Equivalence of $L^p$ norm in a bounded domain
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For $1<p<q<infty$ and a bounded domain $Omega subseteq mathbb R^2$, is the following true? $$C_1Vert f Vert_{L^{p}(Omega)} leq Vert f Vert_{L^{q}(Omega)}leq C_2Vert f Vert_{L^{p}(Omega)} $$
The first inequality is easy by simply using the Holder inequality and $C_1$ depends on the finite measure of $Omega$. How to show the second inequality?
real-analysis functional-analysis measure-theory
asked Dec 10 '18 at 19:04
BourneBourne
548
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add a comment |
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For $1<p<q<infty$ and a bounded domain $Omega subseteq mathbb R^2$, is the following true? $$C_1Vert f Vert_{L^{p}(Omega)} leq Vert f Vert_{L^{q}(Omega)}leq C_2Vert f Vert_{L^{p}(Omega)} $$
The first inequality is easy by simply using the Holder inequality and $C_1$ depends on the finite measure of $Omega$. How to show the second inequality?
real-analysis functional-analysis measure-theory
asked Dec 10 '18 at 19:04
BourneBourne
548
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What are the roles of $p$ and $q$ in the question?
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– Umberto P.
Dec 10 '18 at 19:31
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@UmbertoP. Sorry for the typo. 1 and 2 should be p and q.
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– Bourne
Dec 11 '18 at 0:49
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When your domain has finite total measure (whether bounded or not), then you have ONE of your inequalities, but not the other. When your domain has infinite total measure, then you have neither of the inequalities.
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– GEdgar
Dec 11 '18 at 12:57
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@GEdgar Thanks!
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– Bourne
Dec 11 '18 at 19:00
add a comment |
$begingroup$
For $1<p<q<infty$ and a bounded domain $Omega subseteq mathbb R^2$, is the following true? $$C_1Vert f Vert_{L^{p}(Omega)} leq Vert f Vert_{L^{q}(Omega)}leq C_2Vert f Vert_{L^{p}(Omega)} $$
The first inequality is easy by simply using the Holder inequality and $C_1$ depends on the finite measure of $Omega$. How to show the second inequality?
real-analysis functional-analysis measure-theory
asked Dec 10 '18 at 19:04
BourneBourne
548
$endgroup$
For $1<p<q<infty$ and a bounded domain $Omega subseteq mathbb R^2$, is the following true? $$C_1Vert f Vert_{L^{p}(Omega)} leq Vert f Vert_{L^{q}(Omega)}leq C_2Vert f Vert_{L^{p}(Omega)} $$
The first inequality is easy by simply using the Holder inequality and $C_1$ depends on the finite measure of $Omega$. How to show the second inequality?
real-analysis functional-analysis measure-theory
real-analysis functional-analysis measure-theory
asked Dec 10 '18 at 19:04
BourneBourne
548
asked Dec 10 '18 at 19:04
BourneBourne
548
edited Dec 11 '18 at 0:49
Bourne
asked Dec 10 '18 at 19:04
BourneBourne
548
asked Dec 10 '18 at 19:04
BourneBourne
548
asked Dec 10 '18 at 19:04
BourneBourne
548
548
$begingroup$
What are the roles of $p$ and $q$ in the question?
$endgroup$
– Umberto P.
Dec 10 '18 at 19:31
$begingroup$
@UmbertoP. Sorry for the typo. 1 and 2 should be p and q.
$endgroup$
– Bourne
Dec 11 '18 at 0:49
$begingroup$
When your domain has finite total measure (whether bounded or not), then you have ONE of your inequalities, but not the other. When your domain has infinite total measure, then you have neither of the inequalities.
$endgroup$
– GEdgar
Dec 11 '18 at 12:57
$begingroup$
@GEdgar Thanks!
$endgroup$
– Bourne
Dec 11 '18 at 19:00
add a comment |
$begingroup$
What are the roles of $p$ and $q$ in the question?
$endgroup$
– Umberto P.
Dec 10 '18 at 19:31
$begingroup$
@UmbertoP. Sorry for the typo. 1 and 2 should be p and q.
