Show that a semigroup is strongly continuous on the domain of its generator
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Let $(kappa_t)_{tge0}$ be a semigroup of linear, conservative, contractive and nonnegative operators on $C_0(E)$ with $$(kappa_tf)(x)xrightarrow{tto0}f(x)tag1;;;text{for all }xinmathbb R$$ for all $fin C_0(mathbb R)$. Let $(mathcal D(A),A)$ denote the generator of $(kappa_t)_{tge0}$. Suppose we know that $$kappa_tf-f=int_0^tkappa_s Af:{rm d}s;;;text{for all }finmathcal D(A)tag2.$$
How can we conclude that $(kappa_t)_{tge0}$ is strongly continuous on $mathcal D(A)$?
Clearly, $(2)$ again yields $(1)$, but I don't see why the convergence is uniform with respect to $x$.
functional-analysis semigroup-of-operators
asked Nov 24 at 10:18
0xbadf00d
1,71041429
add a comment |
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0
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Let $(kappa_t)_{tge0}$ be a semigroup of linear, conservative, contractive and nonnegative operators on $C_0(E)$ with $$(kappa_tf)(x)xrightarrow{tto0}f(x)tag1;;;text{for all }xinmathbb R$$ for all $fin C_0(mathbb R)$. Let $(mathcal D(A),A)$ denote the generator of $(kappa_t)_{tge0}$. Suppose we know that $$kappa_tf-f=int_0^tkappa_s Af:{rm d}s;;;text{for all }finmathcal D(A)tag2.$$
How can we conclude that $(kappa_t)_{tge0}$ is strongly continuous on $mathcal D(A)$?
Clearly, $(2)$ again yields $(1)$, but I don't see why the convergence is uniform with respect to $x$.
functional-analysis semigroup-of-operators
asked Nov 24 at 10:18
0xbadf00d
1,71041429
$C_{0}(mathbb{R})$ denotes the space of continuous functions vanishing at infinity, right?
– sharpe
Nov 24 at 10:27
@sharpe Yes, that's correct.
– 0xbadf00d
Nov 24 at 11:26
$(2)$ gives $$|kappa_t f-f|_{infty} leq t |Af|_{infty}$$ which immediately yields the desired convergence.
– saz
Nov 30 at 14:00
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(kappa_t)_{tge0}$ be a semigroup of linear, conservative, contractive and nonnegative operators on $C_0(E)$ with $$(kappa_tf)(x)xrightarrow{tto0}f(x)tag1;;;text{for all }xinmathbb R$$ for all $fin C_0(mathbb R)$. Let $(mathcal D(A),A)$ denote the generator of $(kappa_t)_{tge0}$. Suppose we know that $$kappa_tf-f=int_0^tkappa_s Af:{rm d}s;;;text{for all }finmathcal D(A)tag2.$$
How can we conclude that $(kappa_t)_{tge0}$ is strongly continuous on $mathcal D(A)$?
Clearly, $(2)$ again yields $(1)$, but I don't see why the convergence is uniform with respect to $x$.
functional-analysis semigroup-of-operators
asked Nov 24 at 10:18
0xbadf00d
1,71041429
Let $(kappa_t)_{tge0}$ be a semigroup of linear, conservative, contractive and nonnegative operators on $C_0(E)$ with $$(kappa_tf)(x)xrightarrow{tto0}f(x)tag1;;;text{for all }xinmathbb R$$ for all $fin C_0(mathbb R)$. Let $(mathcal D(A),A)$ denote the generator of $(kappa_t)_{tge0}$. Suppose we know that $$kappa_tf-f=int_0^tkappa_s Af:{rm d}s;;;text{for all }finmathcal D(A)tag2.$$
How can we conclude that $(kappa_t)_{tge0}$ is strongly continuous on $mathcal D(A)$?
Clearly, $(2)$ again yields $(1)$, but I don't see why the convergence is uniform with respect to $x$.
functional-analysis semigroup-of-operators
functional-analysis semigroup-of-operators
asked Nov 24 at 10:18
0xbadf00d
1,71041429
asked Nov 24 at 10:18
0xbadf00d
1,71041429
asked Nov 24 at 10:18
0xbadf00d
1,71041429
asked Nov 24 at 10:18
0xbadf00d
1,71041429
asked Nov 24 at 10:18
0xbadf00d
1,71041429
1,71041429
$C_{0}(mathbb{R})$ denotes the space of continuous functions vanishing at infinity, right?
