Equality case in Hölder's inequality
$begingroup$
Let $p$ and $q$ be dual exponents and let $f in L^p(0,1)$ and $g in L^q(0,1)$ be non-zero functions. Further assume that $||fg||_1=||f||_p||g||_q$. Show that $f=cg^{q-1}$ where $c$ is constant.
By Hölder we know that $||fg||_1 leq||f||_p||g||_q$, but I'm not sure how this helps us. By setting $f=cg^{q-1}$ I was able to verify the equality, but I don't know how to prove this is the only solution.
measure-theory inequality lp-spaces
edited Dec 11 '18 at 12:44
Davide Giraudo
127k16151263
asked Dec 9 '18 at 22:06
TrobjonTrobjon
234
$endgroup$
add a comment |
$begingroup$
Let $p$ and $q$ be dual exponents and let $f in L^p(0,1)$ and $g in L^q(0,1)$ be non-zero functions. Further assume that $||fg||_1=||f||_p||g||_q$. Show that $f=cg^{q-1}$ where $c$ is constant.
By Hölder we know that $||fg||_1 leq||f||_p||g||_q$, but I'm not sure how this helps us. By setting $f=cg^{q-1}$ I was able to verify the equality, but I don't know how to prove this is the only solution.
measure-theory inequality lp-spaces
edited Dec 11 '18 at 12:44
Davide Giraudo
127k16151263
asked Dec 9 '18 at 22:06
TrobjonTrobjon
234
$endgroup$
$begingroup$
This result is standard. See the discussion after Theorem 3.5 in Rudin's RCA.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:35
$begingroup$
I looked at Rudin and I don't understand.
$endgroup$
– Trobjon
Dec 9 '18 at 23:48
add a comment |
$begingroup$
Let $p$ and $q$ be dual exponents and let $f in L^p(0,1)$ and $g in L^q(0,1)$ be non-zero functions. Further assume that $||fg||_1=||f||_p||g||_q$. Show that $f=cg^{q-1}$ where $c$ is constant.
By Hölder we know that $||fg||_1 leq||f||_p||g||_q$, but I'm not sure how this helps us. By setting $f=cg^{q-1}$ I was able to verify the equality, but I don't know how to prove this is the only solution.
measure-theory inequality lp-spaces
edited Dec 11 '18 at 12:44
Davide Giraudo
127k16151263
asked Dec 9 '18 at 22:06
TrobjonTrobjon
234
$endgroup$
Let $p$ and $q$ be dual exponents and let $f in L^p(0,1)$ and $g in L^q(0,1)$ be non-zero functions. Further assume that $||fg||_1=||f||_p||g||_q$. Show that $f=cg^{q-1}$ where $c$ is constant.
By Hölder we know that $||fg||_1 leq||f||_p||g||_q$, but I'm not sure how this helps us. By setting $f=cg^{q-1}$ I was able to verify the equality, but I don't know how to prove this is the only solution.
measure-theory inequality lp-spaces
measure-theory inequality lp-spaces
edited Dec 11 '18 at 12:44
Davide Giraudo
127k16151263
asked Dec 9 '18 at 22:06
TrobjonTrobjon
234
edited Dec 11 '18 at 12:44
Davide Giraudo
127k16151263
asked Dec 9 '18 at 22:06
TrobjonTrobjon
234
edited Dec 11 '18 at 12:44
Davide Giraudo
127k16151263
edited Dec 11 '18 at 12:44
Davide Giraudo
127k16151263
edited Dec 11 '18 at 12:44
Davide Giraudo
127k16151263
127k16151263
asked Dec 9 '18 at 22:06
TrobjonTrobjon
234
asked Dec 9 '18 at 22:06
TrobjonTrobjon
234
asked Dec 9 '18 at 22:06
TrobjonTrobjon
234
234
$begingroup$
This result is standard. See the discussion after Theorem 3.5 in Rudin's RCA.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:35
$begingroup$
I looked at Rudin and I don't understand.
$endgroup$
– Trobjon
Dec 9 '18 at 23:48
add a comment |
$begingroup$
This result is standard. See the discussion after Theorem 3.5 in Rudin's RCA.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:35
$begingroup$
I looked at Rudin and I don't understand.
