It looks not that complicated but I'm stuck in the middle.

$\mathbf{x}=[x_1,x_2, \cdots ,x_n]$.

  1. $g(\mathbf{x})=\exp(-\mathbf{x})$ is a decreasing, and convex function.
  2. $h(\mathbf{x})=x_1+x_2+x_3+\dotsm\;$ is a linear, increasing function, convex/concave.

So, $g(h(\mathbf{x}))$ is a convex function. Am I right so far?

and the problem is that $x_i$ is multiplied to $g(h(\mathbf{x}))$. $x_i$'s are positive and between 0 to 1 real value.

How can I prove $f_i(\mathbf{x})=x_i g(h(\mathbf{x}))$ is concave?

Thanks a lot.


You can do the hessian H of $f_i(x)$ and prove that H is negative semidefinite (if x'Hx ≤ 0 for all x).

the Hessian of $f_i$ is positive semi definite in this case which means that $f_i$ is convex if $x_i$ >0.

  • $\begingroup$ Thank you for your answer. I tried by deriving Hessian H of $f_i(\mathbf{x})$, but I failed to prove $\mathbf{v}'H\mathbf{v}\leq 0$ for all $\mathbf{v}$. Is there other way to prove this? I guess this function will be concave in 0<x<1.. at least quasiconcave. $\endgroup$ Jul 21 '18 at 6:11
  • $\begingroup$ Try: $ f_i( \lambda x +(1- \lambda )y) \leq \lambda f_i(x) +(1-\lambda) f_i(y)$ $g(x)=exp(-x)$ convex and you can prove that easily in $R^n$ So, $f_i( \lambda x+(1-\lambda) y) \leq x_i*(\lambda g_i(x)+ (1- \lambda) g_i(y) )$ $\endgroup$
    – seifedd
    Jul 21 '18 at 19:27
  • $\begingroup$ I hope that this helps !! $\endgroup$
    – seifedd
    Jul 21 '18 at 19:32
  • $\begingroup$ Thank you for your kindness!!, for $x,y\in R^n$ and $f_i(x)=x_i g(x)$, $f_i(\lambda x+ (1-\lambda)y) = (\lambda x_i +(1-\lambda) y_i) g_i(\lambda x + (1-\lambda)y) \leq (\lambda x_i +(1-\lambda) y_i) [\lambda g_i(x) +(1-\lambda)g_i(y)] $ $\endgroup$ Jul 22 '18 at 4:01
  • $\begingroup$ Then, i get $\lambda^2 x_i g_i(x) +(1-\lambda)^2 y_i g_i(y) +\lambda (1-\lambda)[y_i g_i(x) + x_i g_i(y)]=\lambda^2 f_i(x) + (1-\lambda)^2 f_i(y) + \lambda (1-\lambda)[y_i g_i(x) + x_i g_i(y)]$.. Here I stuck again..I don't know how the cross term $\lambda(1-\lambda)x_i f_i(y)$ can be dealt with. ;-( $\endgroup$ Jul 22 '18 at 4:08

Note that $f(x) = x e^{-x}$ is strictly convex for $x \ge 2$ (just compute $f''(x)$).

  • $\begingroup$ you are right. $f''(x) = (2-x)e^{-x}$. My consideration is only in $x\in[0,1]$ :) Thank you for your answer $\endgroup$ Jul 22 '18 at 4:34
  • $\begingroup$ You might want to add that to your question, otherwise this is an answer in the negative. $\endgroup$
    – copper.hat
    Jul 22 '18 at 4:39

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