# Let $A$ be an open convex subset of a normed space. If $f:A \to \mathbb R$ is convex and continuous at $a$, then $f'_+(a, \cdot)$ is Lipschitz

This thread is meant to record a question that I feel interesting during my self-study. I'm very happy to receive your suggestion and comments.

See: SE blog: Answer own Question and MSE meta: Answer own Question. Anyway, it is written as problem.

Let $$X$$ be a normed space, $$A \subset X$$ an open set, $$f: A \to \mathbb{R}$$ a function, and $$a \in A$$ a point. For a "direction" $$v \in X$$, we shall consider the right directional derivative $$f_{+}^{\prime}(a, v)$$, the left directional derivative $$f_{-}^{\prime}(a, v)$$, and the (bilateral) directional derivative $$f^{\prime}(a, v)$$, which are defined by: \begin{aligned} f_{+}^{\prime}(a, v) &=\lim _{t \to 0+} \frac{f(a+t v)-f(a)}{t} \\ f_{-}^{\prime}(a, v) &=\lim _{t \to 0-} \frac{f(a+t v)-f(a)}{t} \\ f^{\prime}(a, v) &=\lim _{t \to 0} \frac{f(a+t v)-f(a)}{t}. \end{aligned}

We shall say that $$f$$ is:

• Gâteaux differentiable at $$a$$ if $$f^{\prime}(a, \cdot) \in X^*$$ (that is, $$f^{\prime}(a, \cdot)$$ is everywhere defined, real-valued, linear and continuous);
• Fréchet differentiable at $$a$$ if there exists $$x^{*} \in X^{*}$$ such that $$\lim _{h \to 0} \frac{f(a+h)-f(a)-x^{*}(h)}{\|h\|}=0 .$$

Theorem: Let $$X$$ be a normed space, $$A \subset X$$ an open set, $$f: A \to \mathbb{R}$$ convex. Fix $$a\in A$$. Then

• (a) $$f_{+}^{\prime}(a, v)$$ exists finite for each $$v \in X$$, and the functional $$p=f_{+}^{\prime}(a, \cdot)$$ is sublinear on $$X$$.
• (b) $$-p(-v) \leq p(v)$$ for each $$v \in X$$.
• (c) The set $$V=\left\{v \in X \mid f^{\prime}(a, v) \text { exists }\right\}=\{v \in X \mid -p(-v)=p(v)\}$$ is linear (that is, $$V$$ is a subspace of $$X$$ ) and the restriction $$\left.p\right|_{V}$$ is linear.
• (d) If $$f$$ is continuous at $$a$$, then $$p$$ is Lipschitz (in particular, $$\left.p\right|_{V} \in V^{*}$$ ).

A direct corollary is as follows.

Let $$X$$ be a normed space, $$A \subset X$$ an open set, $$f: A \to \mathbb{R}$$ convex and continuous at $$a\in A$$. If $$f^{\prime}(a, v)$$ exists for all $$v \in X$$, then $$f$$ is Gâteaux differentiable at $$a$$.

• (a) Fix $$v \in X$$. There is $$\varepsilon>0$$ such that $$a+tv \subset A$$ for all $$t$$ such that $$|t| < \varepsilon$$. Consider the map $$\varphi: (-\varepsilon, \varepsilon) \to \mathbb R, t \mapsto f(a+tv).$$

Then $$\varphi$$ is convex. Let $$-t_1<0. By the chordal slope lemma, we get $$\frac{\varphi(0) - \varphi(-t_1)}{t_1} \le\frac{\varphi(t_1) - \varphi(0)}{t_1} \le \frac{\varphi(t_2) - \varphi(0)}{t_2} \le \frac{\varphi(t_2) - \varphi(t_1)}{t_2-t_1}.$$

