Convexity. Po-Shen Loh. 25 June 2008

Transcription

1 Conveity Po-Shen Loh 5 June 008 Warm-up Sharpening inequalities can help to reduce the number of variables Prove that inequality ( is sharper than (, for all reals, y 0 ( y y ( 6 ( y y ( 6 Then prove (, using the substitution t = y Single-variable inequalities can be quite easy! Solution: The substitution gives something equivalent to ( t 0 Tools Recall that a set S (say in the plane is conve if for any, y S, the line segment with endpoints and y is completely contained in S It turns out that this concept is very useful in the theory of functions Definition We say that a function f( is conve on the interval I when the set {(, y : I, y f(} is conve On the other hand, if the set {(, y : I, y f(} is conve, then we say that f is concave Note that it is possible for f to be neither conve nor concave We say that the conveity/concavity is strict if the graph of f( over the interval I contains no straight line segments Remark Plugging in the definition of set-theoretic conveity, we find the following equivalent definition The function f is conve on the interval I iff for every a, b I, the line segment between the points (a, f(a and (b, f(b is always above or on the curve f Analogously, f is concave iff the line segment always lies below or on the curve This definition is illustrated in Figure After the above remark, the following famous and useful inequality should be quite believable and easy to remember Theorem (Basic version of Jensen s Inequality Let f( be conve on the interval I a, b I, ( a b f(a f(b f On the other hand, if f( is concave on I, then we have the reverse inequality for all a, b I: ( a b f(a f(b f Then for any Remark This can be interpreted as for conve f, the average of f eceeds f of the average, which motivates the general form of Jensen s inequality:

2 Figure : The function in (i is conve, (ii is concave, and (iii is neither In each diagram, the dotted line segments represent a sample line segment as in the definition of conveity However, note that a function that fails to be globally conve/concave can be conve/concave on parts of their domains For eample, the function in (iv is conve on the part where it is solid and concave on the part where it is dotted Theorem (Jensen s Inequality Let f( be conve on the interval I Then for any,, t I, f(average of { i } average of {f( i } If f were concave instead, then the inequality would be reversed Remark We wrote the above inequality without eplicitly stating what average meant This is to allow weighted averages, subject to the condition that the same weight pattern is used on both the LHS and RHS That s great, but how do I prove that a function is conve? If you know calculus, take the second derivative It is a well-known fact that if the second derivative f ( is 0 for all in an interval I, then f is conve on I On the other hand, if f( 0 for all I, then f is concave on I By the above test or by inspection, here are some basic functions that you should safely be able to claim are conve/concave Constant functions f( = c are both conve and concave Powers of : f( = r with r are conve on the interval 0 < <, and with 0 < r are concave on that same interval (Note that f( = is both conve and concave! Reciprocal powers: f( = are conve on the interval 0 < < for all powers r > 0 For r negative odd integers r, f( is concave on the interval < < 0, and for negative even integers r, f( is conve on the interval < < 0 The logarithm f( = log is concave on the interval 0 < <, and the eponential f( = e is conve everywhere 3 f( is conve iff f( is concave 4 You can combine basic conve functions to build more complicated conve functions If f( is conve, then g( = c f( is also conve for any positive constant multiplier c If f( is conve, then g( = f(a b is also conve for any constants a, b R But the interval of conveity will change: for eample, if f( were conve on 0 < < and we had a = 5, b =, then g( would be conve on < < 7 If f( and g( are conve, then their sum h( = f( g( is conve

3 5 If you are brave, you can invoke the oft-used claim by observation, eg, f( = is conve by observation The above statement might actually pass, depending on the grader, but the following desparate statement definitely will not: f( = ( (3 is concave by observation (It is false anyway, so that would ruin your credibility Now prove the following results f( = is conve on < < For any constant c R, f( = c is conve on < < c Solution: ] [ c = c = c c = c c c c Each summand is conve 3 f( = ( is conve Solution: Epand: / / This is sum of two conve functions 4 f( = ( ( r r! is conve on r < < Solution: Forget about the r! factor Think of f( as a product of r linear terms Then by the product rule, f is the sum of r products, each consisting of r linear terms For eample, one of the terms would be ( ( 3( 4 ( r ; this corresponds to the factor missing But then the second derivative f is a sum of even more products, but each product consists of r linear terms (with two factors missing But for > r, every factor is > 0, and we are taking a sum of products of them, so f > 0 Direct applications of Jensen Now use direct applications of Jensen s inequality to prove the following inequalities (India 995, from Kiran Let,, n be positive numbers summing to Prove that n n n n Solution: Done immediately by Jensen, just need to prove that / is conve on the interval 0 < < Use the substitution t =, and prove conveity (on the same interval of ( t/ t = / t ( t But this is the sum of two conve functions, hence conve! (MMO 963 For a, b, c > 0, prove: a b c b c a c a b 3 Solution: Homogeneous, so WLOG a b c = Then LHS is sum over terms of the form /( = /(, which is conve on 0 < < So by Jensen, it is the case when everything is at the average /3 This gives 3/ not the Massively Multiplayer Olympiad 3

