Let $A \in \mathbb {F}^{m \times n}$ with $m< n,$ $b \in \mathbb{F}^m$ and let $x$ be unknown in $\mathbb{F}^n.$ Assume $0<p<1.$ Then $$\arg \min\limits_{x: Ax=b} \|x\|_0 = \lim\limits_{p \to 0} \{\arg \min\limits_{x:Ax=b} \|x\|_p^p\} $$

The reason I'm asking this question is that I am anticipating it as a *possible* objection that could be raised during my final defense. If this question I ask above is, in fact, true, I can take care of it easily by showing that attempts to "approximate" $\|x\|_0$ for an underdetermined system of linear equations via taking examples from a sequence of minimizers decreasing in $p$ to $0$ is itself intractable, because simply finding $\arg \min\limits_{x: Ax=b} \|x\|_p^p$ is itself NP Hard (via reducing from the partition problem as given in Foucart).

My problem, however, is whether this should even be a concern; is the limit above valid, or, if the way I stated this is in error, could it slightly be modified to be valid? I know that $$\lim\limits_{p \to 0} \|x\|_p^p = \|x\|_0$$ (crudely, nonzero numbers proceed to 1 while zeros stay zero) but i do not know how to show this rigorously as my analysis is rusty, and I do not know whether even showing this part would provide aid to my main problem at all.

Any help would be deeply appreciated!

nottrue that $\|x \|_p \to \|x\|_0$, as can be seen by the different scaling behaviour on both sides. What is true is $\|x\|_p^p \to \|x\|_0$. $\endgroup$ – PhoemueX Mar 12 '17 at 9:06