I need to show that this function is convex: $$f(x) = (m-1)e^{-\frac{s}{m+x}} + \left(1-\frac{s\cdot x}{m+x}\right)^{1+1/x},$$ for all values of the parameters $m\geq 1$ and $s\geq 0$, on the domain $x\in(0,u]$, where $$ u = \left\{\begin{array}{ll} 1 & \text{ if }\ s\leq m+1 \\ \frac{m}{s-1} & \text{otherwise.} \end{array} \right. $$

The problem is that this function is not the sum of two convex functions in general (the first term with the exponential can be nonconvex). However, my simulations suggest that the second term somehow compensate, and that $f$ is always convex for $m\geq1$ and $s\geq 0$.

The second derivative of $f$ is very ugly, I don't know if we can do something with it. However, the second derivative has a nice limit in $0$: $$ \lim_{x\to 0} f''(x) = e^{-s/m} \frac{s^2}{12m^4} (4m(s+3)+3s(s-4)), $$ and we can easily show that this is positive when $m\geq 1$ and $s\geq 0$. I don't know it this can help, but at least it comforts me in thinking that $f$ can be convex.

Do you have any thought about what we could use to prove the convexity of this function? Maybe there is an elegant, short proof that does not use the second derivative of $f$ ?

  • $\begingroup$ i would compute the second derivative! $\endgroup$ – Dr. Sonnhard Graubner Mar 22 '18 at 10:53
  • $\begingroup$ As I said, the second derivative is very ugly. And even in very simple cases, I don't find it obvious that it is nonnegative. Consider the case m=s=1. Then, f''(x) has the same sign as $(2x+3)x^2 + (x+1)\log(x+1)[\log(x+1)-4x]$... $\endgroup$ – guigux Mar 23 '18 at 20:04

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