– sharpe
Nov 24 at 10:27
@sharpe Yes, that's correct.
– 0xbadf00d
Nov 24 at 11:26
$(2)$ gives $$|kappa_t f-f|_{infty} leq t |Af|_{infty}$$ which immediately yields the desired convergence.
– saz
Nov 30 at 14:00
add a comment |
$C_{0}(mathbb{R})$ denotes the space of continuous functions vanishing at infinity, right?
– sharpe
Nov 24 at 10:27
@sharpe Yes, that's correct.
– 0xbadf00d
Nov 24 at 11:26
$(2)$ gives $$|kappa_t f-f|_{infty} leq t |Af|_{infty}$$ which immediately yields the desired convergence.
– saz
Nov 30 at 14:00
$C_{0}(mathbb{R})$ denotes the space of continuous functions vanishing at infinity, right?
– sharpe
Nov 24 at 10:27
$C_{0}(mathbb{R})$ denotes the space of continuous functions vanishing at infinity, right?
– sharpe
Nov 24 at 10:27
@sharpe Yes, that's correct.
– 0xbadf00d
Nov 24 at 11:26
@sharpe Yes, that's correct.
– 0xbadf00d
Nov 24 at 11:26
$(2)$ gives $$|kappa_t f-f|_{infty} leq t |Af|_{infty}$$ which immediately yields the desired convergence.
– saz
Nov 30 at 14:00
$(2)$ gives $$|kappa_t f-f|_{infty} leq t |Af|_{infty}$$ which immediately yields the desired convergence.
– saz
Nov 30 at 14:00
add a comment |
1 Answer
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Let $E$ be a locally compact separable metric space and $m$ a nonnegative Radon measure on $E$.
We denote $C_{infty}(E)$ by the space of continuous functions on $E$ vanishing at infinity. $C_{infty}(E)$ becomes a Banach space with respect to the usual sup-norm $|f|_{infty}=sup_{x in E}|f(x)|$.
Let ${kappa_t}_{t>0}$ be a semigroup of linear, contractive operators on $C_{infty}(E)$. Contractive means that $|kappa_t| le 1$ for any $t>0$. Here we define $$|kappa_t|=sup_{f in C_{infty}(E), |f|_{infty}=1}|kappa_t f|.$$
Assume that ${kappa_t}_{t>0}$ satisfies the next condition: for any $xin E$ and any $ f in C_{infty}(E)$,
$${bf (C)} quad lim_{t to 0}kappa_tf(x)=f(x). $$
Claim: ${bf (C)}$ implies that ${kappa_t}_{t>0}$ is strongly continuous on $C_{infty}(E)$. Namely, it holds that for any $f in C_{infty}(E)$,
$$lim_{t to 0}|kappa_tf-f|_{infty}=0.$$
In particular, ${kappa_t}_{t>0}$ is strongly continuous on $D(A)$ because $D(A)$ is a linear subspace of $C_{infty}(E)$.
Sketch of Proof
For $alpha>0$, we define a operator $R_alpha$ on $C_{infty}(E)$ by
$$R_{alpha}f(x)=int_{0}^{infty}e^{-alpha t}kappa_tf(x),dt,quad f in C_{infty}(E).$$
(Here, we used the contractibity of ${kappa_t}$.)
It is easy to see that:
$R_{alpha}-R_{beta}=(beta-alpha)R_{alpha}R_{beta}$ on $C_{infty}(E)$.
Thus, $mathcal{X}:=R_{alpha}(C_{infty}(E))$ does not depend on $alpha>0$.
$lim_{alpha to infty}alpha R_{alpha}f(x)=f(x)$ for any $x in E$, and any $f in C_{infty}(E)$. Thus, $mathcal{X}$ is dense in $C_{infty}(E)$.
