Mixing
product of two Lebesgue spaces
$begingroup$
I want to show that if $fin L_p(R), gin L_q(R)$. Then $h=fg in L_s(R)$ ($1/p+1/q=1/s$). I've said that $|h|^s = |h|^{pq/(p+q)}$. I then define $f=h|h|^{p/(p+q)-1}$ and $g=|h|^{q/(p+q)}$. Then it's trivial that $fgin L_s(R)$. But why are $f,g$ in $L_p, L_q$ respectively?
real-analysis
asked Jan 30 at 21:50
JackJack
1218
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add a comment |
$begingroup$
I want to show that if $fin L_p(R), gin L_q(R)$. Then $h=fg in L_s(R)$ ($1/p+1/q=1/s$). I've said that $|h|^s = |h|^{pq/(p+q)}$. I then define $f=h|h|^{p/(p+q)-1}$ and $g=|h|^{q/(p+q)}$. Then it's trivial that $fgin L_s(R)$. But why are $f,g$ in $L_p, L_q$ respectively?
real-analysis
asked Jan 30 at 21:50
JackJack
1218
$endgroup$
$begingroup$
$f$ and $g$ are given to you. Why are trying to define $f$ and $g$?
$endgroup$
– Kavi Rama Murthy
Jan 30 at 23:31
$begingroup$
see if my revised answer is what you are looking for.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 6:19
add a comment |
$begingroup$
I want to show that if $fin L_p(R), gin L_q(R)$. Then $h=fg in L_s(R)$ ($1/p+1/q=1/s$). I've said that $|h|^s = |h|^{pq/(p+q)}$. I then define $f=h|h|^{p/(p+q)-1}$ and $g=|h|^{q/(p+q)}$. Then it's trivial that $fgin L_s(R)$. But why are $f,g$ in $L_p, L_q$ respectively?
real-analysis
asked Jan 30 at 21:50
JackJack
1218
$endgroup$
I want to show that if $fin L_p(R), gin L_q(R)$. Then $h=fg in L_s(R)$ ($1/p+1/q=1/s$). I've said that $|h|^s = |h|^{pq/(p+q)}$. I then define $f=h|h|^{p/(p+q)-1}$ and $g=|h|^{q/(p+q)}$. Then it's trivial that $fgin L_s(R)$. But why are $f,g$ in $L_p, L_q$ respectively?
real-analysis
real-analysis
asked Jan 30 at 21:50
JackJack
1218
asked Jan 30 at 21:50
JackJack
1218
edited Jan 30 at 23:44
Jack
asked Jan 30 at 21:50
JackJack
1218
asked Jan 30 at 21:50
JackJack
1218
asked Jan 30 at 21:50
JackJack
1218
1218
$begingroup$
$f$ and $g$ are given to you. Why are trying to define $f$ and $g$?
$endgroup$
– Kavi Rama Murthy
Jan 30 at 23:31
$begingroup$
see if my revised answer is what you are looking for.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 6:19
add a comment |
$begingroup$
$f$ and $g$ are given to you. Why are trying to define $f$ and $g$?
$endgroup$
– Kavi Rama Murthy
Jan 30 at 23:31
$begingroup$
see if my revised answer is what you are looking for.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 6:19
$begingroup$
$f$ and $g$ are given to you. Why are trying to define $f$ and $g$?
$endgroup$
– Kavi Rama Murthy
Jan 30 at 23:31
$begingroup$
$f$ and $g$ are given to you. Why are trying to define $f$ and $g$?
$endgroup$
– Kavi Rama Murthy
Jan 30 at 23:31
$begingroup$
see if my revised answer is what you are looking for.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 6:19
$begingroup$
see if my revised answer is what you are looking for.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 6:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think there is a mistype when you said $fg in L_p$. It should be $fg in L_s$. If $fin L_p$ and $g in L_q$ then $int |fg|^{s} leq (int |f|^{p})^{s/p} (int |g|^{q})^{s/q}$ by Holder's inequality (because $frac s p +frac s q=1$). Hence $int |fg|^{s} <infty$.
