Semigroup of Bounded Linear Operators is C0 iff Point Evaluations Continuous

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Theorem

Let $\GF \in \set {\R, \C}$.

Let $X$ be a Banach space over $\GF$.

Let $\family {\map T t}_{t \ge 0}$ be a semigroup of bounded linear operators.

For each $x \in X$, define $x^\wedge : \hointr 0 \infty \to X$ by:

$\map {x^\wedge} t = \map T t x$

for each $t \in \hointr 0 \infty$.


Then $\family {\map T t}_{t \ge 0}$ is a $C_0$ semigroup if and only if $x^\wedge$ is continuous for each $x \in X$.


Proof

Sufficient Condition

Suppose that $x^\wedge$ is continuous for each $x \in X$.

In particular, for each $x \in X$ we have that $x^\wedge$ is continuous at $0$.

That is:

$\ds \lim_{t \mathop \to 0^+} \map T t x = \map T 0 x = x$ for each $x \in X$.

So $\family {\map T t}_{t \ge 0}$ is a $C_0$ semigroup.

$\Box$

Necessary Condition

Suppose that $\family {\map T t}_{t \ge 0}$ is a $C_0$ semigroup.

By Bound on C0 Semigroup, there exists $M \ge 1$ and $\omega \ge 0$ such that:

$\norm {\map T t}_{\map B X} \le M e^{\omega t}$

for each $t \in \hointr 0 \infty$.

Let $x \in X$.

We show that $x^\wedge$ is continuous.

We have that:

$\ds \lim_{t \mathop \to 0^+} \map T t x = x = \map T 0 x$

so $x^\wedge$ is continuous at $0$.

Let $t > 0$ and let $h > 0$.

Then:

\(\ds \norm {\map {x^\wedge} {t + h} - \map {x^\wedge} t}\) \(=\) \(\ds \norm {\map T {t + h} x - \map T t x}\)
\(\ds \) \(=\) \(\ds \norm {\map T t \paren {\map T h x - x} }\) Definition of Semigroup of Bounded Linear Operators
\(\ds \) \(\le\) \(\ds \norm {\map T t}_{\map B X} \norm {\map T h x - x}\) Norm on Bounded Linear Transformation is Submultiplicative
\(\ds \) \(\le\) \(\ds M e^{\omega t} \norm {\map T h x - x}\)
\(\ds \) \(\to\) \(\ds 0\) as $h \to 0^+$, Definition of C0 Semigroup

That is:

$\ds \lim_{h \mathop \to 0^+} \norm {\map {x^\wedge} {t + h} - \map {x^\wedge} t} = 0$

Now let $0 \ge h > -t$.

We have:

\(\ds \norm {\map {x^\wedge} {t + h} - \map {x^\wedge} t}\) \(=\) \(\ds \norm {\map T {t + h} x - \map T t x}\)
\(\ds \) \(=\) \(\ds \norm {\map T {t + h} \paren {x - \map T {-h} x} }\) Definition of Semigroup of Bounded Linear Operators
\(\ds \) \(\le\) \(\ds \norm {\map T {t + h} }_{\map B X} \norm {\map T {-h} x - x}\) Norm on Bounded Linear Transformation is Submultiplicative
\(\ds \) \(\le\) \(\ds M e^{\omega \paren {t + h} } \norm {\map T {-h} x - x}\)

We have:

$\ds \lim_{h \mathop \to 0^-} M e^{\omega \paren {t + h} } = M e^{\omega t}$

Also:

$\ds \lim_{h \mathop \to 0^-} \norm {\map T {-h} x - x} = 0$

from the definition of a $C_0$ semigroup.

From Combination Theorem for Limits of Functions: Product Rule and Squeeze Theorem, we have:

$\ds \lim_{h \mathop \to 0^-} \norm {\map {x^\wedge} {t + h} - \map {x^\wedge} t} = 0$

So:

$\ds \lim_{h \mathop \to 0} \norm {\map {x^\wedge} {t + h} - \map {x^\wedge} t} = 0$

So $x^\wedge$ is continuous for each $x \in X$.

$\blacksquare$


Sources

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