Here is a sketch how to prove the result using the so-called (strong) generator of the semigroup $(T_t)_{t \geq 0}$: Let $$\begin{align} \mathcal{D}(A_s) &:= \left\{u \in C_{\infty}(\mathbb{R}^d); \exists f \in C_{\infty}(\mathbb{R}^d): \lim_{t \to 0} \left\| \frac{T_t u-u}{t} -f \right\|_{\infty} = 0 \right\} \\ A_s u &:= \lim_{t \to 0} \frac{T_t u-u}{t} \qquad (u \in \mathcal{D}(A_s)) \end{align}$$ (The limit is taken w.r.t to the sup-norm.) Then $(A_s,\mathcal{D}(A_s))$ is called the (strong) generator of $(T_t)_{t \geq 0}$. The idea is to show that the weak and the strong generator coincide. The following theorem is quite helpful:
Theorem Let $(A_s,\mathcal{D}(A_s))$ the generator of a Feller semigroup $(T_t)_{t \geq 0}$. Let $(A,\mathcal{D}(A))$ an extension of
$(A_s,\mathcal{D}(A_s))$ such that $$Au = u \Rightarrow u=0 \quad (u \in \mathcal{D}(A)) \tag{1}$$ Then $(A,\mathcal{D}(A))= (A_s,\mathcal{D}(A_s))$.
If we would be able to show that the weak generator $(A,\mathcal{D}(A))$ satisfies $(1)$ we would be finished: For $u \in \mathcal{D}(A) = \mathcal{D}(A_s)$ we have $$\frac{T_t u-u}{t} \to Au \quad \text{uniformly}$$ hence in particular $$\sup_{t>0} \left\| \frac{T_t u-u}{t} \right\|_{\infty} < \infty$$
So - that's it. Here is the remaining part of the proof:
Lemma Let $(A,\mathcal{D}(A))$ the weak generator of a Feller semigroup. Let $u \in \mathcal{D}(A)$, $x_0 \in \mathbb{R}^d$ such
that $u(x_0)=\sup_{x \in \mathbb{R}^d} u(x) \geq 0$. Then $$Au(x_0)
\leq 0$$ (i.e. $A$ satisfies the maximum principle). In particular,
$A$ is dissipative, i.e. $$\forall \lambda>0: \|\lambda \cdot u-Au\|_{\infty} \geq \lambda \cdot \|u\|_{\infty} \tag{2}$$
This shows that the weak generator $(A,\mathcal{D}(A))$ satisfies $(1)$ (put $\lambda=1$ in $(2)$).
Literature René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes (Chapter 7).
Remark The given hint (i.e. applying Banach Steinhaus theorem) was not correct - or at least much more difficult than the creator of this exercise was expecting. One would have to apply Banach Steinhaus to the dual space of $C_{\infty}$ (using the fact that for the dirac measures $\delta_x$ the boundedness is given by the pointwise convergence ... and some more considerations about the (vague) density of dirac measures.)
Here are some basic facts about strongly continuous semigroups that will be useful:
- Let $A$ be the infinitesimal generator of $S(\cdot)$. Then $D(A)$ (the domain of $A$) is dense and $A$ is closed.
- For any $x\in D(A)$ we have $S(t)Ax=\frac{d}{dt}S(t)x$.
- If two strongly continuous semigroups have the same infinitesimal generator, then in fact they are the same semigroup.
Fix $\lambda>0$, $x\in H$ and define $J(t):=\int_0^t S(\tau)x\,d\tau$ ($J$ depends on $x$ as well, but I will omit it for simplicity).
By hypothesis we have $\|J(t)\|\le t\|x\|$, so the integral
$$R(\lambda)x:=\lambda\int_0^\infty e^{-\lambda t}J(t)\,dt$$
makes sense. Let us check that $R(\lambda)=(\lambda I-A)^{-1}$.
