Lecture notes Fundmentl inequlities: techniques nd pplictions Mnh Hong Duong Mthemtics Institute, University of Wrwick Emil: m.h.duong@wrwick.c.uk Februry 8, 207
2 Abstrct Inequlities re ubiquitous in Mthemtics (nd in rel life). For exmple, in optimiztion theory (prticulrly in liner progrmming) inequlities re used to described constrints. In nlysis inequlities re frequently used to derive priori estimtes, to control the errors nd to obtin the order of convergence, just to nme few. Of prticulr importnce inequlities re Cuchy-Schwrz inequlity, Jensen s inequlity for convex functions nd Fenchel s inequlities for dulity. These inequlities re simple nd flexible to be pplicble in vrious settings such s in liner lgebr, convex nlysis nd probbility theory. The im of this mini-course is to introduce to undergrdute students these inequlities together with useful techniques nd some pplictions. In the first section, through vriety of selected problems, students will be fmilir with mny techniques frequently used. The second section discusses their pplictions in mtrix inequlities/nlysis, estimting integrls nd reltive entropy. No dvnced mthemtics is required. This course is tught for Wrwick s tem for Interntionl Mthemtics Competition for University Students, 24th IMC 207. The competition is plnned for students completing their first, second, third or fourth yer of university eduction.
Contents Fundmentl inequlities: Cuchy-Schwrz inequlity, Jensen s inequlity for convex functions nd Fenchel s dul inequlity 4. Cuchy-Schwrz inequlity........................... 4.2 Convex functions................................ 4. Jensen s inequlity............................... 5.4 Convex conjugte nd Fenchel s inequlity.................. 7.5 Some techniques to prove inequlities..................... 8
Fundmentl inequlities: Cuchy-Schwrz inequlity, Jensen s inequlity for convex functions nd Fenchel s dul inequlity In this section, we review three bsic inequlities tht re Cuchy-Schwrz inequlity, Jensen s inequlity for convex functions nd Fenchel s inequlity for dulity. For simplicity of presenttion, we only consider simplest underlying spces such s R n or finite set. These inequlities, however, cn be stted in much more complex situtions. Mny techniques for proving inequlities re presented vi selected exmples nd exercises in Section.5. 4. Cuchy-Schwrz inequlity Let u, v be two vectors of n inner product spce (over R, for simplicity ). The Cuchy- Schwrz inequlity sttes tht u, v u v. Proof. If v = 0, the inequlity is obvious true. Let v 0. For ny λ R, we hve 0 u + λv 2 = u 2 + 2λ u, v + λ 2 v 2. Consider this s qudrtic function of λ. Therefore we hve u, v 2 u 2 v 2, which implies the Cuchy-Schwrz inequlity..2 Convex functions Definition. (Convex function). Let X R n be convex set. A function f : X R is cll convex if for ll x, x 2 X nd t [0, ] f(tx + ( t)x 2 ) tf(x ) + ( t)f(x 2 ). A function f : X R is cll strictly convex if for ll x, x 2 X nd t (0, ) f(tx + ( t)x 2 ) < tf(x ) + ( t)f(x 2 ). Exmple.. Exmples of convex functions: ffine functions: f(x) = x + b, for, b R n, exponentil functions: f(x) = e x for ny R,
5 ( n Eucliden-norm: f(x) = x = x 2 i ) 2. Verifying convexity of differentible function Note tht in Definition., the function f does not require to be differentible. The following criteri cn be used to verify the convexity of f when it is differentible. () If f : X R is differentible, then it is convex if nd only if f(x) f(y) + f(y) (x y) for ll x, y X. (2) If f : X R is twice differentible, then it is convex if nd only if its Hessin 2 f(x) is semi-positive definite for ll x X. Some importnt properties of convex functions Lemm.2. Below re some importnt properties of convex functions. () If f nd g re convex functions, then so re m(x) = mx{f(x), g(x)} nd s(x) = f(x) + g(x). (2) If f nd g re convex functions nd g is non-decresing, then h(x) = g(f(x)) is convex. Proof. These properties cn be directly proved by verifying the definition.. Jensen s inequlity Theorem. (Jensen s inequlity). Let f be convex function, 0 α i ; i =,..., n such tht α i =. Then for ll x,..., x n, we hve ( ) f α i x i α i f(x i ). () Proof. We prove by induction. The cses n =, 2 re obvious. Suppose tht the sttement is true for n =,..., K. Suppose tht α,..., α K re non-negtive numbers such tht K α i =. We need to prove tht ( K ) f α i x i K α i f(x i ).
