Prob 6 - Basics

# Prob 6 - Basics - AN ELEMENTARY DERIVATION OF THE CAUCHY...

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Unformatted text preview: AN ELEMENTARY DERIVATION OF THE CAUCHY, H6LDER, AND MINKOWSKI INEQUALITIES FROM YOUNG’S INEQUALITY ELMER TOLSTED, Pomona College 1. Introduction. Three of the most famous “classical inequalities” are those of Cauchy, Holder, and Minkowski. These inequalities are “pulled out of the hat” so frequently in mathematical proofs that an early acquaintance with them would be useful for most students. We shall deduce these three inequalities from an inequality involving integrals due to W. H. Young. In section 2 we present an elementary geometric proof of Young’s inequality. In section 3 we obtain an important special case of Young’s inequality from which we deduce the Holder and Minkowski in— equalities for ﬁnite sequences in sections 4 and 5. Finally in section 6 we use the inequality in section 3 to prove Holder’s and Minkowski’s inequalities for Riemann integrals of continuous functions. In our presentation Cauchy’s inequality appears simply as a special case of Holder’s inequality. Historically, Cauchy’s inequality was published in 1821, whereas Holder’s generalization did not appear until 1889. Minkowski’s in- equality appeared in 1896, while Young’s inequality, which we use as a point of departure, was not published until 1912. See the References for details. This approach seems desirable for an elementary discussion for several reasons: First: All three of these famous inequalities are achieved easily and quickly. Moreover, the geometrically obvious necessary and sufﬁcient condition for equality to hold in the case of Young’s inequality leads directly to necessary and sufﬁcient conditions for equality in the other cases. Second: The Holder and Minkowski inequalities are demonstrated for real exponents. The usual extra limiting process needed to proceed from rational to real exponents is avoided because of the elementary calculus assumed in applying Young's inequality. Third: There is an appealing novelty in deducing inequalities for sequences from one containing integrals. This is the reverse of the usual procedure in inter- mediate courses in analysis, where sequences are studied ﬁrst and properties of integrals deduced later. Note 1. In his two-volume work, Trigonometric Series, Zygmund begins the section on inequalities by assuming Young’s inequality as geometrically ob- vious. He then deduces from it, in a most elegant and condensed fashion, the above inequalities together with many others for ﬁnite and inﬁnite sequences of complex numbers and for Lebesgue integrals. Note 2. In the book entitled Inequalities, by Hardy, Littlewood, and Polya, there are 404 theorems each containing one or more inequalities which are fre- quently used by mathematicians. The inequalities of Young, Holder, Cauchy, and Minkowski appear as Theorems 156, 11, 7, and 25 respectively. Obviously this approach is considerably different from that of Zygmund. We are indebted to this comprehensive work for much of our bibliography. 2 1964] DERIVATION OF THREE INEQUALITIES 3 Note 3. Analytic Inequalities, by Kazarinoff gives a most readable introduc- tion to inequalities for the undergraduate student. We recommend this 90-page book most highly. The Holder and Minkowski inequalities are deduced on pages 67—74. S(¢“(b), b) R(0, 11) 0(0, ¢(a)) R(0, 17) 0(0, 0) Ha, 0) FIG. 1. b=¢(a), ab=A1+A2. FIG. 2. b<¢(a), ab<A1+A2. 