# Proving strong stability of semigroup

$X$ is the Hilbert space $L^{2}(0,\infty)$ and let $T(t):X\to X$ with $t\ge 0$ be defined by $(T(t)f)(\zeta):=f(t+\zeta)$.

I want to prove that the $C_{0}$-semigroup $(T(t))_{t\ge 0}$ is strongly stable, but not exponentially stable.

We have that $(T(t))_{t\ge 0}$ is strongly stable if $T(t)x\to 0$ as $t\to \infty$, $\forall x\in X$.

By Datko's lemma, if $(T(t))_{t\ge0}$ is exponentially stable, then $\displaystyle \int_{0}^{\infty}\|T(t)x\|^{2}dt<\infty$, $\forall x\in X$.

Suppose $(T(t))_{t\ge 0}$ is exponentially stable, then $\displaystyle \int_{0}^{\infty}\|T(t)x\|^{2}dt=\int_{0}^{\infty}$...

Am I going about this the right way, or could someone point me in the right direction?

By definition,

$$\|T_t f\|^2 = \int_0^{\infty} |(T_t f)(\xi)|^2 \, d\xi = \int_0^{\infty} |f(t+\xi)|^2 \, d\xi = \int_t^{\infty} |f(\eta)|^2 \, d\eta. \tag{1}$$

For $f(x) := \min \left\{\frac{1}{x},1 \right\}$, we have $f \in X$ and, by $(1)$,

$$\|T_t f\|^2 = \frac{1}{t} \qquad \text{for all t \geq 1.}$$

Thus,

$$\int_0^{\infty} \|T_t f\|^2 \, dt \geq \int_1^{\infty} \frac{1}{t} = \infty.$$

This shows that $(T_t)_{t \geq 0}$ cannot be exponentially stable.