“Reverse Hölder” type inequality on a dense subset?
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I want to know if the following is possible.
Fix some measure space $(X,mu)$. Take any $fin L^1$. Recall that $L^1 cap L^2$ is dense in $L^1$, i.e. for every $varepsilon > 0$ there is some $gin L^1 cap L^2 $ such that $Vert f - g Vert_1 < varepsilon$. If $mu(X)<infty$ we can bound the $L^1$ norm of $g$ in terms of the $L^2 $ norm, like so: $Vert g Vert_1 leq mu(X)^{1/2} Vert g Vert_2$; this is an immediate consequence of Hölder's inequality.
I want to know when you can get a similar bound in the other direction, that is, that there exists some $C$ such that $Vert g Vert_2 leq C Vert g Vert_1$. More specifically, (under what circumstances) can we show that there exists some constant $C$ (possibly depending on $Vert f Vert_1 $ and $varepsilon$) such that there exists some some $gin L^1 cap L^2 $ such that $Vert f - g Vert_1 < varepsilon$ and $Vert g Vert_2 leq C Vert g Vert_1$? (In other words, can we show a reversal of the usual interpolation inequality on a dense subset?)
functional-analysis lp-spaces
asked Nov 29 '18 at 16:15
pseudocydoniapseudocydonia
405211
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add a comment |
$begingroup$
I want to know if the following is possible.
Fix some measure space $(X,mu)$. Take any $fin L^1$. Recall that $L^1 cap L^2$ is dense in $L^1$, i.e. for every $varepsilon > 0$ there is some $gin L^1 cap L^2 $ such that $Vert f - g Vert_1 < varepsilon$. If $mu(X)<infty$ we can bound the $L^1$ norm of $g$ in terms of the $L^2 $ norm, like so: $Vert g Vert_1 leq mu(X)^{1/2} Vert g Vert_2$; this is an immediate consequence of Hölder's inequality.
I want to know when you can get a similar bound in the other direction, that is, that there exists some $C$ such that $Vert g Vert_2 leq C Vert g Vert_1$. More specifically, (under what circumstances) can we show that there exists some constant $C$ (possibly depending on $Vert f Vert_1 $ and $varepsilon$) such that there exists some some $gin L^1 cap L^2 $ such that $Vert f - g Vert_1 < varepsilon$ and $Vert g Vert_2 leq C Vert g Vert_1$? (In other words, can we show a reversal of the usual interpolation inequality on a dense subset?)
functional-analysis lp-spaces
asked Nov 29 '18 at 16:15
pseudocydoniapseudocydonia
405211
$endgroup$
add a comment |
$begingroup$
I want to know if the following is possible.
Fix some measure space $(X,mu)$. Take any $fin L^1$. Recall that $L^1 cap L^2$ is dense in $L^1$, i.e. for every $varepsilon > 0$ there is some $gin L^1 cap L^2 $ such that $Vert f - g Vert_1 < varepsilon$. If $mu(X)<infty$ we can bound the $L^1$ norm of $g$ in terms of the $L^2 $ norm, like so: $Vert g Vert_1 leq mu(X)^{1/2} Vert g Vert_2$; this is an immediate consequence of Hölder's inequality.
I want to know when you can get a similar bound in the other direction, that is, that there exists some $C$ such that $Vert g Vert_2 leq C Vert g Vert_1$. More specifically, (under what circumstances) can we show that there exists some constant $C$ (possibly depending on $Vert f Vert_1 $ and $varepsilon$) such that there exists some some $gin L^1 cap L^2 $ such that $Vert f - g Vert_1 < varepsilon$ and $Vert g Vert_2 leq C Vert g Vert_1$? (In other words, can we show a reversal of the usual interpolation inequality on a dense subset?)
functional-analysis lp-spaces
asked Nov 29 '18 at 16:15
pseudocydoniapseudocydonia
405211
$endgroup$
I want to know if the following is possible.
Fix some measure space $(X,mu)$. Take any $fin L^1$. Recall that $L^1 cap L^2$ is dense in $L^1$, i.e. for every $varepsilon > 0$ there is some $gin L^1 cap L^2 $ such that $Vert f - g Vert_1 < varepsilon$. If $mu(X)<infty$ we can bound the $L^1$ norm of $g$ in terms of the $L^2 $ norm, like so: $Vert g Vert_1 leq mu(X)^{1/2} Vert g Vert_2$; this is an immediate consequence of Hölder's inequality.