4 3 Let a, b, c be positive real numbers with a b c Prove that a b c b c a c a b Solution: Let S = abc Then the fractions are of the form S, which is conve by the eercise in the previous section Hence LHS is 3 (S/3 (/3S = S/, and we assumed that S 4 (Ireland 998/7a Prove for all positive real numbers a, b, c: 9 a b c ( a b b c c a Solution: Use substitution S = a b c WLOG, S = Then we need conveity of /( on (0,, which is clear 3 Endpoints of conve functions If f( is conve on the interval a b, then f( attains a maimum, and that value is either f(a or f(b If f( is concave on the interval a b, then f( attains a minimum, and that value is either f(a or f(b Now try these problems Use the above facts to observe that it suffices to manually check just a few possible such assignments Finally, check those values! (Bulgaria, 995 Let n and 0 i for all i =,,, n Show that n ( n ( 3 n Bonus: determine when there is equality Solution: This is actually linear in each variable separately Therefore, we can iteratively unsmooth each variable to an endpoint {0, } Rewrite LHS as ( ( 3 n ( With {0, }-assignments, each term is zero unless i = and i = 0, in which case the term is This is a, 0 pattern in consecutive variables when reading the i cyclically Clearly, the maimum possible number of, 0 patterns is the claimed RHS (USAMO 980/5 Show that for all real numbers 0 a, b, c, a b c b c a c ( a( b( c a b Solution: Consider b, c as constant, and see what happens if we perturb a Then, the LHS is a conve function of a, because first term is linear, nd/3rd terms are of the form, and the third term is a purely linear function of a Hence LHS is maimized at one of the etreme values of a (either 0 or, so we may assume it is one of these Similarly, b, c {0, } This gives 8 assignmets of (a, b, c to check, and they all work 4

5 4 Smoothing and unsmoothing Theorem Let f( be conve on the interval I Suppose a < b are both in I, and suppose ɛ > 0 is a real number for which a ɛ b ɛ Then f(a f(b f(a ɛ f(b ɛ If f is strictly conve, then the inequality is strict This means that: For conve functions f, we can decrease the sum f(a f(b by smoothing a and b together, and increase the sum by unsmoothing a and b apart For concave functions f, we can increase the sum f(a f(b by smoothing a and b together, and decrease the sum by unsmoothing a and b apart In all of the above statements, if the conveity/concavity is strict, then the increasing/decreasing is strict as well This smoothing principle gives another way to draw conclusions about the assignments to the variables which bring the LHS and RHS closest together (ie, sharpening the inequality Hopefully, this process will give us a simpler inequality to prove Warning Be careful to ensure that your smoothing process terminates For eample, if we are trying to prove that (a b c/3 3 abc for a, b, c 0, we can observe that if we smooth any pair of variables together, then the LHS remains constant while the RHS increases Therefore, a naïve smoothing procedure would be: As long as there are two unequal variables, smooth them both together into their arithmetic mean At the end, we will have a = b = c, and the RHS will be eactly equal to the LHS, so we are done Unfortunately, the end may take infinitely long to occur, if the initial values of a, b, c are unfavorable! Instead, one could use a smoothing argument for which each iteration increases the number of a, b, c equal to their arithmetic mean (a b c/3 Now try these problems (Zvezda 998 Prove for all reals a, b, c 0: (a b c 3 a bc b ca c ab Solution: If we smooth, say, a and b together, then LHS is invariant, and RHS is of the form c ab c ab( a b But ab will grow, as will a b Smooth until each element hits the arithmetic mean Therefore, RHS will grow, so we may assume that all are equal This corresponds to (3t /3 3t, which is indeed true 3 Problems Now you re on your own Use any method from this lecture to solve these problems Let m n be positive integers Prove that every graph with m edges and n vertices has at least m n V-shapes, which are defined to be unordered triples of vertices which have eactly two edges between themselves Solution: The number is eactly ( d i, where the di are the degrees Average degree is m/n, and ( ( is conve, so Jensen implies that this is n m/n n(m/n(m/n / m /n 5

