Important inequalities: the dual inequality

How far can the inner product $(x|y)$ rise above $F(x)$? Whatever is left over has to be payable from $y$ alone.

The surplus of one power

The board freezes three extra assumptions: the dimension is $1$, the exponent $p$ stays above $1$, and $F(x)=|x{|}^p/p$.

The highest point of that surplus sits on the bar $|y{|}^q/q$. The exponent $q$ is fixed by $1/p+1/q=1$.

For a known function $F:{\mathbb{R}}^N\to \mathbb{R}$, the dual function $F^{\ast }:{\mathbb{R}}^N\to \mathbb{R}$ records the greatest surplus of the inner product over $F$:

$F^{\ast }(y):={\sup }_{x\in {\mathbb{R}}^N}((x|y)-F(x)).$

Every surplus stays at most that height. Moving $F(x)$ across the comparison gives

$(x|y)\le F(x)+F^{\ast }(y),x,y\in {\mathbb{R}}^N.$

This comparison is (0.25). On the line the pairing $(x|y)$ is the product $xy$. With $F(x)=|x{|}^p/p$ and $F^{\ast }(y)=|y{|}^q/q$,

$xy\le \frac{|x{|}^p}{p}+\frac{|y{|}^q}{q}.$

That real product is Young's inequality.

The dual function writes down, for each $y$, the greatest surplus of the inner product over $F$. The comparison (0.25) is that account. For the power $|x{|}^p/p$ the account is Young's inequality.