I want to know when you can get a similar bound in the other direction, that is, that there exists some $C$ such that $Vert g Vert_2 leq C Vert g Vert_1$. More specifically, (under what circumstances) can we show that there exists some constant $C$ (possibly depending on $Vert f Vert_1 $ and $varepsilon$) such that there exists some some $gin L^1 cap L^2 $ such that $Vert f - g Vert_1 < varepsilon$ and $Vert g Vert_2 leq C Vert g Vert_1$? (In other words, can we show a reversal of the usual interpolation inequality on a dense subset?)
functional-analysis lp-spaces
functional-analysis lp-spaces
asked Nov 29 '18 at 16:15
pseudocydoniapseudocydonia
405211
asked Nov 29 '18 at 16:15
pseudocydoniapseudocydonia
405211
asked Nov 29 '18 at 16:15
pseudocydoniapseudocydonia
405211
asked Nov 29 '18 at 16:15
pseudocydoniapseudocydonia
405211
asked Nov 29 '18 at 16:15
pseudocydoniapseudocydonia
405211
405211
add a comment |
add a comment |
1 Answer
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A "reverse Hölder inequality" exists in the literature. It is studied in the theory of "weights". Essentially a weight $w$ is a positive function for which the measure $w(x) dx$ have the same boudedness properties with respect to some classes of singular integral operators. Look into [D: Theorem 7.4].
The only difference between the condition that you demand and the $A_p$ conditions imposed in weighted theory is that they tend to be "open". I.e. you can have a reverse Hölder inequality for your exponent plus some (potentially small) $epsilon >0$.
[D]: Duoandikoetxea, Javier, Fourier analysis. Transl. from the Spanish and revised by David Cruz-Uribe, Graduate Studies in Mathematics. 29. Providence, RI: American Mathematical Society (AMS). xviii, 222 p. (2001). ZBL0969.42001.
answered Nov 30 '18 at 11:09
Adrián González-PérezAdrián González-Pérez
1,004138
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$begingroup$
Very cool! Thanks for the reference, I'll have a look.
$endgroup$
– pseudocydonia
Nov 30 '18 at 19:32
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Okay, so I am looking at Duoandikoetxea's book, and this looks related but maybe is not 100% what I'm looking for. For one thing the underlying setting is $mathbb{R}^d$ rather than an arbitrary measure space, I guess because the Calderón-Zygmund decomposition is used. Also it's not exactly obvious to me how $w$ being in $A_p$ relates to being in $L_p$.
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– pseudocydonia
Dec 1 '18 at 3:12
add a comment |
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1 Answer
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1 Answer
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$begingroup$
A "reverse Hölder inequality" exists in the literature. It is studied in the theory of "weights". Essentially a weight $w$ is a positive function for which the measure $w(x) dx$ have the same boudedness properties with respect to some classes of singular integral operators. Look into [D: Theorem 7.4].
The only difference between the condition that you demand and the $A_p$ conditions imposed in weighted theory is that they tend to be "open". I.e. you can have a reverse Hölder inequality for your exponent plus some (potentially small) $epsilon >0$.
[D]: Duoandikoetxea, Javier, Fourier analysis. Transl. from the Spanish and revised by David Cruz-Uribe, Graduate Studies in Mathematics. 29. Providence, RI: American Mathematical Society (AMS). xviii, 222 p. (2001). ZBL0969.42001.
answered Nov 30 '18 at 11:09
Adrián González-PérezAdrián González-Pérez
1,004138
$endgroup$
$begingroup$
Very cool! Thanks for the reference, I'll have a look.
$endgroup$
– pseudocydonia
Nov 30 '18 at 19:32
$begingroup$
Okay, so I am looking at Duoandikoetxea's book, and this looks related but maybe is not 100% what I'm looking for. For one thing the underlying setting is $mathbb{R}^d$ rather than an arbitrary measure space, I guess because the Calderón-Zygmund decomposition is used. Also it's not exactly obvious to me how $w$ being in $A_p$ relates to being in $L_p$.
$endgroup$
– pseudocydonia
Dec 1 '18 at 3:12
add a comment |
$begingroup$
A "reverse Hölder inequality" exists in the literature. It is studied in the theory of "weights". Essentially a weight $w$ is a positive function for which the measure $w(x) dx$ have the same boudedness properties with respect to some classes of singular integral operators. Look into [D: Theorem 7.4].
The only difference between the condition that you demand and the $A_p$ conditions imposed in weighted theory is that they tend to be "open". I.e. you can have a reverse Hölder inequality for your exponent plus some (potentially small) $epsilon >0$.
[D]: Duoandikoetxea, Javier, Fourier analysis. Transl. from the Spanish and revised by David Cruz-Uribe, Graduate Studies in Mathematics. 29. Providence, RI: American Mathematical Society (AMS). xviii, 222 p. (2001). ZBL0969.42001.
answered Nov 30 '18 at 11:09
Adrián González-PérezAdrián González-Pérez
1,004138
$endgroup$
$begingroup$
Very cool! Thanks for the reference, I'll have a look.