Wait, let me guess. You have made of mess of things but I think what you really want is a proof of the fact that $h in L_s$ implies there exist $fin L_p, gin L_q$ such that $h=fg$. For this define $f=h|h|^{s/p-1}$ and $g=|h|^{s/q}$. Then $fg=h|h|^{s(frac 1 p+frac 1 q -1)}=h$ and it is obvious that $fin L_p$ and $ gin L_q$.
answered Jan 30 at 23:30
Kavi Rama MurthyKavi Rama Murthy
72.5k53170
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add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
I think there is a mistype when you said $fg in L_p$. It should be $fg in L_s$. If $fin L_p$ and $g in L_q$ then $int |fg|^{s} leq (int |f|^{p})^{s/p} (int |g|^{q})^{s/q}$ by Holder's inequality (because $frac s p +frac s q=1$). Hence $int |fg|^{s} <infty$.
Wait, let me guess. You have made of mess of things but I think what you really want is a proof of the fact that $h in L_s$ implies there exist $fin L_p, gin L_q$ such that $h=fg$. For this define $f=h|h|^{s/p-1}$ and $g=|h|^{s/q}$. Then $fg=h|h|^{s(frac 1 p+frac 1 q -1)}=h$ and it is obvious that $fin L_p$ and $ gin L_q$.
answered Jan 30 at 23:30
Kavi Rama MurthyKavi Rama Murthy
72.5k53170
$endgroup$
add a comment |
$begingroup$
I think there is a mistype when you said $fg in L_p$. It should be $fg in L_s$. If $fin L_p$ and $g in L_q$ then $int |fg|^{s} leq (int |f|^{p})^{s/p} (int |g|^{q})^{s/q}$ by Holder's inequality (because $frac s p +frac s q=1$). Hence $int |fg|^{s} <infty$.
Wait, let me guess. You have made of mess of things but I think what you really want is a proof of the fact that $h in L_s$ implies there exist $fin L_p, gin L_q$ such that $h=fg$. For this define $f=h|h|^{s/p-1}$ and $g=|h|^{s/q}$. Then $fg=h|h|^{s(frac 1 p+frac 1 q -1)}=h$ and it is obvious that $fin L_p$ and $ gin L_q$.
answered Jan 30 at 23:30
Kavi Rama MurthyKavi Rama Murthy
72.5k53170
$endgroup$
add a comment |
$begingroup$
I think there is a mistype when you said $fg in L_p$. It should be $fg in L_s$. If $fin L_p$ and $g in L_q$ then $int |fg|^{s} leq (int |f|^{p})^{s/p} (int |g|^{q})^{s/q}$ by Holder's inequality (because $frac s p +frac s q=1$). Hence $int |fg|^{s} <infty$.
Wait, let me guess. You have made of mess of things but I think what you really want is a proof of the fact that $h in L_s$ implies there exist $fin L_p, gin L_q$ such that $h=fg$. For this define $f=h|h|^{s/p-1}$ and $g=|h|^{s/q}$. Then $fg=h|h|^{s(frac 1 p+frac 1 q -1)}=h$ and it is obvious that $fin L_p$ and $ gin L_q$.
answered Jan 30 at 23:30
Kavi Rama MurthyKavi Rama Murthy
72.5k53170
$endgroup$
I think there is a mistype when you said $fg in L_p$. It should be $fg in L_s$. If $fin L_p$ and $g in L_q$ then $int |fg|^{s} leq (int |f|^{p})^{s/p} (int |g|^{q})^{s/q}$ by Holder's inequality (because $frac s p +frac s q=1$). Hence $int |fg|^{s} <infty$.
Wait, let me guess. You have made of mess of things but I think what you really want is a proof of the fact that $h in L_s$ implies there exist $fin L_p, gin L_q$ such that $h=fg$. For this define $f=h|h|^{s/p-1}$ and $g=|h|^{s/q}$. Then $fg=h|h|^{s(frac 1 p+frac 1 q -1)}=h$ and it is obvious that $fin L_p$ and $ gin L_q$.
answered Jan 30 at 23:30
Kavi Rama MurthyKavi Rama Murthy
72.5k53170
edited Jan 31 at 6:17
answered Jan 30 at 23:30
Kavi Rama MurthyKavi Rama Murthy
72.5k53170
answered Jan 30 at 23:30
Kavi Rama MurthyKavi Rama Murthy
72.5k53170
answered Jan 30 at 23:30
Kavi Rama MurthyKavi Rama Murthy
72.5k53170
72.5k53170
add a comment |
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$begingroup$
$f$ and $g$ are given to you. Why are trying to define $f$ and $g$?
$endgroup$
– Kavi Rama Murthy
Jan 30 at 23:31
$begingroup$
see if my revised answer is what you are looking for.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 6:19