Home Analysis Translation operator

Translation operator

translation-operator

We discuss the properties of the translation operator defined in a Lebesgue space. The latter appears in the study of differential equations, mainly equations with negative memories called also delay equations.

The semigroup property of the translation operator 

Let $p\ge 1$ be a real number and $E=L^p([0,+\infty)$ be the space of all measurable functions such that \begin{align*} \int^{+\infty}_0 |f(s)|^pds <\infty.\end{align*} If we select \begin{align*} \|f\|_p:=\left( \int^{+\infty}_0 |f(s)|^pds <\infty \right)^{\frac{1}{p}}\end{align*}\then $(E,\|\cdot\|)$ is a Banach space of infinite dimension, the Lebesgue space. On the other hand, let us define \begin{align*} (T_t f)(s)=f(t+s),\qquad \forall s,t\ge 0,\; f\in E.\end{align*}

Fact 1: for any $t\ge 0,$ $T_t$ is a linear continuous operator on $E,$ i.e. $T_t\in \mathscr{L}(E)$. Moreover, $\|T_t\|_{\mathscr{L}(E)}\le 1$ for any $t$.

Proof: Let $t\ge 0$ and $f\in E$. The function $s\mapsto (T_t f)(s)=f(t+s)$ is measurable because as composition of two measurables functions, $f$ and the function $s\mapsto t+s$. Moreover, we estimate \begin{align*} \int^{+\infty}_0 |(T_tf)(s)|^pds&= \int^{+\infty}_0 |f(t+s)|^pds \cr & =\int_t^{+\infty}|f(\sigma)|^pd\sigma\cr & \le \int^{+\infty}_0 |f(\sigma)|^pd\sigma=\|f\|_p^p.\end{align*} This shows that $T_f f\in E$, and \begin{align*} \|T_t f\|_p\le \|f\|_p,\qquad \forall t\ge 0,\;f\in E.\end{align*}

Definition: The operator $T_t$ is called the translation operator or the shift operator.

Fact 2: $T_0=Id_E$ and for any $t,s\ge 0,$ $T_{t+s}=T_tT_s$, here $T_tT_s:=T_t\circ T_s$.

Proof: For $f\in E$ and $t,s\ge 0,$ we have \begin{align*} (T_{t+s}f)(s)=f(s+(t+\sigma))=(T_sf)(t+\sigma)=(T_t T_sf)(\sigma).\end{align*}Definition: The familly of operators $(T_t)_{t\ge 0}$ satisying the fact 1 and the fact 2 is called semigroup of operators or the shift semigroup.

The strong continuity of the shift semigroup 

In this section, we show that the translation operator $T_t$ satisfies a nice topological property.

Fract 3: The family $(T_t)_{t\ge 0}$ is strongly continuous at zero. That is for any $f\in E,$ we have $$\lim_{t\to 0}\|T_t f-f\|_p=0.$$

Proof:  For $f\in E,$ we select $u(t):=T_tf$. According to Fact 2, it suffices to prove that $\|u(t)-u(0)\|_p\to 0$ as $t\to 0$. Before solving this question, we first need to recall some facts. First, a function $f$ has a support compact if it vanishes outside a compact. The support of $f$ is the compact set \begin{align*}{\rm supp}(f):=\overline{\{t: f(t)\neq 0\}}.\end{align*} Second, we recall that the set of continuous functions with compact support, $\mathscr{C}_c([0,+\infty))$, is dense in $L^p([0,+\infty))$.

As $f\in E,$ then by density there exists a sequence $(f_n)_n\subset \mathcal{C}_c([0,+\infty)) $ such that $\|f_n-f\|_p\to 0$ as $n\to+\infty$. For any $\varepsilon,$ there exists $m\in \mathbb{N}$ such that $\|f_m-f\|\le \frac{\varepsilon}{3}$. On the other hand, we have\begin{align*}\|u(t)-u(0)\|&=\|T_t f-f\|_p\cr & \le \|T_t f-T_t f_m\|_p + \|T_t f_m-f_m\|_p+\|f_m-f\|_p\cr & \le 2 \|f_m-f\|_p + \|T_t f_m-f_m\|_p \cr &\le \frac{2}{3}\varepsilon+ \|T_t f_m-f_m\|_p .\end{align*}

Now it suffices to show that $\|T_t f_m-f_m\|_p\to 0$ as $t\to 0$. In fact, put $K={\rm supp}(f)$. Then, \begin{align*} \|T_t f_m-f_m\|_p ^p&=\int_K |f_m(t+s)-f(s)|^p ds\cr &\le {\rm meas}(K) \left(\sup_{s\in K}|f(t+s)-f(s)|\right )^p .\end{align*} Here $ {\rm meas}(K) $ is the Lebesgue measure of the set $K$. As the fonction $f$ is uniformy continuous on the compact set, then we have the following uniforme convergence \begin{align*}\lim_{t\to 0} \sup_{s\in K}|f(t+s)-f(s)| =0.\end{align*} Then there exists $\delta>0$ such that for any $t\in (0,\delta)$, we have \begin{align*} \|T_t f_m-f_m\|_p \le \frac{\varepsilon}{3}.\end{align*}Thus, for any $\varepsilon,$ there exists $\delta>0$ such that for any $t,$ \begin{align*} t\in (0,\delta)\Longrightarrow \|u(t)-u(0)\| < \varepsilon.\end{align*}

Previous articleProperties of Real Numbers
Next articleRiemann Integral Exercises