Bound on C0 Semigroup

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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 $C_0$ semigroup.

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


Then 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 $t \in \hointr 0 \infty$.


Proof

We first show that $\norm {\map T t}_{\map B X}$ is bounded on $\closedint 0 \delta$ for some $\delta > 0$.

Aiming for a contradiction, suppose $\norm {\map T t}_{\map B X}$ is unbounded on $\closedint 0 \delta$ for each $\delta > 0$.

Then for each $n \in \N$ we can pick $t_n \in \closedint 0 {\frac 1 n}$ such that $\norm {\map T {t_n} }_{\map B X} \ge n$.

Then, we have:

$\ds \sup_{n \mathop \in \N} \norm {\map T {t_n} }_{\map B X} = \infty$

By definition, each $\map T {t_n}$ is a linear operator.

So we can apply Principle of Condensation of Singularities to obtain:

$\ds \sup_{n \mathop \in \N} \norm {\map T {t_n} x} = \infty$

for some $x \in X$.

But since:

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

and:

$\ds \lim_{n \mathop \to \infty} t_n = 0$ by the Squeeze Theorem

we must have that:

$\ds \sup_{n \mathop \in \N} \norm {\map T {t_n} x} < \infty$

by Convergent Sequence in Normed Vector Space is Bounded.

So we have a contradiction, and there exists $\delta > 0$ such that $\norm {\map T t}_{\map B X}$ is bounded on $\closedint 0 \delta$.

Suppose that:

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

for $t \in \closedint 0 \delta$.

Since $\map T 0 = I$, we have $M \ge 1$.

Now for general $t \ge 0$, write:

$t = N \delta + r$ for $N \in \Z_{\ge 0}$ 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
\(\ds \) \(=\) \(\ds M^{N + 1}\) since $r \in \closedint 0 \delta$

Since $t = N \delta + r$, we have:

$\ds \frac t \delta = N + \frac r \delta \ge N$

so we have:

\(\ds M^{N + 1}\) \(=\) \(\ds M M^N\)
\(\ds \) \(\le\) \(\ds M M^{\frac t \delta}\)
\(\ds \) \(=\) \(\ds M e^{\frac t \delta \log M}\)

Taking:

$\ds \omega = \frac {\log M} \delta$

we have:

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

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

$\blacksquare$


Sources

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