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 . 
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1Why would $g^q$ be convex? take e.g. $g(x)=2+\cos x$! – HorizonsMaths Jul 13 '12 at 12:14
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@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
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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 $q \ge 1$. – HorizonsMaths Jul 13 '12 at 12:36
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@Mercy : I have added script where i came across – Theorem Jul 13 '12 at 12:44
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It says the function $t \mapsto F(t)= \int_\Omega g^q(x)[f^p(x)g^{-q}(x)]^tdx$ is convex. – HorizonsMaths Jul 13 '12 at 12:48
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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. – HorizonsMaths Jul 13 '12 at 13:14
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@Mercy : Mercy can you tell me how can you introduce exponential ? – Theorem Jul 13 '12 at 16:42
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1There'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 $\alpha>0$. – HorizonsMaths Jul 13 '12 at 16:54
