In $\mathbb{R}^n$ we have the statement that a convex function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is differentiable if and only if $\partial f(x) = \{\nabla f(x)\}$. I.e. if and only if the gradient is the unique subgradient.
Do we have a similar statement for functions spaces $\mathcal{F}$? I.e., that if a convex function $f:\mathcal{F}\rightarrow \mathbb{R}$ is Frechet differentiable that this is equivalent to the subdifferential having a unique element, the Frechet derivative?
"Almost" yes if $f$ is continuous. Let $X$ be a Banach space and $f \colon X \to \overline{\mathbb R}$ be a convex proper function.
If $f$ is Gateaux differentiable at $x_0 \in dom(f)$ then $\partial f (x_0)=\{f'(x_0)\}$. If $f$ is continuous at $x_0$ and $\partial f (x_0)$ is a singleton then $f$ is Gateaux differentiable at $x_0$ and $\partial f (x_0)=\{f'(x_0)\}$.
If $f$ is Fréchet differentiable at $x_0 \in dom(f)$ then $\partial f (x_0)=\{f'(x_0)\}$. If $f$ is continuous at $x_0$, $\partial f (x_0)=\{x_0^*\}$ is a singleton and $\partial f$ is continuous at $x_0$ in the sense that, whenever $x_n \to x_0$ and $x_n^* \in \partial f(x_n)$ then $x_n^* \to x_0^*$, then $f$ is Fréchet differentiable at $x_0$ and $\partial f (x_0)=\{f'(x_0)\}$.
You can find the proofs here.
Edit: As gerw mentioned in the comments below, when $f \colon X \to \overline{\mathbb R}$ is proper, convex and lower semicontinuous then $f$ is continuous at every point of its interior domain. In fact, $x_0 \in \operatorname{int} \operatorname{dom} (f)$ if and only if $f$ is continuous at $x_0$, see Proposition 5.1.21 in Applied nonlinear functional analysis: An introduction. Hence for real-valued convex functions, semi continuity is equivalent to continuity.
