# A rather abstract strongly continuous semigroup

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:

1. $\forall t\in\mathbb{R}$, $T(t)$ is a bounded linear operator on $X$;
2. $T(0)=I$;
3. $T(t+\tau)=T(t)T(\tau)$ $\forall t,\tau\in\mathbb{R}$;
4. $\forall x_{0}\in X$, $\|T(t)x_{0}-x_{0}\|_{X}\to 0$ 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?

• In order to show that $T_t$ is a bounded operator, you have to calculate/estimate $\|T_t f\|$ for $f \in X$. So, since $$(T_t f)(\xi) = f(t+\xi),$$ we have $$\|T_t f\|^2 = \int_0^{\infty} |(T_t f)(\xi)|^2 \, d\xi = \int_0^{\infty} |f(t+\xi)|^2 \, d\xi$$ Any idea how to proceed?
– saz
Apr 17 '15 at 16:08
• No, I can't figure out how to bound it. Apr 17 '15 at 20:07

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 $\lim_{t\downarrow 0}\|T(t)g-g\|=0$.
• What about showing that $T(0)=I$. So $(T(0)f)(\zeta)=f(0+\zeta)=f(\zeta)=...$? Apr 25 '15 at 15:51
• @user3482534 : The algebraic properties are not so difficult to verify; as you noted $T(0)=I$ follows immediately. And $T(t)T(\tau)=T(t+\tau)$ follows directly as well. Apr 25 '15 at 16:16
• Sorry for sounding stupid, but how does $f(\zeta)=I$? Apr 25 '15 at 16:36
• It is the operator $T$ evaluated at $0$ which is the identity $I$. In other words $T(0)f = f$. $T(t)$ acts on a function $f$ and gives a new function $g=T(t)f$ given by $g(x) = f(t+x)$. When $t=0$, the function $g$ is $g(x)=f(0+x)=f(x)$. The notation $(T(t)f)(x)$ means: apply the operator $T(t)$ to the function $f$ and then evaluation the resulting function at $x$. It's the notation that is confusing. The operator $T(t)$ is the translation operator where the graph of the function is moved to the left by $t$ units, with the part of the graph to the left of the $y$ axis being discarded. Apr 25 '15 at 19:53