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 | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {a^p} p\) | \(\le\) | \(\ds a b + \frac {b^{-q} } q\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a b\) | \(\ge\) | \(\ds \frac {a^p} p - \frac {b^{-q} } q\) |
$\blacksquare$
Geometric Proof
Since
- $\dfrac 1 p - \dfrac 1 q = 1$
we have $ p - 1 = \dfrac 1 {-q-1}$.
A graph $ y = x^{p - 1} $ on the $ xy $-plane is thus also a graph $ x = y^{-q - 1}$.
The above region corresponds to $\ds \int_0^a x^{p - 1} \rd x$
The above region corresponds to $\ds \int_b^\infty y^{-q - 1} \rd y$.
The above region corresponds to $\ds \int_0^a x^{p - 1} \rd x - \int_b^\infty y^{-q - 1} \rd y$.
Since
- $\dfrac 1 p = \dfrac 1 q + 1 > 1$
we have $0 < p < 1$.
Let $a, b$ be any positive real numbers.
Since $a b$ is the area of the rectangle in the given figure, we have:
- $\ds a b \ge \int_0^a x^{p - 1} \rd x - \int_b^\infty y^{-q - 1} \rd y = \frac {a^p} p - \frac {b^{-q}} q$
$\blacksquare$
Source of Name
This entry was named for William Henry Young.