Image of a transformation becomes a Hilbert space !!
$begingroup$
Let $T: L^2(mathbb R)to O(mathbb C)$ an transformation from the Hilbert space $L^2(mathbb R)$ to the Fréchet spaces $O(mathbb C)$ of all holomorphic functions on $mathbb C$. We assume that $T$ is injective and continuous.
Now, Can we endowed the space $mathcal H:= Im T$ (image of $T$) with the Hilbert inner product
from $left(L^{2}(mathbb{R}), left<.,.right>right)$ for that $mathcal H:= Im T$ becomes a Hilbert space of holomorphic functions ? If yes, how?
I think it is possible if we consider the inner product $(f,g)_{mathcal H}=left<T^{-1}(f),T^{-1}(g)right>_{L^{2}(mathbb{R})}$ ??
Thank you in advance
complex-analysis operator-theory hilbert-spaces harmonic-analysis
asked Dec 3 '18 at 15:25
Z. AlfataZ. Alfata
895514
$endgroup$
add a comment |
$begingroup$
Let $T: L^2(mathbb R)to O(mathbb C)$ an transformation from the Hilbert space $L^2(mathbb R)$ to the Fréchet spaces $O(mathbb C)$ of all holomorphic functions on $mathbb C$. We assume that $T$ is injective and continuous.
Now, Can we endowed the space $mathcal H:= Im T$ (image of $T$) with the Hilbert inner product
from $left(L^{2}(mathbb{R}), left<.,.right>right)$ for that $mathcal H:= Im T$ becomes a Hilbert space of holomorphic functions ? If yes, how?
I think it is possible if we consider the inner product $(f,g)_{mathcal H}=left<T^{-1}(f),T^{-1}(g)right>_{L^{2}(mathbb{R})}$ ??
Thank you in advance
complex-analysis operator-theory hilbert-spaces harmonic-analysis
asked Dec 3 '18 at 15:25
Z. AlfataZ. Alfata
895514
$endgroup$
$begingroup$
I suspect that you need the inverse of $T$ to be continuous in order to guarantee completeness. This is not the case in general.
$endgroup$
– Martin
Dec 3 '18 at 16:29
$begingroup$
@ Martin, we have assume that $T$ is injective and continuous. Then, $T$ becomes an isomorphism continuous from $L^{2}(mathbb{R})$ to $Im T$.
$endgroup$
– Z. Alfata
Dec 3 '18 at 16:53
$begingroup$
Yes but take a Cauchy sequence in the image. You now need to show that it's convergent. Now the you'd like to map that Cauchy sequence back to $L^2$ in order to use completeness of this space. However we hit a wall - the inverse is not necessarily continuous - thus elements close to each other in $H$ need not be close in $L^2$. I'm not saying it's not doable just that we might need continuity of $T^{-1}$ that's all :)
$endgroup$
– Martin
Dec 3 '18 at 17:00
$begingroup$
@ Martin, I think it is possible if we consider the inner product $(f,g)_{mathcal H}=left<T^{-1}(f),T^{-1}(g)right>_{L^{2}(mathbb{R})}$ ??
$endgroup$
– Z. Alfata
Dec 5 '18 at 9:41
add a comment |
$begingroup$
Let $T: L^2(mathbb R)to O(mathbb C)$ an transformation from the Hilbert space $L^2(mathbb R)$ to the Fréchet spaces $O(mathbb C)$ of all holomorphic functions on $mathbb C$. We assume that $T$ is injective and continuous.
Now, Can we endowed the space $mathcal H:= Im T$ (image of $T$) with the Hilbert inner product
from $left(L^{2}(mathbb{R}), left<.,.right>right)$ for that $mathcal H:= Im T$ becomes a Hilbert space of holomorphic functions ? If yes, how?
I think it is possible if we consider the inner product $(f,g)_{mathcal H}=left<T^{-1}(f),T^{-1}(g)right>_{L^{2}(mathbb{R})}$ ??
Thank you in advance
complex-analysis operator-theory hilbert-spaces harmonic-analysis
asked Dec 3 '18 at 15:25
Z. AlfataZ. Alfata
895514
$endgroup$
Let $T: L^2(mathbb R)to O(mathbb C)$ an transformation from the Hilbert space $L^2(mathbb R)$ to the Fréchet spaces $O(mathbb C)$ of all holomorphic functions on $mathbb C$. We assume that $T$ is injective and continuous.
Now, Can we endowed the space $mathcal H:= Im T$ (image of $T$) with the Hilbert inner product
from $left(L^{2}(mathbb{R}), left<.,.right>right)$ for that $mathcal H:= Im T$ becomes a Hilbert space of holomorphic functions ? If yes, how?
I think it is possible if we consider the inner product $(f,g)_{mathcal H}=left<T^{-1}(f),T^{-1}(g)right>_{L^{2}(mathbb{R})}$ ??
Thank you in advance
complex-analysis operator-theory hilbert-spaces harmonic-analysis
complex-analysis operator-theory hilbert-spaces harmonic-analysis
asked Dec 3 '18 at 15:25
Z. AlfataZ. Alfata
895514
asked Dec 3 '18 at 15:25
Z. AlfataZ. Alfata
895514
edited Dec 5 '18 at 9:40
Z. Alfata
asked Dec 3 '18 at 15:25
Z. AlfataZ. Alfata
895514
asked Dec 3 '18 at 15:25
Z. AlfataZ. Alfata
895514
asked Dec 3 '18 at 15:25
Z. AlfataZ. Alfata
895514
895514
$begingroup$
I suspect that you need the inverse of $T$ to be continuous in order to guarantee completeness. This is not the case in general.
$endgroup$
– Martin
Dec 3 '18 at 16:29
$begingroup$
@ Martin, we have assume that $T$ is injective and continuous. Then, $T$ becomes an isomorphism continuous from $L^{2}(mathbb{R})$ to $Im T$.
