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


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

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.


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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:

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

if and only if:

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

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

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