Mixing
How can we prove that a (locally bounded) semigroup is strongly continuous on the closure of its generator?
$begingroup$
Let $E$ be a $mathbb R$-Banach space and $(T(t))_{tge0}$ be a semigroup on $E$, i.e. $T(t)$ is a bounded linear operator on $E$ for all $tge0$, $T(0)=operatorname{id}_E$ and $$T(s+t)=T(s)T(t);;;text{for all }s,tge0.tag1$$ Let $$operatorname{orb}x:[0,infty)to E;,;;;tmapsto T(t)x$$ for $xin E$, $$mathcal D(A):=left{xin E:operatorname{orb}xtext{ is right-differentiable at }0right}$$ and $$Ax:=(operatorname{orb}x)'(0);;;text{for }xinmathcal D(A).$$
How can we show that $(T(t))_{tge0}$ is strongly continuous on $overline{mathcal D(A)}$?
By the semigroup property, it should suffice to show strong continuity at $0$. Moreover, by density it should suffice to consider $xinmathcal D(A)$. Now, my usual reflex would be to obtain the claim from the identity $$T(t)x-x=int_0^tT(s)Ax:{rm d}s;;;text{for all }tge0tag2$$ which is valid for any strongly continuous semigroup and its generator. However, with strong continuity being the property we're asked to prove, I don't see why $(2)$ should hold (actually, I don't see why the Riemann integral should exist in that case).
So, what do we need to do?
I think that we need to assume that $(T(t))_{tge0}$ is locally bounded (e.g. quasicontractive), i.e. $$sup_{sin[0,:t]}left|T(s)right|_{mathfrak L(E)}<infty;;;text{for all }tge0tag3.$$ Under that assumption we obtain $$sup_{sin[0,:t]}left|frac{T(s+h)x-T(s)x}h-T(s)Axright|_Elesup_{sin[0,:t]}left|T(s)right|_{mathfrak L(E)}left|frac{T(h)x-x}h-Axright|_Exrightarrow{hto0+}0tag4$$ for all $tge0$ and hence locally uniform right-differentiability of $operatorname{orb}x$. Maybe we can build up on that.
functional-analysis operator-theory semigroup-of-operators
asked Feb 1 at 23:43
0xbadf00d0xbadf00d
1,71741534
$endgroup$
add a comment |
$begingroup$
Let $E$ be a $mathbb R$-Banach space and $(T(t))_{tge0}$ be a semigroup on $E$, i.e. $T(t)$ is a bounded linear operator on $E$ for all $tge0$, $T(0)=operatorname{id}_E$ and $$T(s+t)=T(s)T(t);;;text{for all }s,tge0.tag1$$ Let $$operatorname{orb}x:[0,infty)to E;,;;;tmapsto T(t)x$$ for $xin E$, $$mathcal D(A):=left{xin E:operatorname{orb}xtext{ is right-differentiable at }0right}$$ and $$Ax:=(operatorname{orb}x)'(0);;;text{for }xinmathcal D(A).$$
How can we show that $(T(t))_{tge0}$ is strongly continuous on $overline{mathcal D(A)}$?
By the semigroup property, it should suffice to show strong continuity at $0$. Moreover, by density it should suffice to consider $xinmathcal D(A)$. Now, my usual reflex would be to obtain the claim from the identity $$T(t)x-x=int_0^tT(s)Ax:{rm d}s;;;text{for all }tge0tag2$$ which is valid for any strongly continuous semigroup and its generator. However, with strong continuity being the property we're asked to prove, I don't see why $(2)$ should hold (actually, I don't see why the Riemann integral should exist in that case).
So, what do we need to do?