Straightforward computation yields that
$$|kappa_t R_{alpha}f-R_{alpha}f|_{infty} le (e^{alpha t}-1)|R_{alpha}f|_{infty}+t|f|$$
for any $t,alpha>0$, and any $f in C_{infty}$. This implies that
$$lim_{t to 0}|kappa_t R_{alpha}f-R_{alpha}f|_{infty} =0$$
for any $alpha>0$, and any $f in C_{infty}$.
Since $mathcal{X}$ is dense in $C_{infty}(E)$, we conclude that ${kappa_t}_{t>0}$ is strongly continuous on $C_{infty}(E)$.
answered Nov 24 at 11:37
sharpe
71312
add a comment |
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1 Answer
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Let $E$ be a locally compact separable metric space and $m$ a nonnegative Radon measure on $E$.
We denote $C_{infty}(E)$ by the space of continuous functions on $E$ vanishing at infinity. $C_{infty}(E)$ becomes a Banach space with respect to the usual sup-norm $|f|_{infty}=sup_{x in E}|f(x)|$.
Let ${kappa_t}_{t>0}$ be a semigroup of linear, contractive operators on $C_{infty}(E)$. Contractive means that $|kappa_t| le 1$ for any $t>0$. Here we define $$|kappa_t|=sup_{f in C_{infty}(E), |f|_{infty}=1}|kappa_t f|.$$
Assume that ${kappa_t}_{t>0}$ satisfies the next condition: for any $xin E$ and any $ f in C_{infty}(E)$,
$${bf (C)} quad lim_{t to 0}kappa_tf(x)=f(x). $$
Claim: ${bf (C)}$ implies that ${kappa_t}_{t>0}$ is strongly continuous on $C_{infty}(E)$. Namely, it holds that for any $f in C_{infty}(E)$,
$$lim_{t to 0}|kappa_tf-f|_{infty}=0.$$
In particular, ${kappa_t}_{t>0}$ is strongly continuous on $D(A)$ because $D(A)$ is a linear subspace of $C_{infty}(E)$.
Sketch of Proof
For $alpha>0$, we define a operator $R_alpha$ on $C_{infty}(E)$ by
$$R_{alpha}f(x)=int_{0}^{infty}e^{-alpha t}kappa_tf(x),dt,quad f in C_{infty}(E).$$
(Here, we used the contractibity of ${kappa_t}$.)
It is easy to see that:
$R_{alpha}-R_{beta}=(beta-alpha)R_{alpha}R_{beta}$ on $C_{infty}(E)$.
Thus, $mathcal{X}:=R_{alpha}(C_{infty}(E))$ does not depend on $alpha>0$.
$lim_{alpha to infty}alpha R_{alpha}f(x)=f(x)$ for any $x in E$, and any $f in C_{infty}(E)$. Thus, $mathcal{X}$ is dense in $C_{infty}(E)$.
Straightforward computation yields that
$$|kappa_t R_{alpha}f-R_{alpha}f|_{infty} le (e^{alpha t}-1)|R_{alpha}f|_{infty}+t|f|$$
for any $t,alpha>0$, and any $f in C_{infty}$. This implies that
$$lim_{t to 0}|kappa_t R_{alpha}f-R_{alpha}f|_{infty} =0$$
for any $alpha>0$, and any $f in C_{infty}$.
Since $mathcal{X}$ is dense in $C_{infty}(E)$, we conclude that ${kappa_t}_{t>0}$ is strongly continuous on $C_{infty}(E)$.
answered Nov 24 at 11:37
sharpe
71312
add a comment |
up vote
0
down vote
Let $E$ be a locally compact separable metric space and $m$ a nonnegative Radon measure on $E$.
We denote $C_{infty}(E)$ by the space of continuous functions on $E$ vanishing at infinity. $C_{infty}(E)$ becomes a Banach space with respect to the usual sup-norm $|f|_{infty}=sup_{x in E}|f(x)|$.