2. Young’s inequality. Let y =¢(x) be a continuous, strictly increasing func- tion for x20, and let (15(0) =0 and §b(a) =b, where a and b are any positive real numbers. (See Figure 1.) Solving this equation for x in terms of y we obtain x=¢‘1(y). We call (12—1 the function inverse to <15, and it too is a continuous, strictly increasing function. Note that ¢>-1(0) =0 and ¢‘1(b)=a, and that the equations y=¢(x) and x=¢“‘(y) have the same graph. From elementary calculus we know that the areas A1 and A2 in Figure 1 are given by (2.1) A1 =foaydx =foa¢(x)dx and (2.2) A2 = bxdy= fob¢-1<y)dy= fob¢~1<x)dx. Since ab, the area of the rectangle OPQR in Figure 1, is equal to the sum of the areas A1 and A2, it follows from (2.1) and (2.2) that (2.3) I ab =fa¢(x)dx+f ¢—1(x)dx. Suppose now that b is not equal to §b(a). (See Figure 2 for the case (Ma) >b and Figure 3 for the case ¢(a) <b.) In each ﬁgure, ab is the area of rectangle 0PTR. But this area is smaller than the sum of areas A1 and A2 by the amount STQ which is shaded in the ﬁgures. 4 MATHEMATICS MAGAZINE [Jam—Feb. T(a, b) sea—1(2)), 1:) 0(6, (15(0)) R(0, b) (20:. 4502)) (o, 0) m, 0) FIG. 3. b>¢(a), ab<A1+Ag. FIG. 4. b=¢(a), ab§A1+A2. Combining these three cases we obtain Young’s inequality which states: Let ¢(x) and ¢‘1(x) be two functions, continuous, vanishing at the origin, strictly increasing, and inverse to each other. Then for a, [720 we have (2.4) ab éfa¢(x)dx+fb¢—1(x)dx. From our previous discussion it is clear that equality holds if and only if (2.5) b = ¢(a). Note 4. Had ¢(x) not been assumed to be strictly increasing, but merely continuous, as suggested in Figure 4, we could not use the simple formula (2.2) for area A2. Fortunately, in our applications of Young’s inequality, the functions ¢(x) will vanish at the origin and be continuous and strictly increasing for x :0. 3. Applications of Young’s inequality. In this section we shall obtain in- equalities from Young’s inequality by choosing particular functions ¢(x). Example 1. Let ¢(x) =x. Then ¢”1(x)=x and (2.4) yields a 1) a2 b2 (1be xdx+ xdx=_+_. o o 2 This is the well-known inequality (3-1) 2ab g a2 + 1,2. The condition b=¢(a) which is necessary and sufﬁcient for equality in (3.1) is in this case (3.2) b = a. Note 5. The usual way of proving (3.1) and (3.2) is simply to use the fact that (a—b)2;0. 1964] DERIVATION OF THREE INEQUALITIES 5 We now get the inequality essential to our purposes by choosing ¢(x) = x“ (a > 0)- In this case the inverse function is ¢"1(x) =x1/“, and (2.4) yields a b aa+1 b(1/a)+1 ab éf xadx +f xl/“dx = + ——————~ - o o a + 1 + 1 If in this last inequality we let r=a+1 and r’= (1/a)+1, then we may write 1 1 (m) ag—w+7w. 1' 1' Since [1/{a+1}]+[1/{(1/a)+1}]=1 we see that r and r’ are related by (1/r)+(1/r’) =1 or equivalently an H: Thus the inequality (3.3) holds under the conditions a>0, b>0, r>1, and (1/7) +(1/r’) = 1- The condition for equality in (3.3) is derived again from (2.5). In this case the condition b=¢(a) becomes as b=w=aH. Note 6. An even more roundabout way of obtaining (3.1) and (3.2) would be to let r=r’=2 in (3.3) and (3.5). Note 7. It is by allowing a to be any positive real exponent in the preceding integration that we are able to prove Holder’s inequality for real rather than rational exponents only in the next section. Note that r and 1” may be any real numbers greater than unity and related by the formula in (3.4). Note 8. Inequality (3.3) appears as Theorem 61 in Hardy, Littlewood, and Polya but is not there credited to any individual. Note 9. If a = O and (or) b = 0 the reader can verify that (3.3) still holds. This fact will enable us to prove, in sections 4, 5, and 6, the Holder and Minkowski inequalities in cases in which some terms of the sequences or some values of the functions involved are zero. One could choose values for r in (3.3) such as 3, 7r, 5, sec 31°, etc., and ob- tain all kinds of nonobvious but also noninteresting inequalities. However, we shall use (3.3) in the next section to deduce Holder’s inequality which is also nonobvious but is not noninteresting. 