$endgroup$
– Trobjon
Dec 9 '18 at 23:48
$begingroup$
This result is standard. See the discussion after Theorem 3.5 in Rudin's RCA.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:35
$begingroup$
This result is standard. See the discussion after Theorem 3.5 in Rudin's RCA.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:35
$begingroup$
I looked at Rudin and I don't understand.
$endgroup$
– Trobjon
Dec 9 '18 at 23:48
$begingroup$
I looked at Rudin and I don't understand.
$endgroup$
– Trobjon
Dec 9 '18 at 23:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I will assume that the function are non-negative. Otherwise, the result may not hold, as we can have when $q$ is an even integer $f=1$ and $g=1$ on $(0,1/2)$ and $-1$ on $(1/2,1)$ or even worse $g^{q-1}$ may not be well-defined if $q$ is not rational and $g$ take negative values.
Since $f$ and $g$ are not zero, by scaling, it suffices to consider the case where
$leftlVert frightrVert_p=leftlVert qrightrVert_q=1$.
Observe that for all positive numbers $a$ and $b$, by convexity of the exponential function,
$$
ab=expleft(frac 1pln left(a^pright)+frac 1qln left(b^qright)right)
leqslant frac 1pexpleft(ln left(a^pright) right)+
frac 1qexpleft(ln left(b^qright) right)
$$
hence
$$
tag{*}
ableqslant frac{a^p}p+frac{b^q}q, quad a,bgeqslant 0
$$
and by strict convexity, equality holds if and only if $a^p=b^q$.
Let
$$
h:=frac{f^p}p+frac{g^q}q-fg.
$$
By $(*)$, $h$ is non-negative and by assumption, $int h=0$ hence $h=0$ almost everywhere. By the equality case in $(*)$, we derive that $f^p=g^q$ almost everywhere.
answered Dec 10 '18 at 10:33
Davide GiraudoDavide Giraudo
127k16151263
$endgroup$
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%2f3033091%2fequality-case-in-h%25c3%25b6lders-inequality%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
$begingroup$
I will assume that the function are non-negative. Otherwise, the result may not hold, as we can have when $q$ is an even integer $f=1$ and $g=1$ on $(0,1/2)$ and $-1$ on $(1/2,1)$ or even worse $g^{q-1}$ may not be well-defined if $q$ is not rational and $g$ take negative values.
Since $f$ and $g$ are not zero, by scaling, it suffices to consider the case where
$leftlVert frightrVert_p=leftlVert qrightrVert_q=1$.
Observe that for all positive numbers $a$ and $b$, by convexity of the exponential function,
$$
ab=expleft(frac 1pln left(a^pright)+frac 1qln left(b^qright)right)
leqslant frac 1pexpleft(ln left(a^pright) right)+
frac 1qexpleft(ln left(b^qright) right)
$$
hence
$$
tag{*}
ableqslant frac{a^p}p+frac{b^q}q, quad a,bgeqslant 0
$$
and by strict convexity, equality holds if and only if $a^p=b^q$.
Let
$$
h:=frac{f^p}p+frac{g^q}q-fg.
$$
By $(*)$, $h$ is non-negative and by assumption, $int h=0$ hence $h=0$ almost everywhere. By the equality case in $(*)$, we derive that $f^p=g^q$ almost everywhere.
answered Dec 10 '18 at 10:33
Davide GiraudoDavide Giraudo
127k16151263
$endgroup$
add a comment |
$begingroup$
I will assume that the function are non-negative. Otherwise, the result may not hold, as we can have when $q$ is an even integer $f=1$ and $g=1$ on $(0,1/2)$ and $-1$ on $(1/2,1)$ or even worse $g^{q-1}$ may not be well-defined if $q$ is not rational and $g$ take negative values.
Since $f$ and $g$ are not zero, by scaling, it suffices to consider the case where
$leftlVert frightrVert_p=leftlVert qrightrVert_q=1$.
