Prove boundedness of operator on $L^p$
$begingroup$
Let $alpha in (0,1)$, $pin(1,infty)$ and $T >0$
The operator is defined as $I^alpha f(t) = frac{1}{Gamma(alpha)}int_{0}^{t} (t-r)^{alpha-1}f(r)dr$.
I want to prove that $I^alpha in B(L^p(0,T))$.
Let $H = L^p(0,T)$.
$||I^alpha f||^p_{H}= frac{1}{Gamma^p(alpha)} int_{0}^{T} |int_{0}^{t} s^{alpha-1}f(t-s)ds|^p dt$.
Well, the $p$-th power is the problem and the inner divergent integral as $alpha-1 <0$. I don't know how to proceed and estimate it properly.
functional-analysis banach-spaces lp-spaces
asked Jan 16 at 15:10
treskovtreskov
145111
$endgroup$
add a comment |
$begingroup$
Let $alpha in (0,1)$, $pin(1,infty)$ and $T >0$
The operator is defined as $I^alpha f(t) = frac{1}{Gamma(alpha)}int_{0}^{t} (t-r)^{alpha-1}f(r)dr$.
I want to prove that $I^alpha in B(L^p(0,T))$.
Let $H = L^p(0,T)$.
$||I^alpha f||^p_{H}= frac{1}{Gamma^p(alpha)} int_{0}^{T} |int_{0}^{t} s^{alpha-1}f(t-s)ds|^p dt$.
Well, the $p$-th power is the problem and the inner divergent integral as $alpha-1 <0$. I don't know how to proceed and estimate it properly.
functional-analysis banach-spaces lp-spaces
asked Jan 16 at 15:10
treskovtreskov
145111
$endgroup$
$begingroup$
What exactly is $B(L^p(0,T))$? Is it the space of all bounded linear automorphism on $L^p$?
$endgroup$
– BigbearZzz
Jan 16 at 15:41
$begingroup$
@BigbearZzz That's correct.
$endgroup$
– treskov
Jan 16 at 15:44
$begingroup$
@Song I looked it up and I can't see why it follows from the theorem you mentioned. There must be easier way to estimate the norm.
$endgroup$
– treskov
Jan 16 at 19:24
add a comment |
$begingroup$
Let $alpha in (0,1)$, $pin(1,infty)$ and $T >0$
The operator is defined as $I^alpha f(t) = frac{1}{Gamma(alpha)}int_{0}^{t} (t-r)^{alpha-1}f(r)dr$.
I want to prove that $I^alpha in B(L^p(0,T))$.
Let $H = L^p(0,T)$.
$||I^alpha f||^p_{H}= frac{1}{Gamma^p(alpha)} int_{0}^{T} |int_{0}^{t} s^{alpha-1}f(t-s)ds|^p dt$.
Well, the $p$-th power is the problem and the inner divergent integral as $alpha-1 <0$. I don't know how to proceed and estimate it properly.
functional-analysis banach-spaces lp-spaces
asked Jan 16 at 15:10
treskovtreskov
145111
$endgroup$
Let $alpha in (0,1)$, $pin(1,infty)$ and $T >0$
The operator is defined as $I^alpha f(t) = frac{1}{Gamma(alpha)}int_{0}^{t} (t-r)^{alpha-1}f(r)dr$.
I want to prove that $I^alpha in B(L^p(0,T))$.
Let $H = L^p(0,T)$.
$||I^alpha f||^p_{H}= frac{1}{Gamma^p(alpha)} int_{0}^{T} |int_{0}^{t} s^{alpha-1}f(t-s)ds|^p dt$.
Well, the $p$-th power is the problem and the inner divergent integral as $alpha-1 <0$. I don't know how to proceed and estimate it properly.
functional-analysis banach-spaces lp-spaces
functional-analysis banach-spaces lp-spaces
asked Jan 16 at 15:10
treskovtreskov
145111
asked Jan 16 at 15:10
treskovtreskov
145111
edited Jan 16 at 20:49
treskov
asked Jan 16 at 15:10
treskovtreskov
145111
asked Jan 16 at 15:10
treskovtreskov
145111
asked Jan 16 at 15:10
treskovtreskov
145111
145111
$begingroup$
What exactly is $B(L^p(0,T))$? Is it the space of all bounded linear automorphism on $L^p$?
$endgroup$
– BigbearZzz
Jan 16 at 15:41
$begingroup$
@BigbearZzz That's correct.
$endgroup$
– treskov
Jan 16 at 15:44
$begingroup$
@Song I looked it up and I can't see why it follows from the theorem you mentioned. There must be easier way to estimate the norm.
$endgroup$
– treskov
Jan 16 at 19:24
add a comment |
$begingroup$
What exactly is $B(L^p(0,T))$? Is it the space of all bounded linear automorphism on $L^p$?
$endgroup$
– BigbearZzz
Jan 16 at 15:41
$begingroup$
@BigbearZzz That's correct.
$endgroup$
– treskov
Jan 16 at 15:44
$begingroup$
@Song I looked it up and I can't see why it follows from the theorem you mentioned. There must be easier way to estimate the norm.
$endgroup$
– treskov
Jan 16 at 19:24
$begingroup$
What exactly is $B(L^p(0,T))$? Is it the space of all bounded linear automorphism on $L^p$?
$endgroup$
– BigbearZzz
Jan 16 at 15:41
$begingroup$
What exactly is $B(L^p(0,T))$? Is it the space of all bounded linear automorphism on $L^p$?
$endgroup$
– BigbearZzz
Jan 16 at 15:41
$begingroup$
@BigbearZzz That's correct.
$endgroup$
– treskov
Jan 16 at 15:44
$begingroup$
@BigbearZzz That's correct.
$endgroup$
– treskov
Jan 16 at 15:44
$begingroup$
@Song I looked it up and I can't see why it follows from the theorem you mentioned. There must be easier way to estimate the norm.
$endgroup$
– treskov
Jan 16 at 19:24
$begingroup$
@Song I looked it up and I can't see why it follows from the theorem you mentioned. There must be easier way to estimate the norm.
$endgroup$
– treskov
Jan 16 at 19:24
add a comment |
1 Answer
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$begingroup$
I managed to solve this.
Let $H=L^p(0,T)$.
Note that $||I^alpha f||^p_{H} = ||g*f||^p_{H}$, where $g(t)=t^{alpha-1}$.
Now it's sufficient to use Young's inequality for convolutions:
For $pin(1,infty)$, $gin L^1(0,T)$ and $fin L^p(0,T)$ the inequality below holds:
$||g*f||_{H} le ||g||_{L^1(0,T)} ||f||_H$.
answered Jan 16 at 20:49
treskovtreskov
145111
$endgroup$
add a comment |
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1 Answer
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active
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votes
1 Answer
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active
oldest
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$begingroup$
I managed to solve this.
Let $H=L^p(0,T)$.
Note that $||I^alpha f||^p_{H} = ||g*f||^p_{H}$, where $g(t)=t^{alpha-1}$.
Now it's sufficient to use Young's inequality for convolutions:
For $pin(1,infty)$, $gin L^1(0,T)$ and $fin L^p(0,T)$ the inequality below holds:
$||g*f||_{H} le ||g||_{L^1(0,T)} ||f||_H$.
answered Jan 16 at 20:49
treskovtreskov
145111
$endgroup$
add a comment |
$begingroup$
I managed to solve this.
Let $H=L^p(0,T)$.
Note that $||I^alpha f||^p_{H} = ||g*f||^p_{H}$, where $g(t)=t^{alpha-1}$.
Now it's sufficient to use Young's inequality for convolutions:
For $pin(1,infty)$, $gin L^1(0,T)$ and $fin L^p(0,T)$ the inequality below holds:
$||g*f||_{H} le ||g||_{L^1(0,T)} ||f||_H$.
answered Jan 16 at 20:49
treskovtreskov
145111
$endgroup$
add a comment |
$begingroup$
I managed to solve this.
Let $H=L^p(0,T)$.
Note that $||I^alpha f||^p_{H} = ||g*f||^p_{H}$, where $g(t)=t^{alpha-1}$.
Now it's sufficient to use Young's inequality for convolutions:
For $pin(1,infty)$, $gin L^1(0,T)$ and $fin L^p(0,T)$ the inequality below holds:
$||g*f||_{H} le ||g||_{L^1(0,T)} ||f||_H$.
answered Jan 16 at 20:49
treskovtreskov
145111
$endgroup$
I managed to solve this.
Let $H=L^p(0,T)$.
Note that $||I^alpha f||^p_{H} = ||g*f||^p_{H}$, where $g(t)=t^{alpha-1}$.
Now it's sufficient to use Young's inequality for convolutions:
For $pin(1,infty)$, $gin L^1(0,T)$ and $fin L^p(0,T)$ the inequality below holds:
$||g*f||_{H} le ||g||_{L^1(0,T)} ||f||_H$.
answered Jan 16 at 20:49
treskovtreskov
145111
answered Jan 16 at 20:49
treskovtreskov
145111
answered Jan 16 at 20:49
treskovtreskov
145111
answered Jan 16 at 20:49
treskovtreskov
145111
145111
add a comment |
add a comment |
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$begingroup$
What exactly is $B(L^p(0,T))$? Is it the space of all bounded linear automorphism on $L^p$?
$endgroup$
– BigbearZzz
Jan 16 at 15:41
$begingroup$
@BigbearZzz That's correct.
$endgroup$
– treskov
Jan 16 at 15:44
$begingroup$
@Song I looked it up and I can't see why it follows from the theorem you mentioned. There must be easier way to estimate the norm.
$endgroup$
– treskov
Jan 16 at 19:24