Hölder s and Minkowski s Inequality

Transcription

1 Hölder s and Minkowski s Inequality James K. Peterson Deartment of Biological Sciences and Deartment of Mathematical Sciences Clemson University Setember 10, 2018

2 Outline 1 Conjugate Exonents 2 Hölder s Inequality 3 Minkowski s Inequality 4 Norms and Vector Saces 5 Homework

3 Conjugate Exonents We say the ositive numbers and q are conjugate exonents if > 1 and 1/ + 1/q = 1. If = 1, we define its conjugate exonent to be q =. Conjugate exonents satisfy some fundamental identities. Clearly, if > 1, q = 1 = 1 = + q q From that it follows and from that using factoring q = + q ( 1) (q 1) = 1 We will use these identies quite a bit.

4 Conjugate Exonents Lemma The α β Lemma: Let α and β be ositive real numbers and and q be conjugate exonents. Then α β α + βq q. To see this is a straightforward integration. We haven t discussed Riemann integration in deth yet, but all of you know how to integrate from your earlier courses. Let u = t 1. Then, t = u 1/( 1) and using the identity ( 1) (q 1) = 1, we have t = u q 1. Now we are going to draw the curve u = t 1 or t = u q 1 in the first quadrant and show you how this inequality makes sense. We will draw the curve u = t 1 as if it was concave u (i.e. like when = 3) even though it could be concave down (i.e. like when = 3/2 ). Whether the curve is concave u or down does not change how the argument goes. So make sure you can see that. A lot of times our ictures are just aids to heling us think through an argument. A laceholder, so to seak!

5 MATH 4540: Analysis Two Conjugate Exonents Here the area of the rectangle α β the area of Region I + the area of Region II + the area marked in green.

6 MATH 4540: Analysis Two Conjugate Exonents Here the area of the rectangle α β the area of Region I + the area of Region II.

7 Conjugate Exonents In the first icture, (1): the area of Region I is the area under the curve t = u q 1 from u = 0 to u = β. This is β 0 u q 1 du. (2): the area of Region II + the area marked in green is the area under the curve u = t 1 from t = 0 to t = α. This is α 0 t 1 dt. In the second icture, (1): the area of Region I is still the area under the curve t = u q 1 from u = 0 to u = β. This is β u q 1 du. 0 (2): the area of Region II is the area under the curve u = t 1 from t = 0 to t = α. This is α t 1 dt. 0 So in both cases α β β 0 α u q 1 du + t 1 dt = βq 0 q + α.

8 Hölder s Inequality Definition Let 1. The collection of all sequence, (a n ) for which a n converges is denoted by the symbol l. (1) l 1 = {(a n ) : a n converges. } (2) l 2 = {(a n ) : a n 2 converges. } We also define l = {(a n ) : su n 1 a n < }. Theorem Hölder s Inequality: Let > 1 and and q be conjugate exonents. If x l and y l q, then ( ) 1/ ( ) 1/q x n y n x n y n q where x = (x n ) and y = (y n ).

9 Hölder s Inequality This inequality is clearly true if either of the two sequences x and y are the zero sequence. So we can assume both x and y have some nonzero terms in them. Then x l, we know 0 < u = ( ) 1/ ( ) 1/q x n <, 0 < v = y n q < Now define new sequences, ˆx and ŷ by ˆx n = x n /u and ŷ n = y n /v. Then, we have ˆx n = ŷ n q = x n u = 1 u y n q v q = 1 v q x n = u u = 1. y n q = v q v q = 1.

10 Hölder s Inequality Now aly the α β Lemma to α = ˆx n and β = ŷ n for any nonzero terms ˆx n and ŷ n. Then ˆx n ŷ n ˆx n / + ŷ n q /q. This is also true, of course, if either ˆx n or ŷ n are zero although the α β lemma does not aly! Now sum over N terms to get N ˆx n ŷ n 1 N ˆx n + 1 q N ŷ n q Since we know x l and y l q, we know N ˆx n ˆx n = 1 N ŷ n q ŷ n q = 1

11 Hölder s Inequality So we have N ˆx n ŷ n q = 1 This is true for all N the artial sums N ˆx n ŷ n are bounded above. Hence, the artial sums converge to this suremum which is denoted by ˆx n ŷ n. We conclude ˆx n ŷ n 1. But ˆx n ŷ n = 1/(u v) x n y n and so we have 1 u v x n y n 1 which imlies the result as ( ) 1/ ( ) 1/q x n y n u v = x n y n q

12 Hölder s Inequality Theorem Hölder s Theorem for = 1 and q = if x l 1 and y l, then x ny n ( ) x n su n 1 y n. We know since y l, y n su k 1 y k. Thus, N x n y n ) x n su y k k 1 Thus the sequence of artial sums N ( ) x ny n is bounded above by x n su k 1 y k. This gives us our result.

