Mixing
self-adjoint bounded generates analytic semigroup
$begingroup$
Engel Nagel A Short Course on Operator Semigroups Corollary II.4.8 states:
(There should be a typo. If $delta=0$ then the spectum is empty, but normal operator has a non-empty spectrum? Anyways,)
Then they proceed,
In particular, Corollary 4.8 shows that the semigroup generated by a self-adjoint operator A that is bounded above, which means that there exists $winmathbb{R}$ such that
$$
( A x | x ) leq w | x | ^ { 2 } quad text { for all } x in D ( A )
$$
is analytic of angle $pi/2$. Moreover, this semigroup is bounded if and only if
$wleq 0$.
I do not see how Corollary 4.8 implies this. If $w=0$, then $(Ax|x)leq 0$, so it is still possible that $0in sigma(A)$. But $0notin {zinmathbb{C}mid |arg(-z)|<delta}$.
Are we meant to consider something like $A+w$, consider ${e}^{t(A+w)}$ and then say ${e}^{tA}$ is well-defined further ${e}^{t(A+w)}={e}^{tA}e^{tw}$ etc.?
functional-analysis semigroup-of-operators
asked Jan 18 at 12:35
user41467user41467
1919
$endgroup$
add a comment |
$begingroup$
Engel Nagel A Short Course on Operator Semigroups Corollary II.4.8 states:
(There should be a typo. If $delta=0$ then the spectum is empty, but normal operator has a non-empty spectrum? Anyways,)
Then they proceed,
In particular, Corollary 4.8 shows that the semigroup generated by a self-adjoint operator A that is bounded above, which means that there exists $winmathbb{R}$ such that
$$
( A x | x ) leq w | x | ^ { 2 } quad text { for all } x in D ( A )
$$
is analytic of angle $pi/2$. Moreover, this semigroup is bounded if and only if
$wleq 0$.
I do not see how Corollary 4.8 implies this. If $w=0$, then $(Ax|x)leq 0$, so it is still possible that $0in sigma(A)$. But $0notin {zinmathbb{C}mid |arg(-z)|<delta}$.
Are we meant to consider something like $A+w$, consider ${e}^{t(A+w)}$ and then say ${e}^{tA}$ is well-defined further ${e}^{t(A+w)}={e}^{tA}e^{tw}$ etc.?
functional-analysis semigroup-of-operators
asked Jan 18 at 12:35
user41467user41467
1919
$endgroup$
$begingroup$
There is a typo (or, more precisely, $mathrm{arg}(0)$ is not well-defined). It should be $sigma(A)subseteq {zinmathbb{C}: |mathrm{arg}(-z)|<delta}cup {0}$.
$endgroup$
– MaoWao
Jan 18 at 12:50
$begingroup$
Ah, I thought it should be $|arg(-z)|leq delta$ so that for $delta=0$ we have self-adjoint negative definite. Your corrected assumption makes the result stronger than my version, which I guess is related to an answer of yours (Pazy's $0in rho(A)$ is not needed). Thanks!
$endgroup$
– user41467
Jan 18 at 13:29
$begingroup$
You have to clarify whether $0in sigma(A)$ is allowed or not. For every $zneq 0$ it does not matter if there is $<$ or $leq$, you can always replace $delta$ by $delta'in (delta,pi/2)$.
$endgroup$
– MaoWao
Jan 18 at 13:33
$begingroup$
I was looking at Yagi before this and thought $sigma(A)subseteq {|arg(−z)|leq delta}$ implicitly excludes $0$. But now I see $0insigma(A)$ is fine. Thanks for the discussion on this typo.
$endgroup$
– user41467
Jan 18 at 13:40
add a comment |
$begingroup$
Engel Nagel A Short Course on Operator Semigroups Corollary II.4.8 states:
(There should be a typo. If $delta=0$ then the spectum is empty, but normal operator has a non-empty spectrum? Anyways,)
Then they proceed,
In particular, Corollary 4.8 shows that the semigroup generated by a self-adjoint operator A that is bounded above, which means that there exists $winmathbb{R}$ such that
$$
( A x | x ) leq w | x | ^ { 2 } quad text { for all } x in D ( A )
$$
is analytic of angle $pi/2$. Moreover, this semigroup is bounded if and only if
$wleq 0$.
