# 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 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 |

By Definition of Strictly Convex Real Function, the equality occurs if and only if:

- $\map \ln {a^p} = \map \ln {b^q}$

That is, if and only if:

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

$\blacksquare$

## Geometric Proof

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.

This page has been identified as a candidate for refactoring of medium complexity.In particular: It occurs to me that the equality proof should be extracted and placed into its own (transcluded) page, as it should then be applied to all proofs.Until this has been finished, please leave
`{{Refactor}}` in the code.
Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Refactor}}` from the code. |

This article needs to be tidied.In particular: including grammarPlease fix formatting and $\LaTeX$ errors and inconsistencies. It may also need to be brought up to our standard house style.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Tidy}}` from the code. |

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

$\blacksquare$

## Proof by Calculus

Without loss of generality assume that $a^p \ge b^q$, otherwise swap $a$ with $b$ and $p$ with $q$.

Define $f : \hointr 0 \infty \to \R$ by:

- $\ds \map f t = \frac {t^p} p + \frac 1 q - t$

We have, from Derivative of Power and Sum Rule for Derivatives:

- $\ds \map {f'} t = t^{p - 1} - 1$

So for $t \ge 1$ we have:

- $\ds \map {f'} t \ge 0$

So, from Real Function with Positive Derivative is Increasing, we have:

- $f$ is increasing on $\hointr 1 \infty$.

That is:

- $\map f t \ge \map f 1$

for all $t \ge 1$.

Since $a^p \ge b^q$, we have:

- $a^p b^{-q} \ge 1$

so:

- $a b^{-q/p} \ge 1$

So:

- $\map f {a b^{-q/p} } \ge \map f 1$

That is:

- $\ds \frac {a^p b^{-q} } p + \frac 1 q - a b^{-q/p} \ge \frac 1 p + \frac 1 q - 1$

By hypothesis, we have:

- $\ds \frac 1 p + \frac 1 q = 1$

So it follows that:

- $\ds \frac {a^p b^{-q} } p + \frac 1 q \ge a b^{-q/p}$

So:

- $\ds \frac {a^p} p + \frac {b^q} q \ge a b^{q \paren {1 - 1/p} }$

Finally, we have:

- $\ds 1 - \frac 1 p = \frac 1 q$

so:

- $\ds q \paren {1 - \frac 1 p} = 1$

giving:

- $\ds \frac {a^p} p + \frac {b^q} q \ge a b$

For the equality case, note that:

- $\map {f'} t > 0$

for $t > 1$.

So, from Real Function with Strictly Positive Derivative is Strictly Increasing, we have:

- $f$ is strictly increasing for $t > 1$.

So:

- $\map f {a b^{-q/p} } = \map f 1$

that is:

- $\ds a b = \frac {a^p} p + \frac {b^q} q$

- $a b^{-q/p} = 1$

That is:

- $b = a^{p/q}$

We have:

- $\ds \frac 1 p + \frac 1 q = 1$

and so:

- $\ds 1 + \frac p q = p$

giving:

- $\ds \frac p q = p - 1$

So we have:

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

as required.

$\blacksquare$

## Also presented as

Statements of Young's inequality will commonly insist that $p, q > 1$.

This is no different to the statement presented here, since if:

- $\ds \frac 1 p + \frac 1 q = 1$

for positive $p$ and $q$ we must have:

- $\ds \frac 1 p \le 1$

and:

- $\ds \frac 1 q \le 1$

So $p, q \ge 1$.

But we cannot have $p = 1$ or $q = 1$, otherwise $\dfrac 1 q = 0$ or $\dfrac 1 p = 0$ respectively.

This is understood to be solved by $q = \infty$ and $p = \infty$ respectively, but there is no real meaning to the right hand side in this case, so we arrive at the usual hypotheses of $p, q > 1$.

## Source of Name

This entry was named for William Henry Young.

## Sources

- 1953: Walter Rudin:
*Principles of Mathematical Analysis*: Exercise $6.10 a$