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Given an operator $T\in B(X,X)$, I'm trying to prove that

$\lim_{n\to\infty} \left( ||T^n||\right)^\frac1n \leq ||T|| $

I can show that $||T^n||\leq||T||^n$, but only for finite $n$. How can I be sure that this holds in the limit? I'm a bit confused as to what kind of continuity that would require.

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You know that for all finite $n$ we have $ \| T^n \|^{1/n} \leq \| T \|.$ Recall that if $a_n\to a$, $b_n \to b$ and $a_n \leq b_n$ then $a\leq b.$ Hence, if $\lim_{n\to\infty} \| T^n \|^{1/n}$ exists, then it is less than or equal to $\| T\|.$ The limit does indeed exist - Gelfand's formula tells us that it is the spectral radius of $T.$

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I'm not sure what you are trying to prove. The inequality $ \|T^n\|^{1/n}\leq\|T\| $ is trivial, since $\|T^n\|\leq\|T\|^n$ by the Banach-norm inequality.

What is not trivial is to show that your limit exists and it agrees with the spectral radius. Every proof I know uses some form of complex analysis, and it appears in every functional analysis book.

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    @MartinArgerami Ahh yes, sorry about that.2012-12-18