Semigroup of Bounded Linear Operators Uniformly Continuous iff Continuous as Map from Non-Negative Reals to Bounded Linear Operators

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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.

Let $\struct {\map B X, \norm {\, \cdot \,}_{\map B X} }$ be the space of bounded linear transformations equipped with the canonical norm.


Then $\family {\map T t}_{t \ge 0}$ is uniformly continuous if and only if:

the mapping $T : \hointr 0 \infty \to \map B X$ is continuous.


Proof

Necessary Condition

If $T$ is continuous, then in particular it is continuous at $0$.

Since $\map T 0 = I$, we therefore have:

$\ds \lim_{t \mathop \to 0^+} \norm {\map T t - I}_{\map B X} = 0$

So $\family {\map T t}_{t \ge 0}$ is uniformly continuous.

$\Box$


Sufficient Condition

Suppose that $\family {\map T t}_{t \mathop \ge 0}$ is uniformly continuous.

We have that:

$\ds \lim_{t \mathop \to 0^+} \norm {\map T t - I}_{\map B X} = 0$

so $T$ is continuous at $0$.

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

Then, we have:

\(\ds \norm {\map T {t + h} - \map T t}_{\map B X}\) \(=\) \(\ds \norm {\map T t \map T h - \map T t}_{\map B X}\) Definition of Semigroup of Bounded Linear Operators
\(\ds \) \(=\) \(\ds \norm {\map T t \paren {\map T h - I} }_{\map B X}\)
\(\ds \) \(\le\) \(\ds \norm {\map T t}_{\map B X} \norm {\map T h - I}_{\map B X}\) Norm on Bounded Linear Transformation is Submultiplicative
\(\ds \) \(\to\) \(\ds 0\) as $h \to 0^+$, since $\norm {\map T t}_{\map B X}$ is a fixed real number and $\norm {\map T h - I}_{\map B X} \to 0$ as $t \to 0^+$

so:

$\ds \lim_{h \to 0^+} \norm {\map T {t + h} - \map T t}_{\map B X} = 0$

Now let $-t < h < 0$.

Then $h + t \in \closedint 0 t$ for all such $h$, so there exists $M > 0$:

$\norm {\map T {t + h} }_{\map B X} \le M$

for all $-t < h < 0$, from Uniformly Continuous Semigroup Bounded on Compact Intervals.

Then we have:

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

So we have:

$\ds \lim_{h \to 0^-} \norm {\map T {t + h} - \map T t}_{\map B X} = 0$

and hence:

$\ds \lim_{h \to 0} \norm {\map T {t + h} - \map T t}_{\map B X} = 0$

for $t > 0$.

So $T : \hointr 0 \infty \to \map B X$ is continuous.

$\blacksquare$


Sources

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