It follows that the map $$(0, \varepsilon) \to \mathbb R, t \mapsto \frac{f(a+tv)-f(a)}{t}$$ is non-decreasing and bounded from below. Then $$f'_+(a, v)$$ exists for all $$v\in X$$. We have $$\frac{f(a+t(\lambda v))-f(a)}{t} = \lambda \frac{f(a+(\lambda t)v)-f(a)}{\lambda t} \quad \forall \lambda >0,$$ so $$p$$ is positively homogeneous. Let $$v_1, v_2 \in X$$. We have $$f(a+t(v_1+v_2)) = f \left ( \frac{(a+2tv_1) + (a+2tv_2)}{2} \right ) \le \frac{1}{2} f(a+2tv_1) + \frac{1}{2} f(a+2tv_2)$$ Then $$\frac{f(a+t(v_1+v_2))-f(a)}{t} \le \frac{f(a+2tv_1)-f(a)}{2t} + \frac{f(a+2tv_2)-f(a)}{2t}.$$

We take the limit $$t \to 0^+$$ and get that $$p$$ is sub-additive.

• (b) As shown above, $$\frac{f(a)-f(a-t_1v)}{t_1} = \frac{\varphi(0) - \varphi(-t_1)}{t_1} \le\frac{\varphi(t_1) - \varphi(0)}{t_1} = \frac{f(a+t_1v) - f(a)}{t_1}.$$ Then $$-\frac{f(a+t_1(-v)) - f(a)}{t_1} \le \frac{f(a+t_1v) - f(a)}{t_1}.$$ The claim then follows by taking the limit $$t_1 \to 0^+$$.

• (c) Notice that $$f'_-(a, v) = -f'_+(a, -v) = -p(-v)$$ and that $$v\in V \iff -v\in V$$.

Let $$\lambda \in \mathbb R$$ and $$v \in V$$. If $$\lambda \ge 0$$, then $$p(-\lambda v) = \lambda p(-v) = -\lambda p(v) = -p(\lambda v)$$ because $$v\in V$$. Now consider $$\lambda<0$$. We have $$p(-\lambda v) = p(-(-\lambda) (-v)) = - p((-\lambda)(-v)) = - p(\lambda v)$$ because $$-\lambda>0$$ and $$-v\in V$$. So $$\lambda v \in V$$.

Let $$v_1, v_2 \in V$$. By (b), we have $$p(v_1+v_2) \ge -p(-v_1-v_2)$$. Because $$p$$ is sub-additive, $$p(v_1+v_2) \le p(v_1)+p(v_2) = -p(-v_1)-p(-v_2) = -[p(-v_1)+p(-v_2)] \le -p(-v_1-v_2)$$. Hence $$p(v_1+v_2) = -p(-v_1-v_2)$$. So $$v_1+v_2 \in V$$. It follows that $$V$$ is a linear subspace.

Let $$\lambda \in \mathbb R$$ and $$v \in V$$. If $$\lambda \ge 0$$, then $$p(\lambda v) =\lambda p(v)$$. Now consider $$\lambda<0$$. We have $$p(\lambda v) =p((-\lambda)(-v)) = -\lambda p(-v) =-\lambda(-p(v)) = \lambda p(v)$$ because $$-\lambda>0$$ and $$-v\in V$$.

Let $$v_1, v_2 \in V$$. We have $$p(v_1+v_2) \le p(v_1)+p(v_2)$$ by sub-additivity. On the other hand, $$p(v_1+v_2)=p(-(-v_1-v_2))=-p(-v_1-v_2) \ge -[p(-v_1)+p(-v_2)] = -[-p(v_1)-p(v_2)] = p(v_1)+p(v_2)$$. Hence $$p(v_1+v_2)=p(v_1)+p(v_2)$$. Hence $$p$$ is linear on $$V$$.

• (d) We need the following Lemma.

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 $$C$$.

It follows that $$f$$ is $$L$$-Lipschitz on $$A$$ for some $$L>0$$. We have \begin{align} |p(v_1)-p(v_2)| &= \left | \lim_{t\to 0^+} \frac{f(a+tv_1) - f(a+tv_2)}{t} \right | \\ &= \lim_{t\to 0^+} \frac{|f(a+tv_1) - f(a+tv_2)|}{t} \\ &\le \lim_{t\to 0^+} \frac{Lt\|v_1-v_2\|}{t} \\ &= L\|v_1-v_2\|. \end{align}

This completes the proof.