An application of Holder's Inequality
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Suppose $1leq p,qleq infty$ and $1/p+1/q=1$. Let $finmathcal{L}^p(E)$. Show that $f=0$ a.e. if and only if
begin{align*}
int_E fcdot gdm=0
end{align*}
for all $gin mathcal{L}^q(E)$. Hint: Choose a smart $g$ so that $int_Efcdot gdm=|f|_p^p$.
functional-analysis holder-inequality
asked Dec 4 '18 at 19:54
TNTTNT
596
$endgroup$
add a comment |
$begingroup$
Suppose $1leq p,qleq infty$ and $1/p+1/q=1$. Let $finmathcal{L}^p(E)$. Show that $f=0$ a.e. if and only if
begin{align*}
int_E fcdot gdm=0
end{align*}
for all $gin mathcal{L}^q(E)$. Hint: Choose a smart $g$ so that $int_Efcdot gdm=|f|_p^p$.
functional-analysis holder-inequality
asked Dec 4 '18 at 19:54
TNTTNT
596
$endgroup$
$begingroup$
In fact, if $f in L^p(E)$ then $$|f|_p = sup_{g in L^q(E)} int_E fg , dm.$$
$endgroup$
– Umberto P.
Dec 4 '18 at 20:20
add a comment |
$begingroup$
Suppose $1leq p,qleq infty$ and $1/p+1/q=1$. Let $finmathcal{L}^p(E)$. Show that $f=0$ a.e. if and only if
begin{align*}
int_E fcdot gdm=0
end{align*}
for all $gin mathcal{L}^q(E)$. Hint: Choose a smart $g$ so that $int_Efcdot gdm=|f|_p^p$.
functional-analysis holder-inequality
asked Dec 4 '18 at 19:54
TNTTNT
596
$endgroup$
Suppose $1leq p,qleq infty$ and $1/p+1/q=1$. Let $finmathcal{L}^p(E)$. Show that $f=0$ a.e. if and only if
begin{align*}
int_E fcdot gdm=0
end{align*}
for all $gin mathcal{L}^q(E)$. Hint: Choose a smart $g$ so that $int_Efcdot gdm=|f|_p^p$.
functional-analysis holder-inequality
functional-analysis holder-inequality
asked Dec 4 '18 at 19:54
TNTTNT
596
asked Dec 4 '18 at 19:54
TNTTNT
596
asked Dec 4 '18 at 19:54
TNTTNT
596
asked Dec 4 '18 at 19:54
TNTTNT
596
asked Dec 4 '18 at 19:54
TNTTNT
596
596
$begingroup$
In fact, if $f in L^p(E)$ then $$|f|_p = sup_{g in L^q(E)} int_E fg , dm.$$
$endgroup$
– Umberto P.
Dec 4 '18 at 20:20
add a comment |
$begingroup$
In fact, if $f in L^p(E)$ then $$|f|_p = sup_{g in L^q(E)} int_E fg , dm.$$
$endgroup$
– Umberto P.
Dec 4 '18 at 20:20
$begingroup$
In fact, if $f in L^p(E)$ then $$|f|_p = sup_{g in L^q(E)} int_E fg , dm.$$
$endgroup$
– Umberto P.
Dec 4 '18 at 20:20
$begingroup$
In fact, if $f in L^p(E)$ then $$|f|_p = sup_{g in L^q(E)} int_E fg , dm.$$
$endgroup$
– Umberto P.
Dec 4 '18 at 20:20
add a comment |
1 Answer
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Hint: try $g=mathrm{sign}(f)cdot|f|^alpha$ for some $alphainmathbb R$.
Second hint: $alpha=p/q$.
answered Dec 4 '18 at 19:59
FedericoFederico
4,984514
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1 Answer
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1 Answer
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$begingroup$
Hint: try $g=mathrm{sign}(f)cdot|f|^alpha$ for some $alphainmathbb R$.
Second hint: $alpha=p/q$.
answered Dec 4 '18 at 19:59
FedericoFederico
4,984514
$endgroup$
add a comment |
$begingroup$
Hint: try $g=mathrm{sign}(f)cdot|f|^alpha$ for some $alphainmathbb R$.
Second hint: $alpha=p/q$.
answered Dec 4 '18 at 19:59
FedericoFederico
4,984514
$endgroup$
add a comment |
$begingroup$
Hint: try $g=mathrm{sign}(f)cdot|f|^alpha$ for some $alphainmathbb R$.
Second hint: $alpha=p/q$.
answered Dec 4 '18 at 19:59
FedericoFederico
4,984514
$endgroup$
Hint: try $g=mathrm{sign}(f)cdot|f|^alpha$ for some $alphainmathbb R$.
Second hint: $alpha=p/q$.
answered Dec 4 '18 at 19:59
FedericoFederico
4,984514
answered Dec 4 '18 at 19:59
FedericoFederico
4,984514
answered Dec 4 '18 at 19:59
FedericoFederico
4,984514
answered Dec 4 '18 at 19:59
FedericoFederico
4,984514
4,984514
add a comment |
add a comment |
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$begingroup$
In fact, if $f in L^p(E)$ then $$|f|_p = sup_{g in L^q(E)} int_E fg , dm.$$
$endgroup$
– Umberto P.
Dec 4 '18 at 20:20