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 $$ \vert f(x) g(x)\vert \le \Vert g\Vert_{L^\infty} \vert f(x)\vert. $$ Also if $f=0$ or $g=0$ the result is obvious. Otherwise notice that (applying Young inequality) we have $$ \frac{\Vert fg\Vert_1}{\Vert f\Vert_p\cdot \Vert g\Vert_q} = \int_X \frac{\vert f\vert}{\Vert f\Vert_p} \cdot \frac{\vert g\vert}{\Vert g\Vert_q} \, d\mu \le \frac 1 p \int_X \left(\frac{\vert f\vert}{\Vert f\Vert_p}\right)^p\, d\mu + \frac 1 q \int_X \left(\frac{\vert g\vert}{\Vert g\Vert_q}\right)^q\, d\mu = \frac 1 p + \frac 1 q = 1 $$ hence the desired inequality holds $$ \int_X \vert f g\vert = \Vert fg\Vert_1 \le \Vert f \Vert_p \cdot \Vert g\Vert_q = \left(\int_X \vert f\vert^p\right)^{\frac 1 p} \left(\int_X \vert g\vert ^q\right)^{\frac 1 q}. $$

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 $$ \Vert x \Vert_p = \left(\sum_k \vert x_k\vert^p\right)^{\frac 1 p} $$ we have $$ \frac{\left\vert \sum_k x_k y_k \right\vert}{\Vert x\Vert_p \cdot \Vert y \Vert_q} \le \frac{\sum_k \vert x_k\vert \vert y_k\vert}{\Vert x\Vert_p\cdot \Vert y\Vert_q} = \sum_k \frac{\vert x_k\vert}{\Vert x\Vert_p} \frac{\vert y_k\vert}{\Vert y\Vert_q} \le \frac 1 p \sum_k \frac{\vert x_k\vert^p}{\Vert x\Vert_p^p} + \frac 1 q \sum_k \frac{\vert y_k\vert^q}{\Vert y\Vert_q^q} = \frac 1 p + \frac 1 q = 1. $$