# A discontinuous midpoint convex function

Let’s recall that a real function $$f: \mathbb R \to \mathbb R$$ is called convex if for all $$x, y \in \mathbb R$$ and $$\lambda \in [0,1]$$ we have $f((1- \lambda) x + \lambda y) \le (1- \lambda) f(x) + \lambda f(y)$ $$f$$ is called midpoint convex if for all $$x, y \in \mathbb R$$ $f \left(\frac{x+y}{2}\right) \le \frac{f(x)+f(y)}{2}$ One can prove that a continuous midpoint convex function is convex. Sierpinski proved the stronger theorem, that a real-valued Lebesgue measurable function that is midpoint convex will be convex.

Can one find a discontinuous midpoint convex function? The answer is positive but requires the Axiom of Choice. Why? Because Robert M. Solovay constructed a model of Zermelo-Fraenkel set theory (ZF), exclusive of the axiom of choice where all functions are Lebesgue measurable. Hence convex according to Sierpinski theorem. And one knows that convex functions defined on $$\mathbb R$$ are continuous.

Referring to my previous article on the existence of discontinuous additive map, let’s use a Hamel basis $$\mathcal B = (b_i)_{i \in I}$$ of $$\mathbb R$$ considered as a vector space on $$\mathbb Q$$. Take $$i_1 \in I$$, define $$f(i_1)=1$$ and $$f(i)=0$$ for $$i \in I\setminus \{i_1\}$$ and extend $$f$$ linearly on $$\mathbb R$$. $$f$$ is midpoint convex as it is linear. As the image of $$\mathbb R$$ under $$f$$ is $$\mathbb Q$$, $$f$$ is discontinuous as explained in the discontinuous additive map counterexample.

Moreover, $$f$$ is unbounded on all open real subsets. By linearity, it is sufficient to prove that $$f$$ is unbounded around $$0$$. Let’s consider $$i_1 \neq i_2 \in I$$. $$G= b_{i_1} \mathbb Z + b_{i_2} \mathbb Z$$ is a proper subgroup of the additive $$\mathbb R$$ group. Hence $$G$$ is either dense of discrete. It cannot be discrete as the set of vectors $$\{b_1,b_2\}$$ is linearly independent. Hence $$G$$ is dense in $$\mathbb R$$. Therefore, one can find a non vanishing sequence $$(x_n)_{n \in \mathbb N}=(q_n^1 b_{i_1} + q_n^2 b_{i_2})_{n \in \mathbb N}$$ (with $$(q_n^1,q_n^2) \in \mathbb Q^2$$ for all $$n \in \mathbb N$$) converging to $$0$$. As $$\{b_1,b_2\}$$ is linearly independent, this implies $$\vert q_n^1 \vert, \vert q_n^2 \vert \underset{n\to+\infty}{\longrightarrow} \infty$$ and therefore $\lim\limits_{n \to \infty} \vert f(x_n) \vert = \lim\limits_{n \to \infty} \vert f(q_n^1 b_{i_1} + q_n^2 b_{i_2}) \vert = \lim\limits_{n \to \infty} \vert q_n^1 \vert = \infty.$

# A discontinuous real convex function

Consider a function $$f$$ defined on a real interval $$I \subset \mathbb R$$. $$f$$ is called convex if: $\forall x, y \in I \ \forall \lambda \in [0,1]: \ f((1-\lambda)x+\lambda y) \le (1-\lambda) f(x) + \lambda f(y)$

Suppose that $$I$$ is a closed interval: $$I=[a,b]$$ with $$a < b$$. For $$a < s < t < u < b$$ one can prove that: $\frac{f(t)-f(s)}{t-s}\le \frac{f(u)-f(s)}{u-s}\le\frac{f(u)-f(t)}{u-t}.$ It follows from those relations that $$f$$ has left-hand and right-hand derivatives at each point of the interior of $$I$$. And therefore that $$f$$ is continuous at each point of the interior of $$I$$.
Is a convex function defined on an interval $$I$$ continuous at all points of the interval? That might not be the case and a simple example is the function: $\begin{array}{l|rcl} f : & [0,1] & \longrightarrow & \mathbb R \\ & x & \longmapsto & 0 \text{ for } x \in (0,1) \\ & x & \longmapsto & 1 \text{ else}\end{array}$

It can be easily verified that $$f$$ is convex. However, $$f$$ is not continuous at $$0$$ and $$1$$.