Observe that for all positive numbers $a$ and $b$, by convexity of the exponential function,
$$
ab=expleft(frac 1pln left(a^pright)+frac 1qln left(b^qright)right)
leqslant frac 1pexpleft(ln left(a^pright) right)+
frac 1qexpleft(ln left(b^qright) right)
$$
hence
$$
tag{*}
ableqslant frac{a^p}p+frac{b^q}q, quad a,bgeqslant 0
$$
and by strict convexity, equality holds if and only if $a^p=b^q$.
Let
$$
h:=frac{f^p}p+frac{g^q}q-fg.
$$
By $(*)$, $h$ is non-negative and by assumption, $int h=0$ hence $h=0$ almost everywhere. By the equality case in $(*)$, we derive that $f^p=g^q$ almost everywhere.
answered Dec 10 '18 at 10:33
Davide GiraudoDavide Giraudo
127k16151263
$endgroup$
add a comment |
$begingroup$
I will assume that the function are non-negative. Otherwise, the result may not hold, as we can have when $q$ is an even integer $f=1$ and $g=1$ on $(0,1/2)$ and $-1$ on $(1/2,1)$ or even worse $g^{q-1}$ may not be well-defined if $q$ is not rational and $g$ take negative values.
Since $f$ and $g$ are not zero, by scaling, it suffices to consider the case where
$leftlVert frightrVert_p=leftlVert qrightrVert_q=1$.
Observe that for all positive numbers $a$ and $b$, by convexity of the exponential function,
$$
ab=expleft(frac 1pln left(a^pright)+frac 1qln left(b^qright)right)
leqslant frac 1pexpleft(ln left(a^pright) right)+
frac 1qexpleft(ln left(b^qright) right)
$$
hence
$$
tag{*}
ableqslant frac{a^p}p+frac{b^q}q, quad a,bgeqslant 0
$$
and by strict convexity, equality holds if and only if $a^p=b^q$.
Let
$$
h:=frac{f^p}p+frac{g^q}q-fg.
$$
By $(*)$, $h$ is non-negative and by assumption, $int h=0$ hence $h=0$ almost everywhere. By the equality case in $(*)$, we derive that $f^p=g^q$ almost everywhere.
answered Dec 10 '18 at 10:33
Davide GiraudoDavide Giraudo
127k16151263
$endgroup$
I will assume that the function are non-negative. Otherwise, the result may not hold, as we can have when $q$ is an even integer $f=1$ and $g=1$ on $(0,1/2)$ and $-1$ on $(1/2,1)$ or even worse $g^{q-1}$ may not be well-defined if $q$ is not rational and $g$ take negative values.
Since $f$ and $g$ are not zero, by scaling, it suffices to consider the case where
$leftlVert frightrVert_p=leftlVert qrightrVert_q=1$.
Observe that for all positive numbers $a$ and $b$, by convexity of the exponential function,
$$
ab=expleft(frac 1pln left(a^pright)+frac 1qln left(b^qright)right)
leqslant frac 1pexpleft(ln left(a^pright) right)+
frac 1qexpleft(ln left(b^qright) right)
$$
hence
$$
tag{*}
ableqslant frac{a^p}p+frac{b^q}q, quad a,bgeqslant 0
$$
and by strict convexity, equality holds if and only if $a^p=b^q$.
Let
$$
h:=frac{f^p}p+frac{g^q}q-fg.
$$
By $(*)$, $h$ is non-negative and by assumption, $int h=0$ hence $h=0$ almost everywhere. By the equality case in $(*)$, we derive that $f^p=g^q$ almost everywhere.
answered Dec 10 '18 at 10:33
Davide GiraudoDavide Giraudo
127k16151263
answered Dec 10 '18 at 10:33
Davide GiraudoDavide Giraudo
127k16151263
answered Dec 10 '18 at 10:33
Davide GiraudoDavide Giraudo
127k16151263
answered Dec 10 '18 at 10:33
Davide GiraudoDavide Giraudo
127k16151263
127k16151263
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%2f3033091%2fequality-case-in-h%25c3%25b6lders-inequality%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
$begingroup$
This result is standard. See the discussion after Theorem 3.5 in Rudin's RCA.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:35
$begingroup$
I looked at Rudin and I don't understand.
$endgroup$
– Trobjon
Dec 9 '18 at 23:48