P-norm and spectral radius inequality Could someone help me with this proof? I have no idea how to begin.
$$\rho(M) \leq  \|M^k\|^{1/k}_p $$    with $M$ an $n\times n$ matrix, $k$ is a natural number. What can you say about $k \to \infty$ ?
Thank you in advance
 A: I'm assuming that $\| M \|_{p}=\displaystyle\max_{x\neq 0 }\frac{\|Mx\|_p}{\|x\|_p}$.
Let $M\in\Bbb C^{n\times n}$ and $v\in \Bbb C^n, v \neq 0$ be such that $Mv=\lambda v$ with $|\lambda|=\rho(M)$. Let $\|\cdot\|$ be any norm on $\Bbb C^n$, then it holds
$$\rho(M)=|\lambda|\frac{\|v\|}{\|v\|}=\frac{\|Mv\|}{\|v\|}\leq \| M \|_{V\to V}:=\max_{x\neq 0 }\frac{\|Mx\|}{\|x\|}.$$
Now, since for every $k$ it holds $\rho(M^k)=\rho(M)^k$, we deduce that
$$\rho(M)^k=\rho(M^k)\leq \| M^k \|_{V\to V}$$
which implies the desired result for $\|\cdot\|=\|\cdot\|_p$.
The case $k\to \infty$ is known as Gelfand's formula. 
The main idea is the following: First, note that $\| M^{m+k}\|_{V\to V}\leq \| M^{m}\|_{V\to V}\| M^{k}\|_{V\to V}$ for all $k,m\geq 1$, then use Fekete's Lemma on $a_m=\ln(\| M^{m}\|_{V\to V})$ to show that
$$\lim_{m\to \infty}\| M^{m}\|_{V\to V}^{1/m}=\inf_{m\geq 1}\| M^{m}\|_{V\to V}^{1/m}.$$
By the argument above, we have that 
$$\rho(M)\leq \lim_{m\to \infty}\| M^{m}\|_{V\to V}^{1/m}$$
For the other direction, note that
$$\rho(M)=\lim_{m\to \infty}\|M^mv\|^{1/m}\leq \lim_{m\to \infty}\| M^{m}\|_{V\to V}^{1/m}\|v\|^{1/m}= \lim_{m\to \infty}\| M^{m}\|_{V\to V}^{1/m}$$
A: I'm working over $\mathbb{C}$ with the definition $$\rho(M) = \sup_{\lambda \in \sigma(A)} |\lambda|$$
We have $\rho(M) \le \|M\|$. Indeed, for $|\lambda| > \|M\|$, the matrix $M - \lambda I$ is invertible with the inverse $$(M - \lambda I)^{-1} = -\frac1\lambda\sum_{n=0}^\infty \frac{M^n}{\lambda^n}$$
Note that the series converges absolutely because of $\lambda > \|M\|$. Therefore $\lambda \notin \sigma(M)$.
Now the spectral mapping theorem implies $\sigma(M^k) = \sigma(M)^k$ so
$$\rho(M)^k = \rho(M^k) \le \|M^k\|$$
and hence $\rho(M) \le \|M^k\|^{1/k}$. 

Prepare for some complex analysis.
Taking the infimum above over $k \in \mathbb{N}$ gives $$\rho(M) \le \inf_{k\in\mathbb{N}} \|M^k\|^{1/k} \le  \liminf_{k\to\infty} \|M^k\|^{1/k}$$
Consider the open ball $\Delta = B\left(0, \frac1{\rho(M)}\right) \subseteq \mathbb{C}$. Let $\phi$ be an arbitrary (bounded) linear functional on the space of matrices. For $\lambda \in \Delta$ the matrix $I - \lambda M$ is invertible so define $f : \Delta \to \mathbb{C}$ as $f(\lambda) = \phi\left((I - \lambda M)^{-1}\right)$. We will show that $f$ is holomorphic on $\Delta$.
\begin{align}
\lim_{\lambda \to \lambda_0} \frac{f(\lambda) - f(\lambda_0)}{\lambda - \lambda_0} &= \phi\left(\lim_{\lambda \to \lambda_0} \frac{(I-\lambda M)^{-1} - (I - \lambda_0 M)^{-1}}{\lambda - \lambda_0}\right)\\
&= \varphi\left(\lim_{\lambda \to \lambda_0} \frac{[(I-\lambda M) - (I-\lambda_0 M)](I-\lambda M)^{-1}(I - \lambda_0 M)^{-1}}{\lambda - \lambda_0}\right)\\
&= -\varphi\left(\lim_{\lambda \to \lambda_0} (I-\lambda M)^{-1}(I - \lambda_0 M)^{-1}\right)\\
&= -\varphi\left( (I-\lambda_0 M)^{-2}\right) \in \mathbb{C}
\end{align}
Hence $f$ has a Taylor series on $\Delta$, meaning there exists a unique sequence $(\alpha_n)_{n=0}^\infty$ in $\mathbb{C}$ such that 
$$f(\lambda) = \sum_{n=0}^\infty \alpha_n \lambda^n , \forall \lambda \in \Delta$$
If $|\lambda| < \frac1{\|M\|}$, then $\|\lambda M\| < 1$ so check that 
$$(I - \lambda M)^{-1} = \sum_{n=0}^\infty \lambda^nM^n$$
Acting on this with $\phi$ gives
$$f(\lambda) = \phi\left((I - \lambda M)^{-1}\right) = \phi\left(\sum_{n=0}^\infty \lambda^nM^n\right) = \sum_{n=0}^\infty \phi(M^n)\lambda^n$$
Since the Taylor series of $f$ is unique, we conclude $\alpha_n = \phi(M^n)$ for all $n \in \mathbb{N}_0$.
Let $\lambda \in \Delta$. We in particular have $\phi(M^n)\lambda^n \xrightarrow{n\to\infty} 0$ so $(\phi(M^n)\lambda^n )_{n=0}^\infty$ is bounded. Since $\phi$ was arbitrary, the Uniform Boundedness Principle implies that $(\lambda^n M^n)_{n=0}^\infty$ is a bounded sequence of matrices. Hence $\exists C(\lambda) > 0$ such that $\|\lambda^n M^n\| \le C(\lambda), \forall n \in \mathbb{N}_0$. Therefore $$\|M^n\|^{1/n}  \le \frac{C(\lambda)^{1/n}}{|\lambda|}, \forall n \in \mathbb{N}$$ so $\limsup_{n\to\infty} \|M^n\|^{1/n} \le \frac1{|\lambda|}$. 
Since $\lambda$ was an arbitrary number such that $\rho(M) < \frac1{|\lambda|}$, we conclude that $$\limsup_{n\to\infty} \|M^n\|^{1/n} \le \rho(M)$$
Finally we get
$$\rho(M) \le \inf_{k\in\mathbb{N}} \|M^k\|^{1/k} \le  \liminf_{k\to\infty} \|M^k\|^{1/k} \le \limsup_{k\to\infty} \|M^k\|^{1/k} \le \rho(M)$$
so $$\rho(M) = \inf_{k\in\mathbb{N}}\|M^k\|^{1/k} = \lim_{k\to\infty}\|M^k\|^{1/k} $$
