A way to show this for $\mathbb{R}^n$ (I'm not sure how to generalize it to a nice subset $\Omega \subset \mathbb{R}^n$) which is maybe a little bit easier is to use Corollary 4.10 of
- Klaus-Jochen Engel, Rainer Nagel: "A Short Course on Operator Semigroups",
which states that if A is the generator of a strongly continuous group, then $A^2$ generates an analytic semigroup of angle $\frac{\pi}{2}$.
The latter part means that the semigroup is defined for parameter values
$$
| \arg z | \lt \frac{\pi}{2}
$$
Since in one dimension
$$
f \mapsto f'
$$
is the generator of the translations, it follows immediatly from this corollary that
$$
f \mapsto f''
$$
generates an analytic semigroup. In several dimensions you'll have to note that the semigroups of translations $T_1, ..., T_n$ along the coordinate axes of a cartesian coordinate system commute, as do the resolvents of the generators $A_1, ..., A_n$.
The definition of the translation groups is:
$$
T_i(t) (f(x)) := f(x_1, ..., x_i + t , ..., x_n)
$$
According to the corollary, every $A^2_i$ generates an analytic semigroup $U_i (z)$, and the $U_i$ also commute. Therefore we can define
$$
U(z) := U_1(z) \cdot \cdot \cdot U_n(z)
$$
and get that it is an analytic semigroup. It's generator contains at least all functions that are twice continuously differentiable and is on this domain given by
$$
A^2 f = (A_1^2 + ... + A^2_n) f = \triangle f
$$
Strongly continuous means
$$
\forall x: \quad \lim_{t\to0}\|T(t)x-x\|_X \to 0,
$$
while continuity in operator norm means $\|T(t)-Id\|_{\mathcal L(X,X)}\to 0$, which is equivalent to
$$
\lim_{t\to0} \sup_{x: \|x\|\le1} \|T(t)x-x\|_X \to 0,
$$
in the second case the convergence has to be uniform with respect to $x$.
In your example, the first property is a consequence of uniform continuity of $x\in X$. To show that the second property is not fulfilled, it is enought to construct for each $\epsilon>0$ a function $x\in X$ such that $\|T(t)x-x\|>\epsilon$. This can be achieved using $x(s):=\max(-1,\min(ns,1))$ for sufficiently large $n$.
Best Answer
Yes, the bi-Laplacian with Dirichlet boundary conditions generates an analytic semigroup on $L^p(\Omega), \; p\in (1,\infty)$. For $p=2$, you can show that the operator is self-adjoint. A general result was proved in Theorem 5.6, pp. 189 in