Let $||A||=(\sum_{i=1}^{n}\sum_{j=1}^{n}{a_{ij}^p})^{1/p}$, and let p=2. Then prove that $\|AB\|\le \|A\|\|B\|$
I have looked at numerous proofs for this, and I don't see one that satisfies me fully. I don't want to use traces. I see proofs bringing an x into the left side, divide by $||Bx||$ then pull things out and cancel out and what not. But these don't involve using p=2. Thanks in advance
Here is one proof:
Note that $\|A\|_F^2 = \sum_k \|Ae_k\|_2^2$. Any set of orthonormal vectors will do, this also shows that $\|A\|_2 \le \|A\|_F$.
Then $\|AB\|_F^2 = \sum_k \|ABe_k\|_2^2 \le \|A\|_2^2 \sum_k \|Be_k\|_2^2 = \|A\|_2^2 \|B\|_F^2 \le \|A\|_F^2 \|B\|_F^2$.