Young's Inequality for Products

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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, for any $a, b \in \R_{\ge 0}$:

$a b \le \dfrac {a^p} p + \dfrac{b^q} q$

Equality occurs if and only if:

$b = a^{p-1}$.


Proof 1

The result follows directly if $a = 0$ or $b = 0$.

Without loss of generality, assume that $a > 0$ and $b > 0$.

Then:

\(\displaystyle a b\) \(=\) \(\displaystyle \map \exp {\map \ln {a b} }\) Exponential of Natural Logarithm
\(\displaystyle \) \(=\) \(\displaystyle \map \exp {\ln a + \ln b}\) Sum of Logarithms
\(\displaystyle \) \(=\) \(\displaystyle \map \exp {\frac 1 p p \ln a + \frac 1 q q \ln b}\) Definition of Multiplicative Identity and Definition of Multiplicative Inverse
\(\displaystyle \) \(=\) \(\displaystyle \map \exp {\frac 1 p \map \ln {a^p} + \frac 1 q \map \ln {b^q} }\) Logarithms of Powers
\(\displaystyle \) \(\le\) \(\displaystyle \frac 1 p \map \exp {\map \ln {a^p} } + \frac 1 q \map \exp {map \ln {b^q} }\) Exponential is Strictly Convex and the hypothesis that $\dfrac 1 p + \dfrac 1 q = 1$
\(\displaystyle \) \(=\) \(\displaystyle \frac {a^p} p + \frac {b^q} q\) Exponential of Natural Logarithm

$\blacksquare$


Proof 2

The blue colored region corresponds to $\displaystyle \int_0^\alpha t^{p-1} \mathrm d t$ and the red colored region to $\displaystyle \int_0^\beta u^{q-1} \mathrm d u$.

In order for $\dfrac 1 p + \dfrac 1 q = 1$ it is necessary for both $p > 1$ and $q > 1$.

\(\displaystyle \frac 1 p + \frac 1 q\) \(=\) \(\displaystyle 1\)
\(\displaystyle \implies \ \ \) \(\displaystyle p + q\) \(=\) \(\displaystyle p q\) multiplying both sides by $p q$
\(\displaystyle \implies \ \ \) \(\displaystyle p + q - p - q + 1\) \(=\) \(\displaystyle p q - p - q + 1\) adding $1 - p - q$ to both sides
\(\displaystyle \implies \ \ \) \(\displaystyle 1\) \(=\) \(\displaystyle \left({p - 1}\right) \left({q - 1}\right)\) elementary algebra
\(\displaystyle \implies \ \ \) \(\displaystyle \frac 1 {p - 1}\) \(=\) \(\displaystyle q - 1\)

Accordingly:

$u = t^{p-1} \iff t = u^{q-1}$


Let $a, b$ be any positive real numbers.

Since $a b$ is the area of the rectangle in the given figure, we have:

$\displaystyle a b \le \int_0^a t^{p-1} \ \mathrm d t + \int_0^b u^{q-1} \ \mathrm d u = \frac {a^p} p + \frac {b^q} q$

Note that even if the graph intersected the side of the rectangle corresponding to $t = a$, this inequality would hold.

Also note that if either $a = 0$ or $b = 0$ then this inequality holds trivially.

$\blacksquare$


Source of Name

This entry was named for William Henry Young.


Sources