Help showing a linear functional is bounded
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Let $(X,M,mu)$ be a $sigma$-finite measure space, and $k:Xtimes X rightarrow mathbb{C}$ (or $mathbb{R}$) be $Xtimes X$ measurable. Suppose there are measurable functions $h,g:Xrightarrow(0,infty)$ and constants $c_1,c_2>0$ such that $$int_X |k(x,y)|g^q(y)dmu(y)leq c_1^qh^q(x)$$ a.e. and
$$int_X |k(x,y)|h^p(x)dmu(x)leq c_2^pg^p(y)$$ a.e.
Show that $(Tf)(x)=int_X k(x,y)f(y)dmu(y)$ defines a bounded linear operator $T:L^p(X)rightarrow L^p(X)$ with $||T||leq c_1c_2.$
I can show that $T$ is linear with not trouble, but I don't see how to get it is bounded. Any help would be most appriciated.
functional-analysis
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add a comment |
$begingroup$
Let $(X,M,mu)$ be a $sigma$-finite measure space, and $k:Xtimes X rightarrow mathbb{C}$ (or $mathbb{R}$) be $Xtimes X$ measurable. Suppose there are measurable functions $h,g:Xrightarrow(0,infty)$ and constants $c_1,c_2>0$ such that $$int_X |k(x,y)|g^q(y)dmu(y)leq c_1^qh^q(x)$$ a.e. and
$$int_X |k(x,y)|h^p(x)dmu(x)leq c_2^pg^p(y)$$ a.e.
Show that $(Tf)(x)=int_X k(x,y)f(y)dmu(y)$ defines a bounded linear operator $T:L^p(X)rightarrow L^p(X)$ with $||T||leq c_1c_2.$
I can show that $T$ is linear with not trouble, but I don't see how to get it is bounded. Any help would be most appriciated.
functional-analysis
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Your domain is a Banach space. So you can apply the closed graph theorem to prove a linear transformation is bounded.
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– GEdgar
Dec 28 '18 at 21:44
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@GEdgar I doubt that's helpful. I especially doubt it's helpful for showing $||T|| le c_1c_2$.
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– mathworker21
Dec 28 '18 at 21:47
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@mathworker21: you are probably right.
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– GEdgar
Dec 28 '18 at 21:57
add a comment |
$begingroup$
Let $(X,M,mu)$ be a $sigma$-finite measure space, and $k:Xtimes X rightarrow mathbb{C}$ (or $mathbb{R}$) be $Xtimes X$ measurable. Suppose there are measurable functions $h,g:Xrightarrow(0,infty)$ and constants $c_1,c_2>0$ such that $$int_X |k(x,y)|g^q(y)dmu(y)leq c_1^qh^q(x)$$ a.e. and
$$int_X |k(x,y)|h^p(x)dmu(x)leq c_2^pg^p(y)$$ a.e.
Show that $(Tf)(x)=int_X k(x,y)f(y)dmu(y)$ defines a bounded linear operator $T:L^p(X)rightarrow L^p(X)$ with $||T||leq c_1c_2.$
I can show that $T$ is linear with not trouble, but I don't see how to get it is bounded. Any help would be most appriciated.
functional-analysis
$endgroup$
Let $(X,M,mu)$ be a $sigma$-finite measure space, and $k:Xtimes X rightarrow mathbb{C}$ (or $mathbb{R}$) be $Xtimes X$ measurable. Suppose there are measurable functions $h,g:Xrightarrow(0,infty)$ and constants $c_1,c_2>0$ such that $$int_X |k(x,y)|g^q(y)dmu(y)leq c_1^qh^q(x)$$ a.e. and
$$int_X |k(x,y)|h^p(x)dmu(x)leq c_2^pg^p(y)$$ a.e.
Show that $(Tf)(x)=int_X k(x,y)f(y)dmu(y)$ defines a bounded linear operator $T:L^p(X)rightarrow L^p(X)$ with $||T||leq c_1c_2.$
I can show that $T$ is linear with not trouble, but I don't see how to get it is bounded. Any help would be most appriciated.
functional-analysis
functional-analysis
asked Dec 28 '18 at 21:34
ScottScott
38918
38918
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Your domain is a Banach space. So you can apply the closed graph theorem to prove a linear transformation is bounded.
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– GEdgar
Dec 28 '18 at 21:44
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@GEdgar I doubt that's helpful. I especially doubt it's helpful for showing $||T|| le c_1c_2$.
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– mathworker21
Dec 28 '18 at 21:47
$begingroup$
@mathworker21: you are probably right.
