0
$\begingroup$

My professor gave us the following question:

Refute (with a simple example): Let 𝑓,𝑔:ℝ→ℝ be two convex functions. The composition ℎ≜𝑓∘𝑔 (that is, ℎ(𝑥)=𝑓(𝑔(𝑥))) is also a convex function.

But from what I can read online the composition of two convex functions is convex as well, what am I missing here?

The composition of two convex functions is convex

$\endgroup$
3
  • 2
    $\begingroup$ $g$ must be non-decreasing in your linked post. $\endgroup$
    – Aig
    Commented Mar 7 at 4:29
  • $\begingroup$ @Aig ohh the title is misleading thanks, so in my counter example I should look for deceasing g but wasn't successful finding one, can u guide me? (I will do the proof) $\endgroup$
    – David
    Commented Mar 7 at 4:31
  • 1
    $\begingroup$ $f(x)=-x$ is convex! $\endgroup$
    – geetha290krm
    Commented Mar 7 at 4:33

1 Answer 1

Reset to default
2
$\begingroup$

Let $f$ and $g$ be $f(x)=-x$, $g(x)=x^2$. Then $f$ and $g$ are convex (since they are twice continuously differentiable and its second derivatives are $\geq 0$). However, $f(g(x))=-x^2$ is not convex.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .