1
$\begingroup$

Consider $1\le p,q < \infty , t\in \mathbb R , f,g>0$ then is $g(x)^q[f^pg^{-q} (x)]^t$ convex in the following context?

Thanks for help . enter image description here

  • 1
    Why would $g^q$ be convex? take e.g. $g(x)=2+\cos x$!2012-07-13
  • 0
    @Mercy : I meant to say that if $g(x) >0 , q>1 $ then $g(x)^q$ is convex . But i think the exp is not convex .2012-07-13
  • 0
    The problem is not the function $x \mapsto x^q$ but the function $g$. But if $g>0$ is convex, so is $g^q$ for $q \ge 1$.2012-07-13
  • 0
    @Mercy : I have added script where i came across2012-07-13
  • 0
    It says the function $t \mapsto F(t)= \int_\Omega g^q(x)[f^p(x)g^{-q}(x)]^tdx$ is convex.2012-07-13
  • 0
    Look at $h_x(t)=g^q(x)[f^p(x)g^{-q}(x)]^t$ as $h_x(t)=g^q(x)\exp(a(x)t)$, with $a(x)=f^p(x)g^{-q}(x)$. For every $x$ the function $t \mapsto h_x(t)$ is convex (a quick way to see this is that $h_x''(t)>0$ for every $t$). Therefore $F(t)=\int_\Omega h_x(t)d\mu(x)$ is also convex.2012-07-13
  • 0
    @Mercy : Mercy can you tell me how can you introduce exponential ?2012-07-13
  • 1
    There's a mistake with $a(x)$, it should be $a(x)=\ln[f^p(x)g^{-q}(x)]$ instead. I just used the identity $\alpha^\beta=\exp(\beta\ln\alpha)$ provided $\alpha>0$.2012-07-13

1 Answers 1