Young's inequality for integral operators

In mathematical analysis, the Young's inequality for integral operators, is a bound on the [math]\displaystyle{ L^p\to L^q }[/math] operator norm of an integral operator in terms of [math]\displaystyle{ L^r }[/math] norms of the kernel itself.

Statement

Assume that [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math] are measurable spaces, [math]\displaystyle{ K : X \times Y \to \mathbb{R} }[/math] is measurable and [math]\displaystyle{ q,p,r\geq 1 }[/math] are such that [math]\displaystyle{ \frac{1}{q} = \frac{1}{p} + \frac{1}{r} -1 }[/math]. If

[math]\displaystyle{ \int_{Y} |K (x, y)|^r \,\mathrm{d} y \le C^r }[/math] for all [math]\displaystyle{ x\in X }[/math]

and

[math]\displaystyle{ \int_{X} |K (x, y)|^r \,\mathrm{d} x \le C^r }[/math] for all [math]\displaystyle{ y\in Y }[/math]

then

[math]\displaystyle{ \int_{X} \left|\int_{Y} K (x, y) f(y) \,\mathrm{d} y\right|^q \, \mathrm{d} x \le C^q \left( \int_{Y} |f(y)|^p \,\mathrm{d} y\right)^\frac{q}{p}. }[/math]

Particular cases

Convolution kernel

If [math]\displaystyle{ X = Y = \mathbb{R}^d }[/math] and [math]\displaystyle{ K (x, y) = h (x - y) }[/math], then the inequality becomes Young's convolution inequality.

See also

Young's inequality for products