# Young's inequality

In mathematics, the standard form of Young's 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 we have

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

Equality holds if and only if "a""p" = "b""q". Young's inequality is a special case of the inequality of weighted arithmetic and geometric means. It is named for William Henry Young.

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

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

which also gives rise to the so-called Young's inequality with "ε" (valid for any "ε" > 0),

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

Generalization using Legendre transforms

If "f" is a convex function and its Legendre transform is denoted by "g", then:$ab le f\left(a\right) + g\left(b\right). ,$This follows immediately from the definition of the Legendre transform. This inequality also holds — in the form "a" &middot;"b" ≤ "f"("a") + "g"("b") — if "f" is a convex function taking a vector argument harv|Arnold|1989|loc=§14.

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 the standard Young inequality 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.

An inequality for "L""p" norms

In real analysis, the following result, first proved in Young (1912) is also called Young's inequality:

Suppose "f" is in "L""p" and "g" is in "L""q" and

:$frac\left\{1\right\}\left\{p\right\} + frac\left\{1\right\}\left\{q\right\} = frac\left\{1\right\}\left\{r\right\} + 1$

with 1 ≤ "p", "q" ,"r" ≤ ∞ and 1/"p" + 1/"q" ≥ 1. Then

: $|f*g| _rle|f|_p|g|_q.$

Here the star denotes convolution, "L""p" is Lebesgue space, and:$|f|_p = Bigl\left(int |f\left(x\right)|^p,dx Bigr\right)^\left\{1/p\right\}$denotes the usual "L""p" norm. This can be proved by use of the Hölder inequality.

An example application is that Young's inequality can be used to show that the Heat Semigroup is a contraction semigroup using the "L""2" norm.

The result can be strengthened to a sharp form, viz: $|f*g| _rle c_\left\{p,q\right\} |f|_p|g|_q.$ where the constant "c""p","q"<1.

Use

Young's inequality is used in the proof of 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.

Proof of the standard form

The proof is trivial if "a" = 0 or "b" = 0. Therefore, assume "a", "b" > 0.

If "a""p" = "b""q", then by the rules for exponentiation and the assumption 1/"p" + 1/"q" = 1,:$ab = a\left(b^q\right)^\left\{1/q\right\} = aa^\left\{p/q\right\} =a^\left\{p/p\right\}a^\left\{p/q\right\} =a^\left\{p\left(1/p+1/q\right)\right\}= a^p = \left\{a^p over p\right\} + \left\{b^q over q\right\},$and we have equality in Young's inequality.

Assume in addition "a""p" ≠ "b""q" for the remaining part of the proof. By the functional equation of the natural logarithm,:$ln ab=ln a +ln b=frac\left\{ln a^p\right\}p + frac\left\{ln b^q\right\}q.$Note that the natural logarithm is strictly increasing, because its first derivative is positive for every positive number, hence ln "a""p" ≠ ln "b""q". Its inverse is the exponential function "f"("x") = exp("x"), which is strictly convex, since its second derivative is positive for every real number. Therefore the exponential function satisfies the defining property of strictly convex functions: for every "t" in the open interval (0,1) and all real numbers "x" and "y" with "x" ≠ "y",

:$f\left(tx+\left(1-t\right)y\right)< t f\left(x\right)+\left(1-t\right)f\left(y\right),.$

Applying this strict inequality for "t" = 1/"p",1 – "t" = 1/"q","x" = ln "a""p" and"y" = ln "b""q" gives

:

which completes the proof.

Proof of the elementary case

Young's inequality with exponent 2 is the special case "p" = "q" = 2. However, it has a more elementary proof, just observe that:$0le \left(a-b\right)^2=a^2+b^2-2ab,$add 2"ab" to every side and divide by 2.

Young's inequality with "ε" follows by applying Young's inequality with exponent 2 to :$a\text{'}=a/sqrt\left\{varepsilon\right\}, ext\left\{ \right\}b\text{'}=sqrt\left\{varepsilon\right\}b.$

References

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