$endgroup$
– Bourne
Dec 11 '18 at 0:49
$begingroup$
When your domain has finite total measure (whether bounded or not), then you have ONE of your inequalities, but not the other. When your domain has infinite total measure, then you have neither of the inequalities.
$endgroup$
– GEdgar
Dec 11 '18 at 12:57
$begingroup$
@GEdgar Thanks!
$endgroup$
– Bourne
Dec 11 '18 at 19:00
$begingroup$
What are the roles of $p$ and $q$ in the question?
$endgroup$
– Umberto P.
Dec 10 '18 at 19:31
$begingroup$
What are the roles of $p$ and $q$ in the question?
$endgroup$
– Umberto P.
Dec 10 '18 at 19:31
$begingroup$
@UmbertoP. Sorry for the typo. 1 and 2 should be p and q.
$endgroup$
– Bourne
Dec 11 '18 at 0:49
$begingroup$
@UmbertoP. Sorry for the typo. 1 and 2 should be p and q.
$endgroup$
– Bourne
Dec 11 '18 at 0:49
$begingroup$
When your domain has finite total measure (whether bounded or not), then you have ONE of your inequalities, but not the other. When your domain has infinite total measure, then you have neither of the inequalities.
$endgroup$
– GEdgar
Dec 11 '18 at 12:57
$begingroup$
When your domain has finite total measure (whether bounded or not), then you have ONE of your inequalities, but not the other. When your domain has infinite total measure, then you have neither of the inequalities.
$endgroup$
– GEdgar
Dec 11 '18 at 12:57
$begingroup$
@GEdgar Thanks!
$endgroup$
– Bourne
Dec 11 '18 at 19:00
$begingroup$
@GEdgar Thanks!
$endgroup$
– Bourne
Dec 11 '18 at 19:00
add a comment |
1 Answer
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It is simply untrue. Let $Omega$ be any bounded open set containing the origin and let $f(x) = dfrac 1{|x|^{2/q}}$.
answered Dec 10 '18 at 19:17
Umberto P.Umberto P.
40.1k13368
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$begingroup$
It is simply untrue. Let $Omega$ be any bounded open set containing the origin and let $f(x) = dfrac 1{|x|^{2/q}}$.
answered Dec 10 '18 at 19:17
Umberto P.Umberto P.
40.1k13368
$endgroup$
add a comment |
$begingroup$
It is simply untrue. Let $Omega$ be any bounded open set containing the origin and let $f(x) = dfrac 1{|x|^{2/q}}$.
answered Dec 10 '18 at 19:17
Umberto P.Umberto P.
40.1k13368
$endgroup$
add a comment |
$begingroup$
It is simply untrue. Let $Omega$ be any bounded open set containing the origin and let $f(x) = dfrac 1{|x|^{2/q}}$.
answered Dec 10 '18 at 19:17
Umberto P.Umberto P.
40.1k13368
$endgroup$
It is simply untrue. Let $Omega$ be any bounded open set containing the origin and let $f(x) = dfrac 1{|x|^{2/q}}$.
answered Dec 10 '18 at 19:17
Umberto P.Umberto P.
40.1k13368
edited Dec 11 '18 at 12:51
answered Dec 10 '18 at 19:17
Umberto P.Umberto P.
40.1k13368
answered Dec 10 '18 at 19:17
Umberto P.Umberto P.
40.1k13368
answered Dec 10 '18 at 19:17
Umberto P.Umberto P.
40.1k13368
40.1k13368
add a comment |
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$begingroup$
What are the roles of $p$ and $q$ in the question?
$endgroup$
– Umberto P.
Dec 10 '18 at 19:31
$begingroup$
@UmbertoP. Sorry for the typo. 1 and 2 should be p and q.
$endgroup$
– Bourne
Dec 11 '18 at 0:49
$begingroup$
When your domain has finite total measure (whether bounded or not), then you have ONE of your inequalities, but not the other. When your domain has infinite total measure, then you have neither of the inequalities.
$endgroup$
– GEdgar
Dec 11 '18 at 12:57
$begingroup$
@GEdgar Thanks!
$endgroup$
– Bourne
Dec 11 '18 at 19:00