Let $\| \cdot \|_0$ be $0$-norm on $\mathbb{R}^n$ defined as $$ \|\cdot \|_0 := \text{the number of nonzero elements in}\,\,x, \forall x\in \mathbb{R}^n $$ Show that $\| \cdot \|_0$ is lower semi-continuous on $\mathbb{R}^n$.

We can use the original definition of lower semi-contnuity as

Def: A function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is lower semi-continuous at $x_*$ if $\forall \epsilon >0$ there exist $\delta >0 $ such that $\forall x$ with $\|x-x_*\|< \delta$,

$$f(x_*)-\epsilon \leq f(x)$$

Or we could show it using the following Lemma:

Lemma: A function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is lower semi-continuous at $x_*$ if and only if for any sequence $(x_n) \rightarrow x_*$ : $f(x_*) \leq \lim \inf f(x_n)$.

I think to show it we need to use the above Lemma.

  • $\begingroup$ your definition is that of a continuous function $\endgroup$ – LinAlg Sep 11 '18 at 22:40
  • $\begingroup$ Sorry, I had made a mistake. The first definition was not the correct one. $\endgroup$ – Saeed Sep 12 '18 at 0:46

What happens if you take $n=1$ and consider $x_n = 1/n$?

$1 = \lim \inf ||1/n||_0 > f(0) = 0$

  • $\begingroup$ What do you mean by that? $\endgroup$ – Saeed Sep 12 '18 at 17:14
  • $\begingroup$ I mean that what you need to show is not necessarily true $\endgroup$ – LinAlg Sep 12 '18 at 18:09
  • $\begingroup$ @Saeed did my hint answer your question? $\endgroup$ – LinAlg Oct 10 '18 at 22:15

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