6 Since K α i =, t lest one of the coefficients α i must be strictly positive. Assume tht α > 0. Then by the conducting ssumptions, we obtin ( K ) ( f α i x i = f α x + ( α ) where we hve used the fct tht K K i=2 α ) i x i α ( K α ) i α f(x ) + ( α )f x i α α f(x ) + ( α ) = K α i f(x i ), i=2 α i α =. K i=2 i=2 α i α f(x i ) Remrk.4 (Jensen s inequlity- probbilistic form). Jensen s inequlity cn lso be stted using probbilistic form. Let (Ω, A, µ) be probbility spce. If g is rel-vlued function tht is µ integrble nd if f is convex function on the rel line, then ( ) f g dµ f g dµ. Ω Ω Exmple.2 (Exmples of Jensen s inequlity). ) For ll rel numbers x,..., x n, it holds ( ) 2 x i n x 2 i. Proof. Since f(x) = x 2 is convex, we hve ( n ) 2 ( x i = f n ) x i n f(x i) = n x 2 i, which is the desired sttement. 2) Arithmetic-Geometric (AM-GM) Inequlity. Let (x i ) i n nd (λ i ) i n be rel number stisfying x i 0, λ i 0, λ i =.
7 Then, with the convention 0 0 =, λ i x i n x λ i i. (2) In prticulr, tking λ =... = λ n = yields n x i n /n x... x n. Proof. By tking the logrithmic both sides, (2) is equivlent to ( λ i ln(x i ) ln λ i x i ). This is exctly Jensen s inequlity pplying to the convex function f(x) = ln(x)..4 Convex conjugte nd Fenchel s inequlity The convex conjugte of function f : R d R is, f : R d R, defined by f (y) = sup x R d {x y f(y)}. Fenchel s inequlity: for x, y R d, we hve f(x) + f (y) x y. Exmple.. Exmples of Fenchel s inequlity ) f(x) = x 2, then f (y) = sup{x y x 2 } = y 2. Fenchel s inequlity reds 2 2 x R d 2 ( x 2 + y 2 ) x y. 2) f(x) = p x p where p >. Then f (y) = sup{x y p x p } = q y q where + =. p q x R d Fenchel s inequlity becomes: for x, y R d nd p, q > such tht + =, we hve p q p x p + q y q x y.
8.5 Some techniques to prove inequlities In prctises, three inequlities introduced in previous sections often do not pper in stndrd forms. It is crucil to recognize them. In this section, through exercises we will lern some techniques to prove inequlities. Exercise. Let i, b i R, b i > 0 for i =,..., n. Prove tht 2 i b i ( n ) 2 i n b i. Proof. By the Cuchy-Schrz inequlity we hve ( ) 2 ( i ) 2 ( 2 )( i = bi i b i ). bi b i Exercise 2 (Problem 6, IMC 205). Prove tht n= Proof. By the AM-GM Inequlity we hve which implies tht Dividing both sides by n(n + ) yields Hence by summing up over n we obtin n(n + ) < 2. 2(n + ) = n + (n + ) > 2 n(n + ), n= 2(n + ) 2 n(n + ) >. ( < 2 n ). n(n + ) n + n(n + ) < 2 n= ( ) n = 2. n + Exercise. Let A, B, C be three ngles of tringle. Prove tht sin A + sin B + sin C 2.
