An inequality for abstract Cauchy problem 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.
 A: 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 $a>0$. 
