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I have been very confused about an argument on Theorem 2.6 from Pazy, Semigroups of Linear Operators and Applications to Partial Differential Equations. Here is the theorem and part of the proof: enter image description here

I just can not prove that red part from the picture just using that the semigroups are $C_0$. Remember that a semigroup $\{T_t\}_{t \geq 0}$ of bounded linear operators on a Banach space is $C_0$ if $\lim_{t \rightarrow 0^+} T(t)x = x,$ for all $x \in X.$ Here there's another discussion about the same doubt, which I still didn't understand.

Here is my attempt:

Fixed $t > 0$, consider the function $\varphi(r) = T(t - r)(S(r)x),$ with $r \leq t$. So, for a $s \geq 0$, we want to describe $\frac{d \varphi}{d r}(s)$. Notice that \begin{align} \frac{\varphi(s + h) - \varphi(s)}{h} & = \frac{1}{h}[T(t - (s + h))(S(s + h)x) - T(t - s)(S(s)x)] \\ & = \frac{1}{h}[T(t - (s + h))(S(s + h)x) - T(t - s)(S(s)x) - T(t - (s+h))(S(s)x) + T(t - (s+h))(S(s)x)] \\ & = \frac{1}{h} [T(t - (s + h))(S(s + h)x) - T(t - (s + h))(S(s)x)] + \frac{1}{h}[T(t - (s+h))(S(s)x) - T(t - s)(S(s)x) ] \\ & \end{align} Lets conclude something about the second quotient above. For $\tilde{h} = - h$, we deduce \begin{align} \frac{1}{h}[T(t - (s+h))(S(s)x) - T(t - s)(S(s)x)] & = -\frac{1}{\tilde{h}}[T(\tilde{h} + (t - s))(S(s)x) - T(t - s)(S(s)x)] \\ & = -\frac{1}{\tilde{h}}[T(\tilde{h})( T(t - s)(S(s)x)) - T(t - s)(S(s)x)], \end{align} which converges to $-A(T(t-s)(S(s)x))$. For me the problem is in the first quotient. Notice that \begin{align} \frac{1}{h} [T(t - (s + h))(S(s + h)x) - T(t - (s + h))(S(s)x)] & = T(t - (s+h)) \frac{(S(s+h)x - S(s)x)}{h} \\ & = T(t - (s+h)) \frac{(S(h)(S(s)x) - S(s)x)}{h}. \end{align} Somehow, I think that this should converge to $T(t-s)(B(S(s)x))$ as $h \rightarrow 0^+$. However, I was not able to obtain this just using that $\{T(t)\}$ is a $C_0$ semigroup. All I know it is $$ \lim_{h \rightarrow 0^+} \frac{(S(h)(S(s)x) - S(s)x)}{h} = B(S(s)x). $$

Any help is very welcome.

ThiagoGM
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2 Answers2

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The way suggested in the question is not good. Lets see another approach. We will first calculate the left derivative. For an $h < 0$ sufficiently small, \begin{align} \varphi(s+h) - \varphi(s) & = T(t - (s+h)) (S(s+h)x) - T(t-s)(S(s)x) \\ & = T(t - (s+h)) (S(s+h)x - S(s)x + S(s)x - h BS(s)x + h BS(s)x) - T(t-s)(S(s)x) \\ & = T(t - (s+h))(S(s+h)x - S(s)x - h BS(s)x) + T(t - (s+h))(h B S(s)x) + T(t - (s+h))(S(s)x) - T(t-s)(S(s)x). \end{align} Thus \begin{align} \frac{\varphi(s+h) - \varphi(s)}{h} = T(t - (s+h))\left(\frac{S(s+h)x - S(s)x}{h} - BS(s)x\right) + T(t - (s+h))(B S(s)x) + \frac{T(t - (s+h))(S(s)x) - T(t-s)(S(s)x)}{h}. \end{align} The first term goes to zero since we have exponential limitation for $||T(t - (s+h))||$. The second converges to $T(t - s)(BS(s)x)$, since $s \mapsto T(s) \hat{x}$ is continous where $\hat{x} = B S(s)x$. Finally, in the third term, taking $\tilde{h} = -h > 0$, one can easily see this term converges to $-AT(t - s) S(s)x$. The right derivative is analogous.

ThiagoGM
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Here is a detailed calculation for the Engel's approach:

For $h>0$, we have

\begin{align} \frac{\varphi(s + h) - \varphi(s)}{h} & = \frac{1}{h} [T(t - (s + h))(S(s + h)x) - T(t - (s + h))(S(s)x)] \\ &\qquad\quad+ \frac{1}{h}[T(t - (s+h))(S(s)x) - T(t - s)(S(s)x) ] \\ & = T(t - s - h)\left(\frac{S(s + h)x - S(s)x}{h} \right) \\ &\qquad\quad+ \frac{1}{h}[T(t - s-h)(S(s)x) - T(t - s-h+h)(S(s)x) ] \\ & = T(t - s - h)\left(\frac{S(s + h)x - S(s)x}{h} \right) \\ &\qquad\quad- T(t - s-h)\left(\frac{T(h)(S(s)x) - S(s)x}{h} \right). \end{align}

Since $\{T(t)\}$ is $C_0$, the map $[0,\infty)\ni t\mapsto T(t)x$ is continuous for every $x\in X$ (Pazy, p. 4). Therefore, for every compact $K\subset[0,\infty)$ and every compact $C\subset X$, the map $G:K\times C\to X$ defined by $G(t,x)=T(t)x$ is uniformly continuous (Engel, p. 37).

Taking $K=[0,t]$ and $C=\left\{\frac{S(s+h)x-S(s)x}{h}\mid h\in(0,1]\right\}\cup\{BS(s)x\}$ (Engel, p. 51) we obtain

$$\begin{aligned} \lim_{h\to 0^+} T(t - s-h) \left(\frac{S(s+h)x - S(s)x}{h}\right) &=\lim_{h\to 0^+} G\left(t-s-h,\frac{S(s+h)x-S(s)x}{h}\right)\\ &= G\left(\lim_{h\to 0^+}(t-s-h),\lim_{h\to 0^+}\frac{S(s+h)x-S(s)x}{h}\right)\\ &= G\left(t-s,BS(s)x\right)\\ &=T(t-s) BS(s)x \end{aligned}$$

and, analogously,

$$\begin{aligned} \lim_{h\to 0^+} T(t - s-h)\left(\frac{T(h)(S(s)x) - S(s)x}{h} \right) &=\lim_{h\to 0^+} G\left(t-s-h,\frac{T(h)(S(s)x) - S(s)x}{h}\right)\\ &= G\left(t-s,AS(s)x\right)\\ &=T(t-s) AS(s)x. \end{aligned}$$

Therefore, the right derivative is $$\lim_{h\to0^+} \frac{\varphi(s + h) - \varphi(s)}{h}=T(t-s) BS(s)x-T(t-s) AS(s)x=0.$$

For $h<0$, we have \begin{align} \frac{\varphi(s + h) - \varphi(s)}{h} & = T(t - s - h)\left(\frac{S(s + h)x - S(s)x}{h} \right) \\ &\qquad\quad+ \frac{1}{h}[T(t - s-h)(S(s)x) - T(t - s)(S(s)x) ] \\ & = T(t - s - h)\left(\frac{S(s + h)x - S(s)x}{h} \right) \\ &\qquad\quad- T(t - s)\left(\frac{T(-h)(S(s)x) - S(s)x}{-h} \right) \\ \end{align} and the argument is similar for the left derivative.

Pedro
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