Proving a function defined in terms of a $C_{0}$-semigroup is continuously differentiable

Question

Suppose that $u\in C([0,\infty))\cap C^{1}([0,\infty))$ is a solution of $$\begin{cases}u'(t)=Lu(t),& t\ge 0, \\ u(0)=x\end{cases}$$

Fix $t>0$ and define the function $$v(s):=T(t-s)u(s),\qquad 0\le s\le t.$$

Since $v(0)=T(t)x$, I have by a proposition that $$v'(0)=\frac{d}{dt}T(t)x=LT(t)x=T(t)Lx,\qquad t\ge 0.$$

Now I want to prove right differentiability for fixed $s_{0}>0$. Let $h>0$. Then $$\begin{aligned}\lim_{h\downarrow 0}\frac{T(t-s_{0}+h)u(s_{0}+h)-T(t-s_{0})u(s_{0})}{h}=T(t-s_{0})\lim_{h\downarrow 0}\frac{T(h)u(s_{0}+h)-u(s_{0})}{h}\end{aligned}$$

At this point I am stuck. I know that $$\lim_{h\downarrow 0}\frac{u(s_{0}+h)-u(s_{0})}{h}=Lu(s_{0}),\qquad \lim_{h\downarrow 0}T(h)=I,$$

But I cannot separate the limit in that way.

Answer 1

Since $T(0)=I$, simply use:

\begin{align} T(h)u(s_0+h)-u(s_0)&=T(h)u(s_0+h) - T(h)u(s_0) + T(h)u(s_0)-T(0)u(s_0)\\ &= T(h)(u(s_0+h) - u(s_0)) + (T(h)-T(0))u(s_0). \end{align}

Now use that the (real) limit of a product is the product of limits and that $\lim_{h\to 0} T(h)=I$ as you mentioned.