$\|S(t)\| \leq Me^{ct} \Rightarrow \|u\|_{C(0,T,H)} \leq \|u_0\|_H+\|f\|_{L^1(0,T,H)}$?

Let $A$ be the infinitesimal generator of a $C_0$-semigroup of contractions $(S(t))$ in a Hilbert space $H$ and $f\in L^1(0,T;H)$. We know that the mild solution of the problem \begin{equation} \begin{cases} u_t=Au+f,\\ u(0)=u_0 \in H, \end{cases} \end{equation} is given by $$u(t)=S(t)u_0+\int_{0}^{t}S(t-s)f(s)ds.\tag{*}$$ Then, since $\|S(t)\|_{\mathcal{L}(H)}\leq 1$ for all $t \geq 0$ it follows that $$\|u\|_{C(0,T,H)} \leq \|u_0\|_H+\|f\|_{L^1(0,T,H)} \tag{**}.$$

My question: Does the inequality $(**)$ holds with the assumption $\|S(t)\| \leq Me^{ct}$ for all $t \geq 0$, with $M,c>0$?

From $(*)$ we have that: $$\|u\|_{C(0,T,H)} \leq Me^{cT}(\|u_0\|_{H}+\|f\|_{L^1(0,T,H)}).$$

Is there any way to remove the term $Me^{cT}$ from the last inequality?

Solution 1:

To see that this bound is the best that you can get in the general setting, consider the case $H = \mathbb{R}$ and $S(t)x = e^t \cdot x$. Then $Ax = x$ and the equation in question is $$u_t - u = f.$$ To see an example where equality is achieved in your bound, consider $f = 0$ and $u_0 = 1$. Then one has $u(t) = e^{t}$.