Mixing
How to bound uniformly this integral?
$begingroup$
could someone help me with this question please?
I need to prove that if $1<p<2<q;$ and $fin{L^p(mathbb{R^2})}cap{L^q(mathbb{R^2})}$ then $$g(z)=int_mathbb{R^2}dfrac{|f(y)|}{|z-y|}dy$$ is uniformly bounded.
I tried to write that as
$$g(z)=int_{|z-y|<R}dfrac{|f(y)|}{|z-y|}dy+int_{|z-y|>R}dfrac{|f(y)|}{|z-y|}dy$$
But I only know how to bound each one if $p=1$ and $q=infty$ after fixing a concrete $R$.
Thanks.
real-analysis functional-analysis
asked Jan 6 at 15:48
mathlifemathlife
629
$endgroup$
add a comment |
$begingroup$
could someone help me with this question please?
I need to prove that if $1<p<2<q;$ and $fin{L^p(mathbb{R^2})}cap{L^q(mathbb{R^2})}$ then $$g(z)=int_mathbb{R^2}dfrac{|f(y)|}{|z-y|}dy$$ is uniformly bounded.
I tried to write that as
$$g(z)=int_{|z-y|<R}dfrac{|f(y)|}{|z-y|}dy+int_{|z-y|>R}dfrac{|f(y)|}{|z-y|}dy$$
But I only know how to bound each one if $p=1$ and $q=infty$ after fixing a concrete $R$.
Thanks.
real-analysis functional-analysis
asked Jan 6 at 15:48
mathlifemathlife
629
$endgroup$
add a comment |
$begingroup$
could someone help me with this question please?
I need to prove that if $1<p<2<q;$ and $fin{L^p(mathbb{R^2})}cap{L^q(mathbb{R^2})}$ then $$g(z)=int_mathbb{R^2}dfrac{|f(y)|}{|z-y|}dy$$ is uniformly bounded.
I tried to write that as
$$g(z)=int_{|z-y|<R}dfrac{|f(y)|}{|z-y|}dy+int_{|z-y|>R}dfrac{|f(y)|}{|z-y|}dy$$
But I only know how to bound each one if $p=1$ and $q=infty$ after fixing a concrete $R$.
Thanks.
real-analysis functional-analysis
asked Jan 6 at 15:48
mathlifemathlife
629
$endgroup$
could someone help me with this question please?
I need to prove that if $1<p<2<q;$ and $fin{L^p(mathbb{R^2})}cap{L^q(mathbb{R^2})}$ then $$g(z)=int_mathbb{R^2}dfrac{|f(y)|}{|z-y|}dy$$ is uniformly bounded.
I tried to write that as
$$g(z)=int_{|z-y|<R}dfrac{|f(y)|}{|z-y|}dy+int_{|z-y|>R}dfrac{|f(y)|}{|z-y|}dy$$
But I only know how to bound each one if $p=1$ and $q=infty$ after fixing a concrete $R$.
Thanks.
real-analysis functional-analysis
real-analysis functional-analysis
asked Jan 6 at 15:48
mathlifemathlife
629
asked Jan 6 at 15:48
mathlifemathlife
629
asked Jan 6 at 15:48
mathlifemathlife
629
asked Jan 6 at 15:48
mathlifemathlife
629
asked Jan 6 at 15:48
mathlifemathlife
629
629
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Fix $R=1$. The first term
$$begin{eqnarray}
int_{|y|<1}dfrac{|f(z-y)|}{|y|}dy&le&left[int_{|y|<1}|f(z-y)|^qdyright]^{1/q}left[int_{|y|<1}frac{1}{|y|^q{^*}}dyright]^{1/q^*}\&le& |f|_qleft[int_{|y|<1}frac{1}{|y|^q{^*}}dyright]^{1/q^*}le M_1<infty
end{eqnarray}$$ by Holder's inequality since $1le q^*<2$ ($q^*$ is a Holder conjugate of $q$) and $ymapsto |y|^{-s}$ is locally integrable for $s<2$. We can check this using polar coordinate $dy = rdr$. The second term is estimated similarly:
$$begin{eqnarray}
int_{|y|ge 1}dfrac{|f(z-y)|}{|y|}dy&le&left[int_{|y|ge 1}|f(z-y)|^pdyright]^{1/p}left[int_{|y|ge 1}frac{1}{|y|^p{^*}}dyright]^{1/p^*}\&le& |f|_pleft[int_{|y|ge 1}frac{1}{|y|^p{^*}}dyright]^{1/p^*}le M_2 <infty.