4. Holder’s inequality. In order to keep printing and visual complexities at a minimum, we shall prove the three classical inequalities for ﬁnite sequences of positive real numbers only. The two sequences involved will be denoted by {01;} :{a1, a2, - - - , an}, and {bi} E{b1, b2, - - - , bu}. Note 10. We could prove our three inequalities for moduli of complex num- 6 MATHEMATICS MAGAZINE [Jam—Feb. bers by inserting appropriate absolute value signs throughout the following proofs. But this would clutter up the pages considerably. The following proofs are also valid for inﬁnite sequences provided all sums involved are ﬁnite. Let r r r1/r rllr s=(al+a2+---+a..) 2(Za.) , (4.1) :r = of + 122' + - - - + 125)”? (2 b3)”,- Replace a and b in (3.3) by (Lg/S and bi/T respectively. Then for i: 1, 2, - - - , n we get the n inequalities (4,2) 2 E g 1(2)’ + i(_b:)" (i=1,2,...,,,,, S T r S r’ T Adding up the right and left hand sides of the n inequalities (4.2) and, using (4.1) and (3.4), we get a1b1+a2b2+---+a.b. i(ai+a2+---+a2) (4-3) —-————-——— é Sr ST r +iI<bl'+b;I+’---+b:.l> r T’ 1 1 *ﬂHTmﬂ. f 7 Finally multiplying both extremes of (4.3) by ST, we have a1b1+a2b2+ - - - +anbn§ST which is Holder’s inequality. Using the Z notation and (4.1), we write H b'lder’s inequality in the usual form, (w 2m§EMWZﬂW where r and r’ are real numbers greater than unity and (1 /r) +(1/r’) = 1. We shall have equality in Holder’s inequality if and only if we have equality in each of the n inequalities (4.2). This happens if and only if (3.5) is satisﬁed 71 times, that is if b; a,- ’_1 (4.5) —T- = (i: 1, 2, - - - , 11). Since S and T are independent of i, we see that (4.5) implies there is a con- stant k (k= T/ST—l) such that r—l , bi = kit,- (1 =1, 2, ' ' ' , In this case, we say that the sequences {oi—1} and {In} are proportional. Conversely, if {arl} and {1).} are proportional, so that there is a constant k for which (4.6) holds, one can show that k= T/S"1 and thus that (4.6) implies 1964] DERIVATION OF THREE INEQUALITIES 7 (4.5). This proves that equality holds in Holder’s inequality if and only if (4.6) is satisﬁed. If we now raise both sides of (4.6) to the r’ power, we have (4.7) 1):, = [flay—1)" = lewd: (i = 1, 2, . . . , 1;). Since 12” is also a constant, we may conclude from (4.7) that Holder’s inequality becomes an equality if and only if the sequences {a2} and {bf} are proportional. If we now let r=r’ = 2 in (4.4) and (4.6), we have Cauchy’s inequality, (4.8) 2 an.- é <2 ai>“”<2 bi)“ with equality holding if and only if (4.9) g = kai. Note 11. So far we have proved Cauchy’s inequality for positive a,- and b;. It is easy to see that the inequality will still hold if some or all of the a.- and b.- are allowed to be negative; for the right hand side will be unaffected by a change in sign of the a,- or bi, and the left side will certainly not increase numerically. Thus Cauchy’s inequality holds when a,- and b.- are any real numbers. See Note 9 for the case in which an or bi assume the value zero. 5. Minkowski’s inequality. For any real number r>1 we may write 2 (a; + bi)’ = Z (M + bi)(ai + In)"1 = Z a.(a.+ bar-1+ Zb.(a.-+ bar—1. Application of Holder’s inequality (4.4) to each of the two sums on the right hand side of (5.1) yields 2 (a£+ bi), g 0211/12 (at. + bi)(r——1)r’]1/n + [Z bill/TE (a.