Suppose $X$ and $A$ are two matrices of compatible dimensions.
Is it possible to prove that $$ \lim_{\|X\|\to 0} \frac{\|X^T A X\|}{\|X\|} = 0, $$
where:
- A is squared, X may not be squared.
- The norms can be any norm (Frobenius, induced, max, etc).
- The norms in the numerator and denominator can be different.
- The submultiplicative property $\|AB\|\le \|A\| \|B\|$ may not be satisfied.
- The limit is taken for any sequence $\{X_n\}$ whose norm goes to zero.
I can do it for some cases, but not in general. Any ideas?
Yes, this follows from homogeneity: for any two norms, and for any positive scalar $\lambda $
$$ \frac{\|(\lambda X)^T A (\lambda X)\| }{|||\lambda X|||} = \lambda \frac{\|X^T A X\| }{|||X|||} \tag{1} $$
Indeed, let $K = \{X: |||X||| =1\} $. Since this is a compact set, the function $X\mapsto \|X^T A X\|$ attains its maximum on it, let this maximum be $M$. The property (1) yields $$ |||X||| = \lambda \implies \frac{\|X^T A X\|}{|||X|||} \le M\lambda $$ or simply put, $$\frac{\|X^T A X\|}{|||X|||} \le M \, |||X|||$$ The claim follows.