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

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


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 of Natural Logarithm          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \exp \left({ \ln a + \ln b }\right)\) \(\displaystyle \) \(\displaystyle \)          Sum 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 Multiplicative Identity and Multiplicative Inverse          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \exp \left({ \frac 1 p \ln \left( {a^p} \right) + \frac 1 q \ln \left( {b^q} \right) }\right)\) \(\displaystyle \) \(\displaystyle \)          Logarithms of Powers          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\le\) \(\displaystyle \) \(\displaystyle \frac 1 p \exp \left( {\ln \left( {a^p} \right)} \right) + \frac 1 q \exp \left( {\ln \left( {b^q} \right)} \right)\) \(\displaystyle \) \(\displaystyle \)          Exponential is Strictly Increasing and 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 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 $q > 1$.

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \frac 1 p + \frac 1 q\) \(=\) \(\displaystyle \) \(\displaystyle 1\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle p + q\) \(=\) \(\displaystyle \) \(\displaystyle p q\) \(\displaystyle \) \(\displaystyle \)          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