Important inequalities: Hölder's inequality

Young's inequality already splits one product by conjugate exponents. Can the absolute value of a sum of products exceed the product of the two Euclidean lengths?

Freeze the exponent at 2

Both arrows are real, $N$ is frozen at $2$, and the exponent is frozen at $2$.

For $x,y\in {\mathbb{C}}^N$,

$(x|y):={\sum }_{k=1}^N\overline{x_k}{y}_k.$

The bar on $x_k$ is the complex conjugate. For these real arrows it leaves each coordinate as it is, and $\Vert x{\Vert }_2$ is the Euclidean length. With the exponent frozen at $2$,

$|(x|y)|\le \Vert x{\Vert }_2\Vert y{\Vert }_2.$

This is the Schwarz inequality.

Let the two exponents be conjugates

Let $p>1$, and let $q$ be fixed by $1/p+1/q=1$.

$\Vert x{\Vert }_p:=({\sum }_{k=1}^N|x_k{|}^p{)}^{1/p},\Vert x{\Vert }_{\infty }:={\max }_{1\le k\le N}|x_k|.$

For $x,y\in {\mathbb{C}}^N$ and real $p,q>1$ with $1/p+1/q=1$,

$|(x|y)|\le \Vert x{\Vert }_p\Vert y{\Vert }_q.$

The $\infty$-length is the largest absolute coordinate, used when a sum of vectors is measured at $p=\infty$ on the Minkowski inequality.

The same pairing, integrated

For real $p,q>1$ with $1/p+1/q=1$, whenever the integrals on the right exist,

$\left|{\int }_{G}f(x)g(x)dx\right|\le ({\int }_G|f(x){|}^{p}dx{)}^{1/p}({\int }_G|g(x){|}^{q}dx{)}^{1/q}.$

When $p=q=2$ this is the Schwarz inequality for integrals. The next board freezes the region as the interval $[0,1]$ and the two functions as $f(t)=at$ and $g(t)=1$.