Is this statement still true with a weaker condition?
$begingroup$
Let $H$ be a complex Hilbert space and let $A:mathrm{dom}(A)to H$ be an unbounded symmetric operator with dense domain. Prove that $A$ is self-adjoint if and only if there is a $lambdainmathbb{C}$ s.t. $lambda I-A:mathrm{dom}(A)to H$ is surjective.
This is an excise, but now I doubt whether the other direction can be achieved. For $A$ self-adjoint, any nonreal $lambda$ will make $lambda I-A$ surjective since the spectrum of $A$ is a subset of $mathbb{R}$. However for the other direction, I can only prove the case $lambda I-A$ and $overline{lambda} I-A$ are both surjective (or at least one surjective one with dense image). I found many other books and they only have similar statement for the case in which both hold. Therefore I am wondering is it really possible that only $lambda I-A$ surjective is sufficient?
Proof for the both-hold case:
Since $A$ is densely defined and symmetric, $A^*$ is an closed extension of $A$. Therefore if we want to show $A$ is self-adjoint, we
just need to show the inclusion from the other direction that $mathrm{dom}(A^*)subsetmathrm{dom}(A)$. Let $xin mathrm{dom}(A^*)$, since $mathrm{Im}(lambda I-A)=H$, there is a $yin mathrm{dom}(A)$
such that $$(lambda I-A)y =(lambda I- A^*)xin H$$ Then for any $zinmathrm{dom}(A)$
we have
$$left<(overline{lambda} I-A)z,xright>=left<z,(lambda I-A^*)x)right>
=left<z,(lambda I-A)yright>=left<(overline{lambda} I-A)z,yright>$$
If $overline{lambda} I-A$ has dense image we will have $x=y$ and so $xinmathrm{dom}(A)$.
I can't think up any amelioration if we remove the condition. Any help will be appreciated.
functional-analysis self-adjoint-operators
asked Dec 6 '18 at 16:07
ApocalypseApocalypse
1378
$endgroup$
add a comment |
$begingroup$
Let $H$ be a complex Hilbert space and let $A:mathrm{dom}(A)to H$ be an unbounded symmetric operator with dense domain. Prove that $A$ is self-adjoint if and only if there is a $lambdainmathbb{C}$ s.t. $lambda I-A:mathrm{dom}(A)to H$ is surjective.
This is an excise, but now I doubt whether the other direction can be achieved. For $A$ self-adjoint, any nonreal $lambda$ will make $lambda I-A$ surjective since the spectrum of $A$ is a subset of $mathbb{R}$. However for the other direction, I can only prove the case $lambda I-A$ and $overline{lambda} I-A$ are both surjective (or at least one surjective one with dense image). I found many other books and they only have similar statement for the case in which both hold. Therefore I am wondering is it really possible that only $lambda I-A$ surjective is sufficient?
Proof for the both-hold case:
Since $A$ is densely defined and symmetric, $A^*$ is an closed extension of $A$. Therefore if we want to show $A$ is self-adjoint, we
just need to show the inclusion from the other direction that $mathrm{dom}(A^*)subsetmathrm{dom}(A)$. Let $xin mathrm{dom}(A^*)$, since $mathrm{Im}(lambda I-A)=H$, there is a $yin mathrm{dom}(A)$
such that $$(lambda I-A)y =(lambda I- A^*)xin H$$ Then for any $zinmathrm{dom}(A)$
we have
$$left<(overline{lambda} I-A)z,xright>=left<z,(lambda I-A^*)x)right>
=left<z,(lambda I-A)yright>=left<(overline{lambda} I-A)z,yright>$$
If $overline{lambda} I-A$ has dense image we will have $x=y$ and so $xinmathrm{dom}(A)$.
I can't think up any amelioration if we remove the condition. Any help will be appreciated.
functional-analysis self-adjoint-operators
asked Dec 6 '18 at 16:07
ApocalypseApocalypse
1378
$endgroup$
$begingroup$
You are right, surjectivity of $lambda-A$ for only one $lambdain mathbb{C}setminusmathbb{R}$ is not sufficient. A counterexample can be found here: math.stackexchange.com/questions/893899/…
$endgroup$
– MaoWao
Dec 6 '18 at 19:00
$begingroup$
Thank you so much for that link!
$endgroup$
– Apocalypse
Dec 7 '18 at 12:20
add a comment |
$begingroup$
Let $H$ be a complex Hilbert space and let $A:mathrm{dom}(A)to H$ be an unbounded symmetric operator with dense domain. Prove that $A$ is self-adjoint if and only if there is a $lambdainmathbb{C}$ s.t. $lambda I-A:mathrm{dom}(A)to H$ is surjective.
This is an excise, but now I doubt whether the other direction can be achieved. For $A$ self-adjoint, any nonreal $lambda$ will make $lambda I-A$ surjective since the spectrum of $A$ is a subset of $mathbb{R}$. However for the other direction, I can only prove the case $lambda I-A$ and $overline{lambda} I-A$ are both surjective (or at least one surjective one with dense image). I found many other books and they only have similar statement for the case in which both hold. Therefore I am wondering is it really possible that only $lambda I-A$ surjective is sufficient?