6 (Ireland 998/7b Prove that if a, b, c are positive real numbers, then a b b c c a ( a b c Solution: Jensen with f(t = /t: f(a f(b ( a b f 3 (T Mildorf s Inequalities, problem 3 Let a,, a n 0 be real numbers summing to Prove that a a a a 3 a n a n 4 Solution: Observe that if any of the a i are zero, then we could increase the LHS by deleting the zero entries (thus reducing the value of n For eample, if n = 4 and the sequence of a i s was (05, 0, 04, 0, then the corresponding LHS with n = 3 and a sequence of (05, 04, 0 would be higher Therefore, we may assume that the sequence of a i has no zeros Net, consider unsmoothing the pair corresponding to a and a 3 Observe that a is only multiplied by a, but a 3 is multiplied by a a 4 > a if n 4, because we just showed all a i > 0 Therefore, we may increase the LHS by pushing a all the way to zero, and giving its mass to a 3 Repeat this process (at most n < times until n 3 Finally, since we only have a, a, and a 3 left, the LHS is simply equal to a (a a 3 = a ( a because they sum to Clearly, this has maimum value /4 (There are also two other cases which correspond to ending up at n = or n =, but those are trivial 4 (Hong Kong 000 Let a a n be real numbers such that a a n = 0 Show that a a n na a n 0 Solution: If there are at least two intermediate a i, neither of which are equal to a or a n, then we can unsmooth them and increase LHS So we may assume that there are of the a i equal to some a = a, at most one of the a i equal to some b, and n or n of the a i equal to some c Case (when there is no b: using a (n c = 0, we solve and get = nc/(c a a (n c = nac, eactly what we needed Then Case (when there is one b: using ab(n c = 0, we solve and get = [b(n c]/(c a Then a b (n c = (n acb( a cb It suffices to show that b( a cb ac If b 0, then use b c to get b bc, hence b( a cb ab ac, final inequality using that a 0 and b c On the other hand, if b 0, then use b a to get b ab, hence b( a cb bc ac, final inequality using that c 0 and b a 5 (IMO 984 For, y, z > 0 and y z =, prove that y yz z yz 7/7 Solution: Smooth with the following epression: (y z yz( Now, if /, then we can push y and z together The mushing algorithm is as follows: first, if there is one of them that is greater than /, pick any other one and mush the other two until all are within / Net we will be allowed to mush with any variable taking the place of Pick the middle term to be ; then by contradiction, the other two terms must be on opposite sides of /3 Hence we can mush to get one of them to be /3 Finally, use the /3 for and mush the other two into /3 Plugging in, we get 7/7 6 (MOP 008/6 Prove for real numbers y : y y y y y y 6

7 7 (MOP Team Contest 008 Let,,, n be positive real numbers with i = Prove: n i 8 (MOP 998/5/5 Let a a n a n = 0 be a sequence of real numbers Prove that: n n a k k( ak a k k= k= Solution: Since the inequality is homogeneous, we can normalize the a k so that a = (If they are all zero, it is trivial anyway Now define the random variable X such that P (X k = a k Then STS E[min{X, X }] E[ X], where X, X, X are iid Prove by induction on n Base case is if n =, trivial Now if you go to n by shifting q amount of probability from P (X = n to P (X = n, RHS will increase by eactly q( n n Yet LHS increases by eactly q under the square root Now since the probability shifted from P (X = n, the square root was originally at least q n In the worst case, the LHS increases by q n q q n, which equals the RHS increase A Compactness Definition Let f(,, t be a function with domain D We say that the function attains a maimum on D at some assignment ( i = (c i if f(c,, c t is greater than or equal to every other value f(,, t with (,, t D Remark Not every function has a maimum! Consider, for eample, the function / on the domain 0 < <, or even the function on the domain 0 < < Definition Let D be a subset of R n If D includes its boundary, then we say that D is closed If there is some finite radius r for which D is contained within the ball of radius r around the origin, then we say that D is bounded If D is both closed and bounded, then we say D is compact Theorem Let f be a continuous function defined over a domain D which is compact Then f attains a maimum on D, and also attains a minimum on D Strictly speaking, this is not the proper definition of compactness, but rather is a consequence of the Heine-Borel Theorem 7