Young's Inequality for Products/Proof by Convexity
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Theorem
Let $p, q \in \R_{> 0}$ be strictly positive real numbers such that:
- $\dfrac 1 p + \dfrac 1 q = 1$
Then:
- $\forall a, b \in \R_{\ge 0}: a b \le \dfrac {a^p} p + \dfrac {b^q} q$
Equality occurs if and only if:
- $b = a^{p - 1}$
Proof
The result follows directly if $a = 0$ or $b = 0$.
Without loss of generality, assume that $a > 0$ and $b > 0$.
Recall Exponential is Strictly Convex.
Consider:
- $x := \map \ln {a^p}$
- $y := \map \ln {b^q}$
- $\alpha := p^{-1}$
- $\beta := q^{-1}$
Note that by hypothesis:
- $\alpha + \beta = 1$
Thus by definition of strictly convex real function:
- $(1): \quad x \ne y \implies \map \exp {\alpha x + \beta y} < \alpha \map \exp x + \beta \map \exp y$
On the other hand:
- $(2): \quad x = y \implies \map \exp {\alpha x + \beta y} = \alpha \map \exp x + \beta \map \exp y$
since:
- $\map \exp {\alpha x + \beta y} = \map \exp x$
and:
- $\alpha \map \exp x + \beta \map \exp y = \alpha \map \exp x + \beta \map \exp x = \map \exp x$
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Therefore:
\(\ds a b\) | \(=\) | \(\ds \map \exp {\map \ln {a b} }\) | Exponential of Natural Logarithm | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \exp {\ln a + \ln b}\) | Sum of Logarithms | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \exp {\frac 1 p p \ln a + \frac 1 q q \ln b}\) | Definition of Multiplicative Identity and Definition of Multiplicative Inverse | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \exp {\frac 1 p \map \ln {a^p} + \frac 1 q \map \ln {b^q} }\) | Logarithms of Powers | |||||||||||
\(\ds \) | \(\le\) | \(\ds \frac 1 p \map \exp {\map \ln {a^p} } + \frac 1 q \map \exp {\map \ln {b^q} }\) | by $(1)$ and $(2)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a^p} p + \frac {b^q} q\) | Exponential of Natural Logarithm |
By $(1)$ and $(2)$, the equality:
- $a b = \dfrac {a^p} p + \dfrac {b^q} q$
occurs if and only if:
- $\map \ln {a^p} = \map \ln {b^q}$
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That is, if and only if:
- $b = a^{p - 1}$
$\blacksquare$
Source of Name
This entry was named for William Henry Young.