I think that we need to assume that $(T(t))_{tge0}$ is locally bounded (e.g. quasicontractive), i.e. $$sup_{sin[0,:t]}left|T(s)right|_{mathfrak L(E)}<infty;;;text{for all }tge0tag3.$$ Under that assumption we obtain $$sup_{sin[0,:t]}left|frac{T(s+h)x-T(s)x}h-T(s)Axright|_Elesup_{sin[0,:t]}left|T(s)right|_{mathfrak L(E)}left|frac{T(h)x-x}h-Axright|_Exrightarrow{hto0+}0tag4$$ for all $tge0$ and hence locally uniform right-differentiability of $operatorname{orb}x$. Maybe we can build up on that.
functional-analysis operator-theory semigroup-of-operators
asked Feb 1 at 23:43
0xbadf00d0xbadf00d
1,71741534
$endgroup$
add a comment |
$begingroup$
Let $E$ be a $mathbb R$-Banach space and $(T(t))_{tge0}$ be a semigroup on $E$, i.e. $T(t)$ is a bounded linear operator on $E$ for all $tge0$, $T(0)=operatorname{id}_E$ and $$T(s+t)=T(s)T(t);;;text{for all }s,tge0.tag1$$ Let $$operatorname{orb}x:[0,infty)to E;,;;;tmapsto T(t)x$$ for $xin E$, $$mathcal D(A):=left{xin E:operatorname{orb}xtext{ is right-differentiable at }0right}$$ and $$Ax:=(operatorname{orb}x)'(0);;;text{for }xinmathcal D(A).$$
How can we show that $(T(t))_{tge0}$ is strongly continuous on $overline{mathcal D(A)}$?
By the semigroup property, it should suffice to show strong continuity at $0$. Moreover, by density it should suffice to consider $xinmathcal D(A)$. Now, my usual reflex would be to obtain the claim from the identity $$T(t)x-x=int_0^tT(s)Ax:{rm d}s;;;text{for all }tge0tag2$$ which is valid for any strongly continuous semigroup and its generator. However, with strong continuity being the property we're asked to prove, I don't see why $(2)$ should hold (actually, I don't see why the Riemann integral should exist in that case).
So, what do we need to do?
I think that we need to assume that $(T(t))_{tge0}$ is locally bounded (e.g. quasicontractive), i.e. $$sup_{sin[0,:t]}left|T(s)right|_{mathfrak L(E)}<infty;;;text{for all }tge0tag3.$$ Under that assumption we obtain $$sup_{sin[0,:t]}left|frac{T(s+h)x-T(s)x}h-T(s)Axright|_Elesup_{sin[0,:t]}left|T(s)right|_{mathfrak L(E)}left|frac{T(h)x-x}h-Axright|_Exrightarrow{hto0+}0tag4$$ for all $tge0$ and hence locally uniform right-differentiability of $operatorname{orb}x$. Maybe we can build up on that.
functional-analysis operator-theory semigroup-of-operators
asked Feb 1 at 23:43
0xbadf00d0xbadf00d
1,71741534
$endgroup$
Let $E$ be a $mathbb R$-Banach space and $(T(t))_{tge0}$ be a semigroup on $E$, i.e. $T(t)$ is a bounded linear operator on $E$ for all $tge0$, $T(0)=operatorname{id}_E$ and $$T(s+t)=T(s)T(t);;;text{for all }s,tge0.tag1$$ Let $$operatorname{orb}x:[0,infty)to E;,;;;tmapsto T(t)x$$ for $xin E$, $$mathcal D(A):=left{xin E:operatorname{orb}xtext{ is right-differentiable at }0right}$$ and $$Ax:=(operatorname{orb}x)'(0);;;text{for }xinmathcal D(A).$$
How can we show that $(T(t))_{tge0}$ is strongly continuous on $overline{mathcal D(A)}$?
By the semigroup property, it should suffice to show strong continuity at $0$. Moreover, by density it should suffice to consider $xinmathcal D(A)$. Now, my usual reflex would be to obtain the claim from the identity $$T(t)x-x=int_0^tT(s)Ax:{rm d}s;;;text{for all }tge0tag2$$ which is valid for any strongly continuous semigroup and its generator. However, with strong continuity being the property we're asked to prove, I don't see why $(2)$ should hold (actually, I don't see why the Riemann integral should exist in that case).
So, what do we need to do?
