Young’s theorem

Let f,g:n. Recall that the convolution of f and g at x is


provided the integral is defined.

The following result is due to William Henry Young.

Theorem 1

Let p,q,r[1,] satisfy

1p+1q-1r=1 (1)

with the convention 1/=0. Let fLp(Rn), gLq(Rn). Then:

  1. 1.

    The function yf(x-y)g(y) belongs to L1(n) for almost all x.

  2. 2.

    The function x(fg)(x) belongs to Lr(n).

  3. 3.

    There exists a constant c=cp,q1, depending on p and q but not on f or g, such that


Observe the analogyMathworldPlanetmath with the similar result with convolution replaced by ordinary (pointwise) productPlanetmathPlanetmath, where the requirement is 1/p+1/q=1/ri.e., 1/p+1/q-1/r=0—instead of (1). The cases

  1. 1.

    1/p+1/q=1, r=

  2. 2.

    p=1, q[1,), r=q

are the most widely known; for these we provide a proof, supposing cp,q=1. We shall use the following facts:

  • If xf(x),xg(x) are measurable, then (x,y)f(x-y)g(y) is measurable.

  • For any x, if fLp, then yf(x-y) belongs to Lp as well, and its Lp-norm is the same as f’s.

  • For any y, if fLp, then xf(x-y) belongs to Lp as well, and its Lp-norm is the same as f’s.

Proof of case 1.

Suppose fLp(n), gLq(n) with 1/p+1/q=1. Then


This holds for all xn, therefore fgfpgq as well.

Proof of case 2.

First, suppose q=1. We may suppose f and g are Borel measurable: if they are not, we replace them with Borel measurable functions f~ and g~ which are equal to f and g, respectively, outside of a set of Lebesgue measureMathworldPlanetmath zero; apply the theoremMathworldPlanetmath to f~, g~, and f~g~; and deduce the theorem for f, g, and fg. By Tonelli’s theorem,


thus the function (x,y)f(x-y)g(y) belongs to L1(n×n). By Fubini’s theorem, the function yf(x-y)g(y) belongs to L1(n) for almost all x, and x(fg)(x) belongs to L1(n); plus,


Suppose now q>1; choose q so that 1/q+1/q=1. By the argumentMathworldPlanetmath above, y|f(x-y)||g(y)|q belongs to L1 for almost all x: for those x, put u(y)=|f(x-y)|1/q,v(y)=|f(x-y)|1/q|g(y)|. Then uLq and vLq with 1/q+1/q=1, so uvL1 and uv1uqvq: but uv=|f(x-y)g(y)|, so point 1 of the theorem is proved. By Hölder’s inequalityMathworldPlanetmath,


but we know that |f||g|qL1, so fgLq and point 2 is also proved. Finally,


but 1/q+1/q=1 means q+q=qq and thus 1+q/q=q, so that point 3 is also proved.


  • 1 G. Gilardi. Analisi tre. McGraw-Hill 1994.
  • 2 W. Rudin. Real and complex analysis. McGraw-Hill 1987.
  • 3 W. H. Young. On the multiplication of successions of Fourier constants. Proc. Roy. Soc. Lond. Series A 87 (1912) 331–339.
Title Young’s theorem
Canonical name YoungsTheorem
Date of creation 2013-03-22 18:17:44
Last modified on 2013-03-22 18:17:44
Owner Ziosilvio (18733)
Last modified by Ziosilvio (18733)
Numerical id 15
Author Ziosilvio (18733)
Entry type Theorem
Classification msc 44A35