Prove Hilbert-Schmidt integral operator is bounded
$begingroup$
I'm having some issues understanding a proof that a Hilbert-Schmidt integral operator is bounded. I'll start off by giving the definitions I'm working with.
Definition of integral operator:
$textit{Let k be a function of two variables (x,t)}in Itimes I=I^2textit{ where I is a finite or infinite real interval. We define a linear integral operator K with kernel k(x,y) as}$
$$Ku(x)=int_{I}k(x,y)u(y) dy, xin I$$
Definition of Hilbert-Schmidt integral operator:
$textit{An integral operator on }L^{2}(I)textit{ is called a Hilbert-Schmidt operator if the kernel k is in }L^2(Itimes I)textit{,}$
$textit{that is if}$
$$||k||^2=int_{I}int_{I}|k(x,y)|^2 dxdy<infty$$
The book now proceeds to show that a Hilbert-Schmidt integral operator is bounded with $||K||leq||k||$. In the proof the following is stated:
$textit{Assume that K is a Hilbert-Schmidt operator with kernel k}in L^2(Itimes I). textit{Let u}in L^2(I),$
$textit{then from Fubini's theorem in measure theory, the function }xmapsto int_{I}|k(x,y)|^2 dy$
$textit{is in }L^1(I)textit{ such that }int_{I} was int_{I}|k(x,y)|^2 dxdy<infty.textit{ Thus it is allowed to use the Cauchy-Schwarz inequality such that}$
$$|Ku(x)|=|int_{I}k(x,y)u(y) dy|leq ||k(x,cdot)|| ||u||$$
$textit{since }||overline{u}||=||u||.$
I' having some difficulties understanding this argument. The remaining part of the proof is not that difficult once this argument is clear. Can anyone give some hints as to why this makes sense?
Thanks in advance!
real-analysis functional-analysis
asked Jan 16 at 15:08
JamesJames
11910
$endgroup$
add a comment |
$begingroup$
I'm having some issues understanding a proof that a Hilbert-Schmidt integral operator is bounded. I'll start off by giving the definitions I'm working with.
Definition of integral operator:
$textit{Let k be a function of two variables (x,t)}in Itimes I=I^2textit{ where I is a finite or infinite real interval. We define a linear integral operator K with kernel k(x,y) as}$
$$Ku(x)=int_{I}k(x,y)u(y) dy, xin I$$
Definition of Hilbert-Schmidt integral operator:
$textit{An integral operator on }L^{2}(I)textit{ is called a Hilbert-Schmidt operator if the kernel k is in }L^2(Itimes I)textit{,}$
$textit{that is if}$
$$||k||^2=int_{I}int_{I}|k(x,y)|^2 dxdy<infty$$
The book now proceeds to show that a Hilbert-Schmidt integral operator is bounded with $||K||leq||k||$. In the proof the following is stated:
$textit{Assume that K is a Hilbert-Schmidt operator with kernel k}in L^2(Itimes I). textit{Let u}in L^2(I),$
$textit{then from Fubini's theorem in measure theory, the function }xmapsto int_{I}|k(x,y)|^2 dy$
$textit{is in }L^1(I)textit{ such that }int_{I} was int_{I}|k(x,y)|^2 dxdy<infty.textit{ Thus it is allowed to use the Cauchy-Schwarz inequality such that}$
$$|Ku(x)|=|int_{I}k(x,y)u(y) dy|leq ||k(x,cdot)|| ||u||$$
$textit{since }||overline{u}||=||u||.$
I' having some difficulties understanding this argument. The remaining part of the proof is not that difficult once this argument is clear. Can anyone give some hints as to why this makes sense?
Thanks in advance!
real-analysis functional-analysis
asked Jan 16 at 15:08
JamesJames
11910
$endgroup$
add a comment |
$begingroup$
I'm having some issues understanding a proof that a Hilbert-Schmidt integral operator is bounded. I'll start off by giving the definitions I'm working with.
