Suppose a nonlinear infinitely continous differentiable function $f:\mathbb{D}\mapsto \mathbb{R^+}$, where $\mathbb{D}\subset\left\{X|\text{rank}{X}=2,X\in\mathbb{R}^{3\times 3}\right\}$ is a continous, open, convex set,


How to define a matrix norm $\left\|\cdot\right\|\_p$ in $\mathbb{D}$, such that $\forall x\_1, x\_2\in \mathbb{D}$:
$$\lim\limits\_{\left\|x\_1-x\_2\right\|\_p\to 0}|f(x\_1)-f(x\_2)|=0,$$ $$\lim\limits\_{\left\|x\_1-x\_2\right\|\_p\to 0}\left\|x\_1-x\_2\right\|\_2=0$$ $$\quad\lim\limits\_{\left\|x\_1-x\_2\right\|\_p\to 0}\left\|x\_1-x\_2\right\|\_{\infty}=0,$$ $$\lim\limits\_{\left\|x\_1-x\_2\right\|\_p\to 0}\left\|x\_1-x\_2\right\|\_1=0,$$


and the $9\times 1$ vectors $y\_1 = x\_1(:),y\_2=x\_2(:)$(use Matlab/Octave operator here)
$$\lim\limits\_{\left\|x\_1-x\_2\right\|\_p\to 0}\left\|y\_1-y\_2\right\|\_2=0,$$


Must the calculation of such matrix norm be `NP`-hard?



How to determine the distance between two matrices under the meaning of a matrix function?
(This isn't an answer, but I'm not allowed to comment yet and I think this is a worthwhile question, even if it means violating the rules; i'll delete the post when the time comes)


You claim that usual matrix norms do not work because of the $|f(x\_1)-f(x\_2)|$ criterion, but if $f$ is differentiable, then given any $x\_1$ it is Lipschitz in some neighborhood of $x\_1$ (with a constant that may depend on $x\_1$), and hence $|f(x\_1)-f(x\_2)|\leq C \|x\_1(:)-x\_2(:)\|\_1$. Did I miss something?


