$\{T(t):t\geq 0\}$ is a strongly continuous semi-group, suppose $x\in \mathbb{X}$, $w-\lim_{t\to 0^{+}}\frac{1}{t}(T(t)-I)x=y$, show that $y=Ax$. 
Let $\{T(t):t\geq 0\}$ be a strongly continuous semi-group on $\mathscr{X}$, suppose $x\in \mathbb{X}$, $w\text{-}\lim_{t\to 0^{+}}\frac{1}{t}(T(t)-I)x=y$, show that $x\in D(A)$ and $y=Ax$.

Considering $f\in \mathbb{X}^{*}$, I try to use $e^{-\lambda t}f(T(t)x)$ has right derivate, but I do not know why it is $e^{-\lambda t}f\Big(T(t)(y-\lambda x)\Big)$. And how to complete the proof?
Thanks for any answer!
 A: Since w-$\lim_\limits{t \to 0^+} \frac1t(T(t) - I)x = y$, we have for $f \in X^*$ that $$\lim_\limits{t \to 0^+} \frac1t (f(T(t)x) - f(x)) = \lim_\limits{t \to 0^+} f(\frac1t(T(t) - I)x) = f(y)$$
so that $f(T(t)x)$ has right derivative at $t=0$ given by $f(y)$. 
Since $g = f \circ T(t) \in X^*$, one then sees that 
$$\lim_\limits{h \to 0^+} \frac1h (f(T(t+h)x) - f(T(t)x)) = \lim_\limits{h \to 0^+} \frac1h (g(T(h)x) - g(x)) = g(y) = f(T(t)y)$$
so that the right derivative of $f(T(t)x)$ at $t$ is $f(T(t)y)$.
By following the proof of the product rule for real valued functions one then sees that $e^{-\lambda t} f(T(t)x)$ has right derivative at $t$ given by $e^{-\lambda t} f(T(t)(y - \lambda x))$ as desired.
To finish the proof, notice that since the right derivative is continuous in $t$,
$$f(x) =  - \int_0^\infty e^{-\lambda t} f(T(t)(y - \lambda x)) dt = f\bigg(- \int_0^\infty e^{-\lambda t} (T(t)(y - \lambda x)) dt \bigg).$$
Since $f \in X^*$ was arbitrary, it follows that 
$$x = - \int_0^\infty e^{-\lambda t} (T(t)(y - \lambda x)) dt.$$
Now, for large $\lambda > 0$, $\lambda$ is in the resolvent set of $A$ and 
$$(\lambda - A)^{-1}z = \int_0^\infty e^{-\lambda t} T(t)z dt.$$
In particular, 
$$x = - (\lambda - A)^{-1}(y - \lambda x) \in D(A).$$ Now apply the operator $\lambda - A$ to both sides for the desired result.
