0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.