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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

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

1 Answer 1

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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.

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  • $\begingroup$ I think for $C^2$ class of convex functions, you have the result that they're either monotonous or break in two monotonous parts. $\endgroup$
    – Keen
    Commented Oct 27, 2017 at 13:55
  • $\begingroup$ Fixed to reflect that I'm speaking of class $\mathcal C^2$ over $\mathbb R$, not some weird disconnected subset. $\endgroup$
    – Lærne
    Commented Feb 2, 2018 at 9:16

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