end{eqnarray}$$ This is because $2<p^*<infty$ and
$$
int_{|y|ge 1}frac{1}{|y|^s}dy =int_1^infty r^{1-s}dr<infty
$$ for all $1-s<-1$. Gathering them together, we get an upper bound $M_1+M_2$ independent of $z$ as wanted.
answered Jan 6 at 16:01
SongSong
10.3k627
$endgroup$
$begingroup$
Don't you use that $1<p$?
$endgroup$
– mathlife
Jan 6 at 16:35
$begingroup$
So, $1<p$ is an unnecessary hypothesis?
$endgroup$
– mathlife
Jan 6 at 16:40
$begingroup$
@mathlife Umm .. Now I got what you meant. $p>1$ is necessary to show $p^*<infty$. But even if $p=1$, the argument is still valid using $p^*=infty$ norm. In this sense, $1=p$ is also allowed. In the same spirit, $q=infty$ is also allowed.
$endgroup$
– Song
Jan 6 at 16:44
$begingroup$
Ok, I see. Thanks
$endgroup$
– mathlife
Jan 6 at 16:48
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
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active
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$begingroup$
Fix $R=1$. The first term
$$begin{eqnarray}
int_{|y|<1}dfrac{|f(z-y)|}{|y|}dy&le&left[int_{|y|<1}|f(z-y)|^qdyright]^{1/q}left[int_{|y|<1}frac{1}{|y|^q{^*}}dyright]^{1/q^*}\&le& |f|_qleft[int_{|y|<1}frac{1}{|y|^q{^*}}dyright]^{1/q^*}le M_1<infty
end{eqnarray}$$ by Holder's inequality since $1le q^*<2$ ($q^*$ is a Holder conjugate of $q$) and $ymapsto |y|^{-s}$ is locally integrable for $s<2$. We can check this using polar coordinate $dy = rdr$. The second term is estimated similarly:
$$begin{eqnarray}
int_{|y|ge 1}dfrac{|f(z-y)|}{|y|}dy&le&left[int_{|y|ge 1}|f(z-y)|^pdyright]^{1/p}left[int_{|y|ge 1}frac{1}{|y|^p{^*}}dyright]^{1/p^*}\&le& |f|_pleft[int_{|y|ge 1}frac{1}{|y|^p{^*}}dyright]^{1/p^*}le M_2 <infty.
end{eqnarray}$$ This is because $2<p^*<infty$ and
$$
int_{|y|ge 1}frac{1}{|y|^s}dy =int_1^infty r^{1-s}dr<infty
$$ for all $1-s<-1$. Gathering them together, we get an upper bound $M_1+M_2$ independent of $z$ as wanted.
answered Jan 6 at 16:01
SongSong
10.3k627
$endgroup$
$begingroup$
Don't you use that $1<p$?
$endgroup$
– mathlife
Jan 6 at 16:35
$begingroup$
So, $1<p$ is an unnecessary hypothesis?
$endgroup$
– mathlife
Jan 6 at 16:40
$begingroup$
@mathlife Umm .. Now I got what you meant. $p>1$ is necessary to show $p^*<infty$. But even if $p=1$, the argument is still valid using $p^*=infty$ norm. In this sense, $1=p$ is also allowed. In the same spirit, $q=infty$ is also allowed.