Definition of integral operator:
$textit{Let k be a function of two variables (x,t)}in Itimes I=I^2textit{ where I is a finite or infinite real interval. We define a linear integral operator K with kernel k(x,y) as}$
$$Ku(x)=int_{I}k(x,y)u(y) dy, xin I$$
Definition of Hilbert-Schmidt integral operator:
$textit{An integral operator on }L^{2}(I)textit{ is called a Hilbert-Schmidt operator if the kernel k is in }L^2(Itimes I)textit{,}$
$textit{that is if}$
$$||k||^2=int_{I}int_{I}|k(x,y)|^2 dxdy<infty$$
The book now proceeds to show that a Hilbert-Schmidt integral operator is bounded with $||K||leq||k||$. In the proof the following is stated:
$textit{Assume that K is a Hilbert-Schmidt operator with kernel k}in L^2(Itimes I). textit{Let u}in L^2(I),$
$textit{then from Fubini's theorem in measure theory, the function }xmapsto int_{I}|k(x,y)|^2 dy$
$textit{is in }L^1(I)textit{ such that }int_{I} was int_{I}|k(x,y)|^2 dxdy<infty.textit{ Thus it is allowed to use the Cauchy-Schwarz inequality such that}$
$$|Ku(x)|=|int_{I}k(x,y)u(y) dy|leq ||k(x,cdot)|| ||u||$$
$textit{since }||overline{u}||=||u||.$
I' having some difficulties understanding this argument. The remaining part of the proof is not that difficult once this argument is clear. Can anyone give some hints as to why this makes sense?
Thanks in advance!
real-analysis functional-analysis
asked Jan 16 at 15:08
JamesJames
11910
$endgroup$
I'm having some issues understanding a proof that a Hilbert-Schmidt integral operator is bounded. I'll start off by giving the definitions I'm working with.
Definition of integral operator:
$textit{Let k be a function of two variables (x,t)}in Itimes I=I^2textit{ where I is a finite or infinite real interval. We define a linear integral operator K with kernel k(x,y) as}$
$$Ku(x)=int_{I}k(x,y)u(y) dy, xin I$$
Definition of Hilbert-Schmidt integral operator:
$textit{An integral operator on }L^{2}(I)textit{ is called a Hilbert-Schmidt operator if the kernel k is in }L^2(Itimes I)textit{,}$
$textit{that is if}$
$$||k||^2=int_{I}int_{I}|k(x,y)|^2 dxdy<infty$$
The book now proceeds to show that a Hilbert-Schmidt integral operator is bounded with $||K||leq||k||$. In the proof the following is stated:
$textit{Assume that K is a Hilbert-Schmidt operator with kernel k}in L^2(Itimes I). textit{Let u}in L^2(I),$
$textit{then from Fubini's theorem in measure theory, the function }xmapsto int_{I}|k(x,y)|^2 dy$
$textit{is in }L^1(I)textit{ such that }int_{I} was int_{I}|k(x,y)|^2 dxdy<infty.textit{ Thus it is allowed to use the Cauchy-Schwarz inequality such that}$
$$|Ku(x)|=|int_{I}k(x,y)u(y) dy|leq ||k(x,cdot)|| ||u||$$
$textit{since }||overline{u}||=||u||.$
I' having some difficulties understanding this argument. The remaining part of the proof is not that difficult once this argument is clear. Can anyone give some hints as to why this makes sense?
Thanks in advance!
real-analysis functional-analysis
real-analysis functional-analysis
asked Jan 16 at 15:08
JamesJames
11910
asked Jan 16 at 15:08
JamesJames
11910
asked Jan 16 at 15:08
JamesJames
11910
asked Jan 16 at 15:08
JamesJames
11910
asked Jan 16 at 15:08
JamesJames
11910
11910
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You don't say what exactly are your difficulties, so I will proceed step by ste. By the Cauchy-Schwarz inequalitybegin{align}
|Ku(x)|&=Bigl|int_{I}k(x,y),u(y),dyBigr|\
&leint_{I}|k(x,y)|,|u(y)|,dy\
&leBigl(int_{I}|k(x,y)|^2,dyBigr)^{1/2}Bigl(int_{I}|u(y)|,dyBigr)^{1/2}\
&=Bigl(int_{I}|k(x,y)|^2,dyBigr)^{1/2}|u|.
