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Hölders inequality is $\int |fg|dx \le||f||_p||g||_q$

Define $F(x)= \frac{f(x)}{(g(x))^{q/p}}$ and $\nu dx =g(x)^q$

and $\Phi(t)= |t|^p$, $p\in (1,\infty)$

Now lets find $\Phi(\int F(x) d\nu)= \frac{(\int fg)^p}{||g||_q^{qp}}$ and similarly find $\int \Phi (F(x))$ and apply jensens inequality .

My doubt is in the setting of $F(x)$ and $\nu dx $ Can anyone help me to make this proof correct .


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up vote 2 down vote accepted

We assume WLOG that $f$ and $g$ are non-negative. We should take $F(x):=\frac{f(x)}{(g(x))^{q/p}}\chi_{g\neq 0}$ and we have to make the change of measure, if $\mu$ is the initial one $\nu=\frac{g(x)^q}{\lVert g\rVert^q_q}\mu$, that is, if $h$ is integrable $$\int h(x)\nu(dx)=\int h(x)g(x)^qd\mu.$$ We can apply Jensen's inequality since $\nu$ is a probability measure. We have \begin{align} \Phi(\int F(x)d\nu(x))&=\left|\int F(x)d\nu(x)\right|^p\\ &=\left|\int F(x)\frac{g(x)^q}{\lVert g\rVert^q_q}d\mu(x)\right|^p\\ &=\frac 1{\lVert g\rVert^{qp}_q}\left|\int\frac{f(x)}{(g(x))^{q/p}}\frac{g(x)^q}{\lVert g\rVert^q_q}\chi_{g\neq 0}d\mu(x)\right|^p\\ &=\frac 1{\lVert g\rVert^{2qp}_q}\left|\int fgd\mu(x)\right|^p, \end{align} since $q-q/p=q/q=1$. We also have \begin{align} \int\Phi(F(x))d\nu(x)&=\int|F(x)^p|d\nu\\ &=\int |F(x)|^p\frac{g(x)^q}{\lVert g\rVert^q_q}d\mu(x)\\ &=\int \left|\frac{f(x)}{(g(x))^{q/p}}\chi_{g\neq 0}\right|^p\frac{g(x)^q}{\lVert g\rVert^q_q}d\mu(x)\\ &=\frac 1{\lVert g\rVert_q^q}\int |f(x)|^pd\mu(x). \end{align}

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Sir , can u elaborate more , i didn't understand much. – Theorem Jul 13 '12 at 8:02
thank you sir.... – Theorem Jul 13 '12 at 8:19
You are welcome! – Davide Giraudo Jul 13 '12 at 8:21

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