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Let $X$ be a Banach space. Denote $$ L(X) = \{T:X\rightarrow X \mid T \text{ is linear and bounded} \}. \tag{1} $$ Then, for any $A\in L(X)$, why $$ |\lambda| > \limsup_{n\to\infty} {\|A^n\|}^{\frac{1}{n}} \tag{2} $$ imply $(\lambda I - A)^{-1} \in L(X)$?

I was stumped by this problem when reading a book on functional analysis. I find a lemma say that

If $T \in L(X)$ and $\|T\|<1$, then $(I-T)^{-1}\in L(X)$ and $(I-T)^{-1} = \sum_{k=0}^\infty T^k$.

Besides, the author says to use the Cauchy-Hadamard theorem, I think it should be that

Consider the power series $$ f(z) =\sum_{n=0}^\infty c_n (z-a)^n. $$ Then the radius of convergence $R$ of $f$ at $a$ is $$ \frac{1}{R} =\limsup_{n\to\infty} \left( {|c_n|}^{1/n} \right). $$

But I don’t know how to use them.

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Enhao Lan
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  • $(|\lambda|-\epsilon)^{n} >|A^{n}|$ for large enough $n$ and $\sum A^{n}/\lambda^{n}$ converges in operator norm. – Kavi Rama Murthy May 05 '25 at 11:21
  • If you read some other book on FA, you are likely to see a proof. – Kavi Rama Murthy May 05 '25 at 11:24
  • @geetha290krm So roughly $(I-A/\lambda)^{-1} \in L(X)$ ? Therefore, $(\lambda I - A)^{-1} \in L(X)$ ? Could you give a detail proof ? It's my first time to contact with these, and I feel a bit uncomfortable. Besides, the book I read is a bit brief, without proof. Thanks. – Enhao Lan May 05 '25 at 11:32
  • The term Neumann series might help, when searching for a reference. – Idividedbyzero May 05 '25 at 11:35
  • See also: https://math.stackexchange.com/questions/2077772/weaker-assumption-for-convergence-of-the-neumann-series – Idividedbyzero May 05 '25 at 11:41

1 Answers1

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We’ll first prove the following theorem, sometimes called Von Neumann’s theorem, which is a slightly stronger version of the lemma in your book, and I’m guessing the proof might be exactly the same

Theorem (von Neumann). Let $X$ be Banach and $T\in L(X)$ and suppose $$\sum_{n\geq 0} \|T^n\|<\infty$$ Then $(I-A)^{-1}\in L(X)$.

Proof. Since $X$ is Banach we know that $L(X)$ is also Banach, therefore, the convergence of that sequence implies the convergence of the following sequence in $L(X)$ $$\sum_nT^n=T'\in L(X).$$ So it’s enough to show that $T'=(I-T)^{-1}$ indeed, using that the map $(A,B)\mapsto AB$ is continuous, we know that \begin{align*}T'(I-T) &=\sum_{n}T^n(I-T)\\&=\sum_nT^n-T^{n+1}\\&=\lim_k\sum_{n=0}^k T^n-T^{n+1}\\&=\lim_kI-T^{k+1}\\&=I,\end{align*} where we used that since the original sum converged, we know $\left\|T^k\right\|\xrightarrow{k\to\infty}0$. A similar calculation shows $$(I-T)T'=I.$$ And so $T'=(I-T)^{-1}$.

Now, back to your problem, notice that $\lambda=0$ is vacuously true. Suppose now $\lambda\neq 0$, then we know

$$\limsup_n\sqrt[n]{\|T^n\|}=\limsup_n\sqrt[n]{\left\|\left(\frac A\lambda\right)^n\right\|}<1.$$

And so by the root test $$\sum_{n\geq 0} \|T^n\|<\infty.$$ And so by von Neumann we have

$$\left(I-\frac A\lambda\right)^{-1}\in L(X)\implies (\lambda I-A)^{-1}\in L(X). \tag*{$\square$} $$

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Bruno Andrades
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