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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.

charmd
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4 Answers4

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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 $|f(\mathbf x^*) - f(\mathbf x)| < \epsilon$.)

Misha Lavrov
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  • I don't really see how you can prove that M exists – charmd Oct 21 '18 at 19:03
  • Is there something specific you don't understand about my proof that $M$ exists? – Misha Lavrov Oct 21 '18 at 20:20
  • When do you first define $M$ ? When reading your first bullet point (btw there is a typo: it is $f(\mathbf{x}^) \le \frac{r}{r+1}...$ and not $\ge$), I had the feeling that you assumed that $M$ was defined as $\sup \limits_{||y-x^||=1} |f(y)|$. Is that it ? – charmd Oct 21 '18 at 22:26
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    Yes. That is the definition of $M$. Proof of existence is in the third paragraph, which I've now signposted more carefully. – Misha Lavrov Oct 21 '18 at 22:37
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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.

Kavi Rama Murthy
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    I think to deserve a bunch of upvotes, an answer should also add at least some explanation rather than just stating a result. Basically, this is little more than a link-only answer. – leftaroundabout Oct 19 '18 at 13:31
  • Indeed. I accepted it since it was the only answer, but would have preferred to have a complete solution – charmd Oct 20 '18 at 14:29
  • I did not expect 6 upvotes for my answer. But it is not at all uncommon to find strange voting patterns, more so with downvoting. – Kavi Rama Murthy Oct 20 '18 at 23:16
  • @CharlesMadeline What extra information are you looking for? I will try to include more information if you tell me what is missing in my answer. – Kavi Rama Murthy Oct 21 '18 at 04:33
  • @Kavi Rama Murthy, thanks for your continued interest. I just would have likes to see 3-4 main ppins in the proof (e.g. a set of points with a special property whuch is introduced, or if the proof shows that the restrictions of $f$ to segments are uniformly Lipschitz on compact sets, or something like that). That would be enough for me to accept your answer oc – charmd Oct 21 '18 at 09:36
  • And btw there are no explicit discontinuous linear form on a Banach space but you can still find some in infinite vector space: see my new example on $\ell^2$ when changing the norm – charmd Oct 21 '18 at 09:37
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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.

David C. Ullrich
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  • Indeed, I've changed it a bit: same space and linear form, but with the supremum norm $||\cdot||_{\infty}$. – charmd Oct 21 '18 at 00:15
  • Another example, taken from https://math.stackexchange.com/questions/99206/discontinuous-linear-functional : on $E = \mathcal{C}^1([0,1],\mathbb{R})$, with the supremum norm $||f||:=\sup\limits_{x\in[0,1]} |f(x)|$. Then $L:f \mapsto f'(0)$ is discontinuous – charmd Oct 21 '18 at 09:40
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    @CharlesMadeline We should note of course that the domain of those explicit unbounded linear functionials is not a Banach space... – David C. Ullrich Oct 21 '18 at 13:25
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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.

Acccumulation
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    Unfortunately, the sufficient statement you give is false. For one example (taken from here), let $f(x,y) = \frac{y}{x^2}(1 - \frac{y}{x^2})$ if $0 < y < x^2$ and $f(x,y) = 0$ otherwise. This is continuous away from $(0,0)$, and continuous along every line through $(0,0)$. So it is continuous along all lines everywhere. But it is not continuous at $(0,0)$. – Misha Lavrov Oct 19 '18 at 19:12