Define $X$ as the Hilbert space $L^{2}(0,\infty)$ and let the operators $T(t):X\to X$, $t\ge 0$ be defined by
$(T(t)f)(\zeta):=f(t+\zeta)$
I want to show that $(T(t))_{t\ge 0}$ is a $C_{0}$-semigroup on $X$.
We know that $(T(t))_{t\in\mathbb{R}}$ is a $C_0$-semigroup if the following hold:
- $\forall t\in\mathbb{R}$, $T(t)$ is a bounded linear operator on $X$;
- $T(0)=I$;
- $T(t+\tau)=T(t)T(\tau)$ $\forall t,\tau\in\mathbb{R}$;
- $\forall x_{0}\in X$, $\|T(t)x_{0}-x_{0}\|_{X}\to 0$ when $t\to 0$.
To solve part the first part of the proof I assume that I have to use the properties of the Hilbert space $L^{2}(0,\infty)$. But I do not know. And I also find the definition of $T(t)$ to be a little abstract to work with in terms of plugging in the values to solve the proceeding parts. How should I start?
Start by showing $\|T(t)f\| \le \|f\|$ for all $f \in L^{2}[0,\infty)$ and $t \ge 0$: \begin{align} \|T(t)f\|^{2} & =\int_{0}^{\infty}|f(x+t)|^{2}dx \\ & = \int_{t}^{\infty}|f(x)|^{2}dx \\ & \le \int_{0}^{\infty}|f(x)|^{2}dx = \|f\|^{2}. \end{align} Because of this norm estimate, the problem of showing $\lim_{t\downarrow 0}T(t)f=f$ is reduced to showing this identity on a dense subspace $\mathcal{M}$ of $L^{2}[0,\infty)$. This is because $$ \begin{align} \|T(t)f-f\| & \le \|T(t)f-T(t)g\|+\|T(t)g-g\|+\|g-f\| \\ & \le 2\|f-g\|+\|T(t)g-g\|. \end{align} $$ One dense subspace of $L^{2}[0,\infty)$ which is particularly easy to deal with for this problem is the set of all continuous functions $g$ on $[0,\infty)$ that vanish outside some interval $[0,R]$. For any such $g$, you can use uniform continuity to get $\lim_{t\downarrow 0}\|T(t)g-g\|=0$.