A step in the proof of the Hille-Yosida theorem from Rudin

Question

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!

Answer 1

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 $0$.

Answer 2

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).

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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 $0$.