$\bullet$ $(S(\epsilon)-I)\int_0^T e^{-\lambda t}J(t)\,dt=\int_0^T e^{-\lambda t}\left(\int_0^t(S(\epsilon)-I)S(\tau)x\,d\tau\right)\,dt$
$=\int_0^T e^{-\lambda t}\left(\int_\epsilon^{t+\epsilon}S(\tau)x\,d\tau-\int_0^t S(\tau)x\,d\tau\right)\,dt$
$=\int_0^T e^{-\lambda t}\left( J(t+\epsilon)-J(\epsilon)-J(t)\right)\,dt$
$=e^{\lambda\epsilon}\int_\epsilon^{T+\epsilon}e^{-\lambda t}J(t)\,dt-\int_0^T e^{-\lambda t}J(t)\,dt-\frac{1-e^{-\lambda T}}{\lambda}J(\epsilon)$
and taking the limit as $T\to\infty$ at the beginning and the end of this chain of equalities and dividing by $\epsilon$ we get
$$\frac{S(\epsilon)-I}{\epsilon}\int_0^\infty e^{-\lambda t}J(t)\,dt=\frac{e^{\lambda\epsilon}-1}{\epsilon}\int_0^\infty e^{-\lambda t}J(t)\,dt-\frac{J(\epsilon)}{\lambda\epsilon}-\frac{e^{\lambda\epsilon}}{\epsilon}\int_0^\epsilon e^{-\lambda t}J(t)\,dt$$
But the RHS possesses a limit as $\epsilon\to 0$, namely $R(\lambda)x-\frac{x}{\lambda}$
(the last term tends to $0$ since $e^{-\lambda t}J(t)=o(1)$): thus
$\frac{R(\lambda)x}{\lambda}=\int_0^\infty e^{-\lambda t}J(t)\,dt\in D(A)$ and
$$A\frac{R(\lambda)x}{\lambda}=R(\lambda)x-\frac{x}{\lambda}$$
i.e. $(\lambda I-A)R(\lambda)x=x$.
$\bullet$ Suppose now $x\in D(A)$. The second fact stated at the beginning gives
$\int_0^T e^{-\lambda t}\left(\int_0^t S(\tau)Ax\,d\tau\right)\,dt
=\int_0^T e^{-\lambda t}(S(t)x-x)\,dt$
$=e^{-\lambda T}J(T)+\lambda\int_0^T e^{-\lambda t}J(t)\,dt-\frac{1-e^{-\lambda T}}{\lambda}x$ (in the last equality we integrated by parts).
Sending $T\to\infty$ we obtain
$$R(\lambda)Ax=\lim_{T\to\infty}\int_0^T e^{-\lambda t}\left(\int_0^t S(\tau)Ax\,d\tau\right)\,dt=R(\lambda)x-\frac{x}{\lambda}$$
so $(\lambda I-A)R(\lambda)x=x$. Moreover $R(\lambda):H\to D(A)$ is a bounded operator.
This proves that $\lambda$ belongs to the resolvent set of $A$ and that $R(\lambda)=(\lambda I-A)^{-1}$.
Finally $\|R(\lambda)x\|\le \lambda\int_0^\infty e^{-\lambda t}\|J(t)\|\,dt
\le \lambda\int_0^\infty e^{-\lambda t}t\|x\|\,dt=\frac{\|x\|}{\lambda}$,
so $\|(\lambda I-A)^{-1}\|\le\frac{1}{\lambda}$.
So Hille-Yosida theorem for contraction semigroups implies that $A$ generates a contraction semigroup, which coincides with $S(\cdot)$.
Best Answer
A semigroup on a Banach space is norm continuous if and only if its generator is bounded. So every $C_0$-semigroup generated by an unbounded operator gives you an example.
To be more concrete, you can take the heat semigroup $$ P_t\colon L^p(\mathbb{R}^d)\to L^p(\mathbb{R}^d),\,P_tf(x)=(4\pi t)^{-d/2}\int e^{-\frac{|x-y|}{4t}}f(y)\,dy $$ for any $p\in [1,\infty)$.
Its generator is $\Delta$ on a suitable domain, which is certainly not bounded. Other examples include the translation semigroup $T_t f(x)=f(x+t)$ on $L^p(\mathbb{R})$, the Ornstein-Uhlenbeck semigroup $T_t f(x)=\int f(e^{-t} x+\sqrt{1-e^{-2t}}y)\,dy$ on $C_b(X)$ and many more.