# Does my proof generalize this exercise about continuity of convex lower semi-continuous function?

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 $$f(x) for all $$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 $$C$$.

• 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.
– daw
May 23, 2022 at 21:09
• @daw This may be of your interest. May 24, 2022 at 10:08

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.
• 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).