Is a contraction semigroup infinitesimal operator bounded?

Let $T_t:L\to L$ be a semigroup of linear operators $T_t$ acting on a Banach space $L$. Assume that $$\|T_t\| := \sup\limits_{f\in L}\frac{\|T_tf\|_L}{\|f\|_L} \leq 1$$ for all $t\geq 0$. The infinitesimal operator is given by $$\mathcal Af = \lim\limits_{h\to 0}\frac{T_hf-f}{h}$$ for any $f\in\mathcal D_{\mathcal A}$, i.e. for each $f$ such that this limit exists. $\mathcal A$ is clearly linear, and my questions are

1. if it is necessary bounded on $\mathcal D_{\mathcal A}$?

2. If it's not, could you please give an example when $\mathcal A$ is unbounded?

3. If there is an example when $\|T_t\|'\leq 1$ and $\|T_t\|''\leq 1$ for two different norms $\|\cdot\|'_L$ and $\|\cdot\|''_L$ but $\|\mathcal A\|'<\infty$ while $\|\mathcal A\|'' =\infty$?

Edited: in the view of the answers, it's still unclear with a part 3. If given one norm on $L$ the linear operator $\mathcal A$ is bounded is it necessary that it will be bounded in another norm on $L$? E.g. in which norm Laplacian $\Delta$ is unbounded and why it is impossbile to give a norm on $C^2$ which will make it bounded?

• You should reformulate question 3 or at least give some motivation to it. As it stands, the answer might be yes, and this second norm will be a very artificial one that is constructed using the axiom of choice. But I don't think that that's what you actually want to know. When applying the theory one wants to work in concrete spaces that come with a "natural" norm, and usually there is only one choice that makes sense. Aug 5 '11 at 22:34
• @Florian: thanks a lot. That's a good point - indeed, the Banach space is already given a norm. Anyway, from your answer I got that sometimes there may be possible to construct such a norm (I don't care much about how artificial is it and the axiom of choice).
– Ilya
Aug 9 '11 at 15:08

In fact, the generator $A$ is bounded if and only if the semigroup is norm continuous, that is, $\|T(t)-1\|\to 0$ as $t\to 0$, and most semigroups don't have this property.

In analysis of PDEs, semigroups generated by partial differential operators are of interest, and these are not bounded (consider the heat semigroup, generated by the Laplacian - this is often used to motivate the whole subject of semigroups in the first place).

$\mathcal A$ is not necessary bounded. Consider for example the one-parameter unitary group $e^{i t \mathcal A}$ generated by an unbounded self-adjoint operator $\mathcal A$ on a Hilbert space $\mathcal H$.

I haven't worked on Operator Semigroups for a long time, and I'm not anywhere near my books, so take this with a grain of salt:

Q 1&2, Usually it is the point that A is unbounded (but densly defined and closed). The archetypical example is the semigroup on $L^2(I)$ on some interval I defined by $T(t)f(s) = f(s + t)$ if $s + t \in I$ and $T(t)f(s) = f(s + t) = 0$ otherwise. The generator A is then given by $A f = f'$ which is clearly unbounded on $L^2$.

About your third question, when an operator is bounded and densly defined, it can be extended to a bounded operator on the whole space. So I'd say this is not possible.

• Ok, so we can extend $\mathcal A$ to the whole state space while it be $\|\|'$-bounded. What does it tell us about the second norm?
– Ilya
Aug 5 '11 at 9:23