$endgroup$
– GEdgar
Dec 28 '18 at 21:57
add a comment |
$begingroup$
Your domain is a Banach space. So you can apply the closed graph theorem to prove a linear transformation is bounded.
$endgroup$
– GEdgar
Dec 28 '18 at 21:44
$begingroup$
@GEdgar I doubt that's helpful. I especially doubt it's helpful for showing $||T|| le c_1c_2$.
$endgroup$
– mathworker21
Dec 28 '18 at 21:47
$begingroup$
@mathworker21: you are probably right.
$endgroup$
– GEdgar
Dec 28 '18 at 21:57
$begingroup$
Your domain is a Banach space. So you can apply the closed graph theorem to prove a linear transformation is bounded.
$endgroup$
– GEdgar
Dec 28 '18 at 21:44
$begingroup$
Your domain is a Banach space. So you can apply the closed graph theorem to prove a linear transformation is bounded.
$endgroup$
– GEdgar
Dec 28 '18 at 21:44
$begingroup$
@GEdgar I doubt that's helpful. I especially doubt it's helpful for showing $||T|| le c_1c_2$.
$endgroup$
– mathworker21
Dec 28 '18 at 21:47
$begingroup$
@GEdgar I doubt that's helpful. I especially doubt it's helpful for showing $||T|| le c_1c_2$.
$endgroup$
– mathworker21
Dec 28 '18 at 21:47
$begingroup$
@mathworker21: you are probably right.
$endgroup$
– GEdgar
Dec 28 '18 at 21:57
$begingroup$
@mathworker21: you are probably right.
$endgroup$
– GEdgar
Dec 28 '18 at 21:57
add a comment |
1 Answer
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Here I'm assuming $1 < p < infty$, and $q$ is conjugate to $p$.
Fix $fin L^p(X)$. Note
$$lvert Tf(x)rvert le int_X lvert k(x,y)rvert lvert f(y)rvert, dmu(y) = int_X lvert k(x,y)rvert^{1/q} left(frac{g(y)}{h(x)}right)cdotlvert k(x,y)rvert^{1/p}left(frac{h(x)}{g(y)}right)lvert f(y)rvert, dmu(y)$$ and by Hölder's inequality, the latter integral is bounded by $$left[int_X lvert k(x,y)rvert left(frac{g(y)}{h(x)}right)^{q}, dmu(y)right]^{1/q} left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, lvert f(y)rvert^p dmu(y)right]^{1/p}$$ Thus
$$lvert Tf(x)rvert le c_1left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y)right]^{1/p}quad text{a.e.}$$ which implies $$|Tf|_p le c_1left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p}$$ By Fubini and the hypothesis,
$$int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x) =int_Xleft[int_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, dmu(x),right]lvert f(y)rvert^p dmu(y) le c_2^p|f|_p$$ Consequently, $$left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p} le c_2|f|_p$$ yielding $|Tf|_p le c_1c_2|f|_p$. As $f$ was arbitrary, $T$ is bounded on $L^p(X)$ with $|T| le c_1 c_2$.
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$begingroup$
Here I'm assuming $1 < p < infty$, and $q$ is conjugate to $p$.
Fix $fin L^p(X)$. Note
$$lvert Tf(x)rvert le int_X lvert k(x,y)rvert lvert f(y)rvert, dmu(y) = int_X lvert k(x,y)rvert^{1/q} left(frac{g(y)}{h(x)}right)cdotlvert k(x,y)rvert^{1/p}left(frac{h(x)}{g(y)}right)lvert f(y)rvert, dmu(y)$$ and by Hölder's inequality, the latter integral is bounded by $$left[int_X lvert k(x,y)rvert left(frac{g(y)}{h(x)}right)^{q}, dmu(y)right]^{1/q} left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, lvert f(y)rvert^p dmu(y)right]^{1/p}$$ Thus
$$lvert Tf(x)rvert le c_1left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y)right]^{1/p}quad text{a.e.}$$ which implies $$|Tf|_p le c_1left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p}$$ By Fubini and the hypothesis,
$$int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x) =int_Xleft[int_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, dmu(x),right]lvert f(y)rvert^p dmu(y) le c_2^p|f|_p$$ Consequently, $$left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p} le c_2|f|_p$$ yielding $|Tf|_p le c_1c_2|f|_p$. As $f$ was arbitrary, $T$ is bounded on $L^p(X)$ with $|T| le c_1 c_2$.
$endgroup$
add a comment |
$begingroup$
Here I'm assuming $1 < p < infty$, and $q$ is conjugate to $p$.
