Semigroup of Bounded Linear Operators is C0 iff Point Evaluations Continuous

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

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.

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:

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:

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 $x \in X$.