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:

$\alpha \map \exp x + \beta \map \exp y = \alpha \map \exp x + \beta \map \exp x = \map \exp x$



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:

$\map \ln {a^p} = \map \ln {b^q}$



That is, if and only if:

$b = a^{p - 1}$

$\blacksquare$


Source of Name

This entry was named for William Henry Young.