Is a convex function always continuous?

Question

It is well known that a convex function defined on $\mathbb{R}$ is continuous (it is even left and right differentiable. We can define a convex function for any normed vector space $E$: a function $f : E\mapsto \mathbb{R}$ is said to be convex iff $$f\big(\lambda x + (1-\lambda)y\big) \le \lambda f(x)+(1-\lambda)f(y)$$

I know that such a function is not necessarily continuous if $E$ has infinite dimension: $f$ can be a discontinuous linear form. For instance, if $E = \ell^2(\mathbb{N})$ the space of square summable sequences (endowed with the supremum norm $||\cdot||_{\infty}$ instead of its natural norm), and $f(u) = \sum \limits_{i \ge 1} \frac{u_i}{i}$, then $f$ is linear, thus convex, yet it is known that $f$ is not continuous.

Now my question is: what about finite dimensions? Does there exist a convex function $f : \mathbb{R}^2 \to \mathbb{R}$ which is not continuous?

I know that there are discontinuous functions from $\mathbb{R}^2$ to $\mathbb{R}$ that have derivatives in every direction (that's a good start since this is a necessary condition !) but I don't know any that is convex.

Answer 1

No: all convex functions $f: \mathbb R^2 \to \mathbb R$ are continuous.

Here's a slightly more general statement. Let $f : \mathbb R^n \to \mathbb R$ be a convex function, and let $\mathbf x^* \in \mathbb R^n$. We show that $f$ is continuous at $\mathbf x^*$.

Let $S = \{\mathbf y \in \mathbb R^n : \|\mathbf x^* - \mathbf y\| = 1\}$. Our first goal is to show that there's some $M\in \mathbb R$ such that $f(\mathbf y)\le M$ for all $\mathbf y \in S$.

To prove that $M$ exists: by Jensen's inequality, if $\mathbf x^{(1)}, \dots, \mathbf x^{(m)}$ are arbitrary points in $\mathbb R^n$, and $\mathbf x$ is a point in their convex hull, then $f(\mathbf x)$ is a weighted average of $f(\mathbf x^{(1)}), \dots, f(\mathbf x^{(m)})$, so it is bounded above by $\max\{f(\mathbf x^{(1)}), \dots, f(\mathbf x^{(m)})\}$. From there, it's enough to find finitely many points whose convex hull contains $S$: for example, the vertices of a hypercube circumscribed about $S$.

Now suppose we take some $\mathbf x$ close to $\mathbf x^*$. Let $r = \|\mathbf x^* - \mathbf x\|$; we may assume $r<1$, since ultimately we want to consider $\|\mathbf x^* - \mathbf x\|$ arbitrarily small.

On the line through $\mathbf x$ and $\mathbf x^*$, we can pick points $\mathbf y^-, \mathbf y^+ \in S$ such that they appear in the order $\mathbf y^-, \mathbf x^*, \mathbf x, \mathbf y^+$ on that line. They can be defined by: $$ \mathbf y^- = \mathbf x^* - \frac{\mathbf x - \mathbf x^*}{r} \text{ and } \mathbf y^+ = \mathbf x^* + \frac{\mathbf x - \mathbf x^*}{r}. $$ From this, we have

• $\mathbf x^* = \frac{r}{r+1} \mathbf y^- + \frac{1}{r+1} \mathbf x$, so $f(\mathbf x^*) \le \frac{r}{r+1} f(\mathbf y^-) + \frac{1}{r+1} f(\mathbf x)$, which gives us the lower bound $$f(\mathbf x) - f(\mathbf x^*) \ge r f(\mathbf x^*) - r f(\mathbf y^-) \ge r(f(\mathbf x^*) - M).$$
• $\mathbf x = r \mathbf y^+ + (1-r) \mathbf x^*$, so $f(\mathbf x) \le r f(\mathbf y^+) + (1-r)f(\mathbf x^*)$, which gives us the upper bound $$f(\mathbf x) - f(\mathbf x^*) \le r f(\mathbf y^+) - r f(\mathbf x^*) \le r(M - f(\mathbf x^*)).$$

Putting these together, we get $$ -r(M - f(\mathbf x^*)) \le f(\mathbf x) - f(\mathbf x^*) \le r(M - f(\mathbf x^*)) $$ which is the statement we need to prove continuity. (In the usual $\epsilon$-$\delta$ form: given $\epsilon > 0$, take $\delta = \frac{\epsilon}{M - f(\mathbf x^*)}$. Then if $\|\mathbf x^* - \mathbf x\| < \delta$, the inequalities above tell us that $|f(\mathbf x^*) - f(\mathbf x)| < \epsilon$.)