I think that we need to assume that $(T(t))_{tge0}$ is locally bounded (e.g. quasicontractive), i.e. $$sup_{sin[0,:t]}left|T(s)right|_{mathfrak L(E)}<infty;;;text{for all }tge0tag3.$$ Under that assumption we obtain $$sup_{sin[0,:t]}left|frac{T(s+h)x-T(s)x}h-T(s)Axright|_Elesup_{sin[0,:t]}left|T(s)right|_{mathfrak L(E)}left|frac{T(h)x-x}h-Axright|_Exrightarrow{hto0+}0tag4$$ for all $tge0$ and hence locally uniform right-differentiability of $operatorname{orb}x$. Maybe we can build up on that.
functional-analysis operator-theory semigroup-of-operators
functional-analysis operator-theory semigroup-of-operators
asked Feb 1 at 23:43
0xbadf00d0xbadf00d
1,71741534
asked Feb 1 at 23:43
0xbadf00d0xbadf00d
1,71741534
edited Feb 2 at 0:06
0xbadf00d
asked Feb 1 at 23:43
0xbadf00d0xbadf00d
1,71741534
asked Feb 1 at 23:43
0xbadf00d0xbadf00d
1,71741534
asked Feb 1 at 23:43
0xbadf00d0xbadf00d
1,71741534
1,71741534
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are right that it suffices to show strong continuity at $0$ (by the semigroup property), but it is not true that it is enough to check strong continuity at $0$ for $xin D(A)$. You would need some uniform bound here. On the other hand, right-differentiability at $0$ automatically implies right-continuity at $0$: If $T_t x-x$ does not tend to zero, then there is no chance for the limit $frac 1 t(T_t x-x)$ to exist.
In your situation, strong continuity on $overline{D(A)}$ is equivalent to local boundedness on $overline{D(A)}$. One implication follows directly from the uniform boundedness principle and the semigroup property. For the other implication (the one you ask about), let $xinoverline{D(A)}$ and $(x_n)$ a sequence in $D(A)$ such that $x_nto x$. Then
$$
|T(t)x-x|leq sup_{sin[0,T]}|T(s)|_{mathcal{L}(overline{D(A)})}|x-x_n|+|T_t x_n-x_n|+|x-x_n|.
$$
Letting first $tto 0$ and then $ntoinfty$ yields the desired convergence.
answered Feb 3 at 20:35
MaoWaoMaoWao
3,953618
$endgroup$
$begingroup$
It's not important for the question, but your notation suggests that $T(s)overline{mathcal D(A)}subseteqoverline{mathcal D(A)}$ for all $sge0$. Why is that the case?
$endgroup$
– 0xbadf00d
Feb 4 at 12:16
$begingroup$
Well, $T(s)$ maps $D(A)$ into $D(A)$ (this can be proven without strong continuity), and then continuity of $T(s)$ implies that the same is true for the closure.
$endgroup$
– MaoWao
Feb 4 at 13:36
$begingroup$
From your proof I guess that's sufficient if there is a small $T>0$ such that the operator norms on $[0,T)$ are bounded, right?
$endgroup$
– 0xbadf00d
Feb 4 at 13:41
1
$begingroup$
That is right. You can also directly show that boundedness on $[0,T)$ for some $T>0$ implies boundedness on all bounded intervals: If $I$ is bounded, then there exists $ninmathbb{N}$ such that $t/n<T$ for $tin I$. Then $|T(t)|leq|T(t/n)|^n$ by the semigroup property.
$endgroup$
– MaoWao
Feb 4 at 13:45
$begingroup$
Do we even obtain left-differentiability of the orbits?
$endgroup$
– 0xbadf00d
Feb 9 at 16:38
|
show 3 more comments
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1 Answer
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1 Answer
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$begingroup$
You are right that it suffices to show strong continuity at $0$ (by the semigroup property), but it is not true that it is enough to check strong continuity at $0$ for $xin D(A)$. You would need some uniform bound here. On the other hand, right-differentiability at $0$ automatically implies right-continuity at $0$: If $T_t x-x$ does not tend to zero, then there is no chance for the limit $frac 1 t(T_t x-x)$ to exist.
