# proof of Hölder inequality

First we prove the more general form (in measure spaces).

Let $(X,\mu)$ be a measure space and let $f\in L^{p}(X)$, $g\in L^{q}(X)$ where $p,q\in[1,+\infty]$ and $\frac{1}{p}+\frac{1}{q}=1$.

The case $p=1$ and $q=\infty$ is obvious since

 $|f(x)g(x)|\leq\|g\|_{L^{\infty}}|f(x)|.$

Also if $f=0$ or $g=0$ the result is obvious. Otherwise notice that (applying http://planetmath.org/node/YoungInequalityYoung inequality) we have

 $\frac{\|fg\|_{1}}{\|f\|_{p}\cdot\|g\|_{q}}=\int_{X}\frac{|f|}{\|f\|_{p}}\cdot% \frac{|g|}{\|g\|_{q}}\,d\mu\leq\frac{1}{p}\int_{X}\left(\frac{|f|}{\|f\|_{p}}% \right)^{p}\,d\mu+\frac{1}{q}\int_{X}\left(\frac{|g|}{\|g\|_{q}}\right)^{q}\,d% \mu=\frac{1}{p}+\frac{1}{q}=1$

hence the desired inequality holds

 $\int_{X}|fg|=\|fg\|_{1}\leq\|f\|_{p}\cdot\|g\|_{q}=\left(\int_{X}|f|^{p}\right% )^{\frac{1}{p}}\left(\int_{X}|g|^{q}\right)^{\frac{1}{q}}.$

If $x$ and $y$ are vectors in ${\mathbb{R}}^{n}$ or vectors in $\ell^{p}$ and $\ell^{q}$-spaces we can specialize the previous result by choosing $\mu$ to be the counting measure on ${\mathbb{N}}$.

In this case the proof can also be rewritten, without using measure theory, as follows. If we define

 $\|x\|_{p}=\left(\sum_{k}|x_{k}|^{p}\right)^{\frac{1}{p}}$

we have

 $\frac{\left|\sum_{k}x_{k}y_{k}\right|}{\|x\|_{p}\cdot\|y\|_{q}}\leq\frac{\sum_% {k}|x_{k}||y_{k}|}{\|x\|_{p}\cdot\|y\|_{q}}=\sum_{k}\frac{|x_{k}|}{\|x\|_{p}}% \frac{|y_{k}|}{\|y\|_{q}}\leq\frac{1}{p}\sum_{k}\frac{|x_{k}|^{p}}{\|x\|_{p}^{% p}}+\frac{1}{q}\sum_{k}\frac{|y_{k}|^{q}}{\|y\|_{q}^{q}}=\frac{1}{p}+\frac{1}{% q}=1.$
Title proof of Hölder inequality ProofOfHolderInequality 2013-03-22 13:31:16 2013-03-22 13:31:16 paolini (1187) paolini (1187) 10 paolini (1187) Proof msc 15A60 msc 46E30 proof of Hölder inequality proof of Holder’s inequality