$endgroup$
– pseudocydonia
Nov 30 '18 at 19:32
$begingroup$
Okay, so I am looking at Duoandikoetxea's book, and this looks related but maybe is not 100% what I'm looking for. For one thing the underlying setting is $mathbb{R}^d$ rather than an arbitrary measure space, I guess because the Calderón-Zygmund decomposition is used. Also it's not exactly obvious to me how $w$ being in $A_p$ relates to being in $L_p$.
$endgroup$
– pseudocydonia
Dec 1 '18 at 3:12
add a comment |
$begingroup$
A "reverse Hölder inequality" exists in the literature. It is studied in the theory of "weights". Essentially a weight $w$ is a positive function for which the measure $w(x) dx$ have the same boudedness properties with respect to some classes of singular integral operators. Look into [D: Theorem 7.4].
The only difference between the condition that you demand and the $A_p$ conditions imposed in weighted theory is that they tend to be "open". I.e. you can have a reverse Hölder inequality for your exponent plus some (potentially small) $epsilon >0$.
[D]: Duoandikoetxea, Javier, Fourier analysis. Transl. from the Spanish and revised by David Cruz-Uribe, Graduate Studies in Mathematics. 29. Providence, RI: American Mathematical Society (AMS). xviii, 222 p. (2001). ZBL0969.42001.
answered Nov 30 '18 at 11:09
Adrián González-PérezAdrián González-Pérez
1,004138
$endgroup$
A "reverse Hölder inequality" exists in the literature. It is studied in the theory of "weights". Essentially a weight $w$ is a positive function for which the measure $w(x) dx$ have the same boudedness properties with respect to some classes of singular integral operators. Look into [D: Theorem 7.4].
The only difference between the condition that you demand and the $A_p$ conditions imposed in weighted theory is that they tend to be "open". I.e. you can have a reverse Hölder inequality for your exponent plus some (potentially small) $epsilon >0$.
[D]: Duoandikoetxea, Javier, Fourier analysis. Transl. from the Spanish and revised by David Cruz-Uribe, Graduate Studies in Mathematics. 29. Providence, RI: American Mathematical Society (AMS). xviii, 222 p. (2001). ZBL0969.42001.
answered Nov 30 '18 at 11:09
Adrián González-PérezAdrián González-Pérez
1,004138
answered Nov 30 '18 at 11:09
Adrián González-PérezAdrián González-Pérez
1,004138
answered Nov 30 '18 at 11:09
Adrián González-PérezAdrián González-Pérez
1,004138
answered Nov 30 '18 at 11:09
Adrián González-PérezAdrián González-Pérez
1,004138
1,004138
$begingroup$
Very cool! Thanks for the reference, I'll have a look.
$endgroup$
– pseudocydonia
Nov 30 '18 at 19:32
$begingroup$
Okay, so I am looking at Duoandikoetxea's book, and this looks related but maybe is not 100% what I'm looking for. For one thing the underlying setting is $mathbb{R}^d$ rather than an arbitrary measure space, I guess because the Calderón-Zygmund decomposition is used. Also it's not exactly obvious to me how $w$ being in $A_p$ relates to being in $L_p$.
$endgroup$
– pseudocydonia
Dec 1 '18 at 3:12
add a comment |
$begingroup$
Very cool! Thanks for the reference, I'll have a look.
$endgroup$
– pseudocydonia
Nov 30 '18 at 19:32
$begingroup$
Okay, so I am looking at Duoandikoetxea's book, and this looks related but maybe is not 100% what I'm looking for. For one thing the underlying setting is $mathbb{R}^d$ rather than an arbitrary measure space, I guess because the Calderón-Zygmund decomposition is used. Also it's not exactly obvious to me how $w$ being in $A_p$ relates to being in $L_p$.
$endgroup$
– pseudocydonia
Dec 1 '18 at 3:12
$begingroup$
Very cool! Thanks for the reference, I'll have a look.
$endgroup$
– pseudocydonia
Nov 30 '18 at 19:32
$begingroup$
Very cool! Thanks for the reference, I'll have a look.
$endgroup$
– pseudocydonia
Nov 30 '18 at 19:32
$begingroup$
Okay, so I am looking at Duoandikoetxea's book, and this looks related but maybe is not 100% what I'm looking for. For one thing the underlying setting is $mathbb{R}^d$ rather than an arbitrary measure space, I guess because the Calderón-Zygmund decomposition is used. Also it's not exactly obvious to me how $w$ being in $A_p$ relates to being in $L_p$.
$endgroup$
– pseudocydonia
Dec 1 '18 at 3:12
$begingroup$
Okay, so I am looking at Duoandikoetxea's book, and this looks related but maybe is not 100% what I'm looking for. For one thing the underlying setting is $mathbb{R}^d$ rather than an arbitrary measure space, I guess because the Calderón-Zygmund decomposition is used. Also it's not exactly obvious to me how $w$ being in $A_p$ relates to being in $L_p$.
$endgroup$
– pseudocydonia
Dec 1 '18 at 3:12
add a comment |
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