# Bounded from above convex function on a normed vector space is locally Lipschitz

Disclaimer: This thread is meant to record. See: SE blog: Answer own Question and MSE meta: Answer own Question. Anyway, it is written as problem. Have fun! :)

I'm trying to generalize this result to n.v.s.

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 $$\overline B(a, r - \varepsilon)$$.
• I am struggling to understand what $x \in B(x,r)$ means. Commented May 22, 2022 at 19:03
• @copper.hat Thank you so much! It should be $x \in B(a,r)$. Commented May 22, 2022 at 19:04

1. If $$f(x) \le m$$ for all $$x \in \overline B(a, r)$$, then $$|f(x)| \le |m| + 2|f(a)|$$ for all $$x \in B(x, r)$$.

WLOG, we assume $$a:=0$$. By convexity of $$f$$, we get $$f(0) \le \frac{1}{2} f(x) + \frac{1}{2} f(-x) \quad \forall x\in \overline B(0, r).$$

Notice that $$x \in \overline B(0, r) \iff -x \in \overline B(0, r)$$, so $$f(x) \ge 2f(0)-f(-x) \ge 2f(0)-m \quad \forall x \in \overline B(0, r).$$

It follows that $$|f(x)| \le \max\{|m|, |2f(0)-m|\} \le 2|f(0)|+|m| \quad \forall x \in \overline B(0, r).$$

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

WLOG, we assume $$a:=0$$. Fix $$x,y \in \overline B(0, r - \varepsilon)$$ such that $$x\neq y$$. Consider $$\varphi: \mathbb R \to \mathbb R, t \mapsto \| t(y-x)+x \|.$$

Then $$\varphi$$ is continuous. Let $$T := \{t \in \mathbb R \mid \varphi(t) \le r\}$$. There are $$t_1, t_2 \in T$$ such that $$1 and $$\varphi_1 :=\varphi (t_1)= r - \varepsilon/2$$ and $$\varphi_2 :=\varphi(t_2) = r$$. Then

• $$\|\varphi_1-y\| = \| [t_1(y-x)+x] -y\| = (t_1-1) \|x -y\|$$.
• $$\|x-\varphi_1\| = \|x- [t_1(y-x)+x]\| = t_1 \| x-y \|$$.

It follows that $$y = \frac{\|\varphi_1-y\| x + \|y-x\| \varphi_1}{\|x-\varphi_1\|}.$$

By convexity of $$f$$, we have $$f(y) \le \frac{\|\varphi_1-y\| }{\|x-\varphi_1\|} f(x) + \frac{\|y-x\|}{\|x-\varphi_1\|} f(\varphi_1),$$ which implies $$\frac{f(y)-f(x)}{\|y-x\|} \le \frac{f(\varphi_1)-f(y)}{\|\varphi_1-y\|}.$$

Similarly, we get $$\frac{f(\varphi_1)-f(y)}{\|\varphi_1-y\|} \le \frac{f(\varphi_2)-f(\varphi_1)}{\|\varphi_2-\varphi_1\|}.$$

It follows that $$\frac{f(y)-f(x)}{\|y-x\|} \le \frac{f(\varphi_2)-f(\varphi_1)}{\|\varphi_2-\varphi_1\|} \le \frac{4M}{\varepsilon}.$$

By symmetry, we obtain $$\frac{f(x)-f(y)}{\|x-y\|} \le \frac{f(\varphi_2)-f(\varphi_1)}{\|\varphi_2-\varphi_1\|} \le \frac{4M}{\varepsilon}.$$

Finally, $$\frac{|f(x)-f(y)|}{\|x-y\|} \le \frac{4M}{\varepsilon}.$$

I have found a cleaner approach for 2. as follows.

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

WLOG, we assume $$a:=0$$. Fix $$x,y \in \overline B(0, r - \varepsilon)$$ such that $$x\neq y$$. We fix $$\lambda>0$$ such that $$z_\lambda := y + \lambda \frac{y-x}{\|y-x\|} \in \overline B(0, r).$$

It follows that $$y = t_\lambda x+(1-t_\lambda) z_\lambda \quad \text{with} \quad t_\lambda := \frac{\lambda}{\lambda+\|y-x\|}.$$

By convexity of $$f$$, we get $$f(y) \le t_\lambda f(x)+(1-t_\lambda)f(z_\lambda),$$ which implies $$\frac{f(y)-f(x)}{1-t_\lambda} \le \frac{f(z_\lambda) - f(y)}{t_\lambda}.$$

It follows that $$\frac{f(y)-f(x)}{|y-x|} \le \frac{f(z_\lambda) - f(y)}{\lambda} \le \frac{2M}{\lambda}.$$

We have $$\|z_\lambda\| \le \|y\| + \lambda \le r - \varepsilon+\lambda.$$

For $$z_\lambda \in \overline B(0, r)$$, it suffices to pick $$\lambda>0$$ such that $$r - \varepsilon+\lambda< r$$, i.e., $$\lambda<\varepsilon$$. Hence $$\frac{f(y)-f(x)}{|y-x|} \le \frac{2M}{\lambda} \le \frac{2M}{\varepsilon}.$$

By symmetry, we also have $$\frac{f(x)-f(y)}{|x-y|}\le \frac{2M}{\varepsilon}.$$

This completes the proof.