{1}{q/r-1}left(varepsilon^{-q/r}-(1+varepsilon)^{-q/r}right)=infty.
$$
answered Sep 21 '17 at 9:27
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
$endgroup$
add a comment |
$begingroup$
Consider $(a,b)=(0,1)$, $rin (p,q)$ and
$$
f_varepsilon(x)=(x+varepsilon)^{-frac{1}{r}}.
$$
Then
$$
int_0^1 |,f_varepsilon|^p<int_0^1 x^{-p/r},dx=frac{1}{1-p/r}=frac{r}{r-p}
$$
and hence
$$
|,f_varepsilon|_p<left(frac{r}{r-p}right)^{1/p}.
$$
Meanwhile
$$
lim_{varepsilonto 0}int_0^1 |,f_varepsilon|^q=lim_{varepsilonto 0}int_0^1 (x+varepsilon)^{-frac{q}{r}},dx=lim_{varepsilonto 0}frac{1}{q/r-1}left(varepsilon^{-q/r}-(1+varepsilon)^{-q/r}right)=infty.
$$
answered Sep 21 '17 at 9:27
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
$endgroup$
Consider $(a,b)=(0,1)$, $rin (p,q)$ and
$$
f_varepsilon(x)=(x+varepsilon)^{-frac{1}{r}}.
$$
Then
$$
int_0^1 |,f_varepsilon|^p<int_0^1 x^{-p/r},dx=frac{1}{1-p/r}=frac{r}{r-p}
$$
and hence
$$
|,f_varepsilon|_p<left(frac{r}{r-p}right)^{1/p}.
$$
Meanwhile
$$
lim_{varepsilonto 0}int_0^1 |,f_varepsilon|^q=lim_{varepsilonto 0}int_0^1 (x+varepsilon)^{-frac{q}{r}},dx=lim_{varepsilonto 0}frac{1}{q/r-1}left(varepsilon^{-q/r}-(1+varepsilon)^{-q/r}right)=infty.
$$
answered Sep 21 '17 at 9:27
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
answered Sep 21 '17 at 9:27
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
answered Sep 21 '17 at 9:27
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
answered Sep 21 '17 at 9:27
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
63.4k1385163
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%2f2438323%2fcontinuous-embeddings-of-ca-b-into-itself-with-lp-norms%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