Answer 2

Corollary 10.1.1 of Convex Analysis by Rockafellar says all convex functions from $\mathbb R^{n}$ to $\mathbb R$ are continuous. The proof is very long and it is not worth reproducing the complete proof here. In the infinite dimensional case there are are discontinuous linear functionals.

Answer 3

Yes, if $E$ is an infinite-dimensional real Banach space then a discontinuous linear functional is a discontinuous convex function. But the map $f$ defined by $f(u)=\sum u_i/i$ is certainly continuous on $\ell_2$.

You're not going to be able to write down a formula for a discontinuous linear functional on a Banach space - it takes the Axiom of Choice to show such a thing exists.

Answer 4

Ref: "Introduction to Continuous Optimization" by Polyak.

Obs: Let ${ f : \mathbb{R} ^n \to \mathbb{R} }$ be a convex function. Let ${ x \in \mathbb{R} ^n }$ be in the convex hull of ${ x _1, \ldots, x _k \in \mathbb{R} ^n . }$ Then

$${ f(x) \leq \max _{ 1 \leq i \leq k} f(x _i) . }$$

Pf: Since ${ x }$ is in the convex hull we can write ${ x = \lambda _1 x _1 + \ldots + \lambda _k x _k }$ with ${ \lambda _i \geq 0 , \sum _{i = 1} ^k \lambda _i = 1 . }$ Note that by Jensen's inequality

$${ f(x) = f \left(\sum _{i = 1} ^k \lambda _i x _i \right) \leq \sum _{i = 1} ^k \lambda _i f(x _i) \leq \max _{ 1 \leq i \leq k} f(x _i) }$$

as needed.

Thm: Let ${ f : \mathbb{R} ^n \to \mathbb{R} }$ be a convex function, and ${ x _0 \in \mathbb{R} ^n . }$ Then ${ f }$ is continuous at ${ x _0. }$
Pf: WLOG we can assume ${ x _0 = 0 . }$ Let ${ (x _s) }$ be any sequence in ${ B(0, 1) \setminus \lbrace 0 \rbrace }$ converging to ${ 0 . }$ The goal is to show

$${ \text{To show: } \quad \lim _{s \to \infty} f(x _s) = f(0) . }$$

Consider the segment

$${ \left[ - \frac{x _s}{\lVert x _s \rVert}, 0, x _s, \frac{x _s}{\lVert x _s \rVert} \right] . }$$

Note that by convexity on ${ [0, x _s, \frac{x _s}{\lVert x _s \rVert}] }$

$${ f(x _s) \leq (1 - \lVert x _s \rVert) f(0) + \lVert x _s \rVert f\left( \frac{x _s}{\lVert x _s \rVert} \right) . }$$

Note that the unit sphere is in the convex hull of ${ \lbrace -1, 1 \rbrace ^n , }$ hence

$${ f \left( \frac{x _s}{\lVert x _s \rVert} \right) \leq \max _{x \in \lbrace -1 , 1 \rbrace ^n} f(x) =: M . }$$

Hence

$${ f(x _s) \leq (1 - \lVert x _s \rVert) f(0) + \lVert x _s \rVert M . }$$

Taking ${ \limsup }$ of the inequality, we have

$${ \limsup _{s \to \infty} f(x _s) \leq f(0) . }$$

Note that by convexity on ${ [ - \frac{x _s}{\lVert x _s \rVert}, 0, x _s] }$

$${ f(0) \leq \frac{\lVert x _s \rVert}{1 + \lVert x _s \rVert } f \left( - \frac{x _s}{\lVert x _s \rVert} \right) + \frac{1}{1 + \lVert x _s \rVert} f(x _s). }$$

Hence

$${ f(0) \leq \frac{\lVert x _s \rVert}{1 + \lVert x _s \rVert } M + \frac{1}{1 + \lVert x _s \rVert} f(x _s) . }$$

Taking ${ \liminf }$ of the inequality, we have

$${ f(0) \leq \liminf _{s \to \infty} f(x _s) . }$$

Combining both the inequalities, we have

$${ \limsup _{s \to \infty} f(x _s) = \liminf _{s \to \infty} f(x _s) = f(0) }$$

as needed. ${ \blacksquare }$

Answer 5

The answer to the question in your title is "no". Consider any convex function $f$ defined on (0,1), and extend $f$ to [0,1) by taking $f(0)$ as $1+\sup_{(0,1)}f(x)$. So we do need the condition that the domain of $f$ be open.

As for the question in the body, it is sufficient to show that $f$ is continuous iff given any function $g$ that takes $\mathbb R$ to a line in $\mathbb R^n$, the function $t \rightarrow f(g(t))$ is continuous.