In mathematics, Young's inequality for products is a mathematical inequality about the product of two numbers.^[1] The inequality is named after William Henry Young and should not be confused with Young's convolution inequality.

Young's inequality for products can be used to prove Hölder's inequality. It is also widely used to estimate the norm of nonlinear terms in PDE theory, since it allows one to estimate a product of two terms by a sum of the same terms raised to a power and scaled.

The standard form of the inequality is the following, which can be used to prove Hölder's inequality.

Theorem — If ${\displaystyle a\geq 0}$ and ${\displaystyle b\geq 0}$ are nonnegative real numbers and if ${\displaystyle p>1}$ and ${\displaystyle q>1}$ are real numbers such that ${\displaystyle {\frac {1}{p}}+{\frac {1}{q}}=1,}$ then

$${\displaystyle ab~\leq ~{\frac {a^{p}}{p}}+{\frac {b^{q}}{q}}.}$$

Equality holds if and only if ${\displaystyle a^{p}=b^{q}.}$

A second proof is via Jensen's inequality.

Proof^[3]

The claim is certainly true if ${\displaystyle a=0}$ or ${\displaystyle b=0}$ so henceforth assume that ${\displaystyle a>0}$ and ${\displaystyle b>0.}$ Put ${\displaystyle t=1/p}$ and ${\displaystyle (1-t)=1/q.}$ Because the logarithm function is concave,

$${\displaystyle \ln \left(ta^{p}+(1-t)b^{q}\right)~\geq ~t\ln \left(a^{p}\right)+(1-t)\ln \left(b^{q}\right)=\ln(a)+\ln(b)=\ln(ab)}$$

with the equality holding if and only if ${\displaystyle a^{p}=b^{q}.}$ Young's inequality follows by exponentiating.

Yet another proof is to first prove it with ${\displaystyle b=1}$ an then apply the resulting inequality to ${\displaystyle {\tfrac {a}{b^{q}}}}$. The proof below illustrates also why Hölder conjugate exponent is the only possible parameter that makes Young's inequality hold for all non-negative values. The details follow:

Proof

Let ${\displaystyle 0<\alpha <1}$ and ${\displaystyle \alpha +\beta =1}$. The inequality

$${\displaystyle x~\leq ~\alpha x^{p}+\beta ,\qquad \,for\quad \ all\quad \ x~\geq ~0}$$

holds if and only if ${\displaystyle \alpha ={\tfrac {1}{p}}}$ (and hence ${\displaystyle \beta ={\tfrac {1}{q}}}$). This can be shown by convexity arguments or by simply minimizing the single-variable function.

To prove full Young's inequality, clearly we assume that ${\displaystyle a>0}$ and ${\displaystyle b>0}$. Now, we apply the inequality above to ${\displaystyle x={\tfrac {a}{b^{s}}}}$ to obtain:

$${\displaystyle {\tfrac {a}{b^{s}}}~\leq ~{\tfrac {1}{p}}{\tfrac {a^{p}}{b^{sp}}}+{\tfrac {1}{q}}.}$$

It is easy to see that choosing ${\displaystyle s=q-1}$ and multiplying both sides by ${\displaystyle b^{q+1}}$ yields Young's inequality.

Young's inequality may equivalently be written as

$${\displaystyle a^{\alpha }b^{\beta }\leq \alpha a+\beta b,\qquad \,0\leq \alpha ,\beta \leq 1,\quad \ \alpha +\beta =1.}$$

Where this is just the concavity of the logarithm function. Equality holds if and only if ${\displaystyle a=b}$ or ${\displaystyle \{\alpha ,\beta \}=\{0,1\}.}$ This also follows from the weighted AM-GM inequality.

Generalizations

Theorem^[4] — Suppose ${\displaystyle a>0}$ and ${\displaystyle b>0.}$ If ${\displaystyle 1<p<\infty }$ and ${\displaystyle q}$ are such that ${\displaystyle {\tfrac {1}{p}}+{\tfrac {1}{q}}=1}$ then

$${\displaystyle ab~=~\min _{0<t<\infty }\left({\frac {t^{p}a^{p}}{p}}+{\frac {t^{-q}b^{q}}{q}}\right).}$$

Using ${\displaystyle t:=1}$ and replacing ${\displaystyle a}$ with ${\displaystyle a^{1/p}}$ and ${\displaystyle b}$ with ${\displaystyle b^{1/q}}$ results in the inequality:

$${\displaystyle a^{1/p}\,b^{1/q}~\leq ~{\frac {a}{p}}+{\frac {b}{q}},}$$

which is useful for proving Hölder's inequality.

Proof^[4]

Define a real-valued function ${\displaystyle f}$ on the positive real numbers by

$${\displaystyle f(t)~=~{\frac {t^{p}a^{p}}{p}}+{\frac {t^{-q}b^{q}}{q}}}$$

for every ${\displaystyle t>0}$ and then calculate its minimum.