Proof for the both-hold case:
Since $A$ is densely defined and symmetric, $A^*$ is an closed extension of $A$. Therefore if we want to show $A$ is self-adjoint, we
just need to show the inclusion from the other direction that $mathrm{dom}(A^*)subsetmathrm{dom}(A)$. Let $xin mathrm{dom}(A^*)$, since $mathrm{Im}(lambda I-A)=H$, there is a $yin mathrm{dom}(A)$
such that $$(lambda I-A)y =(lambda I- A^*)xin H$$ Then for any $zinmathrm{dom}(A)$
we have
$$left<(overline{lambda} I-A)z,xright>=left<z,(lambda I-A^*)x)right>
=left<z,(lambda I-A)yright>=left<(overline{lambda} I-A)z,yright>$$
If $overline{lambda} I-A$ has dense image we will have $x=y$ and so $xinmathrm{dom}(A)$.
I can't think up any amelioration if we remove the condition. Any help will be appreciated.
functional-analysis self-adjoint-operators
asked Dec 6 '18 at 16:07
ApocalypseApocalypse
1378
$endgroup$
Let $H$ be a complex Hilbert space and let $A:mathrm{dom}(A)to H$ be an unbounded symmetric operator with dense domain. Prove that $A$ is self-adjoint if and only if there is a $lambdainmathbb{C}$ s.t. $lambda I-A:mathrm{dom}(A)to H$ is surjective.
This is an excise, but now I doubt whether the other direction can be achieved. For $A$ self-adjoint, any nonreal $lambda$ will make $lambda I-A$ surjective since the spectrum of $A$ is a subset of $mathbb{R}$. However for the other direction, I can only prove the case $lambda I-A$ and $overline{lambda} I-A$ are both surjective (or at least one surjective one with dense image). I found many other books and they only have similar statement for the case in which both hold. Therefore I am wondering is it really possible that only $lambda I-A$ surjective is sufficient?
Proof for the both-hold case:
Since $A$ is densely defined and symmetric, $A^*$ is an closed extension of $A$. Therefore if we want to show $A$ is self-adjoint, we
just need to show the inclusion from the other direction that $mathrm{dom}(A^*)subsetmathrm{dom}(A)$. Let $xin mathrm{dom}(A^*)$, since $mathrm{Im}(lambda I-A)=H$, there is a $yin mathrm{dom}(A)$
such that $$(lambda I-A)y =(lambda I- A^*)xin H$$ Then for any $zinmathrm{dom}(A)$
we have
$$left<(overline{lambda} I-A)z,xright>=left<z,(lambda I-A^*)x)right>
=left<z,(lambda I-A)yright>=left<(overline{lambda} I-A)z,yright>$$
If $overline{lambda} I-A$ has dense image we will have $x=y$ and so $xinmathrm{dom}(A)$.
I can't think up any amelioration if we remove the condition. Any help will be appreciated.
functional-analysis self-adjoint-operators
functional-analysis self-adjoint-operators
asked Dec 6 '18 at 16:07
ApocalypseApocalypse
1378
asked Dec 6 '18 at 16:07
ApocalypseApocalypse
1378
asked Dec 6 '18 at 16:07
ApocalypseApocalypse
1378
asked Dec 6 '18 at 16:07
ApocalypseApocalypse
1378
asked Dec 6 '18 at 16:07
ApocalypseApocalypse
1378
1378
$begingroup$
You are right, surjectivity of $lambda-A$ for only one $lambdain mathbb{C}setminusmathbb{R}$ is not sufficient. A counterexample can be found here: math.stackexchange.com/questions/893899/…
$endgroup$
– MaoWao
Dec 6 '18 at 19:00
$begingroup$
Thank you so much for that link!
$endgroup$
– Apocalypse
Dec 7 '18 at 12:20
add a comment |
$begingroup$
You are right, surjectivity of $lambda-A$ for only one $lambdain mathbb{C}setminusmathbb{R}$ is not sufficient. A counterexample can be found here: math.stackexchange.com/questions/893899/…
$endgroup$
– MaoWao
Dec 6 '18 at 19:00
$begingroup$
Thank you so much for that link!
$endgroup$
– Apocalypse
Dec 7 '18 at 12:20
$begingroup$
You are right, surjectivity of $lambda-A$ for only one $lambdain mathbb{C}setminusmathbb{R}$ is not sufficient. A counterexample can be found here: math.stackexchange.com/questions/893899/…
$endgroup$
– MaoWao
Dec 6 '18 at 19:00
$begingroup$
You are right, surjectivity of $lambda-A$ for only one $lambdain mathbb{C}setminusmathbb{R}$ is not sufficient. A counterexample can be found here: math.stackexchange.com/questions/893899/…
$endgroup$
– MaoWao
Dec 6 '18 at 19:00
$begingroup$
Thank you so much for that link!
$endgroup$
– Apocalypse
Dec 7 '18 at 12:20
$begingroup$
Thank you so much for that link!
$endgroup$
– Apocalypse
Dec 7 '18 at 12:20
add a comment |
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$begingroup$
You are right, surjectivity of $lambda-A$ for only one $lambdain mathbb{C}setminusmathbb{R}$ is not sufficient. A counterexample can be found here: math.stackexchange.com/questions/893899/…
$endgroup$
– MaoWao
Dec 6 '18 at 19:00
$begingroup$
Thank you so much for that link!
$endgroup$
– Apocalypse
Dec 7 '18 at 12:20