# Young's Inequality for Increasing Functions/Equality

## Contents

## Theorem

Let $a_0$ and $b_0$ be strictly positive real numbers.

Let $f: \closedint 0 {a_0} \to \closedint 0 {b_0}$ be a strictly increasing bijection.

Let $a$ and $b$ be real numbers such that $0 \le a \le a_0$ and $0 \le b \le b_0$.

Then $b = \map f a$ if and only if:

- $\displaystyle a b = \int_0^a \map f u \rd u + \int_0^b \map {f^{-1} } v \rd v$

where $\displaystyle \int$ denotes the Darboux integral.

## Proof

### Necessary Condition

### Sufficient Condition

By Monotone Real Function is Darboux Integrable, $f$ and $f^{-1}$ are Darboux integrable.

Let $b = \map f a$.

Define:

- $\displaystyle A = \int_0^a \map f u \rd u + \int_0^b \map {f^{-1} } v \rd v$

Consider any subdivision $P = \set {x_0, x_1, \ldots, x_n}$ of the closed real interval $\closedint 0 a$.

Then:

- $f P = \set {\map f {x_0}, \map f {x_1}, \ldots, \map f {x_n} }$

is a subdivision of $\closedint 0 b$.

We have that:

\(\displaystyle \map U {P, f} + \map L {f P, f^{-1} }\) | \(=\) | \(\displaystyle \sum_{k \mathop = 1}^n \paren {x_k - x_{k - 1} } \map f {x_k} + \sum_{k \mathop = 1}^n \paren {\map f {x_k} - \map f {x_{k - 1} } } \map {f^{-1} } {\map f {x_{k - 1} } }\) | because $f$ and $f^{-1}$ are increasing | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{k \mathop = 1}^n \paren {x_k \map f {x_k} - x_{k - 1} \map f {x_{k - 1} } }\) | by Summation is Linear | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a b\) |

Similarly:

- $\map L {P, f} + \map U {f P, f^{-1} } = a b$

Let $\epsilon \in \R_{>0}$ be an arbitrary strictly positive real number.

By the definition of the Darboux integral, there exist subdivisions $P_+$ and $P_-$ of $\closedint 0 a$ such that:

- $a b = \map U {P_+, f} + \map L {f P_+, f^{-1} } < A + \epsilon$
- $a b = \map L {P_-, f} + \map U {f P_-, f^{-1} } > A - \epsilon$

The result follows from Real Plus Epsilon.

$\blacksquare$

## Source of Name

This entry was named for William Henry Young.