When $f\left(\|Ax\|\right)\leq \|f(A)x\|$ is true?

Question

When is the following relation true?

$$f\left(\|Ax\|\right)\leq \|f(A)x\|$$ where $A$ is any matrix and $\|\cdot\|$ is the spectral norm.

Edit note :

The function $$ is any real concave increasing value function on $[0,∞)$

Answer 1

I don't know what you are looking for. In general, the answer is "very rarely". For instance take $f(t)=t^2$. Then you are asking for $$\tag1 \|Ax\|^2\leq\|A^2x\|. $$ This has obvious problems in that it doesn't scale well: suppose you had equality for $x$: then then inequality would fail for $2x$. Also, if $A$ is nilpotent of order two ($A^2=0$) then the right-hand-side is zero while the left-hand-side will be nonzero for certain $x$.

The scaling problems guarantee that even for positive definite $A$ the inequality $(1)$ has problems: for instance if $A$ is a projection, the inequality becomes $\|Ax\|^2\leq \|Ax\|$, so if $Ax\ne0$ you get $\|Ax\|\leq1$, again with obvious scaling problems.

If $f(x)=\sqrt{x}$, even with positive definite $A$ you still run into problems: take $A$ to be a projection, then you want $$ \|Ax\|^{1/2}\leq\|Ax\|, $$ which is the same (when $Ax\ne0$) to $\|Ax\|\geq1$. So again you have to look at specific $x$ and your inequality doesn't scale.