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?


closed as off-topic by Noah Stein, Andrey Rekalo, Chris Godsil, Stefan Kohl, Daniel Moskovich Feb 12 '14 at 14:13

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  • $\begingroup$ I must be missing something here... why doesn't one of the usual matrix norms work? $\endgroup$ – Federico Poloni Feb 9 '14 at 15:41
  • $\begingroup$ What's a continuous set? $\endgroup$ – Wlodek Kuperberg Feb 9 '14 at 17:17
  • $\begingroup$ by continuous, I mean connected $\endgroup$ – LCFactorization Feb 10 '14 at 0:24
  • $\begingroup$ Usual matrix norm does not work because of the first criterion: $|f(x_1)-f(x_2)|$ which is the most critical $\endgroup$ – LCFactorization Feb 10 '14 at 0:25
  • 1
    $\begingroup$ To talk of continuity and differentiability, you have explain what you mean if you don't mean the usual topology on $\mathbb{R}^9$. If you do, then all norms are equivalent. $\endgroup$ – username Feb 11 '14 at 18:47

(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?

  • $\begingroup$ My puzzle mainly lies in the conflicts between theoretical proof and numerical experiments. Probably because the difference between $x_1$ and $x_2$ has not been small enough? $\endgroup$ – LCFactorization Feb 12 '14 at 6:40
  • $\begingroup$ Do you mean that you did numerical experiments with some norm and the difference didn't seem to go to zero? Assuming you didn't have any bugs, it could be that a) the "norm" you used was degenerate, i.e. wasn't actually a norm, and could be $0$ even when $x_1\neq x_2$, or b) the function you were using was not sufficiently well-behaved to yield stable results at numerical precision. Perhaps you could edit your question with more details about what you were trying out? $\endgroup$ – martin Feb 12 '14 at 9:06
  • $\begingroup$ I will need time to make it clear whether I made any mistake. Thank you for suggestions. $\endgroup$ – LCFactorization Feb 12 '14 at 9:27

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