Vector $p$-norm for square matrices is submultiplicative for $1 le p le 2$

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

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!

Answer

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 $\|x\|_q \le M = \|x\|_p$.