# Lower semi-continuous is norm topology but not in weak topology

I know of the following result:

If $V$ is a Banach space and $f$ : $V \to \mathbb{R}$ is convex and lower semi-continuous in norm topology, then $f$ is lower semi-continuous in weak topology.

I'm looking for an example of a non-convex function which is lower semi-continuous in norm topology, but not lower semi-continuous in weak topology.

Let $C$ be a subset such that it is norm closed but it is not weak closed. Then, let us define $f:=1_{C^c}$, defined by $f(x)=1$ if $x\notin C$ and $f(x)=0$ otherwise. Then, for each real number $r$, the set
$V_r:=\{x\:;\: f(x)\leq r\}$
is always norm closed, but however $V_\frac{1}{2}=C$ is not weak closed. Thus, $f$ is not weakly lower semicontinuous.
• What is $C$? Did you mean $U$? – Sahiba Arora Oct 6 '16 at 19:39
• Yes, $C=U$, sorry. – user178826 Oct 6 '16 at 19:40