- 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
$$
|f(x)| \le \max\{|m|, |2f(0)-m|\} \le 2|f(0)|+|m| \quad \forall x \in \overline B(0, r).
$$
- 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,
$$
\frac{|f(x)-f(y)|}{\|x-y\|} \le \frac{4M}{\varepsilon}.
$$
I have found a cleaner approach for 2. as follows.
- 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.
- (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 $-t_1<0<t_1<t_2<\varepsilon$. 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.
Best Answer
By this result, $\partial f (a) \neq \emptyset$. By this result, $f'_+(a)[v]$ exists finite for all $v\in X$. By this result, $\langle x^*, v \rangle \leq f'_+ (a)[v]$ for all $v \in X$. Let's construct an optimal $x^*$. We need the following lemma.
We extend $f$ to the whole $X$ by setting $f(x) := +\infty$ for all $x \notin A$. We define $$ g: a +\mathbb R v \to \mathbb R, a+tv \mapsto f(a)+tf'_+(a)[v]. $$
Then $g(a) = f(a)$. By Lemma, there is a continuous affine extension $\hat g:X \to \mathbb R$ such that $\hat g \le f$. There are $x^* \in X^*$ and $\alpha \in \mathbb R$ such that $\hat g = x^*+\alpha$.
Because $\hat g (a) = f(a)$, $\hat g(x) = \langle x^*, x-a\rangle +f(a)$. Because $\hat g \le f$, then $f(x)-f(a) \ge \langle x^*, x-a\rangle$ for all $x\in X$.
We have $\hat g(a+v) = \langle x^*, v\rangle +f(a)$ and $g(a+v) = f(a)+f'_+(a)[v]$. Then $\langle x^*, v\rangle = f'_+(a)[v]$. This completes the proof.