nLab Young's inequality

Contents

Young’s inequality for products

Proposition

Given

  • a,b >0a, \,b \,\in\, \mathbb{R}_{\gt 0}, non-negative real numbers;

  • p,q >1p, \, q \,\in\, \mathbb{R}_{\gt 1} such that

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

then the following inequality holds:

aba pp+b qq, a b \;\leq\; \frac{a^p}{p} + \frac{b^q}{q} \,,

which is an equality if and only if a p=b qa^p = b^q.

One proof is by convexity of the exponential function: choosing x,y,tx, y, t such that exp(x)=a p\exp(x) = a^p, exp=b q\exp = b^q and t=1pt = \frac1{p}, Young’s inequality is identical to the convexity constraint

exp(tx+(1t)y)texp(x)+(1t)exp(y).\exp(tx + (1-t)y) \leq t\exp(x) + (1-t)\exp(y).

References

See also:

Last revised on December 27, 2022 at 19:10:59. See the history of this page for a list of all contributions to it.