# A counterexample to Krein-Milman theorem

In the theory of functional analysis, the Krein-Milman theorem states that for a separated locally convex topological vector space $$X$$, a compact convex subset $$K$$ is the closed convex hull of its extreme points.

For the reminder, an extreme point of a convex set $$S$$ is a point in $$S$$ which does not lie in any open line segment joining two points of S. A point $$p \in S$$ is an extreme point of $$S$$ if and only if $$S \setminus \{p\}$$ is still convex.

In particular, according to the Krein-Milman theorem, a non-empty compact convex set has a non-empty set of extreme points. Let see what happens if we weaken some hypothesis of Krein-Milman theorem. Continue reading A counterexample to Krein-Milman theorem

# A topological vector space with no non trivial continuous linear form

We consider here the $$L^p$$- spaces of real functions defined on $$[0,1]$$ for which the $$p$$-th power of the absolute value is Lebesgue integrable. We focus on the case $$0 < p < 1$$. We'll prove that those $$L^p$$-spaces are topological vector spaces on which there exists no continuous non-trivial linear forms (i.e. not vanishing identically). Continue reading A topological vector space with no non trivial continuous linear form

# An unbounded convex not containing a ray

We consider a normed vector space $$E$$ over the field of the reals $$\mathbb{R}$$ and a convex subset $$C \subset E$$.

We suppose that $$0 \in C$$ and that $$C$$ is unbounded, i.e. there exists points in $$C$$ at distance as big as we wish from $$0$$.

The following question arises: “does $$C$$ contains a ray?”. It turns out that the answer depends on the dimension of the space $$E$$. If $$E$$ is of finite dimension, then $$C$$ always contains a ray, while if $$E$$ is of infinite dimension $$C$$ may not contain a ray. Continue reading An unbounded convex not containing a ray

# A compact whose convex hull is not compact

We consider a topological vector space $$E$$ over the field of the reals $$\mathbb{R}$$. The convex hull of a subset $$X \subset E$$ is the smallest convex set that contains $$X$$.

The convex hull may also be defined as the intersection of all convex sets containing X or as the set of all convex combinations of points in X.

The convex hull of $$X$$ is written as $$\mbox{Conv}(X)$$. Continue reading A compact whose convex hull is not compact

# A compact convex set whose extreme points set is not close

Let’s remind that an extreme point $$c$$ of a convex set $$C$$ in a real vector space $$E$$ is a point in $$C$$ which does not lie in any open line segment joining two points of $$C$$.

## The specific case of dimension $$2$$

Proposition: when $$C$$ is closed and its dimension is equal to $$2$$, the set $$\hat{C}$$ of its extreme points is closed.
Continue reading A compact convex set whose extreme points set is not close

# An empty intersection of nested closed convex subsets in a Banach space

We consider a decreasing sequence $$(C_n)_{n \in \mathbb{N}}$$ of non empty closed convex subsets of a Banach space $$E$$.

If the convex subsets are closed balls, their intersection is not empty. To see this let $$x_n$$ be the center and $$r_n > 0$$ the radius of the ball $$C_n$$. For $$0 \leq n < m$$ we have $$\Vert x_m-x_n\Vert \leq r_n – r_m$$ which proves that $$(x_n)_{n \in \mathbb{N}}$$ is a Cauchy sequence. As the space $$E$$ is Banach, $$(x_n)_{n \in \mathbb{N}}$$ converges to a limit $$x$$ and $$x \in \bigcap_{n=0}^{+\infty} C_n$$. Continue reading An empty intersection of nested closed convex subsets in a Banach space