9 Proof. Consider the function f(x) = sin x. Since f (x) = sin 2 (x) 0, f is concve. Therefore, sin A + sin B + sin C ( A + B + C ) sin = sin π = 2. Exercise 4 (Problem, IMC 206). Let n be positive integer nd,..., n ; b,..., b n be rel number such tht i + b i > 0 for i =,..., n. Prove tht ( n ) 2 i b i b 2 i b i b i i i + b i. () ( i + b i ) Proof. We notice the similr form of both sides of (). For A, B R we hve Applying (4) for A = i, B = b i, we get LHS = i b i b 2 i i + b i = AB B 2 A + B = B 2B2 A + B ( ) b i 2b2 i = i + b i b i 2b 2 i i + b i. (4) Similrly now pplying (4) for A = b i, B = n b i, we obtin ( n ) 2 2 b i RHS = b i n. i + b i Therefore () is equivlent to ( n b i By Cuchy-Schwrz inequlity we hve ( ) 2 ( b i = which implies (5) s desired. ) 2 n i + b i b i ) 2 ( i + b i i + b i b 2 i i + b i (5) b 2 ) i i + b i Exercise 5 (Problem, IMC 200). Let 0 < < b. Prove tht b (x 2 + )e x2 dx e 2 e b2 ( i + b i ),
0 Proof. By the AM-GM Inequlity x 2 + 2x > 0 for ny 0 < x b, we hve b (x 2 + )e x2 dx b 2xe x2 dx = e x2 b = b (x 2 + )e x2 dx. Exercise 6 (Problem 6, IMC 200). Let n be n integer nd let f n (x) = sin x sin(2x)... sin(2 n x). Prove tht f n (x) 2 f n (π/). Proof. Let g(x) = sin x sin(2x) /2. We hve 2 ( g(x) = sin x sin(2x) /2 4 = 4 sin x sin x sin x ) 2 cos x 2 sin 2 x + cos 2 x ( ) 2 4 = = g(π/). 4 2 Note we hve use the AM-GM inequlity tht sin 2 x + cos 2 x = sin 2 x + sin 2 x + sin 2 x + ( cos x) 2 4 4 sin x sin x sin x cos x. Therefore, let K = 2 [ ( /2) n ], we hve f n (x) = sin x sin(2x)... sin(2 n x) ) ) /2 ( ) /4 = ( sin x sin(2x) /2 ( sin(2x) sin(4x) /2 sin(4x) sin(8x) /2 ) K ( ) K/2... ( sin(2 n x sin(2 n x) /2 sin(2 n x) = g(x) g(2x) /2... g(2 n x) K ( sin(2 n x) g(π/)g(x) g(2π/) /2... g(2 n π/) K = f n (π/) / sin(2 n π/) K/2 ( 2 ) K/2 = f n (π/) fn (π/) 2. This is the desired inequlity. ) K/2 Exercise 7 (IMO 995). Let, b, c be positive rel numbers such tht bc =. Prove tht (b + c) + b (c + ) + c ( + b) 2.
Proof. Let x =, y =, z =. Then x, y, z re positive rel numbers nd xyz =. We b c hve (b + c) = ) = x2 y + z. Similrly b (c + ) = x ( y + z y2 z + x, c ( + b) = z2 x + y. By Cuchy-Schrz inequlity (see lso Exercise ) nd the Arithmetic-Geometric Inequlity we hve x 2 y + z + y2 z + x + z2 x + y (x + y + z)2 2(x + y + z) = x + y + z 2 xyz 2 = 2. Exercise 8 (Problem, The 26th Annul Vojtech Jrnik Interntionl Competition 206). Let, b, c be positive rel number such tht + b + c =. Prove tht ( + bc Proof. By the AM-GM inequlity we hve )( b c)( + c + ) 728 b + bc = + bc + bc + bc 4 4 b c, nd Therefore, ( + bc ( + b + c ) 27 = bc, i.e., bc 27. )( b c)( + c + ) ( 4 b = 64 bc 4 b c )(4 4 b c )(4 64 27 = 728. ) 4 c b Exercise 9. Let x, y R, y > 0. Prove tht e x + y(ln y ) x y. Proof. Let f(x) = e x. Then for y > 0, we hve f (y) = sup x R {x y e x } = y(ln y ). By Fenchel s inequlity we hve f(x) + f (y) = e x + y(ln y ) x y s desired.
2 Exercise 0 (IMO 200). Let, b, c be positive rel numbers. Prove tht 2 + 8bc + b b2 + 8c + c c2 + 8b. Proof. Since the expression on the LHS does not chnge when we replce (, b, c) by (k, kb, kc) for rbitrry k R, we cn ssume tht + b + c =. Since x x is convex for x > 0, pplying Jensen s inequlity we obtin 2 + 8bc + Next we show tht b b2 + 8c + c c2 + 8b (2 + 8bc) + b(b 2 + 8c) + c(c 2 + 8b) = + b + c + 24bc. (6) +b +c +24bc = (+b+c) = +b +c +( 2 b+ 2 c+b 2 c+b 2 +c 2 +c 2 b)+6bc, (7) which is equivlent to show tht ( 2 b + 2 c + b 2 c + b 2 + c 2 + c 2 b) 6bc. This is indeed true becuse of the AM-GM inequlity ( 2 b + 2 c + b 2 c + b 2 + c 2 + c 2 b) 6 6 2 b 2 c b 2 b 2 c c 2 c 2 b = 6bc. The desired inequlity follows from (6) nd (7).