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Does the sum of a convex function and monotonically increasing function (not necessarily convex) yield a convex function?

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$g(x)=2x+\sin x$ is strictly increasing, $f(x)=\frac15x^2$ is strictly convex. Yet, $f''(x)+g''(x)=\frac25-\sin x$, so $f+g$ is not convex.

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The function $f(x)=0$ is a convex function. Thus, you would require that every monotone increasing function is convex.

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  • $\begingroup$ I understand the trivial example. I was more thinking of a convex function with a single minimum. Does that change anything? $\endgroup$ – Cowboy Trader Jun 7 '18 at 10:43
  • $\begingroup$ It doesn't, see the example by @SaucyO'Path $\endgroup$ – Severin Schraven Jun 7 '18 at 10:46
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    $\begingroup$ @CagdasOzgenc you can arbitrarily well approximate the zero function with a single-minimum function. $\endgroup$ – leftaroundabout Jun 8 '18 at 9:46
  • $\begingroup$ @leftaroundabout That's true. I had a bowl shape function in my mind that appears often in optimization, but the accepted answer covers that case as well. $\endgroup$ – Cowboy Trader Jun 8 '18 at 11:59
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For another example, which visibly fails to be convex and is in fact concave everywhere, add the strictly convex function $f(x) = e^{-x}$ and the strictly increasing function $g(x) = -2 e^{-x}$ to get $f(x) + g(x) = -e^{-x}$.

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No. Let $f(x) = x^2$ and $g(x) = \begin{cases}1 & x>0\\ 0 & x\leq 0\end{cases}$.

Then $f$ is convex, $g$ is monotone increasing, but $f+g$ is not continuous on the interior of its domain and so cannot be convex.

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Easiest counterexample I could think of was $f(x)=|x|-e^{-x}$:

Plot

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