0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.