Invariant subspace for Volterra operator The Next CEO of Stack OverflowInvariant spaces for a shift semigroupContinuity domain for momentum operatorBoundedness of Volterra operator with Sobolev normInvariant Subspace Problem: Non-Seperable Hilbert SpaceTent map invariant densityDirect numerical solutions for first kind Volterra integral equationsProof that $L^p$ is complete in Folland's Real AnalysisVolterra Operator PropertyMixed formulation with Volterra integrals?Subspace of $alpha$ Holder continuous functions is Closed
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Invariant subspace for Volterra operator
The Next CEO of Stack OverflowInvariant spaces for a shift semigroupContinuity domain for momentum operatorBoundedness of Volterra operator with Sobolev normInvariant Subspace Problem: Non-Seperable Hilbert SpaceTent map invariant densityDirect numerical solutions for first kind Volterra integral equationsProof that $L^p$ is complete in Folland's Real AnalysisVolterra Operator PropertyMixed formulation with Volterra integrals?Subspace of $alpha$ Holder continuous functions is Closed
$begingroup$
So, am I stupid or isn't this kind of trivial?
I'm having this problem:
Let $T$ be the Volterra operator on $L^2([0,1])$ defined by
beginequation
Tf(s) = int^s_0 , f(t) , textdt.
endequation
For $0 leq alpha leq 1$ set
beginequation
M_alpha = f in L^2([0,1]) : f(t) = 0, 0 leq t leq alpha.
endequation
Prove that each $M_alpha$ is an invariant subspace for $T$.
My solution is this:
$M_alpha$ is an invariant subspace for $T$ if $f(t) in M_alpha Rightarrow Tf(s) in M_alpha$. So if $f(t) in M_alpha$, then
beginequation*
beginaligned
Tf(s) &= int^s_0 , f(t) , textdt\
&= begincasesdisplaystyleint^s_0 , 0 , textdt &quad textif 0 leq t leq alpha\displaystyleint^s_0 , f(t) , textdt &quad textif alpha < t leq 1\endcases\
&= begincases0 &quad textif 0 leq t leq alpha\displaystyleint^s_0 , f(t) textdt &quad textif alpha < t leq 1\endcases\
&in M_alpha
endaligned
endequation*
Am I wrong or is it this simple?
functional-analysis hilbert-spaces
asked Mar 19 at 21:59
Emil TonklingaEmil Tonklinga
62
$endgroup$
add a comment |
$begingroup$
So, am I stupid or isn't this kind of trivial?
I'm having this problem:
Let $T$ be the Volterra operator on $L^2([0,1])$ defined by
beginequation
Tf(s) = int^s_0 , f(t) , textdt.
endequation
For $0 leq alpha leq 1$ set
beginequation
M_alpha = f in L^2([0,1]) : f(t) = 0, 0 leq t leq alpha.
endequation
Prove that each $M_alpha$ is an invariant subspace for $T$.
My solution is this:
$M_alpha$ is an invariant subspace for $T$ if $f(t) in M_alpha Rightarrow Tf(s) in M_alpha$. So if $f(t) in M_alpha$, then
beginequation*
beginaligned
Tf(s) &= int^s_0 , f(t) , textdt\
&= begincasesdisplaystyleint^s_0 , 0 , textdt &quad textif 0 leq t leq alpha\displaystyleint^s_0 , f(t) , textdt &quad textif alpha < t leq 1\endcases\
&= begincases0 &quad textif 0 leq t leq alpha\displaystyleint^s_0 , f(t) textdt &quad textif alpha < t leq 1\endcases\
&in M_alpha
endaligned
endequation*
Am I wrong or is it this simple?
functional-analysis hilbert-spaces
asked Mar 19 at 21:59
Emil TonklingaEmil Tonklinga
62
$endgroup$
add a comment |
$begingroup$
So, am I stupid or isn't this kind of trivial?
I'm having this problem:
Let $T$ be the Volterra operator on $L^2([0,1])$ defined by
beginequation
Tf(s) = int^s_0 , f(t) , textdt.
endequation
For $0 leq alpha leq 1$ set
beginequation
M_alpha = f in L^2([0,1]) : f(t) = 0, 0 leq t leq alpha.
endequation
Prove that each $M_alpha$ is an invariant subspace for $T$.
My solution is this:
$M_alpha$ is an invariant subspace for $T$ if $f(t) in M_alpha Rightarrow Tf(s) in M_alpha$. So if $f(t) in M_alpha$, then
beginequation*
beginaligned
Tf(s) &= int^s_0 , f(t) , textdt\
&= begincasesdisplaystyleint^s_0 , 0 , textdt &quad textif 0 leq t leq alpha\displaystyleint^s_0 , f(t) , textdt &quad textif alpha < t leq 1\endcases\
&= begincases0 &quad textif 0 leq t leq alpha\displaystyleint^s_0 , f(t) textdt &quad textif alpha < t leq 1\endcases\
&in M_alpha
endaligned
endequation*
Am I wrong or is it this simple?
functional-analysis hilbert-spaces
asked Mar 19 at 21:59
Emil TonklingaEmil Tonklinga
62
$endgroup$
So, am I stupid or isn't this kind of trivial?
I'm having this problem:
Let $T$ be the Volterra operator on $L^2([0,1])$ defined by
beginequation
Tf(s) = int^s_0 , f(t) , textdt.
endequation
For $0 leq alpha leq 1$ set
beginequation
M_alpha = f in L^2([0,1]) : f(t) = 0, 0 leq t leq alpha.
endequation
Prove that each $M_alpha$ is an invariant subspace for $T$.
My solution is this:
$M_alpha$ is an invariant subspace for $T$ if $f(t) in M_alpha Rightarrow Tf(s) in M_alpha$. So if $f(t) in M_alpha$, then
beginequation*
beginaligned
Tf(s) &= int^s_0 , f(t) , textdt\
&= begincasesdisplaystyleint^s_0 , 0 , textdt &quad textif 0 leq t leq alpha\displaystyleint^s_0 , f(t) , textdt &quad textif alpha < t leq 1\endcases\
&= begincases0 &quad textif 0 leq t leq alpha\displaystyleint^s_0 , f(t) textdt &quad textif alpha < t leq 1\endcases\
&in M_alpha
endaligned
endequation*
Am I wrong or is it this simple?
functional-analysis hilbert-spaces
functional-analysis hilbert-spaces
asked Mar 19 at 21:59
Emil TonklingaEmil Tonklinga
62
asked Mar 19 at 21:59
Emil TonklingaEmil Tonklinga
62
asked Mar 19 at 21:59
Emil TonklingaEmil Tonklinga
62
asked Mar 19 at 21:59
Emil TonklingaEmil Tonklinga
62
asked Mar 19 at 21:59
Emil TonklingaEmil Tonklinga
62
62
add a comment |
add a comment |
0
active
oldest
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