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

Question

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! :)

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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 $\overline B(a, r - \varepsilon)$.

Answer 1

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

2. 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 $1<t_1<t_2$ 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, $$ \frac{|f(x)-f(y)|}{\|x-y\|} \le \frac{4M}{\varepsilon}. $$

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

2. 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.