end{align}
Then
$$
|Ku(x)|^2leBigl(int_{I}|k(x,y)|^2,dyBigr)|u|^2
$$
and
$$
|Ku|^2=int_I|Ku(x)|^2dxleBigl(int_IBigl(int_{I}|k(x,y)|^2,dyBigr)dxBigr)|u|^2=|k|^2|u|^2,
$$
that is,
$$|Ku|le|k|,|u|.$$
answered Jan 16 at 16:30
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
$begingroup$
I have gotten the same steps. My main problem is the argument of why $int_{I}|k(x,y)|^2 dy$ makes sense. We know $k(x,y)in L^2(Itimes I)$, but how can we know $int_{I}|k(x,y)|^2 dy$ is finite?
$endgroup$
– James
Jan 16 at 16:38
$begingroup$
You do not care. It may be infinite for some $x$, but Fubini guarantees that it is finite a.e. and integrable as a function of $x$. Moreover, since we are dealing with positive functions, Fubini applies, even if the integral were infinite.
$endgroup$
– Julián Aguirre
Jan 16 at 16:41
$begingroup$
Thank you for the help! Why does Fubini's theorem apply even if the integral is infinite?
$endgroup$
– James
Jan 16 at 16:56
$begingroup$
In that case you get an equality $infty=infty$.
$endgroup$
– Julián Aguirre
Jan 16 at 22:42
$begingroup$
Oh yeah, of course. Thanks for the help, it makes much more sense now!
$endgroup$
– James
Jan 17 at 13:45
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You don't say what exactly are your difficulties, so I will proceed step by ste. By the Cauchy-Schwarz inequalitybegin{align}
|Ku(x)|&=Bigl|int_{I}k(x,y),u(y),dyBigr|\
&leint_{I}|k(x,y)|,|u(y)|,dy\
&leBigl(int_{I}|k(x,y)|^2,dyBigr)^{1/2}Bigl(int_{I}|u(y)|,dyBigr)^{1/2}\
&=Bigl(int_{I}|k(x,y)|^2,dyBigr)^{1/2}|u|.
end{align}
Then
$$
|Ku(x)|^2leBigl(int_{I}|k(x,y)|^2,dyBigr)|u|^2
$$
and
$$
|Ku|^2=int_I|Ku(x)|^2dxleBigl(int_IBigl(int_{I}|k(x,y)|^2,dyBigr)dxBigr)|u|^2=|k|^2|u|^2,
$$
that is,
$$|Ku|le|k|,|u|.$$
answered Jan 16 at 16:30
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
$begingroup$
I have gotten the same steps. My main problem is the argument of why $int_{I}|k(x,y)|^2 dy$ makes sense. We know $k(x,y)in L^2(Itimes I)$, but how can we know $int_{I}|k(x,y)|^2 dy$ is finite?
$endgroup$
– James
Jan 16 at 16:38
$begingroup$
You do not care. It may be infinite for some $x$, but Fubini guarantees that it is finite a.e. and integrable as a function of $x$. Moreover, since we are dealing with positive functions, Fubini applies, even if the integral were infinite.
$endgroup$
– Julián Aguirre
Jan 16 at 16:41
$begingroup$
Thank you for the help! Why does Fubini's theorem apply even if the integral is infinite?
$endgroup$
– James
Jan 16 at 16:56
$begingroup$
In that case you get an equality $infty=infty$.
$endgroup$
– Julián Aguirre
Jan 16 at 22:42
$begingroup$
Oh yeah, of course. Thanks for the help, it makes much more sense now!