I do not see how Corollary 4.8 implies this. If $w=0$, then $(Ax|x)leq 0$, so it is still possible that $0in sigma(A)$. But $0notin {zinmathbb{C}mid |arg(-z)|<delta}$.
Are we meant to consider something like $A+w$, consider ${e}^{t(A+w)}$ and then say ${e}^{tA}$ is well-defined further ${e}^{t(A+w)}={e}^{tA}e^{tw}$ etc.?
functional-analysis semigroup-of-operators
asked Jan 18 at 12:35
user41467user41467
1919
$endgroup$
Engel Nagel A Short Course on Operator Semigroups Corollary II.4.8 states:
(There should be a typo. If $delta=0$ then the spectum is empty, but normal operator has a non-empty spectrum? Anyways,)
Then they proceed,
In particular, Corollary 4.8 shows that the semigroup generated by a self-adjoint operator A that is bounded above, which means that there exists $winmathbb{R}$ such that
$$
( A x | x ) leq w | x | ^ { 2 } quad text { for all } x in D ( A )
$$
is analytic of angle $pi/2$. Moreover, this semigroup is bounded if and only if
$wleq 0$.
I do not see how Corollary 4.8 implies this. If $w=0$, then $(Ax|x)leq 0$, so it is still possible that $0in sigma(A)$. But $0notin {zinmathbb{C}mid |arg(-z)|<delta}$.
Are we meant to consider something like $A+w$, consider ${e}^{t(A+w)}$ and then say ${e}^{tA}$ is well-defined further ${e}^{t(A+w)}={e}^{tA}e^{tw}$ etc.?
functional-analysis semigroup-of-operators
functional-analysis semigroup-of-operators
asked Jan 18 at 12:35
user41467user41467
1919
asked Jan 18 at 12:35
user41467user41467
1919
asked Jan 18 at 12:35
user41467user41467
1919
asked Jan 18 at 12:35
user41467user41467
1919
asked Jan 18 at 12:35
user41467user41467
1919
1919
$begingroup$
There is a typo (or, more precisely, $mathrm{arg}(0)$ is not well-defined). It should be $sigma(A)subseteq {zinmathbb{C}: |mathrm{arg}(-z)|<delta}cup {0}$.
$endgroup$
– MaoWao
Jan 18 at 12:50
$begingroup$
Ah, I thought it should be $|arg(-z)|leq delta$ so that for $delta=0$ we have self-adjoint negative definite. Your corrected assumption makes the result stronger than my version, which I guess is related to an answer of yours (Pazy's $0in rho(A)$ is not needed). Thanks!
$endgroup$
– user41467
Jan 18 at 13:29
$begingroup$
You have to clarify whether $0in sigma(A)$ is allowed or not. For every $zneq 0$ it does not matter if there is $<$ or $leq$, you can always replace $delta$ by $delta'in (delta,pi/2)$.
$endgroup$
– MaoWao
Jan 18 at 13:33
$begingroup$
I was looking at Yagi before this and thought $sigma(A)subseteq {|arg(−z)|leq delta}$ implicitly excludes $0$. But now I see $0insigma(A)$ is fine. Thanks for the discussion on this typo.
$endgroup$
– user41467
Jan 18 at 13:40
add a comment |
$begingroup$
There is a typo (or, more precisely, $mathrm{arg}(0)$ is not well-defined). It should be $sigma(A)subseteq {zinmathbb{C}: |mathrm{arg}(-z)|<delta}cup {0}$.
$endgroup$
– MaoWao
Jan 18 at 12:50
$begingroup$
Ah, I thought it should be $|arg(-z)|leq delta$ so that for $delta=0$ we have self-adjoint negative definite. Your corrected assumption makes the result stronger than my version, which I guess is related to an answer of yours (Pazy's $0in rho(A)$ is not needed). Thanks!
$endgroup$
– user41467
Jan 18 at 13:29
$begingroup$
You have to clarify whether $0in sigma(A)$ is allowed or not. For every $zneq 0$ it does not matter if there is $<$ or $leq$, you can always replace $delta$ by $delta'in (delta,pi/2)$.
