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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 $\alpha C$.

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Hints:

(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.

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