Bounds on matrix norm of Hadamard powers Suppose $A$ and $B$ are non-negative $n\times n$ matrices with $\|A\|= \|B\| = 1.$ In particular, therefore, $0\leq A_{ij}, B_{ij}\leq 1$ for all $i,j.$
Suppose also that $\|A-B\|\leq \delta.$
I want a bound of the form $$\left|\log\|A^{\circ r}\|-\log\|B^{\circ r}\|\right|\leq f(n,\delta,r),\ \ \ \ \  r>0$$ where $A^{\circ r}, B^{\circ r}$ are the Hadamard powers of exponent $r$: $(A^{\circ r})_{ij} = (A_{ij})^r.$
I want the function $f(n,\delta,r)$ to go to zero for any fixed $n,r$ as $\delta\to 0$ as also for any fixed $n,\delta$ as $r\to 1.$
 A: Since $\|A\| = \|B\| = 1$, we have that $|A_{ij}|, |B_{ij}| \le 1$,
Since $\|A - B \| \le \delta$, we have $|A_{ij} - B_{ij}| \le \delta$.  Hence $ |A^r_{ij} - B_{ij}^r| \le g(r,\delta)$, where $g(r,\delta) = r\delta$ if $r \ge 1$ (by the mean value theorem), and $\delta^{1/r}$ if $r < 1$ (since $x^r + y^r \ge (x+y)^r$ if $x,y \ge 0$).
Hence $\|A^{\circ r}-B^{\circ r}\| \le \|A^{\circ r}-B^{\circ r}\|_F \le n g(r,\delta)$, where here $\|\cdot\|_F$ denotes the Frobenius norm.
$$ |\log\|A^{\circ r}\| - \log\|B^{\circ r}\|| = \left| \log\left(\frac{\|A^{\circ r}\|}{\|B^{\circ r}\|} \right) \right| $$
$$ \le \left| \log\left(\frac{\|B^{\circ r}\|+\|A^{\circ r}-B^{\circ r}\|}{\|B^{\circ r}\|} \right) \right| $$
$$ \le \frac{\|A^{\circ r}-B^{\circ r}\|}{\|B^{\circ r}\|}$$
(Here we used $\log(1+x) \ge x$ for $x > -1$.)
Now $\|B\|_F \ge \|B\| \ge 1$, hence there exists $i,j$ such that $|B_{ij}|\ge \frac1n$.  So $\|B^{\circ r}\| \ge \frac1{n^r}$.  Hence
$$ |\log\|A^{\circ r}\| - \log\|B^{\circ r}\|| \le n^{1+\frac1r} g(r,\delta) .$$
Next, for any $0 \le x\le 1$, we have $|x-x^r| \le \frac{r-1}{r^r}$ (do this by solving for the derivative equal to zero).  Hence, going via the Frobenius norm,  $\|A - A^{\circ r}\| \le n \frac{(r-1)}{r^r}$.  So arguing as before
$$ | \log \|A\| - \log\|A^{\circ r}\| | \le \frac{\|A - A^{\circ r}\|}{\|A\|} ,$$
that is $\log\|A^{\circ r}\| \le n \frac{(r-1)}{r^r}$.
Hence $f(n,\delta,r) \le \min\{2n \frac{(r-1)}{r^r},n^{1+\frac1r} g(r,\delta)\}$.
Maybe I made some mistakes here or there, but I think the idea is correct.  Certainly these estimates are by no means the best possible.