In your situation, strong continuity on $overline{D(A)}$ is equivalent to local boundedness on $overline{D(A)}$. One implication follows directly from the uniform boundedness principle and the semigroup property. For the other implication (the one you ask about), let $xinoverline{D(A)}$ and $(x_n)$ a sequence in $D(A)$ such that $x_nto x$. Then
$$
|T(t)x-x|leq sup_{sin[0,T]}|T(s)|_{mathcal{L}(overline{D(A)})}|x-x_n|+|T_t x_n-x_n|+|x-x_n|.
$$
Letting first $tto 0$ and then $ntoinfty$ yields the desired convergence.
answered Feb 3 at 20:35
MaoWaoMaoWao
3,953618
$endgroup$
$begingroup$
It's not important for the question, but your notation suggests that $T(s)overline{mathcal D(A)}subseteqoverline{mathcal D(A)}$ for all $sge0$. Why is that the case?
$endgroup$
– 0xbadf00d
Feb 4 at 12:16
$begingroup$
Well, $T(s)$ maps $D(A)$ into $D(A)$ (this can be proven without strong continuity), and then continuity of $T(s)$ implies that the same is true for the closure.
$endgroup$
– MaoWao
Feb 4 at 13:36
$begingroup$
From your proof I guess that's sufficient if there is a small $T>0$ such that the operator norms on $[0,T)$ are bounded, right?
$endgroup$
– 0xbadf00d
Feb 4 at 13:41
1
$begingroup$
That is right. You can also directly show that boundedness on $[0,T)$ for some $T>0$ implies boundedness on all bounded intervals: If $I$ is bounded, then there exists $ninmathbb{N}$ such that $t/n<T$ for $tin I$. Then $|T(t)|leq|T(t/n)|^n$ by the semigroup property.
$endgroup$
– MaoWao
Feb 4 at 13:45
$begingroup$
Do we even obtain left-differentiability of the orbits?
$endgroup$
– 0xbadf00d
Feb 9 at 16:38
|
show 3 more comments
$begingroup$
You are right that it suffices to show strong continuity at $0$ (by the semigroup property), but it is not true that it is enough to check strong continuity at $0$ for $xin D(A)$. You would need some uniform bound here. On the other hand, right-differentiability at $0$ automatically implies right-continuity at $0$: If $T_t x-x$ does not tend to zero, then there is no chance for the limit $frac 1 t(T_t x-x)$ to exist.
In your situation, strong continuity on $overline{D(A)}$ is equivalent to local boundedness on $overline{D(A)}$. One implication follows directly from the uniform boundedness principle and the semigroup property. For the other implication (the one you ask about), let $xinoverline{D(A)}$ and $(x_n)$ a sequence in $D(A)$ such that $x_nto x$. Then
$$
|T(t)x-x|leq sup_{sin[0,T]}|T(s)|_{mathcal{L}(overline{D(A)})}|x-x_n|+|T_t x_n-x_n|+|x-x_n|.
$$
Letting first $tto 0$ and then $ntoinfty$ yields the desired convergence.
answered Feb 3 at 20:35
MaoWaoMaoWao
3,953618
$endgroup$
$begingroup$
It's not important for the question, but your notation suggests that $T(s)overline{mathcal D(A)}subseteqoverline{mathcal D(A)}$ for all $sge0$. Why is that the case?
$endgroup$
– 0xbadf00d
Feb 4 at 12:16
$begingroup$
Well, $T(s)$ maps $D(A)$ into $D(A)$ (this can be proven without strong continuity), and then continuity of $T(s)$ implies that the same is true for the closure.
$endgroup$
– MaoWao
Feb 4 at 13:36
$begingroup$
From your proof I guess that's sufficient if there is a small $T>0$ such that the operator norms on $[0,T)$ are bounded, right?
