Existence of a semigroup of bounded operators which is not $C_0$

Question

Let $X$ be any Banach space. Then we can define a $C_0 $ semi group of bounded operators on $X$. But my question is that can we define a semi group of bounded operators which is not $C_0$?

Answer 1

Consider the space $X = C_b[0,\infty)$ of bounded continuous functions on $[0,\infty)$ endowed with the uniform norm $$\|f\|_{\infty} := \sup_{x \geq 0} |f(x)|.$$ Define

$$T_t f(x) := f(x+t),\qquad x \geq 0, t \geq 0.$$

Obviously, $(T_t)_{t \geq 0}$ has the semigroup property (i.e. $T_t T_s = T_{t+s}$) and $\|T_t\| \leq 1$. This means that $(T_t)_{t \geq 0}$ defines a semigroup of bounded operators on $X$. On the other hand, the strong continuity

$$\|T_t f - f\|_{\infty} = \sup_{x \geq 0} |f(t+x)-f(x)| \to 0$$

holds if, and only if, $f$ is uniformly continuous. Since there exist $f \in X$ which are not uniformly continuous (e.g. $f(x) := \sin(x^2)$), we conclude that $(T_t)_{t \geq 0}$ is not $C_0$.

Answer 2

Here's a more elaborate example to give you a bounded group that is not $C_{0}$. To do this, I'll define a function $F : \mathbb{R}\rightarrow\mathbb{R}$ such that $F(x+y)=F(x)+F(y)$. Then $E(x)=e^{iF(x)}$ satisfies $E(x+y)=e^{iF(x+y)}=e^{iF(x)}e^{iF(y)}=E(x)E(y)$. Automatically $F(0)=0$ because $F(0)=F(0+0)=F(0)+F(0)$, which gives $E(0)=1$. I'll describe how to construct $F$ so that $E$ is not continuous at $0$, but this is intimately tied up with the axiom of choice. The construction of $F$ is very similar to the classical construction of a non-measurable set.

To define such a function $F$, first observe that $F(0)=0$ is necessary, as well as $F(-x)=F(x)$ because $F(0)=F(x+(-x))=F(x)+F(-x)$. If $p$, $q$ are positive integers, then $$ \begin{align} qF(\frac{p}{q}x) & =F(\frac{p}{q}x)+\cdots+F(\frac{p}{q}x) \;\;\;\mbox{ ($q$ times) }\\ & =F(\frac{p}{q}x+\cdots+\frac{p}{q}x) \;\;\;\mbox{ ($q$ times) }\\ & =F(q\frac{p}{q}x) =F(px) \\ & =F(x)+\cdots +F(x) \;\;\;\mbox{ ($p$ times) } \\ & =pF(x). \end{align} $$ So it is necessary that $$ r F(x)=F(rx),\;\;\; r \in \mathbb{Q}. $$ In other words, $F$ is necessarily linear over $\mathbb{Q}$, but $F$ does not need to be linear over $\mathbb{R}$, as shall be demonstrated.

Define an equivalence relation on $\mathbb{R}\setminus\{0\}$ by $x\sim y$ iff $x/y\in\mathbb{Q}$. It is easy to check that $\sim$ is an equivalence relation on $\mathbb{R}\setminus\{0\}$, which is verified by checking that

1. $x \sim x$;
2. If $x\sim y$, then $y\sim x$.
3. If $x\sim y$ and $y\sim z$, then $x\sim z$.

Denote the equivalence class containing $\alpha\in\mathbb{R}\setminus\{0\}$ by $[\alpha]$. Then $\mathbb{R}\setminus\{0\}=\bigcup_{\alpha\in\Lambda}[\alpha]$ is a disjoint union where $\Lambda$ is some subset of $\mathbb{R}$. $F$ is uniquely determined on $[\alpha]$ by $F(\alpha)$, which may be chosen arbitrarily. There are uncountably many elements of $\Lambda$. To create a discontinuous additive function $F : \mathbb{R}\rightarrow \mathbb{R}$: Let $F(0)=0$, choose any distinct elements $\{ \alpha_{j}\}_{j=1}^{\infty}$, and choose rationals $r_{j}$ so that $r_{j}\alpha_{j} \in [2^{-j-1},2^{-j}]$. Define $F(\alpha_{j})$ so that $F(r_{j}\alpha_{j})=r_{j}F(\alpha_{j})=1$. Then, in any interval $[0,\epsilon)$ for $\epsilon > 0$, there is a point where $F(x)=1$, which keeps $F$ from being continuous at $0$. You can define $F(\alpha)=\alpha$ for all of the other $\alpha\in\Lambda$. Then $F(x)=x$ except for a countable number of points, but you can create functions about as wild you like.

For $F$ as described, $E(x)=e^{iF(x)}$ has modulus $1$ for all $x\in\mathbb{R}$, and there is a sequence $\{ x_{j}\}$ converging to $0$ on which $E(x_{j})=e^{i} \ne 1$. Now you have a bounded group $G(t)=E(t)I$ on any complex Banach space $X$ which is not $C_{0}$.