Young's inequality

In mathematics, Young's inequality is either of two inequalities: one about the product of two numbers,^[1] and one about the convolution of two functions.^[2] 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

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 .^[3] 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

Standard version for increasing functions

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^[4]^[5] 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.