Let $A$ be a square matrix of dimension $n\times n$ and consider the following norm for $1< p<\infty$:

$$\|A\|_{p} = \max_{x \neq 0} \frac{\|Ax\|_p}{\|x\|_p}.$$
Let $\psi_p(x):=\big(|x_1|^{p-1}\operatorname{sign}(x_1),\ldots,|x_n|^{p-1}\operatorname{sign}(x_n)\big)$ and write $p'$ the Hölder conjugate of $p$. Then for every $x$ we have

$$\max_{x\neq 0 }\frac{\|Ax\|_p}{\|x\|_p} = \max_{x\neq 0 }\frac{\langle \psi_p(Ax),Ax\rangle}{\|Ax\|_p^{p-1}\|x\|_p}=\max_{x\neq 0 } \frac{\langle \psi_p(Ax),Ax\rangle}{\|\psi_p(Ax)\|_{p'}\|x\|_p}\leq \max_{x,y\neq 0} \frac{\langle y,Ax\rangle}{\|y\|_{p'}\|x\|_p} \\ \leq \max_{x,y\neq 0} \frac{\|y\|_{p'}\|Ax\|_p}{\|y\|_{p'}\|x\|_p}=\max_{x\neq 0 } \frac{\|Ax\|_p}{\|x\|_p},$$
where we used the Hölder inequality for vectors in the last inequality. It follows that
$$\|A\|_p = \max_{x,y\neq 0} \frac{\langle y,Ax\rangle}{\|y\|_{p'}\|x\|_p} = \max_{x,y\neq 0} \frac{\langle A^*y,x\rangle}{\|y\|_{p'}\|x\|_p}=\|A^*\|_{p'}.$$
Now, since $\|A\|_p$ is a norm induced by a vector norm, it is submultiplicative and the spectral radius $\rho(A)$ of $A$ satisfy $\rho(A)\leq \|A\|_p$. So, we get
$$\|AB\|_2^2 = \rho\big((AB)^*(AB)\big) \leq \|(AB)^*(AB)\|_p \\ \leq \|B^*A^*\|_p\|AB\|_p\leq \|B^*\|_p\|B\|_p\|A^*\|_p\|A\|_p =\big(\|A\|_p\|A\|_{p'}\big)\big(\|B\|_p\|B\|_{p'}\big).$$

i.e.take Holder's inequality and replace $fg$ with $f \circ g$. – usul Feb 27 '14 at 21:46