Let $A\in\mathcal L(E)$. Show that if $\lambda>\|A\|$ then $(\lambda I-A)^{-1}=\int_0^\infty e^{-t(\lambda I-A)}\mathrm dt$.

Here $E$ is a Banach space. I have proved the case for $E$ finite-dimensional using matrices but I dont know how to solve the case for $E$ being infinite dimensional.

My first attempt was to restrict $A$ to arbitrary finite subspaces $H\subset E$, but then in general $A|_H\notin\mathcal L(H)$, that is, $H$ is not necessarily invariant under the action of $A$.

At most I can assume that $A|_H\in\mathcal L(H,A(H))$, but then I cant use the same strategy that used before for the case of $E$ being finite-dimensional, what used the eigenvalues of $A$.

Some help will be appreciated, thank you.

I had an idea (I dont know if it would work): if I prove that $e^{t(\lambda-A)}\to \infty$ seems then easy to show that $e^{-t(\lambda-A)}\to 0$.


1 Answer 1


Hint: justify that

$$\left({\lambda} I-A\right) \int_{0}^{c}{e}^{{-t} \left({\lambda} I-A\right)} d t = \int_{0}^{c}-\frac{d}{d t} {e}^{{-t} \left({\lambda} I-A\right)} d t = I-{e}^{{-c} \left({\lambda} I-A\right)}$$


$${\left({\lambda} I-A\right)}^{{-1}}-\int_{0}^{c}{e}^{{-t} \left({\lambda} I-A\right)} d t = {\left({\lambda} I-A\right)}^{{-1}} {e}^{{-c} \left({\lambda} I-A\right)}$$

Then prove and use that ${e}^{{-c} \left({\lambda} I-A\right)} = {e}^{{-c} {\lambda}} {e}^{c A}$ and

$$\left\|{e}^{c A}\right\| \leqslant {e}^{c \left\|A\right\|}$$

  • $\begingroup$ oh,,, right!!! I didnt see that $\lambda I$ and $A$ commute trivially... $\endgroup$
    – Masacroso
    Aug 30, 2017 at 16:19

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