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How to prove that $log(1+e^{-x})$ is a convex function?

[from comment]: I have tried with the basic definition of convex function..... like $f(ax+by) \leq af(x)+bf(y)$... but was not able to solve further....

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  • $\begingroup$ What have you attempted so far? Did you sketch the graph? $\endgroup$
    – Joffan
    Feb 4 '15 at 18:24
  • $\begingroup$ I have tried with the basic definition of convex function..... like f(ax+by) <= af(x)+bf(y).. but not able to solve furhter..... $\endgroup$
    – tourism
    Feb 4 '15 at 18:26
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Hint: it is a fact that if a function is twice differentiable and has positive second derivative everywhere, then it is convex.

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    $\begingroup$ can i prove it with the basic definition of convex function ? $\endgroup$
    – tourism
    Feb 4 '15 at 18:24
  • $\begingroup$ I don't think it will be easy.. You have to check that $$\log(1+e^{-(tx+(1-t)y)})\leq t \log(1-e^{-x})+(1-t)\log(1+e^{-y}).$$ Maybe trying to reverse engineering it gets you something. $\endgroup$
    – Ivo Terek
    Feb 4 '15 at 18:26
  • $\begingroup$ I have tried it by first expanding log(1+x) series and then to simplify I have taken the log of both sides... but it's giving me 0=0. $\endgroup$
    – tourism
    Feb 4 '15 at 18:31
  • $\begingroup$ It is not an equivalence, since a convex function is continuous for sure but not necessarily twice differentiable. Consider, for instance, $f(x)=|x|$ or $f(x)=|x|^3$. $\endgroup$ Feb 4 '15 at 18:39
  • $\begingroup$ Oh, I see. Thanks for pointing. I have rephrased it. $\endgroup$
    – Ivo Terek
    Feb 4 '15 at 18:42
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$$\frac{d}{dx}\log(1+e^{-x}) = \frac{-e^{-x}}{1+e^{-x}} = -\frac{1}{e^x+1} $$ is an increasing function, since $e^x$ is an increasing function.

This gives that $\log(1+e^{-x})$ is a convex function.

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