I'm trying to prove that the vector $p$-norm for square matrices is submultiplicative for values of $p$ between $1$ and $2$. The vector $p$-norm for a square matrix $A$ is defined as
$\displaystyle \|A\|_p:=\left(\sum_{i=1}^n \sum_{j=1}^n |a_{ij}|^p \right)^{\frac{1}{p}}$
For $p=1$ and $p=2$ the result follows easily from the Cauchy-Schwarz inequality. Now for $1<p<2$ my idea is to use Hölder's inequality, but I'm having trouble with one step.
My approach is as follows:
Let $A=(a_{ij})$ and $B=(b_{ij})$ be two square $n \times n$ matrices and let $q=\frac{p}{p-1}$. Then
$\displaystyle \left(\|AB\|_p\right)^p=\sum_{i=1}^n \sum_{j=1}^n \left| \sum_{k=1}^n a_{ik}b_{kj} \right|^p \le \sum_{i=1}^n \sum_{j=1}^n \left( \sum_{k=1}^n \left| a_{ik}\right|^p\right) \left( \sum_{k=1}^n \left| b_{ik}\right|^q\right)^\frac{p}{q}$
Now, since $p<2$, we have that $q>2$ and as such $\frac{p}{q}<1$. Here comes my doubt: Can I assure that $\left( \sum_{k=1}^n \left| b_{ik}\right|^q\right)^\frac{p}{q} \le \sum_{k=1}^n \left| b_{ik}\right|^p$ ? Because if so, then the result would follow immediately.
Any help would be greatly appreciated!
The answer to your question follows from this: let $\|x\|_p = \left(\sum_{i=1}^n |x_i|^p \right)^{1/p}$. Then is $\|x\|_q \le \|x\|_p$ if $q \ge p$?
This is a very well known result, But here is a proof.
Answer: yes. Let $M = \|x\|_p$. So $\|\frac1Mx\|_p = 1$. Therefore $|\frac{x_i}M|^p \le 1 \Rightarrow |\frac{x_i}M| \le 1 \Rightarrow |\frac{x_i}M|^q \le |\frac{x_i}M|^p$. Hence $$ \|\frac1M x\|_q^q = \sum_{i=1}^n |\frac{x_i}M|^q \le \sum_{i=1}^n |\frac{x_i}M|^p = \|\frac1Mx\|_p^p = 1.$$ Therefore $\|x\|_q \le M = \|x\|_p$.