Uniformly Continuous Semigroup Bounded on Compact Intervals

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 uniformly continuous semigroup.

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

Let $T > 0$ be a real number.

Then there exists $M > 0$ such that:

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

for $t \in \closedint 0 T$.

Proof

Since $\family {\map T t}_{t \ge 0}$ is a uniformly continuous semigroup, we have:

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

From Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence and Modulus of Limit: Normed Vector Space, we have:

$\norm {\map T t}_{\map B X} \to \norm I_{\map B X}$ as $t \to 0^+$

In particular:

$\norm {\map T t}_{\map B X} \to 1$

since $\norm I_{\map B X} = 1$ from Identity Mapping on Normed Vector Space is Bounded Linear Operator.

Then there exists $\delta > 0$ such that:

$\ds \norm {\map T t}_{\map B X} \le \frac 3 2$ for $t \in \closedint 0 \delta$.

For $t_0 \in \closedint 0 T$, write $t_0 = N \delta + r$ for $N \in \N$ and $r \in \hointr 0 \delta$.

Then we have:

\(\ds \norm {\map T t}_{\map B X}\)

\(=\)

\(\ds \norm {\map T {N \delta + r} }_{\map B X}\)

\(\ds \)

\(=\)

\(\ds \norm {\map T {N \delta} \map T r}_{\map B X}\)

Definition of Semigroup of Bounded Linear Operators

\(\ds \)

\(=\)

\(\ds \norm {\paren {\map T \delta}^N \map T r}_{\map B X}\)

\(\ds \)

\(\le\)

\(\ds \norm {\paren {\map T \delta}^N}_{\map B X} \norm {\map T r}_{\map B X}\)

Norm on Bounded Linear Transformation is Submultiplicative

\(\ds \)

\(\le\)

\(\ds \norm {\map T \delta}_{\map B X}^N \norm {\map T r}_{\map B X}\)

Bound on Norm of Power of Element in Normed Algebra

\(\ds \)

\(=\)

\(\ds \paren {\frac 3 2}^{N + 1}\)

since $r \in \closedint 0 \delta$

Now note that if $t_0 \in \closedint 0 T$ and $t_0 = N \delta + r$ for $N \in \Z_{\ge 0}$ and $r \in \hointr 0 \delta$, we have:

$\ds N \le \frac T \delta$

So we have:

$\ds \norm {\map T t}_{\map B X} \le \paren {\frac 3 2}^{\frac T \delta + 1}$

for $t \in \closedint 0 T$.

$\delta$ is independent of $t$, so we have the result.

$\blacksquare$