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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 $C$.
Akira
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1 Answers1

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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.

Akira
  • 18,439