- + b.)"‘“"]”"_ Factoring the right hand side and using (3.4) we have (5.2) 2 (a.- + W s [2 (a.- + b.>']”"[<2 a2>1”+ <2 bbwl. Dividing both sides of (5.2) by the ﬁrst factor on the right hand side gives us (5.1) r 1/r [2 (a.- + b.>']"“”" s (2 ab” + (2 b.) Noting that 1— (1 /r’) = 1 /r, we write this as (5.3) [2 (a.- + bo'l‘" s (2 a2)” + (Z bi)” (r a 1). which is M inkowski’s inequality. Note 12. If r= 1, (5.3) obviously becomes an equality. If r <1, the inequality is reversed. See Theorem 25 of Hardy, Littlewood, and Polya. 8 MATHEMATICS MAGAZINE [Jan—Feb. In order to ﬁnd necessary and sufﬁcient conditions for equality in Minkow- ski’s inequality we note that Holder’s inequality was applied to each of the two sums on the right hand side of (5.1). To ensure equality in each of these applica- tions of Holder’s inequality we must satisfy two sets of conditions of the type (4.6), namely: r—l 7—1 (a.- + b.)"1= k. a.- (5.4) r—l r—1 (ai+b‘)r—1=k2 bi. (i=1,2, ...,n) where the role of k in (4.6) is played by 121“1 and k;“1 respectively. Extracting the (r—l)st roots in (5.4) gives us the equivalent conditions a.- + I); = kid; I); = 031 — Dd; (5.5) or a. + b.- = kzb. b.- = {1/(k2 — 1)}a.. We can satisfy the 2 sets of equations in (5.5) simultaneously if and only if 121—1: {1/(k2— 1) In this case we may write (5.5) as one set of 1; equations bi=kdg 1,2,---,n), where (5.7) k=k1-1= 1 - kz—l Since (5.4) implies (5.6) we now know that (5.6) is necessary for equality in Minkowski's inequality. Question: Does (5.6) imply (5.5) and (5.4)? Answer: Yes, for if we are given two proportional sequences {0;} and {b,-} and hence a value for k such that (5.6) is satisﬁed, we may use (5.7) to compute k1=k+1 and k2: (l/k) +1 and be sure that (5.5) and hence (5.4) are satisﬁed. Thus (5.6) is also sufﬁcient to ensure equality in (5.3). Note 13. For r=2 we can give elementary geometric interpretations for the Holder (Cauchy) and Minkowski inequalities. Let n=3, and let A and B be vectors having the components {(11, (12, as} and {b1, b2, b3} respectively. We know from elementary vector analysis that the lengths of vectors A, B, and (A +B) are given respectively by 1.4] =V{:a§}, |Bl =4/{ibﬁ}, and mm =V{:<..+..2}. Thus the Minkowski inequality, Hie-W}éwiath-W 1964] DERIVATION OF THREE INEQUALITIES 9 asserts the obvious geometric fact that (5.8) lA+BI§IAI+IB|. This is the so called triangle inequality. We also know from elementary vector analysis that the scalar or “dot” product of two vectors A and B, deﬁned as IA! IBI cos 0, is equal to gmb‘. (Here 0 is the angle between the vectors A and B.) Since cos Bé 1 we know that (5.9) IAIIBIcosﬁélAIIBI. This is equivalent to Cauchy’s inequality, iaébiéVW} mi}- It is geometrically obvious that we shall have equality in (5.8) and (5.9) if and only if the vectors A and B have the same direction. We know that two vec- tors have the same direction if and only if their components are proportional, which in this case means that 1': ka‘ (7:: 1:2: We recognize this last equation as (4.9) and (5.6). This geometric argument has thus led us to our necessary and sufﬁcient condition for equality to hold in both the Cauchy and the Minkowski inequalities for the case when r=2 and n=3. 