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Difference between revisions of "H�lder's Inequality"

m (proofreading)
m (Hölder's inequality moved to Hölder's Inequality: capitalization change)
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Revision as of 13:36, 11 July 2006

Hölder's inequality: If $p,q>1$, $1/p+1/q=1$, $f\in L^p, g\in L^q$ then $fg\in L^1$ and $||fg||_1\leq ||f||_p||g||_q$.

Proof: If $||f||_p=0$ then $f=0$ a.e. and there is nothing to prove. Case $||g||_q=0$ is similar. On the other hand, we may assume that $f(x),g(x)\in\mathbb{R}$ for all $x$. Let $a=\frac{|f(x)|^p}{||f||_p^p}, b=\frac{|g(x)|^q}{||g||_q^q},\alpha=1/p,\beta=1/q$. Young's Inequality gives us

$\frac{|f(x)|}{||f||_p}\frac{|g(x)|}{||g||_q}\leq \frac{1}{p}\frac{|f(x)|^p}{||f||_p^p}+\frac{1}{q}\frac{|g(x)|^q}{||g||_q^q}.$

These functions are measurable, so by integrating we get

$\frac{||fg||_1}{||f||_p||g||_q}\leq\frac{1}{p}\frac{||f(x)||^p}{||f||_p^p}+\frac{1}{q}\frac{||g(x)||^q}{||g||_q^q}=\frac{1}{p}+\frac{1}{q}=1$
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