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Suppose that $f:[0,1] \mapsto [0,\infty)$ is an increasing function with $f(0) = 0$. Then, is it true that $x\mapsto xf(x)$ is convex? It appears to be the case when $f$ is convex. However, I couldn't prove it for general $f$. I also couldn't find a counterexample. Any ideas?

themanwhooptimallystopped
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    $f(x)=\dfrac{\ln\left(1+2x^2\right)}{5+3x^3}$ is a function monotonically increasing on $[0,1]$ but $g(x)=xf(x)$ is not a convex function on $[0,1]$. I think it is possible to find simpler counterexamples. – Antonio Aug 21 '21 at 18:37
  • Thanks @Antonio, I've just found a simpler counterexample ( f(x) = 1-exp(-kx), k>2 ) here: https://math.stackexchange.com/a/1060458/960429 – themanwhooptimallystopped Aug 21 '21 at 18:42

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$f(x) = 1-\exp(-kx)$ for $k>2$ is a counterexample.

Theo Bendit
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themanwhooptimallystopped
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