$endgroup$
– Song
Jan 6 at 16:44
$begingroup$
Ok, I see. Thanks
$endgroup$
– mathlife
Jan 6 at 16:48
add a comment |
$begingroup$
Fix $R=1$. The first term
$$begin{eqnarray}
int_{|y|<1}dfrac{|f(z-y)|}{|y|}dy&le&left[int_{|y|<1}|f(z-y)|^qdyright]^{1/q}left[int_{|y|<1}frac{1}{|y|^q{^*}}dyright]^{1/q^*}\&le& |f|_qleft[int_{|y|<1}frac{1}{|y|^q{^*}}dyright]^{1/q^*}le M_1<infty
end{eqnarray}$$ by Holder's inequality since $1le q^*<2$ ($q^*$ is a Holder conjugate of $q$) and $ymapsto |y|^{-s}$ is locally integrable for $s<2$. We can check this using polar coordinate $dy = rdr$. The second term is estimated similarly:
$$begin{eqnarray}
int_{|y|ge 1}dfrac{|f(z-y)|}{|y|}dy&le&left[int_{|y|ge 1}|f(z-y)|^pdyright]^{1/p}left[int_{|y|ge 1}frac{1}{|y|^p{^*}}dyright]^{1/p^*}\&le& |f|_pleft[int_{|y|ge 1}frac{1}{|y|^p{^*}}dyright]^{1/p^*}le M_2 <infty.
end{eqnarray}$$ This is because $2<p^*<infty$ and
$$
int_{|y|ge 1}frac{1}{|y|^s}dy =int_1^infty r^{1-s}dr<infty
$$ for all $1-s<-1$. Gathering them together, we get an upper bound $M_1+M_2$ independent of $z$ as wanted.
answered Jan 6 at 16:01
SongSong
10.3k627
$endgroup$
$begingroup$
Don't you use that $1<p$?
$endgroup$
– mathlife
Jan 6 at 16:35
$begingroup$
So, $1<p$ is an unnecessary hypothesis?
$endgroup$
– mathlife
Jan 6 at 16:40
$begingroup$
@mathlife Umm .. Now I got what you meant. $p>1$ is necessary to show $p^*<infty$. But even if $p=1$, the argument is still valid using $p^*=infty$ norm. In this sense, $1=p$ is also allowed. In the same spirit, $q=infty$ is also allowed.
$endgroup$
– Song
Jan 6 at 16:44
$begingroup$
Ok, I see. Thanks
$endgroup$
– mathlife
Jan 6 at 16:48
add a comment |
$begingroup$
Fix $R=1$. The first term
$$begin{eqnarray}
int_{|y|<1}dfrac{|f(z-y)|}{|y|}dy&le&left[int_{|y|<1}|f(z-y)|^qdyright]^{1/q}left[int_{|y|<1}frac{1}{|y|^q{^*}}dyright]^{1/q^*}\&le& |f|_qleft[int_{|y|<1}frac{1}{|y|^q{^*}}dyright]^{1/q^*}le M_1<infty
end{eqnarray}$$ by Holder's inequality since $1le q^*<2$ ($q^*$ is a Holder conjugate of $q$) and $ymapsto |y|^{-s}$ is locally integrable for $s<2$. We can check this using polar coordinate $dy = rdr$. The second term is estimated similarly:
$$begin{eqnarray}
int_{|y|ge 1}dfrac{|f(z-y)|}{|y|}dy&le&left[int_{|y|ge 1}|f(z-y)|^pdyright]^{1/p}left[int_{|y|ge 1}frac{1}{|y|^p{^*}}dyright]^{1/p^*}\&le& |f|_pleft[int_{|y|ge 1}frac{1}{|y|^p{^*}}dyright]^{1/p^*}le M_2 <infty.
end{eqnarray}$$ This is because $2<p^*<infty$ and
$$
int_{|y|ge 1}frac{1}{|y|^s}dy =int_1^infty r^{1-s}dr<infty
$$ for all $1-s<-1$. Gathering them together, we get an upper bound $M_1+M_2$ independent of $z$ as wanted.
