# are all continuous convex functions increasing?

Hi

My question is about convex functions

I've been reading Boyd's convex optimization and a question came to me. I guess if $f$ is a convex function then $f$ is increasing .

I guess it is conceptually but I didn't find a mathemathical theorem to prove it .

What constraints that we assume makes the gusse be true ?

I would appreciate any counter examples or constraints :)

And I mean by an increasing definition that $f$ is increasing if : $a \ge b$ then $f(a) \ge f(b)$

And also assume that $f$ is continuous

• No, they can be decreasing then increasing.
– user65203
Commented Oct 27, 2017 at 12:55
• Consider $e^{-x}.$
– zhw.
Commented Oct 27, 2017 at 14:03

No consider a continuous convex increasing function $f$. Then $g:x\mapsto f(-x)$ is a convex decreasing function. This is because a mirror transformation on points $a$ and $b$ will keep the mirrored graph of $f$ below the mirrored segment $[(a,f(a)),(b,f(b))]$, but obviously a mirrored increasing graph becomes decreasing. For example consider $e^{-x}$.
You can even be non-monotonic. Consider something like $x^2$.
For functions of class $\mathcal C^2(\mathbb R)$, what convexity means though, is that your tangent will all raise. So once it becomes increasing, the function cannot become decreasing later; that would give you a smaller derivative.
• I think for $C^2$ class of convex functions, you have the result that they're either monotonous or break in two monotonous parts.
• Fixed to reflect that I'm speaking of class $\mathcal C^2$ over $\mathbb R$, not some weird disconnected subset. Commented Feb 2, 2018 at 9:16