13 Minkowski s Inequality Theorem Minkowski s Inequality Let 1 and let x and y be in l, Then, ( n 1 x n + y n ) 1 ( n 1 x n ) 1 + ( n 1 y n ) 1 (1): = We know x n + y n x n + y n by the triangle inequality. So we have x n + y n su n 1 x n + su n 1 y n. Thus, the right hand is an uer bound for all the terms of the left. We then can say su n 1 x n + y n su n 1 x n + su n 1 which is the result for =. (2): = 1 Again, we know x n + y n x n + y n by the triangle inequality. Sum the first N terms on both sides to get

14 Minkowski s Inequality The right hand side is an uer bound for the artial sums on the left. Hence, we have (3) 1 < < We have x n + y n x n + y n x n + y n = x n + y n x n + y n 1 N x n + y n x n x n + y n 1 + y n x n + y n 1 N N x n x n + y n 1 + y n x n + y n 1 Let a n = x n, b n = x n + y n 1, c n = y n and d n = x n + y n 1. Hölder s Inequality alies just fine to finite sequences: i.e. sequences in R N.

15 Minkowski s Inequality So we have N a n b n a n ) 1 ( N b q n ) 1 q But b q n = x n + y n q( 1) = x n + y n using the conjugate exonents identities we established. So we have found N x n x n + y n 1 ) 1 ( x n N x n + y n ) 1 q We can aly the same reasoning to the terms c n and d n to find N y n x n + y n 1 ) 1 ( y n N x n + y n ) 1 q

16 Minkowski s Inequality We can use the inequalities we just figured out to get the next estimate N x n + y n ( N + ) 1 ( x n N y n ) 1 x n + y n ) 1 q x n + y n ) 1 q Now factor out the common term to get ( N ( N ) 1 ( x n + y n x n N + Rewrite again as x n + y n ) 1 1 q ) 1 ) y n ( N x n ) 1 + x n + y n ) 1 q y n ) 1

17 Minkowski s Inequality But 1 1/q = 1/, so we have x n + y n ) 1 ( x n ) 1 + x n ) 1 + ( y n ) 1 y n ) 1 This says the right hand side is an uer bound for the artial sums on the left. Hence, we know ( x n + y n ) 1 ( x n ) 1 + ( y n ) 1

18 Norms and Vector Saces If x l, we can define a new function, called a norm, on l like this: x = ( x n ) 1, x = su x n, = n 1 The Minkowski Inequality can then be rehrased as x + y x + y 1 < and so we can use as a way to measure both size of x and the distance between x and y. Letting d(x, y) denote the distance between x and y, we have d(x, y) = x y = x + ( y) x + y = x + y We note if d(x, y) = 0, then in the case 1 <, we have x n y n = 0. But here you are adding u non-negative terms, so this must imly x n y n = 0 for all n. This tells us x n = y n for all n; i..e x = y.

19 Norms and Vector Saces On the other hand, when =, we would have su n 1 x n y n = 0. But this says, x n y n 0 for all n. Hence, x n = y n for all n telling us x = y again. Also, note the Minkowski Inequality tells us that l is a vector sace as if x and y are in l, their sum x + y defined by x + y is the sequence (x n + y n ) has x n + y n < because of Minkowski s Inequality. A function like on the set l is called a norm and the set l is called a Normed Linear Sace or Normed Vector Sace. A vector in l is the sequence x and its magnitude or size in x. This is not a finite dimensional vector sace and is another examle of such things to add to your collection along with (C[0, 1], d 1 ) and so forth.

20 Homework Homework Prove a geometric series x with common ratio r in ( 1, 1) is in l 1 and find x If x R 2, the usual Euclidean norm is 2. Note x 3 and so on is also a norm on R 2. Comute x 1, x 2, x 3,..., x 10 for x = [2, 5]. What do you think haens as? 7.3 The Hölder s inequality tells us that in R 2, < x, y > /( x y q ) is in [ 1, 1]. So we can use this to define the angle between x and y. For = q = 2 this is our usual angle, but for = 3, q = 3/2 and so on the calculation changes. Calculate this angle for = 2, = 3, = 4 and = 5 where the q value is the exonent conjugate to for the two vectors x = [ 1, 3] and y = [2, 4].