$endgroup$
– James
Jan 17 at 13:45
add a comment |
$begingroup$
You don't say what exactly are your difficulties, so I will proceed step by ste. By the Cauchy-Schwarz inequalitybegin{align}
|Ku(x)|&=Bigl|int_{I}k(x,y),u(y),dyBigr|\
&leint_{I}|k(x,y)|,|u(y)|,dy\
&leBigl(int_{I}|k(x,y)|^2,dyBigr)^{1/2}Bigl(int_{I}|u(y)|,dyBigr)^{1/2}\
&=Bigl(int_{I}|k(x,y)|^2,dyBigr)^{1/2}|u|.
end{align}
Then
$$
|Ku(x)|^2leBigl(int_{I}|k(x,y)|^2,dyBigr)|u|^2
$$
and
$$
|Ku|^2=int_I|Ku(x)|^2dxleBigl(int_IBigl(int_{I}|k(x,y)|^2,dyBigr)dxBigr)|u|^2=|k|^2|u|^2,
$$
that is,
$$|Ku|le|k|,|u|.$$
answered Jan 16 at 16:30
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
$begingroup$
I have gotten the same steps. My main problem is the argument of why $int_{I}|k(x,y)|^2 dy$ makes sense. We know $k(x,y)in L^2(Itimes I)$, but how can we know $int_{I}|k(x,y)|^2 dy$ is finite?
$endgroup$
– James
Jan 16 at 16:38
$begingroup$
You do not care. It may be infinite for some $x$, but Fubini guarantees that it is finite a.e. and integrable as a function of $x$. Moreover, since we are dealing with positive functions, Fubini applies, even if the integral were infinite.
$endgroup$
– Julián Aguirre
Jan 16 at 16:41
$begingroup$
Thank you for the help! Why does Fubini's theorem apply even if the integral is infinite?
$endgroup$
– James
Jan 16 at 16:56
$begingroup$
In that case you get an equality $infty=infty$.
$endgroup$
– Julián Aguirre
Jan 16 at 22:42
$begingroup$
Oh yeah, of course. Thanks for the help, it makes much more sense now!
$endgroup$
– James
Jan 17 at 13:45
add a comment |
$begingroup$
You don't say what exactly are your difficulties, so I will proceed step by ste. By the Cauchy-Schwarz inequalitybegin{align}
|Ku(x)|&=Bigl|int_{I}k(x,y),u(y),dyBigr|\
&leint_{I}|k(x,y)|,|u(y)|,dy\
&leBigl(int_{I}|k(x,y)|^2,dyBigr)^{1/2}Bigl(int_{I}|u(y)|,dyBigr)^{1/2}\
&=Bigl(int_{I}|k(x,y)|^2,dyBigr)^{1/2}|u|.
end{align}
Then
$$
|Ku(x)|^2leBigl(int_{I}|k(x,y)|^2,dyBigr)|u|^2
$$
and
$$
|Ku|^2=int_I|Ku(x)|^2dxleBigl(int_IBigl(int_{I}|k(x,y)|^2,dyBigr)dxBigr)|u|^2=|k|^2|u|^2,
$$
that is,
$$|Ku|le|k|,|u|.$$
answered Jan 16 at 16:30
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
You don't say what exactly are your difficulties, so I will proceed step by ste. By the Cauchy-Schwarz inequalitybegin{align}
|Ku(x)|&=Bigl|int_{I}k(x,y),u(y),dyBigr|\
&leint_{I}|k(x,y)|,|u(y)|,dy\
&leBigl(int_{I}|k(x,y)|^2,dyBigr)^{1/2}Bigl(int_{I}|u(y)|,dyBigr)^{1/2}\
&=Bigl(int_{I}|k(x,y)|^2,dyBigr)^{1/2}|u|.
end{align}
Then
$$
|Ku(x)|^2leBigl(int_{I}|k(x,y)|^2,dyBigr)|u|^2
$$
and
$$
|Ku|^2=int_I|Ku(x)|^2dxleBigl(int_IBigl(int_{I}|k(x,y)|^2,dyBigr)dxBigr)|u|^2=|k|^2|u|^2,
$$
that is,
$$|Ku|le|k|,|u|.$$
answered Jan 16 at 16:30
Julián AguirreJulián Aguirre
69.5k24297
answered Jan 16 at 16:30
Julián AguirreJulián Aguirre
69.5k24297
answered Jan 16 at 16:30
Julián AguirreJulián Aguirre
69.5k24297
answered Jan 16 at 16:30
Julián AguirreJulián Aguirre
69.5k24297
69.5k24297
$begingroup$
I have gotten the same steps. My main problem is the argument of why $int_{I}|k(x,y)|^2 dy$ makes sense. We know $k(x,y)in L^2(Itimes I)$, but how can we know $int_{I}|k(x,y)|^2 dy$ is finite?
