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 by Convexity

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

$\blacksquare$


Geometric Proof

Holder's Ineq.jpg

In the above diagram, the $\color {blue} {\text {blue} }$ colored region corresponds to $\ds \int_0^\alpha t^{p - 1} \rd t$ and the $\color {red} {\text {red} }$ colored region to $\ds \int_0^\beta u^{q - 1} \rd u$.

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.





It remains to show that $ab=\frac{a^p}p+\frac{b^q}q$ if and only if $b=a^{p-1}$. One direction is nearly trivial: If $b=a^{p-1}$ then

\(\ds \frac{a^p}p+\frac{b^q}q\) \(=\) \(\ds \frac{a^p}p+\frac{(a^{p-1})^q}{q}\) From the assumption
\(\ds \) \(=\) \(\ds \frac{a^p}p+\frac{(a^{p/q})^q}q\) Due to $1/p+1/q=1$
\(\ds \) \(=\) \(\ds a^p\left(\frac 1 p + \frac 1 q \right)\) Algebra
\(\ds \) \(=\) \(\ds a\cdot a^{p-1}\) Algebra and $1/p+1/q=1$
\(\ds \) \(=\) \(\ds ab\) From the assumption

The converse is less trivial, and we prove it by the contrapositive. So assume that $b\ne a^{p-1}$ and we will argue that $ab\ne \frac{a^p}p+\frac{b^q}q$. We separate this into cases, and begin with the case that $b< a^{p-1}$. Note that this is precisely what is diagrammed in the image above (identifying $a=\alpha,b=\beta$). Because the sum of the colored regions strictly exceeds the area of the rectangle, we have that $ab<\frac{a^p}p+\frac{b^q}q$. In the case that $b > a^{p-1}$ one draws a diagram similar to the one above, although integrated with respect to $u$ rather than $t$.

This shows that $b\ne a^{p-1}$ implies $ab\ne \frac{a^p}p+\frac{b^q}q$ which is the contrapositive of the claim that $ab=\frac{a^p}p+\frac{b^q}q$ implies $b=a^{p-1}$.

$\blacksquare$


Source of Name

This entry was named for William Henry Young.


Sources