A step in the proof of the Hille-Yosida theorem from Rudin I'm getting stuck on perhaps a simple step in the Hille-Yosida theorem from 13.37 in Rudin's functional analysis.
I wonder if someone has had this same difficulty before or knows how to get around it -
Setup: $A$ is a densely defined operator with domain $\mathcal{D}(A)$ in a Banach space $X$ and there are constants $C, \gamma >0$ such that for every $\lambda > \gamma$ and $m\in \mathbb{N}$,
$$
\| (\lambda I - A)^{-m}\| \leq C(\lambda - \gamma)^{-m}.
$$
The claim is then that $A$ is the infinitesimal generator of a semi-group of operators.
For small $\varepsilon$, the bounded operator $S(\varepsilon)$ is defined to be $(I-\varepsilon A)^{-1}:X \to \mathcal{D}(A)$.
It follows from the definitions that $AS(\varepsilon) = \varepsilon^{-1}(S(\varepsilon)-I)$ from which one can show that $e^{tAS(\varepsilon)}$ converges weakly to a bounded $Q(t)$. Moreover, $\{Q(t) \}$ gives a semi-group.
Thus it has an infinitesimal generator $\tilde{A}$.
Using the resolvent formulas for $AS(\varepsilon)$ and $\tilde{A}$, we have for all $x$ and $\lambda$ sufficiently large,
$$
(\lambda I - \tilde{A})^{-1}x = \int_0^\infty e^{-\lambda t} Q(t)x dt
$$
and
$$
(\lambda I - AS(\varepsilon))^{-1}x = \int_0^\infty e^{-\lambda t} e^{tAS(\varepsilon)}x dt.
$$
One can easily justify the limit
$$
\lim_{\varepsilon \to 0} \int_0^\infty e^{-\lambda t} e^{tAS(\varepsilon)}x dt = \int_0^\infty e^{-\lambda t} Q(t)x dt.
$$
Question: In order to compare $\tilde{A}$ and $A$, how does one see that
$$
\lim_{\varepsilon \to 0} (\lambda I - AS(\varepsilon))^{-1}x = (\lambda I - A)^{-1}x?
$$
THANK YOU!
 A: First, multiply your condition on $A$ by $\lambda^m$ and substitute $\lambda=\frac{1}{\epsilon}$; we have $$\|(I-\epsilon A)^{-m}\|\leq C\left(1-\epsilon\gamma\right)^{-m}\to1 \tag{1}$$ as $\epsilon\to0$.  In particular, when $m=1$, we see that $S(\epsilon)$ is uniformly bounded.
Second, note that the argument to show that $$\epsilon AS(\epsilon)=S(\epsilon)-1$$ also shows $$\epsilon AS(\epsilon)=S(\epsilon)-1=\epsilon S(\epsilon)A \tag{2}$$ and so $S(\epsilon)$ commutes with $A$.
Third, note that $$X(X^{-1}-Y^{-1})Y=(Y-X)^{-1}\tag{3}$$  Then use (3) to compute the difference between your left and right side: \begin{align*} \require{cancel}
(\lambda-AS(\epsilon))^{-1}x-(\lambda-A)^{-1}x&=(\lambda-AS(\epsilon))^{-1}\cdot(\cancel{\lambda}-A-(\cancel{\lambda}-AS(\epsilon)))\cdot{}\\
&\phantom{{}={}}\quad\quad\quad(\lambda-A)^{-1}x \\
&=(\lambda-AS(\epsilon))^{-1}\cdot A(S(\epsilon)-1)\cdot(\lambda-A)^{-1}x \\
&=(\lambda-AS(\epsilon))^{-1}\cdot(A\cdot\epsilon AS(\epsilon))\cdot(\lambda-A)^{-1}x \\
&=(\lambda-AS(\epsilon))^{-1}\cdot\epsilon S(\epsilon)\cdot A^2(\lambda-A)^{-1}x \tag{4}
\end{align*} where (4) applies the commutation relation from (2).
The first term is a bounded operator (since $\lambda$ is sufficiently large) and the last is fixed in $\epsilon$.  By (1), the middle is going to $0$, and so (4) as a whole tends to $0$.
