Young's inequality

In mathematics, Young's inequality is either of two inequalities: one about the product of two numbers, and one about the convolution of two functions. They are named after William Henry Young.

Young's inequality for products can be used to prove Hölder's inequality. It is also used widely 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.

Young's inequality for products

Standard version for conjugate Hölder exponents

In its standard form, the inequality states that if a and b are nonnegative real numbers and p and q are positive real numbers such that 1/p + 1/q = 1, then

$ab \le \frac{a^p}{p} + \frac{b^q}{q}.$

Equality holds if and only if a^p = b^q. This form of Young's inequality is a special case of the inequality of weighted arithmetic and geometric means and can be used to prove Hölder's inequality.

Proof

The claim is certainly true if a = 0 or b = 0. Therefore, assume a > 0 and b > 0 in the following. Put t = 1/p, and (1 − t) = 1/q. Then since the logarithm function is strictly concave

$\log(t a^p + (1-t) b^q) \ge t \log(a^p) + (1-t) \log(b^q) = \log(a) +\log(b) = \log(ab)$

with equality if and only if a^p = b^q. Young's inequality follows by exponentiating.

Elementary case

An elementary case of Young's inequality is the inequality with exponent 2,

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

which also gives rise to the so-called Young's inequality with ε (valid for every ε > 0), sometimes called the Peter–Paul inequality . 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"

$ab \le \frac{a^2}{2\varepsilon} + \frac{\varepsilon b^2}{2}.$

Proof

Young's inequality with exponent 2 is the special case 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 a and b we can write:

$0\le (a-b)^2$

Work out the square of the right hand side:

$0\le a^2+b^2-2ab$

Add '2ab' to both sides:

$2ab\le a^2+b^2$

Divide both sides by 2 and we have Young's inequality with exponent 2:

$ab\le \frac{a^2}{2}+ \frac{b^2}{2}$

Young's inequality with ε follows by substituting $a'$ and $b'$ as below into Young's inequality with exponent 2:

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

Matricial generalization

T. Ando proved a generalization of Young's inequality for complex matrices ordered by Loewner ordering. It states that for any pair A, B of complex matrices of order n there exists a unitary matrix U such that

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

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

Standard version for increasing functions

File:Young.png

The area of the rectangle a,b can't be larger than sum of the areas under the functions $f$ (red) and $f^{-1}$ (yellow)

For the standard version of the inequality, let f denote a real-valued, continuous and strictly increasing function on [0, c] with c > 0 and f(0) = 0. Let f⁻¹ denote the inverse function of f. Then, for all a ∈ [0, c] and b ∈ [0, f(c)],

$ab \le \int_0^a f(x)\,dx + \int_0^b f^{-1}(x)\,dx$

with equality if and only if b = f(a).

Generalization using Fenchel-Legendre transforms

If f is a convex function and its Legendre transform (convex conjugate) is denoted by g, then

$ab \le f(a) + g(b). \,$

This follows immediately from the definition of the Legendre transform.

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

$\langle u, v \rangle \le f^\star(u) + f(v).$

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

Examples

• The Legendre transform of f(a) = a^p/p is g(b) = b^q/q with q such that 1/p + 1/q = 1, and thus Young's inequality for conjugate Hölder exponents mentioned above is a special case.
• The Legendre transform of f(a) = e^a – 1 is g(b) = 1 − b + b ln b, hence ab ≤ e^a − b + b ln b for all non-negative a and b. This estimate is useful in large deviations theory under exponential moment conditions, because b ln b appears in the definition of relative entropy, which is the rate function in Sanov's theorem.

Young's inequality for convolutions

In real analysis, the following result is also called Young's inequality:

Suppose f is in L^p(R^d) and g is in L^q(R^d) and

$\frac{1}{p} + \frac{1}{q} = \frac{1}{r} +1$

with 1 ≤ p, q, r ≤ ∞. Then

$\|f*g\| _r\le\|f\|_p\|g\|_q.$

Here the star denotes convolution, L^p is Lebesgue space, and

$\|f\|_p = \Bigl(\int |f(x)|^p\,dx \Bigr)^{1/p}$

denotes the usual L^p norm.

An example application is that Young's inequality can be used to show that the heat semigroup is a contracting semigroup using the L² norm (i.e. the Weierstrass transform does not enlarge the L² norm).

In case p, q > 1 Young's inequality can be strengthened to a sharp form, via

$\|f*g\| _r\le c_{p,q} \|f\|_p\|g\|_q.$

where the constant c_p,q < 1.

See also

• Convex conjugate
• Legendre transformation

External links

• Young's Inequality at PlanetMath
• Weisstein, Eric W., "Young's Inequality", MathWorld.