answered Jan 6 at 16:01
SongSong
10.3k627
$endgroup$
Fix $R=1$. The first term
$$begin{eqnarray}
int_{|y|<1}dfrac{|f(z-y)|}{|y|}dy&le&left[int_{|y|<1}|f(z-y)|^qdyright]^{1/q}left[int_{|y|<1}frac{1}{|y|^q{^*}}dyright]^{1/q^*}\&le& |f|_qleft[int_{|y|<1}frac{1}{|y|^q{^*}}dyright]^{1/q^*}le M_1<infty
end{eqnarray}$$ by Holder's inequality since $1le q^*<2$ ($q^*$ is a Holder conjugate of $q$) and $ymapsto |y|^{-s}$ is locally integrable for $s<2$. We can check this using polar coordinate $dy = rdr$. The second term is estimated similarly:
$$begin{eqnarray}
int_{|y|ge 1}dfrac{|f(z-y)|}{|y|}dy&le&left[int_{|y|ge 1}|f(z-y)|^pdyright]^{1/p}left[int_{|y|ge 1}frac{1}{|y|^p{^*}}dyright]^{1/p^*}\&le& |f|_pleft[int_{|y|ge 1}frac{1}{|y|^p{^*}}dyright]^{1/p^*}le M_2 <infty.
end{eqnarray}$$ This is because $2<p^*<infty$ and
$$
int_{|y|ge 1}frac{1}{|y|^s}dy =int_1^infty r^{1-s}dr<infty
$$ for all $1-s<-1$. Gathering them together, we get an upper bound $M_1+M_2$ independent of $z$ as wanted.
answered Jan 6 at 16:01
SongSong
10.3k627
edited Jan 6 at 16:39
answered Jan 6 at 16:01
SongSong
10.3k627
answered Jan 6 at 16:01
SongSong
10.3k627
answered Jan 6 at 16:01
SongSong
10.3k627
10.3k627
$begingroup$
Don't you use that $1<p$?
$endgroup$
– mathlife
Jan 6 at 16:35
$begingroup$
So, $1<p$ is an unnecessary hypothesis?
$endgroup$
– mathlife
Jan 6 at 16:40
$begingroup$
@mathlife Umm .. Now I got what you meant. $p>1$ is necessary to show $p^*<infty$. But even if $p=1$, the argument is still valid using $p^*=infty$ norm. In this sense, $1=p$ is also allowed. In the same spirit, $q=infty$ is also allowed.
$endgroup$
– Song
Jan 6 at 16:44
$begingroup$
Ok, I see. Thanks
$endgroup$
– mathlife
Jan 6 at 16:48
add a comment |
$begingroup$
Don't you use that $1<p$?
$endgroup$
– mathlife
Jan 6 at 16:35
$begingroup$
So, $1<p$ is an unnecessary hypothesis?
$endgroup$
– mathlife
Jan 6 at 16:40
$begingroup$
@mathlife Umm .. Now I got what you meant. $p>1$ is necessary to show $p^*<infty$. But even if $p=1$, the argument is still valid using $p^*=infty$ norm. In this sense, $1=p$ is also allowed. In the same spirit, $q=infty$ is also allowed.
$endgroup$
– Song
Jan 6 at 16:44
$begingroup$
Ok, I see. Thanks
$endgroup$
– mathlife
Jan 6 at 16:48
$begingroup$
Don't you use that $1<p$?
$endgroup$
– mathlife
Jan 6 at 16:35
$begingroup$
Don't you use that $1<p$?
$endgroup$
– mathlife
Jan 6 at 16:35
$begingroup$
So, $1<p$ is an unnecessary hypothesis?
$endgroup$
– mathlife
Jan 6 at 16:40
$begingroup$
So, $1<p$ is an unnecessary hypothesis?
$endgroup$
– mathlife
Jan 6 at 16:40
$begingroup$
@mathlife Umm .. Now I got what you meant. $p>1$ is necessary to show $p^*<infty$. But even if $p=1$, the argument is still valid using $p^*=infty$ norm. In this sense, $1=p$ is also allowed. In the same spirit, $q=infty$ is also allowed.
$endgroup$
– Song
Jan 6 at 16:44
$begingroup$
@mathlife Umm .. Now I got what you meant. $p>1$ is necessary to show $p^*<infty$. But even if $p=1$, the argument is still valid using $p^*=infty$ norm. In this sense, $1=p$ is also allowed. In the same spirit, $q=infty$ is also allowed.
$endgroup$
– Song
Jan 6 at 16:44
$begingroup$
Ok, I see. Thanks
$endgroup$
– mathlife
Jan 6 at 16:48
$begingroup$
Ok, I see. Thanks
$endgroup$
– mathlife
Jan 6 at 16:48
add a comment |
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