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Consider the operators $Q(t)\in \mathcal{B}(L^{2}(\mathbb{R}))$ given by $(Q(t)f)(s)=f(s+t)\quad \forall f\in L^{2}(\mathbb{R})$. Each $Q(t)$ is clearly unitary and the following are satisfied:

  1. $Q(0)=$Identity on $L^{2}(\mathbb{R})$.

  2. $Q(s+t)=Q(s)Q(t)$

We now must show that $\lim_{t\to0}\|Q(t)f-f\|=0\quad\forall f\in L^{2}(\mathbb{R})$.

I'd be grateful for some hints on how to prove this, as always!

Arundhathi
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1 Answers1

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You have to make use of this lemma :

Let $T(\cdot) : R_+ → B(X)$ semigroup of operators. Then the following assertions are equivalent.

(a) $T(\cdot)$ is strongly continuous (and thus a $C_0$–semigroup).

(b) It holds $\lim_{t \rightarrow 0_+} T(t)x = x$ for all $x \in X$.

(c) There are a number $t_0 > 0$ and a dense subspace $D \subseteq X$ such that $\sup_{0\leq t \leq t_0} \|T(t)\| < \infty$ and $\lim_{t \rightarrow 0_+} T(t)x = x$ for all $x \in D$.

Take $D = C_c^{\infty}(\mathbb{R})$ i.e. compactly supported smooth functions on $\mathbb{R}$, and $X = L^2(\mathbb{R})$, and the semigroup is translation semigroup in question.

HINT : Utilise (c).

DiffeoR
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