Reverse Young's Inequality for Products

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Theorem

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

$\dfrac 1 p - \dfrac 1 q = 1$

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

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


Then:

$a b \ge \dfrac {a^p} p - \dfrac {b^{-q} } q$


Proof

Define:

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


Then:

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


Thus Young's Inequality for Products can be applied:

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

$\blacksquare$


Source of Name

This entry was named for William Henry Young.