# Lower semicontinuous of Minkowski functional

Let $$P(x)$$ is Minkowski functional of $$C$$ which is closed convex set and $$\theta \in C$$. After proving $$C=\{ x: P(x)\le 1\}$$ and for any $$\alpha >0$$, $$\alpha C$$ is closed, how to get $$P$$ is lower semicontinuous ?

I think the lower semicontinuous is $$\liminf\limits_{x\rightarrow x_0} P(x)=P(x_0).$$ I can't see any connection between the lower semicontinuous and closure of $$\alpha C$$.

(1). Show that for any $$λ>0$$, the set $$\{ x \in X \colon P(x) \leq λ\}=λC$$.
(2). Conclude that the sublevel sets $$L_λ= \{ x \in X \colon P(x) \leq λ\}$$ are closed, for every $$λ>0$$. This is equivalent to saying that $$P$$ is lower semicontinuous.
• I can prove the (1). But I can't get the (2). Namely, I can't prove that the closure of sublevel sets is equal to $P$ is lower semicontinuous. Mar 19 '21 at 6:03