Theorem — If ${\displaystyle 0\leq p_{i}\leq 1}$ with ${\displaystyle \sum _{i}p_{i}=1}$ then

$${\displaystyle \prod _{i}{a_{i}}^{p_{i}}~\leq ~\sum _{i}p_{i}a_{i}.}$$

Equality holds if and only if all the ${\displaystyle a_{i}}$s with non-zero ${\displaystyle p_{i}}$s are equal.

An elementary case of Young's inequality is the inequality with exponent ${\displaystyle 2,}$

$${\displaystyle ab\leq {\frac {a^{2}}{2}}+{\frac {b^{2}}{2}},}$$

which also gives rise to the so-called Young's inequality with ${\displaystyle \varepsilon }$ (valid for every ${\displaystyle \varepsilon >0}$), sometimes called the Peter–Paul inequality. ^[5] This name refers to the fact that tighter control of the second term is achieved at the cost of losing some control of the first term – one must "rob Peter to pay Paul"

$${\displaystyle ab~\leq ~{\frac {a^{2}}{2\varepsilon }}+{\frac {\varepsilon b^{2}}{2}}.}$$

Proof: Young's inequality with exponent ${\displaystyle 2}$ is the special case ${\displaystyle p=q=2.}$ However, it has a more elementary proof.

Start by observing that the square of every real number is zero or positive. Therefore, for every pair of real numbers ${\displaystyle a}$ and ${\displaystyle b}$ we can write:

$${\displaystyle 0\leq (a-b)^{2}}$$

Work out the square of the right hand side:

$${\displaystyle 0\leq a^{2}-2ab+b^{2}}$$

Add ${\displaystyle 2ab}$ to both sides:

$${\displaystyle 2ab\leq a^{2}+b^{2}}$$

Divide both sides by 2 and we have Young's inequality with exponent ${\displaystyle 2:}$

$${\displaystyle ab\leq {\frac {a^{2}}{2}}+{\frac {b^{2}}{2}}}$$

Young's inequality with ${\displaystyle \varepsilon }$ follows by substituting ${\displaystyle a'}$ and ${\displaystyle b'}$ as below into Young's inequality with exponent ${\displaystyle 2:}$

$${\displaystyle a'=a/{\sqrt {\varepsilon }},\;b'={\sqrt {\varepsilon }}b.}$$

T. Ando proved a generalization of Young's inequality for complex matrices ordered by Loewner ordering.^[6] It states that for any pair ${\displaystyle A,B}$ of complex matrices of order ${\displaystyle n}$ there exists a unitary matrix ${\displaystyle U}$ such that

$${\displaystyle U^{*}|AB^{*}|U\preceq {\tfrac {1}{p}}|A|^{p}+{\tfrac {1}{q}}|B|^{q},}$$

where ${\displaystyle {}^{*}}$ denotes the conjugate transpose of the matrix and ${\displaystyle |A|={\sqrt {A^{*}A}}.}$

For the standard version^[7][8] of the inequality, let ${\displaystyle f}$ denote a real-valued, continuous and strictly increasing function on ${\displaystyle [0,c]}$ with ${\displaystyle c>0}$ and ${\displaystyle f(0)=0.}$ Let ${\displaystyle f^{-1}}$ denote the inverse function of ${\displaystyle f.}$ Then, for all ${\displaystyle a\in [0,c]}$ and ${\displaystyle b\in [0,f(c)],}$

$${\displaystyle ab~\leq ~\int _{0}^{a}f(x)\,dx+\int _{0}^{b}f^{-1}(x)\,dx}$$

with equality if and only if ${\displaystyle b=f(a).}$

With ${\displaystyle f(x)=x^{p-1}}$ and ${\displaystyle f^{-1}(y)=y^{q-1},}$ this reduces to standard version for conjugate Hölder exponents.

For details and generalizations we refer to the paper of Mitroi & Niculescu.^[9]

By denoting the convex conjugate of a real function ${\displaystyle f}$ by ${\displaystyle g,}$ we obtain

$${\displaystyle ab~\leq ~f(a)+g(b).}$$

This follows immediately from the definition of the convex conjugate. For a convex function ${\displaystyle f}$ this also follows from the Legendre transformation.

More generally, if ${\displaystyle f}$ is defined on a real vector space ${\displaystyle X}$ and its convex conjugate is denoted by ${\displaystyle f^{\star }}$ (and is defined on the dual space ${\displaystyle X^{\star }}$), then

$${\displaystyle \langle u,v\rangle \leq f^{\star }(u)+f(v).}$$

where ${\displaystyle \langle \cdot ,\cdot \rangle :X^{\star }\times X\to \mathbb {R} }$ is the dual pairing.

• Convex conjugate – Generalization of the Legendre transformation
• Integral of inverse functions – Mathematical theorem, used in calculus
• Legendre transformation – Mathematical transformation
• Young's convolution inequality – Mathematical inequality about the convolution of two functions