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 at 19:54
TNT
596
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up vote
1
down vote
favorite
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 at 19:54
TNT
596
In fact, if $f in L^p(E)$ then $$|f|_p = sup_{g in L^q(E)} int_E fg , dm.$$
– Umberto P.
Dec 4 at 20:20
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 at 19:54
TNT
596
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 at 19:54
TNT
596
asked Dec 4 at 19:54
TNT
596
asked Dec 4 at 19:54
TNT
596
asked Dec 4 at 19:54
TNT
596
asked Dec 4 at 19:54
TNT
596
596
In fact, if $f in L^p(E)$ then $$|f|_p = sup_{g in L^q(E)} int_E fg , dm.$$
– Umberto P.
Dec 4 at 20:20
add a comment |
In fact, if $f in L^p(E)$ then $$|f|_p = sup_{g in L^q(E)} int_E fg , dm.$$
– Umberto P.
Dec 4 at 20:20
In fact, if $f in L^p(E)$ then $$|f|_p = sup_{g in L^q(E)} int_E fg , dm.$$
– Umberto P.
Dec 4 at 20:20
In fact, if $f in L^p(E)$ then $$|f|_p = sup_{g in L^q(E)} int_E fg , dm.$$
– Umberto P.
Dec 4 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 at 19:59
Federico
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1 Answer
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1 Answer
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active
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up vote
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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 at 19:59
Federico
4,203512
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up vote
2
down vote
Hint: try $g=mathrm{sign}(f)cdot|f|^alpha$ for some $alphainmathbb R$.
Second hint: $alpha=p/q$.
answered Dec 4 at 19:59
Federico
4,203512
add a comment |
up vote
2
down vote
up vote
2
down vote
Hint: try $g=mathrm{sign}(f)cdot|f|^alpha$ for some $alphainmathbb R$.
Second hint: $alpha=p/q$.
answered Dec 4 at 19:59
Federico
4,203512
Hint: try $g=mathrm{sign}(f)cdot|f|^alpha$ for some $alphainmathbb R$.
Second hint: $alpha=p/q$.
answered Dec 4 at 19:59
Federico
4,203512
answered Dec 4 at 19:59
Federico
4,203512
answered Dec 4 at 19:59
Federico
4,203512
answered Dec 4 at 19:59
Federico
4,203512
4,203512
add a comment |
add a comment |
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In fact, if $f in L^p(E)$ then $$|f|_p = sup_{g in L^q(E)} int_E fg , dm.$$
– Umberto P.
Dec 4 at 20:20