6. Integral inequalities. We shall now state and prove the Cauchy, Holder, and Minkowski inequalities for Riemann integrals of continuous functions. If f(x) and g(x) are continuous, nonnegative functions on the closed interval céxéd, then the following inequalities are true. Cauchy-Schwarz: (6.1) [Ldf(x)g(x)dx:l2 _S_ [Ldf2(x)dx]l:£dg2(x)dx]. Equality holds if and only if (6-2) g(x) E kf(x)- H b'lder: . (6.3) df(x)g(x)dx é [Urumqi/Iagr'<x)dx]w, where (1 /r) +(1/r’) = 1 and r and r’ are real numbers greater than 1. Equality holds if and only if (6.4) we a arm—1. M inkowski: (6.5) [awe + g(x>}'dx]m é [dfr(x>dx:|m + [ agr<x)dx]"', 10 MATHEMATICS MAGAZINE [Jam—Feb. where r is any real number greater than or equal to 1. The necessary and sufﬁ- cient condition for equality is (6.2). Note 14. Cauchy’s inequality for integrals is called the Cauchy—Schwarz inequality and is again the special case of Holder’s inequality for which 7 = r’ = 2. It was ﬁrst proved by Cauchy for ﬁnite sums, by Buniakowski for classical integrals, and by Schwarz for Lebesgue integrals. Note 15. Since f(x) and g(x) are continuous nonnegative functions so are fr(x) and g"(x); hence all the above Riemann integrals exist. The proofs that follow are valid for Lebesgue integrals provided all integrals exist. Equality will hold for Lebesgue integrals if the conditions for equality stated in (6.2) and (6.4) hold for almost all x (instead of for all x) on céxéd. Our proof of (6.3) is strictly analogous to the proof of Holder’s inequality in section 4. Assume that neither f(x) nor g(x) is identically zero on céxéd, and let (6.6) S = (fcdf’(x)dx)llr and T = (Ldg"(x)dx>url. Since 5750 and T750, we may choose a=f(x)/S and b=g(x)/T in (3.3) obtain- ing ﬁg g(x) f’(x) 1 g"(x) S T +— S’ r’ T" (chSd). 1 (6.7) _S_ — 1’ Since S and T are deﬁnite integrals, they are constants; hence we may factor them out from under the integral signs when we integrate both sides of (6.7). This yields d d 1.[ﬁmm-+ltﬂyww Sr r’ T" 1 d -§ﬂﬂmmw§— (6.8) r =§m+§m=L Multiplying (6.8) by ST yields ff(x)g(x)dx§ST which, in view of (6.6), is Holder's inequality (6.3). Question: Are we sure that integrating both sides of an inequality like (6.7) really preserves the inequality yielding the inequality (6.8)? Answer: Yes, for if we regard the left side of (6.7) as a continuous function q5(x) and the right hand side of (6.7) as a continuous function 50(x), we may call upon the following theorem from elementary calculus: Let ¢(x) and 30(x) be continuous functions satisfying the inequality d>(x) éiﬂx) for all x on céxéd. Then we have a v d (6.9) f¢(x)dx éf ¢(x)dx. 1964] DERIVATION OF THREE INEQUALITIES 11 Moreover, equality holds in (6.9) if and only if (6.10) ¢(x) = Mac) for all x on 6 § x é d. This theorem is obvious if interpreted in terms of areas. From (6.9) and (6.10) we now know that our integrals are equal when and only when their integrands are equal for all x on c __<_x _S_ d. Thus we have equality in Holder’s inequality if and only if equality holds in (6.7) for all x on céx gd. It then follows from (3.5) that go) = [M T S r—l (6.11) :| for allxonoéxéd is a necessary and sufﬁcient condition for equality in (6.3). Since S and T are constants, (6.11) is equivalent to (6.4). To prove Minkowski’s inequality for integrals, the reader must go through the same procedures with integrals as we did with sums in section 5. Substitut— ing k f(x) for g(x) in both sides of (6.5) and showing that each side reduces to (1 +k) [fff’(x)dx]‘/T is the easy way to show that g(x) = kf(x) is sufﬁcient to en— sure equality. An argument analogous to that leading to (5.6) would show that g(x) Ek f(x) is also necessary for equality in Minkowski’s inequality. Note 16. If we allow f(x) and g(x) to be discontinuous, q5(x) and ¢(x) in (6.9) may also be discontinuous. In this case the integrals in (6.9) will remain equal even though ¢(x) ¢¢(x) at a denumerable set of values of x on céxéd. Thus we are assured of equality in (6.1), (6.3), (6.5) if the condition stated for equality in