will maximum of convex set function intersect with Boundary?

Question

For a real-valued convex function with the domain is a closed convex set too.

Will the pre-image of the maximum value intersect with the Boundary of the domain?

Answer 1

Suppose $f$ is convex (with a convex domain of course) and a $\max$ occurs at an interior point $x^*$, then $f$ is constant.

Pick any $x \in C \setminus \{x^*\}$, then there is some $y$ such that $x^* \in (x,y)$, that is there is some $t \in (0,1)$ such that $x^* = tx+ (1-t)y$ and since $f(x^*) \le tf(x)+(1-t)f(y) \le f(x^*)$ we see that $f(x)=f(x^*)$. Hence $f$ is a constant.

If $x^*$ is a $\max$ then either it occurs at the boundary or at the interior. If it occurs in the interior it is a constant.

If the domain is closed then it will contain the boundary and hence if $f$ is constant then it will attain the $\max$ on the boundary.

Hence if the domain is closed the $\max$ is attained at a boundary point.

Answer 2

This result is not so simple to prove. I will write you a sketch of the proof, and only in the case when the domain $D$ is compact. In this way it is ensured the existence of a maximum of $f$. Denote by $\partial D$ the boundary of $D$.

Let $f: D \to \Bbb R$ be a convex function, and $D$ is a convex compact set ($D$ is a subset of some $\Bbb R^n$).

Then we need two lemmas:

Lemma 1: let $f:[a,b] \to \Bbb R$ be a convex function. Then the maximum of $f$ is attained at $x=a$ or at $x=b$.

proof: by contradiction, suppose that the maximum of $f$ is attained in the interior, and that it is stricty larger than $f(a)$ and $f(b)$. Try to show that this contradicts convexity.

Lemma 2: let $x \in D$. Then there exists $y \in \partial D$ such that $f(x) \le f(y)$.

proof: if $x \in \partial D$ ,then pick $y=x$. Otherwise, if $x \notin \partial D$, consider any line passing through $x$. This will intersect $\partial D$ at two points $a,b$. In other words, the interior point $x$ belongs to the segment $[a;b]$ and $a,b$ belong to the boundary of $D$. Then $f$ is convex on $[a;b]$ and you can apply Lemma 1.

Then you can conclude that the maximum of $f$ is achieved at some point on the boundary. Indeed, if the maximum if attained at some $x \in D$, then you can find $y \in \partial D$ with $f(y) \ge f(x)$, so that $y \in \partial D$ is maximum point of $f$.

Answer 3

Suppose $x^* $ is local maximizer of $f$, but there is an open ball around $x^*$ contained entirely in $X$, so that $x^*$ is an interior point. Strict convexity implies $$ f(x') - f(x^*) > \nabla f(x^*)(x'-x^*) $$ for all $x' \in X$; if $f$ isn't differentiable, substitute a selection from the subgradient into the argument above and below for the gradient, which always exists on the interior for convex functions (Rockafellar's Convex Analysis).

But take $$ x' = x^* - \varepsilon \dfrac{\nabla f(x^*)}{||\nabla f(x^*)||^2} $$ where $\varepsilon>0$ is small enough that $x'$ is in the ball around $x^*$. Substituting that in above yields $$ f(x')-f(x^*) = \varepsilon\dfrac{\nabla f(x^*)'\nabla f(x^*)}{||\nabla f(x^*)||^2} =\varepsilon> 0 $$ so that $f(x') > f(x^*)$ which is a contradiction. So a strictly convex function cannot have a maximizer at an interior point.

One problem is that a function that is convex can still have pretty ugly properties. For example, the function on $X=\{x \in \mathbb{R}^N : ||x|| \le 1\}$ that takes the value zero when $x$ is interior and 1 on the boundary is convex but not strictly convex. On the interior, it is constant, but jumps on the boundary. So the maximizers are the boundary, but that's exactly where the function is discontinuous, and the useful properties of the convexity inequality like interior continuity, a.e. differentiability, and the properties of the subgradient fail. So the weak convexity inequality is not as powerful as you might hope in general.