Young's Inequality for Products

Theorem

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

$\dfrac 1 p + \dfrac 1 q = 1$

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

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

Proof 1

The result is obvious if $a=0$ or $b=0$, so assume WLOG that $a > 0$ and $b > 0$.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle ab\)	\(=\)	\(\displaystyle \)	\(\displaystyle \exp \left({\ln\left({ab}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	exponential and logarithm functions are inverses
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \exp \left({ \ln a + \ln b }\right)\)	\(\displaystyle \)	\(\displaystyle \)	"sum property" of logarithms
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \exp \left({ \frac 1 p p \ln a + \frac 1 q q \ln b }\right)\)	\(\displaystyle \)	\(\displaystyle \)	definitions of the multiplicative identity and inverse
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \exp \left({ \frac 1 p \ln \left( {a^p} \right) + \frac 1 q \ln \left( {b^q} \right) }\right)\)	\(\displaystyle \)	\(\displaystyle \)	"power property" of logarithms
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle \)	\(\displaystyle \frac 1 p \exp \left( {\ln \left( {a^p} \right)} \right) + \frac 1 q \exp \left( {\ln \left( {b^q} \right)} \right)\)	\(\displaystyle \)	\(\displaystyle \)	exponential function is convex and the hypothesis that $\dfrac 1 p + \dfrac 1 q = 1$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac{a^p} p + \frac{b^q} q\)	\(\displaystyle \)	\(\displaystyle \)	exponential and logarithm functions are inverses

$\blacksquare$

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 $q > 1$.

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

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

• Walter Rudin: Principles of Mathematical Analysis (1953): Exercise $6.10 a$