When is a $C_0$-semigroup with infinitesimal generator $A$ of the form $t\mapsto e^{tA}$?

Let

• $E$ be a $\mathbb R$-Banach space
• $T:[0,\infty)\to\mathfrak L(E)$ be a $C^0$-semigroup, i.e.
1. $T(0)=\operatorname{id}_E$
2. $T(s+t)=T(s)T(t)$ for all $s,t\ge 0$
3. $t\mapsto T(t)x$ is continuous from $[0,\infty)$ to $E$ for all $x\in E$
• $A$ be the infinitesimal generator of $T$

I've read that if$^1$ $\mathcal D(A)=E$ and $A$ is bounded, then $$T(t)=e^{tA}:=\sum_{n\in\mathbb N_0}\frac{t^n}{n!}A^n\;\;\;\text{for }t\ge 0\tag 1$$ and I guess it's easy to prove that the series on the rigt-hand side of $(1)$ exists in $E$.

Is there any chance that $T$ is of the form $(1)$ even when $A$ is unbounded on $\mathcal D(A)$?

I guess the answer is yes. If $E$ is a $\mathbb R$-Hilbert space, $(e_n)_{n\in\mathbb N}\subseteq\mathcal D(A)$ is an orthonormal basis of $E$ and $$Ae_n=\lambda_ne_n\;\;\;\text{for all }n\in\mathbb N\tag 2$$ for some $(\lambda_n)_{n\in\mathbb N}\subseteq\mathbb R$, then a formal application of the series representation of the exponential function yields $$e^{tA}x=\sum_{n\in\mathbb N}e^{t\lambda_n}\langle x,e_n\rangle_Ee_n\;\;\;\text{for all }t\ge 0\text{ and }x\in E\;.\tag 3$$ The right-hand side of $(3)$ exists in $E$ iff $$\sum_{n\in\mathbb N}\left|e^{t\lambda_n}\langle x,e_n\rangle_E\right|^2<\infty\;.\tag 4$$ Since $$\sum_{n\in\mathbb N}\left|e^{t\lambda_n}\langle x,e_n\rangle_E\right|^2=\sum_{n\in\mathbb N}e^{2t\lambda_n}\left|\langle x,e_n\rangle_E\right|^2\le\sum_{n\in\mathbb N}\left|\langle x,e_n\rangle_E\right|^2=\left\|x\right\|_E\;,\tag 5$$ if $\lambda_n\le 0$ for all $n\in\mathbb N$, $T$ is of the form $(1)$ whenever $A$ admits a complete set of eigenfunctions with non-positive eigenvalues.

Are there other similar results we can obtain?

$^1$ In fact, if $A$ is bounded on $\mathcal D(A)$, it can always be extended to a bounded linear operator on $E$.

• Concerning your comment: Be careful. If In fact, if $A$ is bounded on $D(A)$ wrt the graph norm (which is always the case if the operator is closed), then it certainly need not have a bounded extension to the whole of $H$. – Delio Mugnolo Jun 22 '17 at 11:44
• You might want to look at the Hille–Yosida theorem. – Harald Hanche-Olsen Jun 22 '17 at 11:56

I don't know if you application of the Spectral Theorem really can be regarded as a direct extension of the exponential formula for bounded operators; but if you want to proceed in this direction, you should look at the literature about Functional Calculus, a topic dedicated to study the possibility of defining bounded operators by plugging a suitable unbounded operator $A$ into a function $f$ with nice properties -- the nicert of $f$, the rougher the properties of $A$. Since $f=\exp$ is a very nice function, $t\mapsto \exp(tA)$ can be defined for a very large class of operators $A$, see e.g. Haase's monograph.