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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?

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Why would $g^q$ be convex? take e.g. $g(x)=2+\cos x$! –  Mercy Jul 13 '12 at 12:14
    
@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 . –  Theorem Jul 13 '12 at 12:33
    
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 $q \ge 1$. –  Mercy Jul 13 '12 at 12:36
    
@Mercy : I have added script where i came across –  Theorem Jul 13 '12 at 12:44
    
It says the function $t \mapsto F(t)= \int_\Omega g^q(x)[f^p(x)g^{-q}(x)]^tdx$ is convex. –  Mercy Jul 13 '12 at 12:48

1 Answer 1

Take $f^p(x)=g^q(x)=2+\cos x$, then $g^q[f^pg^{-q}]^t$ is not convex.

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