Let $E$ be a complex Hilbert space, $\mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.
If $A\in\mathcal{L}(E)$, why $\displaystyle\lim_{n\to+\infty}\|A^n\|^{1/n}$ always exists?
Thank you.
Asked
Active
Viewed 193 times
3
Martin Sleziak
- 53,687
Student
- 4,914
-
1This is Gelfand's formula, i.e. that the limit above is equal to the spectral radius $r(A)$. The proof is essentially to show that $\limsup |A^n|^{1/n} \leq r(A) \leq \liminf |A^n|^{1/n}$. It extends to all elements of a Banach algebra. – I was suspended for talking Jan 17 '18 at 09:41
-
Thank you, I know that this limit is the spectral radius of $A$ but I want to know why this limit is always existing. – Student Jan 17 '18 at 09:44
1 Answers
8
Because of the norm inequality $\|AB\|\le\|A\|\|B\|$, the function $f(n)=\log\|A^n\|$ is subadditive (Subadditivity): $$f(m+n)=\log\|A^{m+n}\|=\log\|A^m\cdot A^n\|\le\log\|A^m\|+\log\|A^n\|=f(m)+f(n),$$ thus by Fekete's lemma, $\displaystyle\lim_{n\to\infty}\frac{f(n)}n$ exists and is equal to $\displaystyle\inf_{n\ge1}\frac{f(n)}n$.
-
1I will ad this link as a useful resource for Fekete's lemma and some related topics: Existence of a limit associated to an almost subadditive sequence. – Martin Sleziak Jan 19 '18 at 16:56
-
@Martin Sleziak That resource may be very useful, indeed. Fekete's lemma and a distant cousin (Jensen's inequality) share the fate to be extremely helpful in some situations, but much less well-known than they deserve. – Jan 19 '18 at 19:29