When does a convex set have a unique outward normal direction?

Question

Let $C$ be a closed, nonempty, and convex set (in a real Hilbert space $\mathcal{X}$), and let $c\in C$ be a point on its boundary. When will the normal cone $N_Cc$ have a unique (nonzero) direction? My definition of the normal cone at $c$ is $N_Cc=\{x\in\mathcal{X} | (\forall d\in C) \langle x|d-c\rangle\leq 0\}$. I already know this holds for many simple sets like balls and half-spaces, but I want a more general result.

This excerpt from Rockafellar/Wets describes precisely the notion I'm looking for:

When $x$ is any point on a curved boundary of the set $C$, the [normal cone] reduces to a ray which corresponds to the outward normal direction indicated classically.

However, the book provides no definition of a "curved" boundary. I'm looking for a rigorous characterization of this class of set. Further references for geometry/convex analysis are greatly appreciated!

Answer 1

Here is a reference that might help you. https://ximera.osu.edu/mklynn2/multivariable/content/03_14_gradient/gradient

The first Theorem in the section ''The gradient and level sets'' states:

Consider a function $f:\mathbb{R}^{n}\to\mathbb{R}$, and suppose $f$ is of class $C^{1}$. For some constant $c$, consider the level set $S = \{x\ |\ f(x)=c\}$. Then, for any point, $x\in S$, the gradient $\nabla f(x)$ is perpendicular to S.

If the function $f$ is a convex function then the set $K = \{ x\ |\ f(x)\leq c \}$ is a convex set and $S$ is the boundary of $K$, i.e., $S = \partial K$. The (unique) gradient $\nabla f(x)$ is the normal at the point $x\in S$.

Answer 2

Ler $p\in \partial C$. By the Hahn Banach Theorem, there exists a hyperplane $H$ meeting $C$ exactly at $p$. Locally around $p$, $\partial C$ is the graph of a function $\varphi $ from $H$ to the line $H^{\perp}$. The function $\varphi $ is convex, and its local behavior control the geometry of $C$ near $p$. For instance if it is derivable at $p$, then the normal cone is the line $H^{\perp}$.