$endgroup$
– James
Jan 16 at 16:38
$begingroup$
You do not care. It may be infinite for some $x$, but Fubini guarantees that it is finite a.e. and integrable as a function of $x$. Moreover, since we are dealing with positive functions, Fubini applies, even if the integral were infinite.
$endgroup$
– Julián Aguirre
Jan 16 at 16:41
$begingroup$
Thank you for the help! Why does Fubini's theorem apply even if the integral is infinite?
$endgroup$
– James
Jan 16 at 16:56
$begingroup$
In that case you get an equality $infty=infty$.
$endgroup$
– Julián Aguirre
Jan 16 at 22:42
$begingroup$
Oh yeah, of course. Thanks for the help, it makes much more sense now!
$endgroup$
– James
Jan 17 at 13:45
add a comment |
$begingroup$
I have gotten the same steps. My main problem is the argument of why $int_{I}|k(x,y)|^2 dy$ makes sense. We know $k(x,y)in L^2(Itimes I)$, but how can we know $int_{I}|k(x,y)|^2 dy$ is finite?
$endgroup$
– James
Jan 16 at 16:38
$begingroup$
You do not care. It may be infinite for some $x$, but Fubini guarantees that it is finite a.e. and integrable as a function of $x$. Moreover, since we are dealing with positive functions, Fubini applies, even if the integral were infinite.
$endgroup$
– Julián Aguirre
Jan 16 at 16:41
$begingroup$
Thank you for the help! Why does Fubini's theorem apply even if the integral is infinite?
$endgroup$
– James
Jan 16 at 16:56
$begingroup$
In that case you get an equality $infty=infty$.
$endgroup$
– Julián Aguirre
Jan 16 at 22:42
$begingroup$
Oh yeah, of course. Thanks for the help, it makes much more sense now!
$endgroup$
– James
Jan 17 at 13:45
$begingroup$
I have gotten the same steps. My main problem is the argument of why $int_{I}|k(x,y)|^2 dy$ makes sense. We know $k(x,y)in L^2(Itimes I)$, but how can we know $int_{I}|k(x,y)|^2 dy$ is finite?
$endgroup$
– James
Jan 16 at 16:38
$begingroup$
I have gotten the same steps. My main problem is the argument of why $int_{I}|k(x,y)|^2 dy$ makes sense. We know $k(x,y)in L^2(Itimes I)$, but how can we know $int_{I}|k(x,y)|^2 dy$ is finite?
$endgroup$
– James
Jan 16 at 16:38
$begingroup$
You do not care. It may be infinite for some $x$, but Fubini guarantees that it is finite a.e. and integrable as a function of $x$. Moreover, since we are dealing with positive functions, Fubini applies, even if the integral were infinite.
$endgroup$
– Julián Aguirre
Jan 16 at 16:41
$begingroup$
You do not care. It may be infinite for some $x$, but Fubini guarantees that it is finite a.e. and integrable as a function of $x$. Moreover, since we are dealing with positive functions, Fubini applies, even if the integral were infinite.
$endgroup$
– Julián Aguirre
Jan 16 at 16:41
$begingroup$
Thank you for the help! Why does Fubini's theorem apply even if the integral is infinite?
$endgroup$
– James
Jan 16 at 16:56
$begingroup$
Thank you for the help! Why does Fubini's theorem apply even if the integral is infinite?
$endgroup$
– James
Jan 16 at 16:56
$begingroup$
In that case you get an equality $infty=infty$.
$endgroup$
– Julián Aguirre
Jan 16 at 22:42
$begingroup$
In that case you get an equality $infty=infty$.
$endgroup$
– Julián Aguirre
Jan 16 at 22:42
$begingroup$
Oh yeah, of course. Thanks for the help, it makes much more sense now!
$endgroup$
– James
Jan 17 at 13:45
$begingroup$
Oh yeah, of course. Thanks for the help, it makes much more sense now!
$endgroup$
– James
Jan 17 at 13:45
add a comment |
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