Relation between $L^p$ spaces.
$begingroup$
We say $fin L_{p}(U)$ if $int_{U}|f|^{p}<infty$.
Can we say $uin L_{4} iff u^{2} in L_{2}$, and $u in L_{2} iff u^{1/2} in L_{4}$
functional-analysis lebesgue-measure lp-spaces
asked Dec 24 '18 at 14:44
MontyMonty
34613
$endgroup$
add a comment |
$begingroup$
We say $fin L_{p}(U)$ if $int_{U}|f|^{p}<infty$.
Can we say $uin L_{4} iff u^{2} in L_{2}$, and $u in L_{2} iff u^{1/2} in L_{4}$
functional-analysis lebesgue-measure lp-spaces
asked Dec 24 '18 at 14:44
MontyMonty
34613
$endgroup$
add a comment |
$begingroup$
We say $fin L_{p}(U)$ if $int_{U}|f|^{p}<infty$.
Can we say $uin L_{4} iff u^{2} in L_{2}$, and $u in L_{2} iff u^{1/2} in L_{4}$
functional-analysis lebesgue-measure lp-spaces
asked Dec 24 '18 at 14:44
MontyMonty
34613
$endgroup$
We say $fin L_{p}(U)$ if $int_{U}|f|^{p}<infty$.
Can we say $uin L_{4} iff u^{2} in L_{2}$, and $u in L_{2} iff u^{1/2} in L_{4}$
functional-analysis lebesgue-measure lp-spaces
functional-analysis lebesgue-measure lp-spaces
asked Dec 24 '18 at 14:44
MontyMonty
34613
asked Dec 24 '18 at 14:44
MontyMonty
34613
edited Dec 28 '18 at 19:56
Monty
asked Dec 24 '18 at 14:44
MontyMonty
34613
asked Dec 24 '18 at 14:44
MontyMonty
34613
asked Dec 24 '18 at 14:44
MontyMonty
34613
34613
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes, that is correct, because of the identity $lvert x^nrvert=lvert xrvert^n$, which holds for all $xinBbb C$ and $ninBbb Z$. For $nnotin Bbb Z$ there is the definition issue of what $x^n$ is when $xnotin[0,infty)$, in which case $f^n$ is arguably not defined.
answered Dec 24 '18 at 14:46
Saucy O'PathSaucy O'Path
6,0491627
$endgroup$
$begingroup$
Yes, so can use this to work with functions $u^{2}$ in $L_{2}$ ? i.e see my edit.
$endgroup$
– Monty
Dec 24 '18 at 14:58
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Yes, that is correct, because of the identity $lvert x^nrvert=lvert xrvert^n$, which holds for all $xinBbb C$ and $ninBbb Z$. For $nnotin Bbb Z$ there is the definition issue of what $x^n$ is when $xnotin[0,infty)$, in which case $f^n$ is arguably not defined.
answered Dec 24 '18 at 14:46
Saucy O'PathSaucy O'Path
6,0491627
$endgroup$
$begingroup$
Yes, so can use this to work with functions $u^{2}$ in $L_{2}$ ? i.e see my edit.
$endgroup$
– Monty
Dec 24 '18 at 14:58
add a comment |
$begingroup$
Yes, that is correct, because of the identity $lvert x^nrvert=lvert xrvert^n$, which holds for all $xinBbb C$ and $ninBbb Z$. For $nnotin Bbb Z$ there is the definition issue of what $x^n$ is when $xnotin[0,infty)$, in which case $f^n$ is arguably not defined.
answered Dec 24 '18 at 14:46
Saucy O'PathSaucy O'Path
6,0491627
$endgroup$
$begingroup$
Yes, so can use this to work with functions $u^{2}$ in $L_{2}$ ? i.e see my edit.
$endgroup$
– Monty
Dec 24 '18 at 14:58
add a comment |
$begingroup$
Yes, that is correct, because of the identity $lvert x^nrvert=lvert xrvert^n$, which holds for all $xinBbb C$ and $ninBbb Z$. For $nnotin Bbb Z$ there is the definition issue of what $x^n$ is when $xnotin[0,infty)$, in which case $f^n$ is arguably not defined.
answered Dec 24 '18 at 14:46
Saucy O'PathSaucy O'Path
6,0491627
$endgroup$
Yes, that is correct, because of the identity $lvert x^nrvert=lvert xrvert^n$, which holds for all $xinBbb C$ and $ninBbb Z$. For $nnotin Bbb Z$ there is the definition issue of what $x^n$ is when $xnotin[0,infty)$, in which case $f^n$ is arguably not defined.
answered Dec 24 '18 at 14:46
Saucy O'PathSaucy O'Path
6,0491627
answered Dec 24 '18 at 14:46
Saucy O'PathSaucy O'Path
6,0491627
answered Dec 24 '18 at 14:46
Saucy O'PathSaucy O'Path
6,0491627
answered Dec 24 '18 at 14:46
Saucy O'PathSaucy O'Path
6,0491627
6,0491627
$begingroup$
Yes, so can use this to work with functions $u^{2}$ in $L_{2}$ ? i.e see my edit.
$endgroup$
– Monty
Dec 24 '18 at 14:58
add a comment |
$begingroup$
Yes, so can use this to work with functions $u^{2}$ in $L_{2}$ ? i.e see my edit.
$endgroup$
– Monty
Dec 24 '18 at 14:58
$begingroup$
Yes, so can use this to work with functions $u^{2}$ in $L_{2}$ ? i.e see my edit.
$endgroup$
– Monty
Dec 24 '18 at 14:58
$begingroup$
Yes, so can use this to work with functions $u^{2}$ in $L_{2}$ ? i.e see my edit.
$endgroup$
– Monty
Dec 24 '18 at 14:58
add a comment |
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