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I'm trying to solve below question (Proposition 0.7.) in this lecture note.

Let $C$ be an open convex subset of a normed space $X$ and $f: C \to \mathbb{R}$ convex.

(a) If $f$ is u.s.c., then $f$ is continuous on $C$.

(b) If $X$ is a Banach space and $f$ l.s.c., then $f$ is continuous on $C$.

In my below proof for (b), I don't need to impose that $X$ is a Banach space. I suspect I made some subtle mistakes. Could you have a check on my attempt?


We need the following useful lemma.

Let $(X, \| \cdot\|)$ be a normed vector space, $C$ its open convex subset, and $f:C \to \mathbb R$ convex. Then the following statements are equivalent.

  • (i) $f$ is locally Lipschitz on $C$;
  • (ii) $f$ is continuous on $C$;
  • (iii) $f$ is continuous at some point of $C$;
  • (iv) $f$ is locally bounded on $C$;
  • (v) $f$ is upper bounded on a nonempty open subset of $C$.

(a) Fix $\varepsilon>0$ and $a \in C$. By upper semi-continuity of $f$, there is $r>0$ such that $B(a,r) \subset C$ and $f(x) <f(a) + \varepsilon$ for all $x \in B(a,r)$. This means $f$ is upper bounded on $B(a,r)$. The claim then follows from our Lemma (v $\implies$ ii).

(b) If $f$ is l.s.c., then $-f$ is u.s.c. By (a), $-f$ is continuous on $C$. It follows that $f$ is continuous on $C$.

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    $\begingroup$ If $f$ is convex then $-f$ is not convex in general. I am not sure whether completeness is important for the result, I would expect it is not. $\endgroup$
    – daw
    May 23, 2022 at 21:09
  • $\begingroup$ @daw This may be of your interest. $\endgroup$
    – Akira
    May 24, 2022 at 10:08

1 Answer 1

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As @daw pointed out in a comment, $f$ is convex does not imply $-f$ is convex. Below is the proof by the author of the note.


(b)

  • If $C=X$, then $F_{n} := \{x \in C \mid f(x) \leq n\}$.

  • If $C \neq X$, then $$ \begin{align} F_{n} &:= \left\{x \in C \,\middle\vert\, f(x) \leq n, \operatorname{dist}(x, X \setminus C) \geq \frac{1}{n}\right\} \\ &= \underbrace{\{x \in C\mid f(x) \le n\}}_{\text{closed in }C} \bigcap \underbrace{\left \{x \in X \mid \operatorname{dist}(x, X \setminus C) \geq \frac{1}{n} \right\}}_{\text{contained in } C \text{ and closed in } X}. \end{align} $$ The sets $(F_{n})_{n \in \mathbb{N}}$ are closed in $C$; but they are also closed in $X$ since $\overline{F_{n}} \subset C$. By the Baire Category Theorem, there exists $k \in \mathbb{N}$ such that $F_{k}$ has a nonempty interior. This implies that $f$ is upper bounded on a nonempty open set. The claim then follows from our Lemma (v $\implies$ ii).

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