Reverse Young's Inequality for Products

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Theorem

Let $p, q \in \R_{> 0}$ be strictly positive real numbers satisfying:

$\displaystyle \frac 1 p - \frac 1 q = 1$

Let $a \in \R_{\ge 0}$ be a positive real number.

Let $b \in \R_{> 0}$ be a strictly positive real number.


Then:

$\displaystyle a b \ge \frac {a^p} p - \frac {b^{-q}} q$


Proof

Define:

\((1):\quad\) \(\displaystyle u\) \(=\) \(\displaystyle \frac 1 p\)
\(\displaystyle v\) \(=\) \(\displaystyle \frac q p\)
\(\displaystyle x\) \(=\) \(\displaystyle \left({a b}\right)^p\)
\(\displaystyle y\) \(=\) \(\displaystyle b^{-p}\)


Then:

\(\displaystyle \frac 1 u + \frac 1 v\) \(=\) \(\displaystyle p + \frac p q\)
\(\displaystyle \) \(=\) \(\displaystyle p \left({1 + \frac 1 q}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle p \left({\frac 1 p}\right)\) because $\dfrac 1 p - \dfrac 1 q = 1$, by hypothesis
\(\displaystyle \) \(=\) \(\displaystyle 1\)


Thus Young's Inequality for Products can be applied:

\(\displaystyle x y\) \(\le\) \(\displaystyle \frac {x^u} u + \frac {y^v} v\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left({a b}\right)^p b^{-p}\) \(\le\) \(\displaystyle \frac {\left({\left({a b}\right)^p}\right)^{1/p} } {1/p} + \frac {\left({b^{-p} }\right)^{q/p} } {q/p}\) substituting for $u, v, x, y$ from $(1)$ above
\(\displaystyle \implies \ \ \) \(\displaystyle a^p\) \(\le\) \(\displaystyle pab + p \frac { b^{-q} } q\) algebraic simplification
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {a^p} p\) \(\le\) \(\displaystyle ab + \frac { b^{-q} } q\)
\(\displaystyle \implies \ \ \) \(\displaystyle ab\) \(\ge\) \(\displaystyle \frac {a^p} p - \frac { b^{-q} } q\)

$\blacksquare$


Source of Name

This entry was named for William Henry Young.