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 $p,q\in [1,+\mathrm{\infty }]$ and $\frac{1}{p}+\frac{1}{q}=1$.

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

$$|f(x)g(x)|\le {\parallel g\parallel }_{L^{\mathrm{\infty }}}|f(x)|.$$

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

$$\frac{{\parallel fg\parallel }_1}{{\parallel f\parallel }_p\cdot {\parallel g\parallel }_q}={\int }_X\frac{|f|}{{\parallel f\parallel }_p}\cdot \frac{|g|}{{\parallel g\parallel }_q}𝑑\mu \le \frac{1}{p}{\int }_X{\left(\frac{|f|}{{\parallel f\parallel }_p}\right)}^p𝑑\mu +\frac{1}{q}{\int }_X{\left(\frac{|g|}{{\parallel g\parallel }_q}\right)}^q𝑑\mu =\frac{1}{p}+\frac{1}{q}=1$$

hence the desired inequality holds

$${\int }_X|fg|={\parallel fg\parallel }_1\le {\parallel f\parallel }_p\cdot {\parallel g\parallel }_q={\left({\int }_X{|f|}^p\right)}^{\frac{1}{p}}{\left({\int }_X{|g|}^q\right)}^{\frac{1}{q}}.$$

If $x$ and $y$ are vectors in ${ℝ}^n$ or vectors in ${\mathrm{\ell }}^p$ and ${\mathrm{\ell }}^q$-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

$${\parallel x\parallel }_p={\left(\sum _k{|x_k|}^p\right)}^{\frac{1}{p}}$$

we have

$$\frac{\left|{\sum }_k{x}_k{y}_k\right|}{{\parallel x\parallel }_p\cdot {\parallel y\parallel }_q}\le \frac{{\sum }_k|x_k||y_k|}{{\parallel x\parallel }_p\cdot {\parallel y\parallel }_q}=\sum _k\frac{|x_k|}{{\parallel x\parallel }_p}\frac{|y_k|}{{\parallel y\parallel }_q}\le \frac{1}{p}\sum _k\frac{{|x_k|}^p}{{\parallel x\parallel }_p^p}+\frac{1}{q}\sum _k\frac{{|y_k|}^q}{{\parallel y\parallel }_q^q}=\frac{1}{p}+\frac{1}{q}=1.$$