Definitions (we could assume $f$ is convex for this problem too, if needbe):
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For a function $f\colon\mathbb{R}^n\to\left]-\infty,+\infty\right]$, the subdifferential at a point $x\in\mathbb{R}^n$, $\partial f(x)$, is the set of all $u\in\mathbb{R}^n$ satisfying
$$ (\forall y\in\mathbb{R}^n)\quad \langle u~|~y-x\rangle+f(x)\leq f(y).$$ -
The directional derivative of $f$ at $x$ in the direction $y$ is
$$f'(x;y)=\lim_{\alpha \searrow 0} \frac{f(x+\alpha y)-f(x)}{\alpha}$$ -
$f$ is differentiable if $f'(x;\cdot)$ is linear. In this case, the slope provided by the Riesz Representation Theorem is the gradient of $f$. In other words, differentiability means that there exists a vector $\nabla f(x)$ such that
$$\langle\nabla f(x)~|~\cdot\rangle=f'(x;\cdot).$$
Question:
If $\partial f$ is always single-valued, does this mean that $f$ is differentiable?
I have been able to show that, regardless of the uniqueness of a subgradient $u$, we always have $\langle u~|~\cdot\rangle\leq f'(x;\cdot)$. If I could get a matching upper bound, we would be done. Proof sketch of the lower bound is below.
Let $x\in\textrm{dom} f$ and $\alpha\in\left]0,1\right[$. Then
\begin{align}
&\langle u~|~x+\alpha y – x\rangle\leq f(x+\alpha y)-f(x)\\\\
\Leftrightarrow\quad & \langle u~|~y\rangle\leq \frac{f(x+\alpha y)-f(x)}{\alpha}.
\end{align}
Taking the limit as $\alpha$ goes to $0$ shows that $\langle u~|~\cdot\rangle\leq f'(x;\cdot)$.
Best Answer
If $\partial f(x)$ is single-valued, then $x$ is in the interior of $dom (f)$:
To see this, assume $x$ is a boundary point of $dom (f)$. As $\partial f(x) \ne \emptyset$, $f(x)<\infty$ follows, so $x\in dom f$. Since we are in a finite-dimensional setting, there is supporting hyperplane of $cl(dom(f))$ at $x$. I.e., there is $y \in \mathbb R^n$ with $y\ne0$ such that $$ y^T( x'- x) \le 0\quad \forall x'\in dom f. $$ Then if $g\in \partial f(x)$ then $g+ty\in \partial f(x)$ for all $t>0$.
Since $x$ is in the interior of $dom (f)$, and hence $f$ is continuous at $x$. Then a standard result shows that $f$ is differentiable: Take a direction $h\ne0$, then we have $f(x) + f'(x;th) \le f(x+th)$ for $t>0$. Separating $int(epi(f))$ and $\{ (x+th,f(x) + f'(x;th)): \ t>0\}$ one can show that $f'(x;h) \le g^Th$ , where $\partial f(x) = \{g\}$.
This result only works in finite-dimensions. It $f:X\to\mathbb R$ is such that $f(x) = |l(x)|^2$ with $l$ linear and discontinous, then $\partial f(0)=\{0\}$, but $f$ is not Gateaux differentiable at zero.