Bijection+Closability?Adjoint of resolvent of self-adjoint, densely-defined operator on a Hilbert spaceTrue/False: Self-adjoint compact operatorDefinition of resolvent setImage of a dense set through unbounded operatorIf $B$ is bounded with $0< Bleqslant 1$ and $T$ is closed, is $BT$ closable?whether the inverse operator of identity plus projection is boundedContinuity of the inverse OperatorClosed operator and dense rangeIs the closure of a closable projection continuous?Is weak derivative a bounded operator from $H^k(mathbbR^n)$ to $L^2(mathbbR^n)$?
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Bijection+Closability?
Adjoint of resolvent of self-adjoint, densely-defined operator on a Hilbert spaceTrue/False: Self-adjoint compact operatorDefinition of resolvent setImage of a dense set through unbounded operatorIf $B$ is bounded with $0< Bleqslant 1$ and $T$ is closed, is $BT$ closable?whether the inverse operator of identity plus projection is boundedContinuity of the inverse OperatorClosed operator and dense rangeIs the closure of a closable projection continuous?Is weak derivative a bounded operator from $H^k(mathbbR^n)$ to $L^2(mathbbR^n)$?
$begingroup$
I was wondering whether a closable operator $A$ with domain $D(A)subset H$ ($H$ is a Hilbert space) which is also bijective has an everywhere defined bounded inverse?
Thanks in advance.
Math.
functional-analysis operator-theory
asked Mar 12 at 20:21
MathMath
995
$endgroup$
add a comment |
$begingroup$
I was wondering whether a closable operator $A$ with domain $D(A)subset H$ ($H$ is a Hilbert space) which is also bijective has an everywhere defined bounded inverse?
Thanks in advance.
Math.
functional-analysis operator-theory
asked Mar 12 at 20:21
MathMath
995
$endgroup$
$begingroup$
We may also assume that $D(A)$ is dense.
$endgroup$
– Math
Mar 13 at 7:58
$begingroup$
If you have such an operator and remove one point from its domain, doesn't it still satisfy the conditions? Maybe you should ask about a "closed densely defined" operator".
$endgroup$
– Keith McClary
Mar 13 at 19:32
add a comment |
$begingroup$
I was wondering whether a closable operator $A$ with domain $D(A)subset H$ ($H$ is a Hilbert space) which is also bijective has an everywhere defined bounded inverse?
Thanks in advance.
Math.
functional-analysis operator-theory
asked Mar 12 at 20:21
MathMath
995
$endgroup$
I was wondering whether a closable operator $A$ with domain $D(A)subset H$ ($H$ is a Hilbert space) which is also bijective has an everywhere defined bounded inverse?
Thanks in advance.
Math.
functional-analysis operator-theory
functional-analysis operator-theory
asked Mar 12 at 20:21
MathMath
995
asked Mar 12 at 20:21
MathMath
995
asked Mar 12 at 20:21
MathMath
995
asked Mar 12 at 20:21
MathMath
995
asked Mar 12 at 20:21
MathMath
995
995
$begingroup$
We may also assume that $D(A)$ is dense.
$endgroup$
– Math
Mar 13 at 7:58
$begingroup$
If you have such an operator and remove one point from its domain, doesn't it still satisfy the conditions? Maybe you should ask about a "closed densely defined" operator".
$endgroup$
– Keith McClary
Mar 13 at 19:32
add a comment |
$begingroup$
We may also assume that $D(A)$ is dense.
$endgroup$
– Math
Mar 13 at 7:58
$begingroup$
If you have such an operator and remove one point from its domain, doesn't it still satisfy the conditions? Maybe you should ask about a "closed densely defined" operator".
$endgroup$
– Keith McClary
Mar 13 at 19:32
$begingroup$
We may also assume that $D(A)$ is dense.
$endgroup$
– Math
Mar 13 at 7:58
$begingroup$
We may also assume that $D(A)$ is dense.
$endgroup$
– Math
Mar 13 at 7:58
$begingroup$
If you have such an operator and remove one point from its domain, doesn't it still satisfy the conditions? Maybe you should ask about a "closed densely defined" operator".
$endgroup$
– Keith McClary
Mar 13 at 19:32
$begingroup$
If you have such an operator and remove one point from its domain, doesn't it still satisfy the conditions? Maybe you should ask about a "closed densely defined" operator".
$endgroup$
– Keith McClary
Mar 13 at 19:32
add a comment |
0
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$begingroup$
We may also assume that $D(A)$ is dense.
$endgroup$
– Math
Mar 13 at 7:58
$begingroup$
If you have such an operator and remove one point from its domain, doesn't it still satisfy the conditions? Maybe you should ask about a "closed densely defined" operator".
$endgroup$
– Keith McClary
Mar 13 at 19:32