# When will the supporting hyperplane of a convex set coincide with the tangent?

Due to the supporting hyperplane theorem, a convex set $C$ in a separable topological space has supporting hyperplance at each of its boundary points. The theorem only guarantees its existence, now I want to discuss the uniqueness of the supporting hyperplane.

A special situation where uniqueness holds is when the supporting hyperplane at $p\in\partial C$ coincides with the tangent pf $\partial C$ at $p$. For example, when the $\partial C$ is piecewise linear, then the supporting hyperplanes must coincide with tangent planes except at vertices of $C$.

There are sufficient conditions, say $C$ being strictly convex, which guarantees the uniqueness of the supporting hyperplane. But strict convexity does not lead to the conclusion that the supporting hyperplane coincide with tangents. Moreover. since the supporting hyperplane theorem is no more than Hahn-Banach theorem, I was wondering

(i)Is there a necessary condition (on $\partial C$, say some sort of algebraic regularity; conditions on $C$ are also fine but less interesting) about the boundary of convex set $C$ to make the supporting hyperplane unique?

(ii)Is there a necessary condition (on $\partial C$, say some sort of algebraic regularity) about the boundary of convex set $C$ such that the supporting hyperplane of $C$ at any point of $p\in\partial C$ coincide with the tangent plane of $\partial C$ at $p$?

For my purpose, I only want to know the case $C\subset\mathbb{R}^n$ with the usual topology. ($C$ does not have to be closed convex set but only convex, closedness is too strong for my purpose.)

Additionally, how will the situation changes if $C\subset H$ for a general (infinite dimensional) Hilbert space? Or even more general, in a Banach space $B$.

• Keyword: smooth point (= point de lissité). – Francois Ziegler Mar 28 '17 at 14:21
• The finite-dimensional case is answered in detail in Rockafellar's book Convex Analysis. Basically, it is the case that the supporting hyperplane is unique at a point $p \in \partial C$ if and only if it is the tangent hyperplane at $p$ if and only if $\partial C$ is differentiable at $p$. – Deane Yang Mar 28 '17 at 14:44
• @DeaneYang After a careful look into the relevant sections of Rockafellar's treatise, the relevant result claimed in the book seems to be Theorem 18.8 and those in Sec 25. However, I did not recognize where he proved "iff", so if would be more explanatory if you could add your citations. Moreover, in Sec 18, Rockafellar actually showed that closed convex set in R^n is always envelope of tangent half-spaces, which does not serve as a necessary condition to my OP since I did not put closedness in OP. Appreciated. – Henry.L Mar 28 '17 at 15:34
• The way I recall it is this: Rotating the body if needed, the boundary near a given point is the graph of a convex function. Each element of the subdifferential of $f$ at that point defines a supporting hyperplane. $f$ is differentiable at the point if and only if there is only one element in the subdifferential and therefore only one supporting hyperplane. I don't have a copy of the book handy, but this is what I recall. – Deane Yang Mar 28 '17 at 19:04
• @DeaneYang Thanks for the clarification,I need some time to figure out which result yo are referring to, you are also welcomed to compose an answer! – Henry.L Mar 28 '17 at 20:05

I think it goes like this: Assume $C$ has nonempty interior. Rotating the convex set if needed, the boundary $\partial C$ near any given point $p \in \partial C$ can be written as the graph of a convex function. More or less by definition, a hyperplane is a supporting hyperplane at $p$ if and only if it corresponds to an element of the subdifferential of $f$. Moreover, $f$ is differentiable at $p$ if and only if the subdifferential contains only one element. In that case, the unique supporting hyperplane is the tangent hyperplane of $\partial C$ at $p$.

In summary, a supporting hyperplane is the tangent hyperplane at $p \in\partial C$ if and only if $f$ is differentiable at $p$ if and only if the subdifferential of $f$ at $p$ contains only one element if and only if there is only one supporting hyperplane at $p$.

This, I believe, follows by results stated and proved in the book Convex Analysis by Rockafellar.

EDIT: As HenryL points out, this holds only if the supporting hyperplane actually touches the boundary. If $C$ is noncompact, then it is possible this does not happen.

• If any upvoters happen to find the results relevant to this answer please comment, I am still searching for it, thanks! – Henry.L Mar 29 '17 at 17:08
• The claim " a hyperplane is a supporting hyperplane at $p$ if and only if it corresponds to an element of the subdifferential of $f$" is in Theorem 23.2(p.215 of [Rockafellar]); The claim "$f$ is differentiable at $p$ if and only if the subdifferential (at $p$) contains only one element." correspond to Theorem 25.1(p.242 of [Rockafellar]), however, it seems to require $f$ to be finite to make this claim hold. Therefore your claim is a partial answer which excludes the possible boundary at $\infty$. – Henry.L Apr 5 '17 at 12:58
• That's a good point. – Deane Yang Apr 5 '17 at 14:07
• Let $H(C,p)$ = family of supporting hyperplanes of $C$ at $p$, then, for any $r > 0$, $H(C,p) = H(C\cap \overline{ B(p,r)}, p)$, where $B(p,r)$ is the ball of radius $r$ centered at $p$. So, without loss of generality, one may assume $C$ is bounded. – VictorZurkowski Apr 5 '17 at 16:11
• I have no doubt that locally, $\partial C$ can be represented as the graph of a function defined on a supporting hyperplane, but the supporting hyperplane has to be chosen with some care. For example, when $C=\{(x,y) | 0 \le x \text{ and } x^2 \le y \}$, one cannot represent $\partial C$ near $(0,0)$ as the graph of a function defined on either $y=0$ or $x=0$ (both of which are supporting hyperplanes of $C$ at $p=(0,0)$ ). – VictorZurkowski Apr 5 '17 at 16:24