Gelfand formula for the spectral radious.

I am reading ''class notes on operator theory on hilbert spaces'' by John Petrovic (http://homepages.wmich.edu/~petrovic/courses/Math6780/m678notesch1.pdf) And in page 37 it proves the Gelfand formula for the spectral radius: $$r(T)=\lim_{n\to\infty}||T^n||^{1/n}$$ I have some questions about the proof:

• First, it proves that $r(T)\leq||T^n||^{1/n}$ from which it deduces that: $$r(T)\leq\liminf_{n\to\infty}||T^n||^{1/n}$$
• In the second part it proves that if $\lambda\neq0$ with $\frac{1}{\lambda}\in\rho(T)$ then: $$|\lambda|||T^n||^{1/n}\leq M^{1/n}$$ for some constant $M>0$. This implies that: $$|\lambda|\limsup_{n\to\infty}||T^n||^{1/n}\leq 1$$ From this, the proof says: Since this is true for any $\lambda$ such that $\frac{1}{\lambda}\in\rho(T)$ it holds all the more whenever $\frac{1}{|\lambda|} > r(T)$. It follows that $\limsup\limits_{n\to\infty} ||T^n||^{1/n}≤ r(T)$ and the theorem is proved.

I have two questions:

• Why we can't avoid the $\limsup$ and $\liminf$? I mean, in the first part it proves $r(T)\leq||T^n||^{1/n}$ so $r(T)\leq\lim\limits_{n\to\infty}||T^n||^{1/n}$ and in the same way in the second part we can prove that $r(T)\geq\lim\limits_{n\to\infty}||T^n||^{1/n}$, right?
• When you have: $$|\lambda|\limsup_{n\to\infty}||T^n||^{1/n}\leq 1$$ I don't understand the last part of the proof (Since this is true for any $\lambda\neq 0$ such that $\frac{1}{\lambda}\in\rho(T)$ it holds all the more whenever $\frac{1}{|\lambda|} > r(T)$. It follows that $\limsup\limits_{n\to\infty} ||T^n||^{1/n}≤ r(T)$ and the theorem is proved.) Maybe it is trivial, but I don't see what is happening there.

Thanks for your help.

To your first point: You have to use $\liminf$, simply because you have not proven yet that the $\lim$ exists. A priori, we can not assume the series $\lVert T^n \rVert^\frac{1}{n}$ to be convergent.

Two your second point. Well, we know that the spectrum is bounded. The lowest possible such bound is $r(T)$ by definition. This means that every complex number that exceeds this bound is not in the spectrum. So if $\lambda \in \mathbb{C}$ satisfies $\frac{1}{\lvert \lambda \rvert} > r(T)$, then $\frac{1}{\lambda}$ is in $\rho(T)$. But this case has already been treated in the proof.