$endgroup$
– Z. Alfata
Dec 3 '18 at 16:53
$begingroup$
Yes but take a Cauchy sequence in the image. You now need to show that it's convergent. Now the you'd like to map that Cauchy sequence back to $L^2$ in order to use completeness of this space. However we hit a wall - the inverse is not necessarily continuous - thus elements close to each other in $H$ need not be close in $L^2$. I'm not saying it's not doable just that we might need continuity of $T^{-1}$ that's all :)
$endgroup$
– Martin
Dec 3 '18 at 17:00
$begingroup$
@ Martin, I think it is possible if we consider the inner product $(f,g)_{mathcal H}=left<T^{-1}(f),T^{-1}(g)right>_{L^{2}(mathbb{R})}$ ??
$endgroup$
– Z. Alfata
Dec 5 '18 at 9:41
add a comment |
$begingroup$
I suspect that you need the inverse of $T$ to be continuous in order to guarantee completeness. This is not the case in general.
$endgroup$
– Martin
Dec 3 '18 at 16:29
$begingroup$
@ Martin, we have assume that $T$ is injective and continuous. Then, $T$ becomes an isomorphism continuous from $L^{2}(mathbb{R})$ to $Im T$.
$endgroup$
– Z. Alfata
Dec 3 '18 at 16:53
$begingroup$
Yes but take a Cauchy sequence in the image. You now need to show that it's convergent. Now the you'd like to map that Cauchy sequence back to $L^2$ in order to use completeness of this space. However we hit a wall - the inverse is not necessarily continuous - thus elements close to each other in $H$ need not be close in $L^2$. I'm not saying it's not doable just that we might need continuity of $T^{-1}$ that's all :)
$endgroup$
– Martin
Dec 3 '18 at 17:00
$begingroup$
@ Martin, I think it is possible if we consider the inner product $(f,g)_{mathcal H}=left<T^{-1}(f),T^{-1}(g)right>_{L^{2}(mathbb{R})}$ ??
$endgroup$
– Z. Alfata
Dec 5 '18 at 9:41
$begingroup$
I suspect that you need the inverse of $T$ to be continuous in order to guarantee completeness. This is not the case in general.
$endgroup$
– Martin
Dec 3 '18 at 16:29
$begingroup$
I suspect that you need the inverse of $T$ to be continuous in order to guarantee completeness. This is not the case in general.
$endgroup$
– Martin
Dec 3 '18 at 16:29
$begingroup$
@ Martin, we have assume that $T$ is injective and continuous. Then, $T$ becomes an isomorphism continuous from $L^{2}(mathbb{R})$ to $Im T$.
$endgroup$
– Z. Alfata
Dec 3 '18 at 16:53
$begingroup$
@ Martin, we have assume that $T$ is injective and continuous. Then, $T$ becomes an isomorphism continuous from $L^{2}(mathbb{R})$ to $Im T$.
$endgroup$
– Z. Alfata
Dec 3 '18 at 16:53
$begingroup$
Yes but take a Cauchy sequence in the image. You now need to show that it's convergent. Now the you'd like to map that Cauchy sequence back to $L^2$ in order to use completeness of this space. However we hit a wall - the inverse is not necessarily continuous - thus elements close to each other in $H$ need not be close in $L^2$. I'm not saying it's not doable just that we might need continuity of $T^{-1}$ that's all :)
$endgroup$
– Martin
Dec 3 '18 at 17:00
$begingroup$
Yes but take a Cauchy sequence in the image. You now need to show that it's convergent. Now the you'd like to map that Cauchy sequence back to $L^2$ in order to use completeness of this space. However we hit a wall - the inverse is not necessarily continuous - thus elements close to each other in $H$ need not be close in $L^2$. I'm not saying it's not doable just that we might need continuity of $T^{-1}$ that's all :)
$endgroup$
– Martin
Dec 3 '18 at 17:00
$begingroup$
@ Martin, I think it is possible if we consider the inner product $(f,g)_{mathcal H}=left<T^{-1}(f),T^{-1}(g)right>_{L^{2}(mathbb{R})}$ ??
$endgroup$
– Z. Alfata
Dec 5 '18 at 9:41
$begingroup$
@ Martin, I think it is possible if we consider the inner product $(f,g)_{mathcal H}=left<T^{-1}(f),T^{-1}(g)right>_{L^{2}(mathbb{R})}$ ??
$endgroup$
– Z. Alfata
Dec 5 '18 at 9:41
add a comment |
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$begingroup$
I suspect that you need the inverse of $T$ to be continuous in order to guarantee completeness. This is not the case in general.
$endgroup$
– Martin
Dec 3 '18 at 16:29
$begingroup$
@ Martin, we have assume that $T$ is injective and continuous. Then, $T$ becomes an isomorphism continuous from $L^{2}(mathbb{R})$ to $Im T$.
$endgroup$
– Z. Alfata
Dec 3 '18 at 16:53
$begingroup$
Yes but take a Cauchy sequence in the image. You now need to show that it's convergent. Now the you'd like to map that Cauchy sequence back to $L^2$ in order to use completeness of this space. However we hit a wall - the inverse is not necessarily continuous - thus elements close to each other in $H$ need not be close in $L^2$. I'm not saying it's not doable just that we might need continuity of $T^{-1}$ that's all :)
$endgroup$
– Martin
Dec 3 '18 at 17:00
$begingroup$
@ Martin, I think it is possible if we consider the inner product $(f,g)_{mathcal H}=left<T^{-1}(f),T^{-1}(g)right>_{L^{2}(mathbb{R})}$ ??
$endgroup$
– Z. Alfata
Dec 5 '18 at 9:41