A: Some more details to Jacob Manaker's solution
(edit: this refers to a previous version of Jacob’s answer; I now added some further details so this is a complete version of his previous solution. For an alternative solution look below):
First note that
$$
(\lambda\Bbb 1 - AS(\varepsilon))^{-1} x = \int_0^\infty e^{-\lambda t} T(t,\varepsilon) dt
$$
with $T(t,\varepsilon)=e^{tAS(\varepsilon)}$. In equation (10) Rudin shows $$
\|T(t,\varepsilon)\|_{op}\leq C\exp\Big(\frac{\gamma t}{1-\varepsilon\gamma}\Big) \leq C\exp\Big(\frac{\gamma t}{1-\varepsilon_0\gamma}\Big).
$$
This implies for $\|x\| \leq 1$ that
$$
\|(\lambda\Bbb 1 - AS(\varepsilon))^{-1}x\| \leq \int_0^\infty C\exp\Big(\frac{\gamma t}{1-\varepsilon_0\gamma}-\lambda t\Big) dt <\infty.
$$
Hence the $(\lambda\Bbb 1 - AS(\varepsilon))^{-1}$ are uniformly continuous.
For $x\in\mathcal D(A)$ we have
$$
(\Bbb 1 - \varepsilon A) S(\varepsilon)x = x = S(\varepsilon)(\Bbb 1 - \varepsilon A)x
$$
thus
$$
AS(\varepsilon)x = S(\varepsilon)Ax.
$$
(Similar to: $S(\varepsilon)=\varepsilon^{-1}R_A(\varepsilon^{-1})$ and in Banachalgebras resolvents commute with the respective element. We just have to pay attention as $A$ is not defined on the whole $X$ and we therefore aren't really in a Banachalgebra)
We can conclude
$$
(\lambda\Bbb 1 - AS(\varepsilon))(\lambda \Bbb 1 - A)x = (\lambda\Bbb 1 - A)(\lambda\Bbb 1 - AS(\varepsilon))x.
$$
This allows Jacob's calculation
\begin{align*}
&\big((\lambda\Bbb 1 - AS(\varepsilon))^{-1}-(\lambda\Bbb 1 - A)^{-1}\big) x \\
=& \big((\lambda\Bbb 1 - AS(\varepsilon))^{-1}(\lambda\Bbb 1 - A)^{-1}\big)\big((\lambda\Bbb 1-A) - (\lambda\Bbb 1 - AS(\varepsilon))\big)x\\
=&(\lambda \Bbb 1 - AS(\varepsilon))^{-1}(\lambda \Bbb 1 - A)^{-1} A^2 \cdot \varepsilon \cdot S(\varepsilon) x\\
=& (\lambda \Bbb 1 - AS(\varepsilon))^{-1}(\lambda \Bbb 1 - A)^{-1} \varepsilon \cdot S(\varepsilon) A^2x.
\end{align*}
Now $A^2 x$ is a fixed element and $S(\varepsilon)A^2x \to A^2x$ as $\varepsilon\to 0$. By equicontinuity of $(\lambda\Bbb 1 - AS(\varepsilon))^{-1}$ we conclude that the whole expression tends to 0. We now have the claim for all $x\in\mathcal D(A)$. We can now work with approximating sequences and equicontinuity of the $(\lambda \Bbb 1 - AS(\varepsilon))^{-1}$ and conclude the claim for all $x\in X$ (like Rudin does several times in the proof).

An alternative solution:
The operator $(\lambda\Bbb 1 - A)\colon \mathcal D(A) \to X$ is invertible, hence there exists an $y\in \mathcal D(A)$ such that $x = (\lambda \Bbb 1-A)y$.
We now wish to show
$$
(\lambda\Bbb 1 - AS(\varepsilon))^{-1}(\lambda\Bbb 1 -A)y = (\lambda \Bbb 1 - AS(\varepsilon))^{-1}x \to (\lambda\Bbb 1 - A)^{-1} x =y.
$$
We calculate
\begin{align*}
(\lambda\Bbb 1 - AS(\varepsilon))^{-1}(\lambda\Bbb 1 - A)y &= (\lambda\Bbb 1 - AS(\varepsilon))^{-1}\big[(\lambda\Bbb 1 - AS(\varepsilon))y + (AS(\varepsilon)-A)y\big]\\
&= y + (\lambda\Bbb 1 - AS(\varepsilon))^{-1}(S(\varepsilon)A-A)y\\
&= y + (\lambda\Bbb 1 - AS(\varepsilon))^{-1}(S(\varepsilon)-\Bbb 1)Ay.
\end{align*}
Now $Ay$ is fixed and thus $(S(\varepsilon)-\Bbb 1)Ay \to 0$ as $\varepsilon \to 0$. The $(\lambda\Bbb 1 - AS(\varepsilon))^{-1}$ are equicontinuous and therefore the last term tends to $0$.