Let ${kappa_t}_{t>0}$ be a semigroup of linear, contractive operators on $C_{infty}(E)$. Contractive means that $|kappa_t| le 1$ for any $t>0$. Here we define $$|kappa_t|=sup_{f in C_{infty}(E), |f|_{infty}=1}|kappa_t f|.$$
Assume that ${kappa_t}_{t>0}$ satisfies the next condition: for any $xin E$ and any $ f in C_{infty}(E)$,
$${bf (C)} quad lim_{t to 0}kappa_tf(x)=f(x). $$
Claim: ${bf (C)}$ implies that ${kappa_t}_{t>0}$ is strongly continuous on $C_{infty}(E)$. Namely, it holds that for any $f in C_{infty}(E)$,
$$lim_{t to 0}|kappa_tf-f|_{infty}=0.$$
In particular, ${kappa_t}_{t>0}$ is strongly continuous on $D(A)$ because $D(A)$ is a linear subspace of $C_{infty}(E)$.
Sketch of Proof
For $alpha>0$, we define a operator $R_alpha$ on $C_{infty}(E)$ by
$$R_{alpha}f(x)=int_{0}^{infty}e^{-alpha t}kappa_tf(x),dt,quad f in C_{infty}(E).$$
(Here, we used the contractibity of ${kappa_t}$.)
It is easy to see that:
$R_{alpha}-R_{beta}=(beta-alpha)R_{alpha}R_{beta}$ on $C_{infty}(E)$.
Thus, $mathcal{X}:=R_{alpha}(C_{infty}(E))$ does not depend on $alpha>0$.
$lim_{alpha to infty}alpha R_{alpha}f(x)=f(x)$ for any $x in E$, and any $f in C_{infty}(E)$. Thus, $mathcal{X}$ is dense in $C_{infty}(E)$.
Straightforward computation yields that
$$|kappa_t R_{alpha}f-R_{alpha}f|_{infty} le (e^{alpha t}-1)|R_{alpha}f|_{infty}+t|f|$$
for any $t,alpha>0$, and any $f in C_{infty}$. This implies that
$$lim_{t to 0}|kappa_t R_{alpha}f-R_{alpha}f|_{infty} =0$$
for any $alpha>0$, and any $f in C_{infty}$.
Since $mathcal{X}$ is dense in $C_{infty}(E)$, we conclude that ${kappa_t}_{t>0}$ is strongly continuous on $C_{infty}(E)$.
answered Nov 24 at 11:37
sharpe
71312
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $E$ be a locally compact separable metric space and $m$ a nonnegative Radon measure on $E$.
We denote $C_{infty}(E)$ by the space of continuous functions on $E$ vanishing at infinity. $C_{infty}(E)$ becomes a Banach space with respect to the usual sup-norm $|f|_{infty}=sup_{x in E}|f(x)|$.
Let ${kappa_t}_{t>0}$ be a semigroup of linear, contractive operators on $C_{infty}(E)$. Contractive means that $|kappa_t| le 1$ for any $t>0$. Here we define $$|kappa_t|=sup_{f in C_{infty}(E), |f|_{infty}=1}|kappa_t f|.$$
Assume that ${kappa_t}_{t>0}$ satisfies the next condition: for any $xin E$ and any $ f in C_{infty}(E)$,
$${bf (C)} quad lim_{t to 0}kappa_tf(x)=f(x). $$
Claim: ${bf (C)}$ implies that ${kappa_t}_{t>0}$ is strongly continuous on $C_{infty}(E)$. Namely, it holds that for any $f in C_{infty}(E)$,
$$lim_{t to 0}|kappa_tf-f|_{infty}=0.$$
In particular, ${kappa_t}_{t>0}$ is strongly continuous on $D(A)$ because $D(A)$ is a linear subspace of $C_{infty}(E)$.
Sketch of Proof
For $alpha>0$, we define a operator $R_alpha$ on $C_{infty}(E)$ by
$$R_{alpha}f(x)=int_{0}^{infty}e^{-alpha t}kappa_tf(x),dt,quad f in C_{infty}(E).$$
(Here, we used the contractibity of ${kappa_t}$.)
It is easy to see that:
$R_{alpha}-R_{beta}=(beta-alpha)R_{alpha}R_{beta}$ on $C_{infty}(E)$.
Thus, $mathcal{X}:=R_{alpha}(C_{infty}(E))$ does not depend on $alpha>0$.
$lim_{alpha to infty}alpha R_{alpha}f(x)=f(x)$ for any $x in E$, and any $f in C_{infty}(E)$. Thus, $mathcal{X}$ is dense in $C_{infty}(E)$.
