# In a normed vector space, the convex function $f:C \to \mathbb R$ is locally Lipschitz if and only if $f$ is upper bounded on an open subset of $C$

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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 $$C$$.

We need the following useful lemma:

Let $$(X, \| \cdot\|)$$ be a n.v.s. Recall that $$B(x, r)$$ (resp. $$\overline B(x, r)$$) denotes the open (resp. closed) ball of radius $$r$$ and center $$x$$. Fix $$a \in X, r>0, \varepsilon \in (0, r)$$, and $$m, M \in \mathbb R$$. Let $$f: \overline B(a, r) \to \mathbb R$$ be convex.

1. If $$f(x) \le m$$ for all $$x \in \overline B(a, r)$$, then $$|f(x)| \le |m| + 2|f(a)|$$ for all $$x \in B(a, r)$$.
2. If $$|f(x)| \le M$$ for all $$x \in \overline B(a, r)$$, then $$f$$ is $$\frac{4M}{\varepsilon}$$-Lipschitz on $$\overline B(a, r - \varepsilon)$$.

The implications $$(i) \Rightarrow(i i) \Rightarrow(i i i) \Rightarrow(v)$$ and $$(i i) \Rightarrow(i v) \Rightarrow(v)$$ are obvious. It remains to show that $$(v)$$ implies $$(i)$$.

Assume $$f$$ is upper bounded by $$m \in \mathbb R$$ on a nonempty open subset $$D$$ of $$C$$, i.e., $$f(x) \le m$$ for all $$x \in D$$. WLOG, we assume $$D := B(0, R)$$ for some $$R>0$$. By our Lemma, $$f$$ is locally Lipschitz on $$D$$.

In particular, $$f$$ is continuous at $$0$$. Then there are $$m>0$$ and $$R>r>0$$ such that $$|f(x)| \le m$$ for all $$x \in B(0, r)$$. We fix $$a\in C$$. Then $$f(x) = f \left ( \frac{2(x-a)+2a}{2} \right ) \le \frac{1}{2} f(2(x-a)) + \frac{1}{2}f(2a).$$

It follows that $$f(x) \le \frac{m+f(2a)}{2}$$ for all $$x\in B(b, r/2)$$. By our Lemma again, $$f$$ is locally Lipschitz at $$a$$. This completes the proof.