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 $a, b \in \R_{\ge 0}$:
- $a b \le \dfrac {a^p} p + \dfrac {b^q} q$
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:
| \(\ds a b\) | \(=\) | \(\ds \map \exp {\map \ln {a b} }\) | 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 | |||||||||||
| \(\ds \) | \(\le\) | \(\ds \frac 1 p \map \exp {\map \ln {a^p} } + \frac 1 q \map \exp {\map \ln {b^q} }\) | Exponential is Strictly Convex and the hypothesis that $\dfrac 1 p + \dfrac 1 q = 1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {a^p} p + \frac {b^q} q\) | Exponential of Natural Logarithm |
$\blacksquare$
Proof 2
In order for $\dfrac 1 p + \dfrac 1 q = 1$ it is necessary for both $p > 1$ and $q > 1$.
| \(\ds \frac 1 p + \frac 1 q\) | \(=\) | \(\ds 1\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds p + q\) | \(=\) | \(\ds p q\) | multiplying both sides by $p q$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds p + q - p - q + 1\) | \(=\) | \(\ds p q - p - q + 1\) | adding $1 - p - q$ to both sides | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 1\) | \(=\) | \(\ds \paren {p - 1} \paren {q - 1}\) | elementary algebra | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac 1 {p - 1}\) | \(=\) | \(\ds q - 1\) |
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:
- $\ds a b \le \int_0^a t^{p - 1} \rd t + \int_0^b u^{q - 1} \rd u = \frac {a^p} p + \frac {b^q} q$
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
- 1953: Walter Rudin: Principles of Mathematical Analysis: Exercise $6.10 a$