$endgroup$
– 0xbadf00d
Feb 4 at 13:41
1
$begingroup$
That is right. You can also directly show that boundedness on $[0,T)$ for some $T>0$ implies boundedness on all bounded intervals: If $I$ is bounded, then there exists $ninmathbb{N}$ such that $t/n<T$ for $tin I$. Then $|T(t)|leq|T(t/n)|^n$ by the semigroup property.
$endgroup$
– MaoWao
Feb 4 at 13:45
$begingroup$
Do we even obtain left-differentiability of the orbits?
$endgroup$
– 0xbadf00d
Feb 9 at 16:38
|
show 3 more comments
$begingroup$
You are right that it suffices to show strong continuity at $0$ (by the semigroup property), but it is not true that it is enough to check strong continuity at $0$ for $xin D(A)$. You would need some uniform bound here. On the other hand, right-differentiability at $0$ automatically implies right-continuity at $0$: If $T_t x-x$ does not tend to zero, then there is no chance for the limit $frac 1 t(T_t x-x)$ to exist.
In your situation, strong continuity on $overline{D(A)}$ is equivalent to local boundedness on $overline{D(A)}$. One implication follows directly from the uniform boundedness principle and the semigroup property. For the other implication (the one you ask about), let $xinoverline{D(A)}$ and $(x_n)$ a sequence in $D(A)$ such that $x_nto x$. Then
$$
|T(t)x-x|leq sup_{sin[0,T]}|T(s)|_{mathcal{L}(overline{D(A)})}|x-x_n|+|T_t x_n-x_n|+|x-x_n|.
$$
Letting first $tto 0$ and then $ntoinfty$ yields the desired convergence.
answered Feb 3 at 20:35
MaoWaoMaoWao
3,953618
$endgroup$
You are right that it suffices to show strong continuity at $0$ (by the semigroup property), but it is not true that it is enough to check strong continuity at $0$ for $xin D(A)$. You would need some uniform bound here. On the other hand, right-differentiability at $0$ automatically implies right-continuity at $0$: If $T_t x-x$ does not tend to zero, then there is no chance for the limit $frac 1 t(T_t x-x)$ to exist.
In your situation, strong continuity on $overline{D(A)}$ is equivalent to local boundedness on $overline{D(A)}$. One implication follows directly from the uniform boundedness principle and the semigroup property. For the other implication (the one you ask about), let $xinoverline{D(A)}$ and $(x_n)$ a sequence in $D(A)$ such that $x_nto x$. Then
$$
|T(t)x-x|leq sup_{sin[0,T]}|T(s)|_{mathcal{L}(overline{D(A)})}|x-x_n|+|T_t x_n-x_n|+|x-x_n|.
$$
Letting first $tto 0$ and then $ntoinfty$ yields the desired convergence.
answered Feb 3 at 20:35
MaoWaoMaoWao
3,953618
answered Feb 3 at 20:35
MaoWaoMaoWao
3,953618
answered Feb 3 at 20:35
MaoWaoMaoWao
3,953618
answered Feb 3 at 20:35
MaoWaoMaoWao
3,953618
3,953618
$begingroup$
It's not important for the question, but your notation suggests that $T(s)overline{mathcal D(A)}subseteqoverline{mathcal D(A)}$ for all $sge0$. Why is that the case?
$endgroup$
– 0xbadf00d
Feb 4 at 12:16
$begingroup$
Well, $T(s)$ maps $D(A)$ into $D(A)$ (this can be proven without strong continuity), and then continuity of $T(s)$ implies that the same is true for the closure.
$endgroup$
– MaoWao
Feb 4 at 13:36
$begingroup$
From your proof I guess that's sufficient if there is a small $T>0$ such that the operator norms on $[0,T)$ are bounded, right?
$endgroup$
– 0xbadf00d
Feb 4 at 13:41
1
$begingroup$
That is right. You can also directly show that boundedness on $[0,T)$ for some $T>0$ implies boundedness on all bounded intervals: If $I$ is bounded, then there exists $ninmathbb{N}$ such that $t/n<T$ for $tin I$. Then $|T(t)|leq|T(t/n)|^n$ by the semigroup property.