Fix $fin L^p(X)$. Note
$$lvert Tf(x)rvert le int_X lvert k(x,y)rvert lvert f(y)rvert, dmu(y) = int_X lvert k(x,y)rvert^{1/q} left(frac{g(y)}{h(x)}right)cdotlvert k(x,y)rvert^{1/p}left(frac{h(x)}{g(y)}right)lvert f(y)rvert, dmu(y)$$ and by Hölder's inequality, the latter integral is bounded by $$left[int_X lvert k(x,y)rvert left(frac{g(y)}{h(x)}right)^{q}, dmu(y)right]^{1/q} left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, lvert f(y)rvert^p dmu(y)right]^{1/p}$$ Thus
$$lvert Tf(x)rvert le c_1left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y)right]^{1/p}quad text{a.e.}$$ which implies $$|Tf|_p le c_1left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p}$$ By Fubini and the hypothesis,
$$int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x) =int_Xleft[int_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, dmu(x),right]lvert f(y)rvert^p dmu(y) le c_2^p|f|_p$$ Consequently, $$left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p} le c_2|f|_p$$ yielding $|Tf|_p le c_1c_2|f|_p$. As $f$ was arbitrary, $T$ is bounded on $L^p(X)$ with $|T| le c_1 c_2$.
$endgroup$
add a comment |
$begingroup$
Here I'm assuming $1 < p < infty$, and $q$ is conjugate to $p$.
Fix $fin L^p(X)$. Note
$$lvert Tf(x)rvert le int_X lvert k(x,y)rvert lvert f(y)rvert, dmu(y) = int_X lvert k(x,y)rvert^{1/q} left(frac{g(y)}{h(x)}right)cdotlvert k(x,y)rvert^{1/p}left(frac{h(x)}{g(y)}right)lvert f(y)rvert, dmu(y)$$ and by Hölder's inequality, the latter integral is bounded by $$left[int_X lvert k(x,y)rvert left(frac{g(y)}{h(x)}right)^{q}, dmu(y)right]^{1/q} left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, lvert f(y)rvert^p dmu(y)right]^{1/p}$$ Thus
$$lvert Tf(x)rvert le c_1left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y)right]^{1/p}quad text{a.e.}$$ which implies $$|Tf|_p le c_1left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p}$$ By Fubini and the hypothesis,
$$int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x) =int_Xleft[int_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, dmu(x),right]lvert f(y)rvert^p dmu(y) le c_2^p|f|_p$$ Consequently, $$left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p} le c_2|f|_p$$ yielding $|Tf|_p le c_1c_2|f|_p$. As $f$ was arbitrary, $T$ is bounded on $L^p(X)$ with $|T| le c_1 c_2$.
$endgroup$
Here I'm assuming $1 < p < infty$, and $q$ is conjugate to $p$.
Fix $fin L^p(X)$. Note
$$lvert Tf(x)rvert le int_X lvert k(x,y)rvert lvert f(y)rvert, dmu(y) = int_X lvert k(x,y)rvert^{1/q} left(frac{g(y)}{h(x)}right)cdotlvert k(x,y)rvert^{1/p}left(frac{h(x)}{g(y)}right)lvert f(y)rvert, dmu(y)$$ and by Hölder's inequality, the latter integral is bounded by $$left[int_X lvert k(x,y)rvert left(frac{g(y)}{h(x)}right)^{q}, dmu(y)right]^{1/q} left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, lvert f(y)rvert^p dmu(y)right]^{1/p}$$ Thus
$$lvert Tf(x)rvert le c_1left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y)right]^{1/p}quad text{a.e.}$$ which implies $$|Tf|_p le c_1left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p}$$ By Fubini and the hypothesis,
$$int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x) =int_Xleft[int_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, dmu(x),right]lvert f(y)rvert^p dmu(y) le c_2^p|f|_p$$ Consequently, $$left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p} le c_2|f|_p$$ yielding $|Tf|_p le c_1c_2|f|_p$. As $f$ was arbitrary, $T$ is bounded on $L^p(X)$ with $|T| le c_1 c_2$.
answered Dec 28 '18 at 23:41
kobekobe
34.9k22248
34.9k22248
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$begingroup$
Your domain is a Banach space. So you can apply the closed graph theorem to prove a linear transformation is bounded.
$endgroup$
– GEdgar
Dec 28 '18 at 21:44
$begingroup$
@GEdgar I doubt that's helpful. I especially doubt it's helpful for showing $||T|| le c_1c_2$.
$endgroup$
– mathworker21
Dec 28 '18 at 21:47
$begingroup$
@mathworker21: you are probably right.
$endgroup$
– GEdgar
Dec 28 '18 at 21:57