# Is a convex function always continuous?

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.

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

• I don't really see how you can prove that M exists Commented Oct 21, 2018 at 19:03
• Is there something specific you don't understand about my proof that $M$ exists? Commented Oct 21, 2018 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 ? Commented Oct 21, 2018 at 22:26
• Yes. That is the definition of $M$. Proof of existence is in the third paragraph, which I've now signposted more carefully. Commented Oct 21, 2018 at 22:37

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.

• 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. Commented Oct 19, 2018 at 13:31
• Indeed. I accepted it since it was the only answer, but would have preferred to have a complete solution Commented Oct 20, 2018 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. Commented Oct 20, 2018 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. Commented Oct 21, 2018 at 4:33
• @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 Commented Oct 21, 2018 at 9:36

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.

• Indeed, I've changed it a bit: same space and linear form, but with the supremum norm $||\cdot||_{\infty}$. Commented Oct 21, 2018 at 0:15
• Another example, taken from math.stackexchange.com/questions/99206/… : 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 Commented Oct 21, 2018 at 9:40
• @CharlesMadeline We should note of course that the domain of those explicit unbounded linear functionials is not a Banach space... Commented Oct 21, 2018 at 13:25

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.

• 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 $(0,0)$. Commented Oct 19, 2018 at 19:12