As @Célestin pointed out in a comment and as in here, the limit being uniform means
$\forall \varepsilon > 0, \exists \delta>0, \forall t \text{ s.t. } |t| < \delta, \forall v \text{ s.t. } \|v\|=1$, we have
$$
\left | \frac{f(a+t v)-f(a)}{t} - f^{\prime}(a)(v) \right | < \varepsilon.
$$
I also present the proof of the equivalence below.
Assume the Fréchet differential of $f$ at $a$ is $x^* \in X^*$. It is well-known that $f'(a) = x^*$. Assume the contrary that there is $\varepsilon>0$ such that for each $n \in \mathbb N$, there is $(t_n, v_n)$ such that $|t_n| < 1/n, \|v_n\|=1$, and
$$
\left | \frac{f(a+t_n v_n)-f(a)}{t_n} - f^{\prime}(a)(v_n) \right | \ge \varepsilon.
$$
We have $t_nv_n \to 0$, so by definition of Fréchet derivative, we get
$$
\lim _{n} \frac{f(a+t_n v_n)-f(a)-x^{*}(t_nv_n)}{\|t_nv_n\|} = \lim_n \frac{f(a+t_nv_n)-f(a)- x^*(t_nv_n)}{t_n} \frac{t_n}{|t_n|} =0 .
$$
Then
$$
\lim_n \left [\frac{f(a+t_nv_n)-f(a)}{t_n} - x^*(v_n) \right ]=0.
$$
This is a contradiction.
Assume that the Gâteaux differential of $f$ at $a$ is $f'(a) \in X^*$. Let $(h_n) \subset X$ such that $h_n \to 0$. We want to prove
$$
\lim_n \frac{f(a+h_n)-f(a)-f'(a)(h_n)}{\|h_n\|} = 0 .
$$
Let $v_n := h_n / \|h_n\|$ and $t_n :=\|h_n\|$. Then $t_n \to 0^+$ and $\|h_n\|=1$. The problem reduces to prove
$$
\lim_n \left [\frac{f(a+t_n v_n)-f(a)}{t_n} - f'(a)(v_n) \right ] = 0 .
$$
On the other hand, then the limit
$$
\lim_{t \to 0} \frac{f(a+t v_n)-f(a)}{t}=f^{\prime}(a)(v_n)
$$
is uniform for all $n$, i.e., $\forall \varepsilon > 0, \exists \delta>0, \forall t_m \text{ s.t. } |t_m| < \delta, \forall v_n$, we have
$$
\left | \frac{f(a+t_m v_n)-f(a)}{t_m} - f^{\prime}(a)(v_n) \right | < \varepsilon.
$$
This in turn implies $\forall t_m \text{ s.t. } |t_m| < \delta$, we have
$$
\left | \frac{f(a+t_m v_m)-f(a)}{t_m} - f^{\prime}(a)(v_m) \right | < \varepsilon.
$$
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
Below, we use $x^*(v)$ and $\langle x^*, v \rangle$ interchangeably. Notice that $f'_-(a)[v] = -f'_+(a)[-v]$, so (ii) is equivalent to (iii).
Let's prove (i) implies (ii). Let $x^* \in \partial f(a)$, i.e., $$ f(x)-f(a) \ge \langle x^*, x-a \rangle \quad \forall x\in A. $$ Then $f(a+tv)-f(a) \ge t\langle x^*, v \rangle$ and thus $$ \frac{f(a+t v)-f(a)}{t} \ge \langle x^*, v \rangle \quad \forall t>0. $$ The claim then follows by taking the limit $t \to 0^+$.
Let's prove (ii) implies (i). Notice that $f$ is convex, so the map $$ \varphi:(0, +\infty) \to \mathbb R, t \mapsto \frac{f(a+t v)-f(a)}{t} $$ is increasing. Assume $\langle x^*, v \rangle \le f_{+}^{\prime}(a)[v]$ for each $v \in X$. Then $\langle x^*, v \rangle \le \varphi (t)$ for all $t>0$. We pick $t=1$ and $v=x-a$. Then $$ \langle x^*, x-a \rangle \le \frac{f(a+1(x-a))-f(a)}{1}. $$ This completes the proof.