Prove $\lim_{n\rightarrow\infty}||A^n||^{\frac{1}{n}}=\inf_n||A^n||^{\frac{1}{n}}$

Question

Let $X$ be a normed space and $A\in B(X)$ (i.e.A is continuous linear map in $X$), prove that $$\lim_{n\rightarrow\infty}||A^n||^{\frac{1}{n}}=\inf_n||A^n||^{\frac{1}{n}}$$

It is easy to see that $||A^n||^{\frac{1}{n}}\leq ||A||$

One method: 4.1.13 from Pedersen's "Analysis Now" – Ben Grossmann Nov 03 '16 at 13:18 — Ben Grossmann, Nov 03 '16 at 13:18
You know that $|CD| \le |C||D|$. So $|A^{n}| \le |A|^{n}$. – Disintegrating By Parts Nov 03 '16 at 14:01 — Disintegrating By Parts, Nov 03 '16 at 14:01

score 5 · Accepted Answer · answered Nov 03 '16 at 13:24

If $\| A^n \| = 0$ for some $n$, then the result is obvious since both sides are zero. Otherwise, define $(a_n)$ by $a_n = \log \|A^n\|$. Then this sequence is subadditive:

$$ a_{m+n} = \log \|A^{m+n}\| \leq \log (\|A^m\|\|A^n\|) = a_m + a_n. $$

Then by the Fekete's lemma we have

$$ \lim_{n\to\infty} \frac{a_n}{n} = \inf_{n\geq 1} \frac{a_n}{n} \in [-\infty, \infty). $$

Taking exponential to both sides yields the desired identity.

score -1 · Answer 2 · answered Nov 03 '16 at 13:14

-1

2n'th element is less or equals when nth element of this sequence => if this limit exists, it equals to the infimum of the sequence

answered Nov 03 '16 at 13:14

kotomord

1,814

Prove $\lim_{n\rightarrow\infty}||A^n||^{\frac{1}{n}}=\inf_n||A^n||^{\frac{1}{n}}$

2 Answers2