Possibly a basic question, but this is not really my area. Suppose that $f,g\in L^2(X,\mu)$ for some probability measure space $(X,\mu)$. Let us write $f=(f_1,f_2)^T:X\to \mathbb{R}^2$, and similarly for $g$. Note $fg^T$ defines a $2\times 2$ matrix. If we equip such matrices with the operator norm (or the Frobenius norm), can we say that $\|fg^T\|_=\int_X \|fg^T\|_{op}\, d\mu\le \|f\|_2\|g\|_2$? Or even $\le C\|f\|_2\|g\|_2$ for some constant independent of $f$ and $g$?
Obviously if it holds for one norm on the 2 by 2 matrices, then it holds for all norms since they are equivalent.
Thanks!
Edit: I think it definitely holds since this is a Bochner space with values in a finite dimensional (hence Banach) space. But a direct proof would be nice!
Explicitly, $$ (fg^T)_{ij} = f_i g_j.$$ Pick as a norm on $d\times d$ matrices $$ \pmb\|(a_{ij})_{i,j}\pmb\|_1 = \sum_{i,j=1}^d |a_{ij}|.$$ Then since for each $i,j$, $$ \int |f_i g_j| d\mu \le \|f_i\|_{L^2}\|g_j\|_{L^2},$$ Summing in $i,j$, we have $$ \int \pmb\| fg^T\pmb\|_1 d\mu \le \sum_{i,j=1}^d \|f_i\|_{L^2}\|g_j\|_{L^2} = \sum_{i=1}^d\|f_i\|_{L^2} \sum_{j=1}^d\|g_j\|_{L^2} \le d \|f\|_{L^2}\|g\|_{L^2} $$ by using the inequality $\|(a_1,\dots, a_d)\|_{\ell^1} \le \sqrt{d} \|(a_1,\dots, a_d)\|_{\ell^2} $ twice.