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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.

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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.

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  • $\begingroup$ What is $C$? Did you mean $U$? $\endgroup$ – Sahiba Arora Oct 6 '16 at 19:39
  • $\begingroup$ Yes, $C=U$, sorry. $\endgroup$ – user178826 Oct 6 '16 at 19:40

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