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:

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


Then:

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


Thus Young's Inequality for Products can be applied:

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

$\blacksquare$


Source of Name

This entry was named for William Henry Young.