Young's inequality for products

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File:Young.png
The area of the rectangle a,b can't be larger than sum of the areas under the functions f (red) and f1 (yellow)

In mathematics, Young's inequality for products is a mathematical inequality about the product of two numbers.[1] The inequality is named after William Henry Young and should not be confused with Young's convolution inequality. Young's inequality for products can be used to prove Hölder's inequality. It is also widely used 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.

Standard version for conjugate Hölder exponents

The standard form of the inequality is the following, which can be used to prove Hölder's inequality.

Theorem — If a0 and b0 are nonnegative real numbers and if p>1 and q>1 are real numbers such that 1p+1q=1, then abapp+bqq. Equality holds if and only if ap=bq.

Proof[2]

Since 1p+1q=1, p1=1q1. A graph y=xp1 on the xy-plane is thus also a graph x=yq1. From sketching a visual representation of the integrals of the area between this curve and the axes, and the area in the rectangle bounded by the lines x=0,x=a,y=0,y=b, and the fact that y is always increasing for increasing x and vice versa, we can see that 0axp1dx upper bounds the area of the rectangle below the curve (with equality when bap1) and 0byq1dy upper bounds the area of the rectangle above the curve (with equality when bap1). Thus, 0axp1dx+0byq1dyab, with equality when b=ap1 (or equivalently, ap=bq). Young's inequality follows from evaluating the integrals. (See below for a generalization.)

A second proof is via Jensen's inequality.

Proof[3]

The claim is certainly true if a=0 or b=0 so henceforth assume that a>0 and b>0. Put t=1/p and (1t)=1/q. Because the logarithm function is concave, ln(tap+(1t)bq)tln(ap)+(1t)ln(bq)=ln(a)+ln(b)=ln(ab) with the equality holding if and only if ap=bq. Young's inequality follows by exponentiating.

Yet another proof is to first prove it with b=1 an then apply the resulting inequality to abq. The proof below illustrates also why Hölder conjugate exponent is the only possible parameter that makes Young's inequality hold for all non-negative values. The details follow:

Proof

Let 0<α<1 and α+β=1. The inequality xαxp+β,forallx0 holds if and only if α=1p (and hence β=1q). This can be shown by convexity arguments or by simply minimizing the single-variable function. To prove full Young's inequality, clearly we assume that a>0 and b>0. Now, we apply the inequality above to x=abs to obtain: abs1papbsp+1q. It is easy to see that choosing s=q1 and multiplying both sides by bq+1 yields Young's inequality.

Young's inequality may equivalently be written as aαbβαa+βb,0α,β1,α+β=1. Where this is just the concavity of the logarithm function. Equality holds if and only if a=b or {α,β}={0,1}. This also follows from the weighted AM-GM inequality.

Generalizations

Theorem[4] — Suppose a>0 and b>0. If 1<p< and q are such that 1p+1q=1 then ab=min0<t<(tpapp+tqbqq). Using t:=1 and replacing a with a1/p and b with b1/q results in the inequality: a1/pb1/qap+bq, which is useful for proving Hölder's inequality.

Proof[4]

Define a real-valued function f on the positive real numbers by f(t)=tpapp+tqbqq for every t>0 and then calculate its minimum.

Theorem — If 0pi1 with ipi=1 then iaipiipiai. Equality holds if and only if all the ais with non-zero pis are equal.

Elementary case

An elementary case of Young's inequality is the inequality with exponent 2, aba22+b22, which also gives rise to the so-called Young's inequality with ε (valid for every ε>0), sometimes called the Peter–Paul inequality. [5] 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" aba22ε+εb22. Proof: Young's inequality with exponent 2 is the special case p=q=2. However, it has a more elementary proof. Start by observing that the square of every real number is zero or positive. Therefore, for every pair of real numbers a and b we can write: 0(ab)2 Work out the square of the right hand side: 0a22ab+b2 Add 2ab to both sides: 2aba2+b2 Divide both sides by 2 and we have Young's inequality with exponent 2: aba22+b22 Young's inequality with ε follows by substituting a and b as below into Young's inequality with exponent 2: a=a/ε,b=εb.

Matricial generalization

T. Ando proved a generalization of Young's inequality for complex matrices ordered by Loewner ordering.[6] It states that for any pair A,B of complex matrices of order n there exists a unitary matrix U such that U*|AB*|U1p|A|p+1q|B|q, where * denotes the conjugate transpose of the matrix and |A|=A*A.

Standard version for increasing functions

For the standard version[7][8] 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 f1 denote the inverse function of f. Then, for all a[0,c] and b[0,f(c)], ab0af(x)dx+0bf1(x)dx with equality if and only if b=f(a). With f(x)=xp1 and f1(y)=yq1, this reduces to standard version for conjugate Hölder exponents. For details and generalizations we refer to the paper of Mitroi & Niculescu.[9]

Generalization using Fenchel–Legendre transforms

By denoting the convex conjugate of a real function f by g, we obtain abf(a)+g(b). This follows immediately from the definition of the convex conjugate. For a convex function f this also follows from the Legendre transformation. More generally, if f is defined on a real vector space X and its convex conjugate is denoted by f (and is defined on the dual space X), then u,vf(u)+f(v). where ,:X×X is the dual pairing.

Examples

The convex conjugate of f(a)=ap/p is g(b)=bq/q with q such that 1p+1q=1, and thus Young's inequality for conjugate Hölder exponents mentioned above is a special case. The Legendre transform of f(a)=ea1 is g(b)=1b+blnb, hence abeab+blnb for all non-negative a and b. This estimate is useful in large deviations theory under exponential moment conditions, because blnb appears in the definition of relative entropy, which is the rate function in Sanov's theorem.

See also

Notes

  1. Young, W. H. (1912), "On classes of summable functions and their Fourier series", Proceedings of the Royal Society A, 87 (594): 225–229, Bibcode:1912RSPSA..87..225Y, doi:10.1098/rspa.1912.0076, JFM 43.1114.12, JSTOR 93236
  2. Pearse, Erin. "Math 209D - Real Analysis Summer Preparatory Seminar Lecture Notes" (PDF). Retrieved 17 September 2022.
  3. Bahouri, Chemin & Danchin 2011.
  4. 4.0 4.1 Jarchow 1981, pp. 47–55.
  5. Tisdell, Chris (2013), The Peter Paul Inequality, YouTube video on Dr Chris Tisdell's YouTube channel,
  6. T. Ando (1995). "Matrix Young Inequalities". In Huijsmans, C. B.; Kaashoek, M. A.; Luxemburg, W. A. J.; et al. (eds.). Operator Theory in Function Spaces and Banach Lattices. Springer. pp. 33–38. ISBN 978-3-0348-9076-2.
  7. Hardy, G. H.; Littlewood, J. E.; Pólya, G. (1952) [1934], Inequalities, Cambridge Mathematical Library (2nd ed.), Cambridge: Cambridge University Press, ISBN 0-521-05206-8, MR 0046395, Zbl 0047.05302, Chapter 4.8
  8. Henstock, Ralph (1988), Lectures on the Theory of Integration, Series in Real Analysis Volume I, Singapore, New Jersey: World Scientific, ISBN 9971-5-0450-2, MR 0963249, Zbl 0668.28001, Theorem 2.9
  9. Mitroi, F. C., & Niculescu, C. P. (2011). An extension of Young's inequality. In Abstract and Applied Analysis (Vol. 2011). Hindawi.

References

External links