Consider the following abstract Cauchy problem: $x'(t)=Ax(t)$, in $(0,T)$, with $x(0)=x\_0$, where $A$ generates an **analytic** $C\_0$-semigroup on a Banach space $X$. How we can prove an inequality of this type:
$$\int\_0^T \|x'(t)\|\_X dt \le C\int\_0^T \|x(t)\|\_X dt +C\|x(T)\|\_X,$$
for some constant $C$, which is independent on $x\_0$, for all $x\_0$ bounded by a constant $M$.


In the scalar case $x'(t)=a x(t)$, $x(0)=\alpha$, we can get such estimate. I'm wondering if we can generalize this for general equations, maybe under some extra assumptions.



An inequality for abstract Cauchy problem
The inequality in question does not hold in general. To show this, we shall consider a case when the initial condition $x\_0$ is given by a highly oscillating function $f\_0\colon\mathbb R\to\mathbb R$. The oscillations of $x\_0=f\_0$ will affect $\|x'(t)\|$ much more than they will affect $\|x(t)\|$, which will lead to the violation of the inequality in question. 


Let $T=1$, $X=L^1(\mathbb R)$, and $Af=f''/2$ for $f\in X\cap C^2(\mathbb R)$. Then $A$ generates the analytic $C\_0$ (heat) semigroup $(S^t)\_{t\ge0}$ defined by the formula 
$$(S^t f)(x)=Ef(x+B\_t)=\int\_{\mathbb R}f(x+y)\frac{dy}{\sqrt t}\,\phi\Big(\frac y{\sqrt t}\Big)
$$
for $f\in X$, real $t>0$, and real $x$, where $B$ is a standard Brownian motion and $\phi$ is the standard normal pdf. 


Take now $f\_0\in X$ defined by the formula $f\_0(x):=\phi(x)\cos ax$ for some real $a>0$ and all real $x$. 
For real $t\in(0,1]$ and real $x$, let 
$$u(t,x):=(S^tf\_0)(x)=\frac1{\sqrt{2 \pi } \sqrt{t+1}}\,
\exp\Big\{-\frac{a^2 t+x^2}{2 t+2}\Big\} \cos\frac{a x}{t+1}, 
$$
so that 
$$u'\_t(t,x)=\frac1{2 \sqrt{2 \pi } (t+1)^{5/2}}\, \\ 
\times\exp\Big\{-\frac{a^2 t+x^2}{2 t+2}\Big\} \\ 
\times\left(2 a x \sin \frac{a x}{t+1}-(a^2+t-x^2+1)
 \cos \frac{a x}{t+1}\right), 
$$
\begin{equation}
 |u(t,x)|\le e^{-(a^2 t+x^2)/4},
\end{equation}
\begin{equation}
 \|u(t,\cdot)\|\_X=\|u(t,\cdot)\|\_1\ll e^{-a^2t/4}, 
\end{equation}
\begin{equation}
 \int\_0^1\|u(t,\cdot)\|\_X\,dt\ll1/a^2,
\end{equation}
so that the right-hand side of your inequality is $\ll1/a^2\to0$; here and in the sequel $U\ll V$ (or, equivalently, $V\gg U$) means $U=O(V)$, and $a\to\infty$. 


On the other hand, the left-hand side of your inequality is $\gg1$ and hence the inequality does not hold for large enough $a>0$. Indeed, for any natural $k$, any $t\in(0,1)$, and any $x\in
J\_k:=[\frac{\pi(3/4+2k)(1+t)}{a},\frac{\pi(1+2k)(1+t)}{a}]$ we have $\cos\frac{a x}{t+1}\le-1/\sqrt2$ and $\sin\frac{a x}{t+1}\ge0$; so, if we additionally assume that $2\pi(1+2k)\le a$, then $0<x<1$ and 
$$|u'\_t(t,x)|\ge 
\exp\Big\{-\frac{a^2 t+1}{2}\Big\} \, a^2. 
$$
Since we have $\gg a$ intervals $J\_k$ with natural $k$ such that $2\pi(1+2k)\le a$, and each of these intervals is of length $\gg1/a$, we see that 
\begin{equation}
 \|u'\_t(t,\cdot)\|\_X=\|u'\_t(t,\cdot)\|\_1\gg e^{-a^2t/2} \, a^2, 
\end{equation}
whence the left-hand side of your inequality is 
\begin{equation}
 \int\_0^1\|u'\_t(t,\cdot)\|\_X\,dt\gg1 
\end{equation}
(actually, it seems to be about $0.64$ for large $a>0$). 


Thus, your inequality does not hold for large enough $a>0$. 


