1
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – saz
    Apr 17 '15 at 16:08
  • $\begingroup$ No, I can't figure out how to bound it. $\endgroup$
    – Jason Born
    Apr 17 '15 at 20:07
1
$\begingroup$

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 $\lim_{t\downarrow 0}\|T(t)g-g\|=0$.

$\endgroup$
4
  • $\begingroup$ What about showing that $T(0)=I$. So $(T(0)f)(\zeta)=f(0+\zeta)=f(\zeta)=...$? $\endgroup$
    – Jason Born
    Apr 25 '15 at 15:51
  • $\begingroup$ @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. $\endgroup$ Apr 25 '15 at 16:16
  • $\begingroup$ Sorry for sounding stupid, but how does $f(\zeta)=I$? $\endgroup$
    – Jason Born
    Apr 25 '15 at 16:36
  • $\begingroup$ 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. $\endgroup$ Apr 25 '15 at 19:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.