$endgroup$
– MaoWao
Feb 4 at 13:45
$begingroup$
Do we even obtain left-differentiability of the orbits?
$endgroup$
– 0xbadf00d
Feb 9 at 16:38
|
show 3 more comments
$begingroup$
It's not important for the question, but your notation suggests that $T(s)overline{mathcal D(A)}subseteqoverline{mathcal D(A)}$ for all $sge0$. Why is that the case?
$endgroup$
– 0xbadf00d
Feb 4 at 12:16
$begingroup$
Well, $T(s)$ maps $D(A)$ into $D(A)$ (this can be proven without strong continuity), and then continuity of $T(s)$ implies that the same is true for the closure.
$endgroup$
– MaoWao
Feb 4 at 13:36
$begingroup$
From your proof I guess that's sufficient if there is a small $T>0$ such that the operator norms on $[0,T)$ are bounded, right?
$endgroup$
– 0xbadf00d
Feb 4 at 13:41
1
$begingroup$
That is right. You can also directly show that boundedness on $[0,T)$ for some $T>0$ implies boundedness on all bounded intervals: If $I$ is bounded, then there exists $ninmathbb{N}$ such that $t/n<T$ for $tin I$. Then $|T(t)|leq|T(t/n)|^n$ by the semigroup property.
$endgroup$
– MaoWao
Feb 4 at 13:45
$begingroup$
Do we even obtain left-differentiability of the orbits?
$endgroup$
– 0xbadf00d
Feb 9 at 16:38
$begingroup$
It's not important for the question, but your notation suggests that $T(s)overline{mathcal D(A)}subseteqoverline{mathcal D(A)}$ for all $sge0$. Why is that the case?
$endgroup$
– 0xbadf00d
Feb 4 at 12:16
$begingroup$
It's not important for the question, but your notation suggests that $T(s)overline{mathcal D(A)}subseteqoverline{mathcal D(A)}$ for all $sge0$. Why is that the case?
$endgroup$
– 0xbadf00d
Feb 4 at 12:16
$begingroup$
Well, $T(s)$ maps $D(A)$ into $D(A)$ (this can be proven without strong continuity), and then continuity of $T(s)$ implies that the same is true for the closure.
$endgroup$
– MaoWao
Feb 4 at 13:36
$begingroup$
Well, $T(s)$ maps $D(A)$ into $D(A)$ (this can be proven without strong continuity), and then continuity of $T(s)$ implies that the same is true for the closure.
$endgroup$
– MaoWao
Feb 4 at 13:36
$begingroup$
From your proof I guess that's sufficient if there is a small $T>0$ such that the operator norms on $[0,T)$ are bounded, right?
$endgroup$
– 0xbadf00d
Feb 4 at 13:41
$begingroup$
From your proof I guess that's sufficient if there is a small $T>0$ such that the operator norms on $[0,T)$ are bounded, right?
$endgroup$
– 0xbadf00d
Feb 4 at 13:41
1
1
$begingroup$
That is right. You can also directly show that boundedness on $[0,T)$ for some $T>0$ implies boundedness on all bounded intervals: If $I$ is bounded, then there exists $ninmathbb{N}$ such that $t/n<T$ for $tin I$. Then $|T(t)|leq|T(t/n)|^n$ by the semigroup property.
$endgroup$
– MaoWao
Feb 4 at 13:45
$begingroup$
That is right. You can also directly show that boundedness on $[0,T)$ for some $T>0$ implies boundedness on all bounded intervals: If $I$ is bounded, then there exists $ninmathbb{N}$ such that $t/n<T$ for $tin I$. Then $|T(t)|leq|T(t/n)|^n$ by the semigroup property.
$endgroup$
– MaoWao
Feb 4 at 13:45
$begingroup$
Do we even obtain left-differentiability of the orbits?
$endgroup$
– 0xbadf00d
Feb 9 at 16:38
$begingroup$
Do we even obtain left-differentiability of the orbits?
$endgroup$
– 0xbadf00d
Feb 9 at 16:38
|
show 3 more comments
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