$endgroup$
– MaoWao
Jan 18 at 13:33
$begingroup$
I was looking at Yagi before this and thought $sigma(A)subseteq {|arg(−z)|leq delta}$ implicitly excludes $0$. But now I see $0insigma(A)$ is fine. Thanks for the discussion on this typo.
$endgroup$
– user41467
Jan 18 at 13:40
$begingroup$
There is a typo (or, more precisely, $mathrm{arg}(0)$ is not well-defined). It should be $sigma(A)subseteq {zinmathbb{C}: |mathrm{arg}(-z)|<delta}cup {0}$.
$endgroup$
– MaoWao
Jan 18 at 12:50
$begingroup$
There is a typo (or, more precisely, $mathrm{arg}(0)$ is not well-defined). It should be $sigma(A)subseteq {zinmathbb{C}: |mathrm{arg}(-z)|<delta}cup {0}$.
$endgroup$
– MaoWao
Jan 18 at 12:50
$begingroup$
Ah, I thought it should be $|arg(-z)|leq delta$ so that for $delta=0$ we have self-adjoint negative definite. Your corrected assumption makes the result stronger than my version, which I guess is related to an answer of yours (Pazy's $0in rho(A)$ is not needed). Thanks!
$endgroup$
– user41467
Jan 18 at 13:29
$begingroup$
Ah, I thought it should be $|arg(-z)|leq delta$ so that for $delta=0$ we have self-adjoint negative definite. Your corrected assumption makes the result stronger than my version, which I guess is related to an answer of yours (Pazy's $0in rho(A)$ is not needed). Thanks!
$endgroup$
– user41467
Jan 18 at 13:29
$begingroup$
You have to clarify whether $0in sigma(A)$ is allowed or not. For every $zneq 0$ it does not matter if there is $<$ or $leq$, you can always replace $delta$ by $delta'in (delta,pi/2)$.
$endgroup$
– MaoWao
Jan 18 at 13:33
$begingroup$
You have to clarify whether $0in sigma(A)$ is allowed or not. For every $zneq 0$ it does not matter if there is $<$ or $leq$, you can always replace $delta$ by $delta'in (delta,pi/2)$.
$endgroup$
– MaoWao
Jan 18 at 13:33
$begingroup$
I was looking at Yagi before this and thought $sigma(A)subseteq {|arg(−z)|leq delta}$ implicitly excludes $0$. But now I see $0insigma(A)$ is fine. Thanks for the discussion on this typo.
$endgroup$
– user41467
Jan 18 at 13:40
$begingroup$
I was looking at Yagi before this and thought $sigma(A)subseteq {|arg(−z)|leq delta}$ implicitly excludes $0$. But now I see $0insigma(A)$ is fine. Thanks for the discussion on this typo.
$endgroup$
– user41467
Jan 18 at 13:40
add a comment |
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$begingroup$
There is a typo (or, more precisely, $mathrm{arg}(0)$ is not well-defined). It should be $sigma(A)subseteq {zinmathbb{C}: |mathrm{arg}(-z)|<delta}cup {0}$.
$endgroup$
– MaoWao
Jan 18 at 12:50
$begingroup$
Ah, I thought it should be $|arg(-z)|leq delta$ so that for $delta=0$ we have self-adjoint negative definite. Your corrected assumption makes the result stronger than my version, which I guess is related to an answer of yours (Pazy's $0in rho(A)$ is not needed). Thanks!
$endgroup$
– user41467
Jan 18 at 13:29
$begingroup$
You have to clarify whether $0in sigma(A)$ is allowed or not. For every $zneq 0$ it does not matter if there is $<$ or $leq$, you can always replace $delta$ by $delta'in (delta,pi/2)$.
$endgroup$
– MaoWao
Jan 18 at 13:33
$begingroup$
I was looking at Yagi before this and thought $sigma(A)subseteq {|arg(−z)|leq delta}$ implicitly excludes $0$. But now I see $0insigma(A)$ is fine. Thanks for the discussion on this typo.
$endgroup$
– user41467
Jan 18 at 13:40