# Show $l_o$-norm is a semi-continuous function?

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.

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

What happens if you take $n=1$ and consider $x_n = 1/n$?
$1 = \lim \inf ||1/n||_0 > f(0) = 0$