If $T$ is a semigroup of bounded linear operators on a Banach space, can we conclude that each $T(t)$ has non-zero operator norm?

Question

Let $E$ be a Banach space and $T:[0,\infty)\to\mathfrak L(E)$ be a semigroup, i.e.

1. $T(0)=\operatorname{id}_E$
2. $T(s+t)=T(s)T(t)$ for all $s,t\ge 0$

It's clear that $$\left\|T(0)\right\|_{\mathfrak L(E)}=1\;.\tag 1$$ Moreover, if there would be some $t_0\ge 0$ with $$\left\|T(t_0)\right\|_{\mathfrak L(E)}=0\;,\tag 2$$ then $T(t_0)=0$ and hence $$T(t)=0\;\;\;\text{for all }t\ge t_0\tag 3$$ by property (2). Can we exploit property (2) and the submultiplicativity of $\left\|\;\cdot\;\right\|_{\mathfrak L(E)}$ to show that $$\left\|T(t)\right\|_{\mathfrak L(E)}\ne 0\tag 4$$ for all $t\ge 0$?

Answer 1

If $T$ is continuous with respect to the operator norm, then by the openness of the set of invertible operators and continuity $T(t)$ is invertible for all sufficiently small $t$, and by the semigroup property it follows that $T(t)$ is invertible for all $t \geqslant 0$.

If we only require $T$ to be continuous with respect to the strong operator topology, then it is possible that there is a $t_0$ such that $T(t) = 0$ for all $t > t_0$. For example, let $E$ be a suitable Banach space of [equivalence classes of] functions $[0,+\infty) \to \mathbb{C}$ vanishing identically [almost everywhere] on $[1,+\infty)$, and $T(t)f \colon x \mapsto f(x+t)$. Suitable such spaces include

• $\{f \in C([0,+\infty)) : x \geqslant 1 \implies f(x) = 0\}$ with the maximum norm, or
• $\{f \in L^p([0,+\infty),\lambda) : f = 0 \text{ a.e. on } [1,+\infty)\}$, where $\lambda$ is the Lebesgue measure, for $1 \leqslant p < +\infty$ with the corresponding $L^p$ norm.

Then $T$ is a semigroup of continuous operators ($\lVert T(t)\rVert \leqslant 1$ for all $t$) with $T(t) = 0$ for $t \geqslant 1$, and $T$ is continuous with respect to the strong operator topology.

For the case of continuous functions, that follows immediately from the uniform continuity of the functions in $E$, for the $L^p$ subspaces, it's a consequence of the continuity of the map $t \mapsto f(\,\cdot\, + t)$ from $\mathbb{R}$ to $L^p(\mathbb{R},\lambda)$ for every fixed $f \in L^p$.

Answer 2

Given the Hilbert spaces: $$\mathbb{I}:=[0,1]:\quad\mathcal{H}:=\ell^2(\mathbb{R})\quad\mathcal{H}_0:=\ell^2(\mathbb{I})$$

Regard the translation: $$\tau_t:\mathcal{H}\to\mathcal{H}:\quad(\tau_tx)(t'):=x(t+t')$$

Consider the semigroup: $$T:\mathbb{R}\to\mathcal{B}(\mathcal{H}):\quad T:=\tau$$

Check strong continuity: $$\ldots$$ Regard the embedding: $$\iota:\mathcal{H}_0\hookrightarrow\mathcal{H}:\quad(\iota x)(t'):=x(t')$$ $$\pi:\mathcal{H}\twoheadrightarrow\mathcal{H}_0:\quad(\pi x)(t'):=x(t')$$

Construct the semigroup: $$T_0:\mathbb{R}\to\mathcal{B}(\ell^2\mathbb{I}):\quad T_0:=\pi T\iota$$

Check the nilpotence: $$t\geq1:\quad T_0(t)=\pi T(t)\iota=\pi\tau(t)\iota=0$$ For details see: Engel & Nagel