each case holds for all but a denumerable set of values of x on 6 § x é d. Note 17. We could have proved each of the three inequalities for integrals from the corresponding inequality for sequences, by approximating the integrals by step functions and using the corresponding inequality for sequences on the step functions. To write out such proofs in detail is much more lengthy and cumbersome than our method. Note 18. A very condensed derivation of the Holder, Cauchy-Schwarz, and Minkowski inequalities from Young’s inequality is presented on pages 20—22 of Volume 1, Metric and {Vormed Spaces, of the two-volume work, Elements of the Theory of Functions and Functional Analysis, by Kolmogorov and Fomin. These inequalities are there crucial in proving that the distance, p(x, y), between points x and y of the authors’ numerous examples of metric spaces satisﬁes the triangle inequality: 906, y) + My, 2) a pr. 2)- References 1. V. Buniakowski, Sur quelques inégalités concernant les intégrales ordinaires et les intégrales aux differences ﬁnies. Mémoires de l’Acad. de St. Pe’tersbourg (VII), 1, 1859, No. 9, p. 4. 2. A. L. Cauchy, Cours d'analyse de l’Ecole Royale Polytechnique. Ire partie. Analyse Algé- brique, Paris, 1821, p. 373. 3. G. H. Hardy, J. E. Littlewood, and G. Polya, Inequalities, Cambridge, 1934. 12 MATHEMATICS MAGAZINE [Jam—Feb. 4 O Holder, Uber einen Mittelwertsatz, Gottingen Nachrichten, 1889, pp. 38—47. 5 6 A N. D. Kazarinoff, Analytic Inequalities, Holt, Rinehart and Winston, New York, 1961. H. . N. Kolmogorov, and S. V. Fomin, Functional Analysis, Vol. 1, Metric and Normed Spaces, Graylock, 1957, pp. 16—23. 7. Minkowski, Geometrie der Zahlen, I, Leipzig, 1896, pp. 115—117. 8. H. A. Schwarz, Uber ein die Flachen kleinsten Flacheninhalts betreffendes Problem der Variationsrechnung, (Werke, I, 224-269) Acta soc. scient. F enn., 15 (1885) 315—362. 9. W. H. Young, On classes of summable functions and their Fourier series, Proc. Royal Soc., (A) 87 (1912) 225—229. 10. A. Zygmund, Trigonometric Series, Vol. I, Cambridge, 1959, pp. 16—19. A REMARK CONCERNING THE DEFINITION OF A FIELD A. H. LIGHTSTONE, Carleton University (Canada) A convenient and well-known method of deﬁning a ﬁeld utilizes the notion of an abelian group. Thus, by a ﬁeld we mean any algebraic system,say (S, +, -, 0, 1), where + and - are binary operators on S, 065 and IES, which possesses the following properties. (i) (S, +, 0) is an abelian group. (ii) (S— {0}, -’, 1) is an abelian group, where -’ is - restricted to S— (iii) The following propositions are true about (S, +, -, 0, 1): (a) 0 ¢ 1 (b) VxVyVle'(y + Z) = x'y + 9W] (c) VxVsz[(y + z)-x = y-x + z-x]. At ﬁrst sight, it might appear that (c) is unnecessary. The purpose of this note is to demonstrate that (c) is not a consequence of the other postulates. Consider the algebraic system ({0, 1}, +, -, 0, 1), where + and - are de- ﬁned by the following tables: + 0 1 - 0 1 0 0 1 0 0 1 1 1 0 1 0 1 . Clearly, ({0, 1}, +, 0) is an abelian group, and ({1}, 1), is an abelian group. Furthermore, the left-hand distributive law holds, since VxVyy=y] holds in the given algebraic system. However, the right-hand distributive law fails, since (1+1)-1=1 while 1-1+1 - 1 =0. It follows that (c) is not a consequence of the remaining ﬁeld postulates. To clarify the intended meaning of postulates (i), (ii), and (iii), we spell out these statements in terms of propositions about the algebraic system (81+! 'v 0; ...
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Prob 6 - Basics - AN ELEMENTARY DERIVATION OF THE CAUCHY...

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