Straightforward computation yields that
$$|kappa_t R_{alpha}f-R_{alpha}f|_{infty} le (e^{alpha t}-1)|R_{alpha}f|_{infty}+t|f|$$
for any $t,alpha>0$, and any $f in C_{infty}$. This implies that
$$lim_{t to 0}|kappa_t R_{alpha}f-R_{alpha}f|_{infty} =0$$
for any $alpha>0$, and any $f in C_{infty}$.
Since $mathcal{X}$ is dense in $C_{infty}(E)$, we conclude that ${kappa_t}_{t>0}$ is strongly continuous on $C_{infty}(E)$.
answered Nov 24 at 11:37
sharpe
71312
Let $E$ be a locally compact separable metric space and $m$ a nonnegative Radon measure on $E$.
We denote $C_{infty}(E)$ by the space of continuous functions on $E$ vanishing at infinity. $C_{infty}(E)$ becomes a Banach space with respect to the usual sup-norm $|f|_{infty}=sup_{x in E}|f(x)|$.
Let ${kappa_t}_{t>0}$ be a semigroup of linear, contractive operators on $C_{infty}(E)$. Contractive means that $|kappa_t| le 1$ for any $t>0$. Here we define $$|kappa_t|=sup_{f in C_{infty}(E), |f|_{infty}=1}|kappa_t f|.$$
Assume that ${kappa_t}_{t>0}$ satisfies the next condition: for any $xin E$ and any $ f in C_{infty}(E)$,
$${bf (C)} quad lim_{t to 0}kappa_tf(x)=f(x). $$
Claim: ${bf (C)}$ implies that ${kappa_t}_{t>0}$ is strongly continuous on $C_{infty}(E)$. Namely, it holds that for any $f in C_{infty}(E)$,
$$lim_{t to 0}|kappa_tf-f|_{infty}=0.$$
In particular, ${kappa_t}_{t>0}$ is strongly continuous on $D(A)$ because $D(A)$ is a linear subspace of $C_{infty}(E)$.
Sketch of Proof
For $alpha>0$, we define a operator $R_alpha$ on $C_{infty}(E)$ by
$$R_{alpha}f(x)=int_{0}^{infty}e^{-alpha t}kappa_tf(x),dt,quad f in C_{infty}(E).$$
(Here, we used the contractibity of ${kappa_t}$.)
It is easy to see that:
$R_{alpha}-R_{beta}=(beta-alpha)R_{alpha}R_{beta}$ on $C_{infty}(E)$.
Thus, $mathcal{X}:=R_{alpha}(C_{infty}(E))$ does not depend on $alpha>0$.
$lim_{alpha to infty}alpha R_{alpha}f(x)=f(x)$ for any $x in E$, and any $f in C_{infty}(E)$. Thus, $mathcal{X}$ is dense in $C_{infty}(E)$.
Straightforward computation yields that
$$|kappa_t R_{alpha}f-R_{alpha}f|_{infty} le (e^{alpha t}-1)|R_{alpha}f|_{infty}+t|f|$$
for any $t,alpha>0$, and any $f in C_{infty}$. This implies that
$$lim_{t to 0}|kappa_t R_{alpha}f-R_{alpha}f|_{infty} =0$$
for any $alpha>0$, and any $f in C_{infty}$.
Since $mathcal{X}$ is dense in $C_{infty}(E)$, we conclude that ${kappa_t}_{t>0}$ is strongly continuous on $C_{infty}(E)$.
answered Nov 24 at 11:37
sharpe
71312
edited Nov 24 at 12:28
answered Nov 24 at 11:37
sharpe
71312
answered Nov 24 at 11:37
sharpe
71312
answered Nov 24 at 11:37
sharpe
71312
71312
add a comment |
add a comment |
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$C_{0}(mathbb{R})$ denotes the space of continuous functions vanishing at infinity, right?
– sharpe
Nov 24 at 10:27
@sharpe Yes, that's correct.
– 0xbadf00d
Nov 24 at 11:26
$(2)$ gives $$|kappa_t f-f|_{infty} leq t |Af|_{infty}$$ which immediately yields the desired convergence.
– saz
Nov 30 at 14:00