Integral inequalities

Transcription

1 Integrl inequlities Constntin P. Niculescu Bsic remrk: If f : [; ]! R is (Riemnn) integrle nd nonnegtive, then f(t)dt : Equlity occurs if nd only if f = lmost everywhere (.e.) When f is continuous, f =.e. if nd only if f = everywhere. Importnt Consequence: Monotony of integrl, f g implies f(t)dt g(t)dt: Lecture presented on Decemer 6, 28, t the Adus Slm School of Mthemticl Sciences, Lhore.

2 In Proility Theory, integrle functions re rndom vriles. Most importnt inequlities refer to: M(f) = f(t)dt (men vlue of f) = V r(f) = M (f M(f)) 2 (vrince of f) f 2 (x)dx f(x) dx! 2 : Theorem Cheyshev s inequlity: If f; g : [; ]! R hve the sme monotony, then f(t)g(t)dt f(t)dt! if f; g hve opposite monotony, then the inequlity should e reversed. g(t)dt! ; Appliction: Let f : [; ]! R e di erentile function hving ounded derivtive. Then V r(f) ( )2 2 sup x f (x) 2 :

3 Theorem 2 (The Men Vlue Theorem). Let f : [; ]! R e continuous function nd g : [; ]! R e nonnegtive integrle function. Then there is c 2 [; ] such tht f(x)g(x) dx = f(c) g(x) dx: Theorem 3 (Boundedness). If f : [; ]! R is integrle, then f is ounded, jfj is integrle nd f(t)dt sup t jf(t)j dt jf(t)j : Remrk 4 If f is integrle, then jf(x)j dx f(x)dx 3 sup x f (x) :

4 Remrk 5 Suppose tht f is continuously di erentile on [; ] nd f() = f() = : Then sup jf(t)j t 2 f (t) dt: Theorem 6 (Cuchy-Schwrz inequlity). If f; g : [; ]! R re integrle, then f(t)g(t)dt f 2 (t)dt! =2 with equlity i f nd g re proportionl.e. g2 (t)dt! =2

5 Specil Inequlities Young s inequlity. Let f : [; ]! [; f()] e strictly incresing continuous function such tht f() = : Using the de nition of derivtive show tht F (x) = x f(x) f(t) dt + f (t) dt xf(x) is di erentile on [; ] nd F (x) = for ll x 2 [; ]: Find from here tht x y xy f(t) dt + f (t) dt: for ll x nd y f(). Figure : The geometric mening of Young s inequlity.

6 Specil cse (corresponding for f(x) = x p nd f (x) = x q ) : For ll ; p; q 2 (; ) nd =p + =q = ; then p p + q q p p + q q if p; q 2 (; ) nd p + q = ; if p 2 ( ; )nfg nd p + q = : The equlity holds if (nd only if) p = q : Theorems 7 nd 8 elow refer to ritrry mesure spces (X; ; ): Theorem 7 (The Rogers-Hölder inequlity for p > ). Let p; q 2 (; ) with =p + =q = ; nd let f 2 L p () nd g 2 L q (): Then fg is in L () nd we hve nd X fg d X jfgj d () X jfgj d kfk L p kgk L q : (2)

7 Thus X fg d kfk L p kgk L q : (3) The ove result extends in strightforwrd mnner for the pirs p = ; q = nd p = ; q = : In the complementry domin, p 2 ( ; )nfg nd =p + =q = ; the inequlity sign should e reversed. For p = q = 2; we retrieve the Cuchy-Schwrz inequlity. Proof. If f or g is zero -lmost everywhere, then the second inequlity is trivil. Otherwise, using the Young inequlity, we hve jf(x)j kfk L p jg(x)j kgk L q p jf(x)jp kfk p L p + q jg(x)jq kgk q L q for ll x in X, such tht fg 2 L (). Thus kfk L p kgk L q X jfgj d nd this proves (2). The inequlity (3) is immedite.

8 Remrk 8 (Conditions for equlity). The sic oservtion is the fct tht f nd X f d = imply f = -lmost everywhere. Consequently we hve equlity in () if, nd only if, f(x)g(x) = e i jf(x)g(x)j for some rel constnt nd for -lmost every x: Suppose tht p; q 2 (; ): In order to get equlity in (2) it is necessry nd su cient to hve jf(x)j kfk L p jg(x)j kgk L q = p jf(x)jp kfk p L p + q jg(x)jq kgk q L q lmost everywhere. The equlity cse in Young s inequlity shows tht this is equivlent to jf(x)j p = kfk p L p = jg(x)j q = kgk q L q lmost everywhere, tht is, A jf(x)j p = B jg(x)j q lmost everywhere for some nonnegtive numers A nd B.

9 If p = nd q = ; we hve equlity in (2) if, nd only if, there is constnt such tht jg(x)j lmost everywhere, nd jg(x)j = for lmost every point where f(x) 6= : Theorem 9 (Minkowski s inequlity). For p < nd f; g 2 L p () we hve jjf + gjj L p jjfjj L p + jjgjj L p : (4) Proof. For p = ; this follows immeditely from jf + gj jfj + jgj. For p 2 (; ) we hve jf + gj p (jfj + jgj) p (2 sup fjfj ; jgjg) p 2 p (jfj p + jgj p ) which shows tht f + g 2 L p ():

10 According to the Rogers-Holder inequlity, jjf + gjj p L p = X jf + gjp d X jf + gjp jfj d + X jfjp d =p X =p X jf + gj(p )q jf + gjp jgj d =q d + =q d + X jgjp d jf + gj(p )q X = (jjfjj L p + jjgjj L p) jjf + gjj p=q L p ; where =p + =q = ; nd it remins to oserve tht p p=q = : Remrk If p = ; we otin equlity in (4) if, nd only if, there is positive mesurle function ' such tht f(x)'(x) = g(x) lmost everywhere on the set fx : f(x)g(x) 6= g : If p 2 (; ) nd f is not lmost everywhere, then we hve equlity in (4) if, nd only if, g = f lmost everywhere, for some :

11 Lndu s inequlity. Let f : [; )! R e twice f di erentile function. Put M k = sup (k) x (x) for k = ; ; 2: If f nd f re ounded, then f is lso ounded nd M 2 q M M 2 : Proof. Notice tht f(x) = f(x )+ x x f (t) f (x ) dt+f (x )(x x ): The cse of functions on the entire rel line. Extension to the cse of functions with Lipschitz derivtive.

12 Inequlities involving convex functions Hermite-Hdmrd inequlity: Let f : [; ]! R e convex function. Then + f f() + f() f(x) dx (HH) 2 2 with equlity only for ne functions. The geometric mening. The cse of ritrry proility mesures. See [2]. Jensen s inequlity: If ' : [; ]! [; ] is n integrle function nd f : [; ]! R is continuous convex function, then f '(x) dx! f ('(x)) dx: (J) The cse of ritrry proility mesures.

13 An ppliction of the Jensen inequlity: Hrdy s inequlity: Suppose tht f 2 L p (; ); f ; where p 2 (; ): Put F (x) = x x f(t) dt; x > : Then kf k L p p p kfk L p with equlity if, nd only if, f = lmost everywhere. The ove inequlity yields the norm of the verging opertor f! F; from L p (; ) into L p (; ): The constnt p=(p ) is est possile (though untinted). The optimlity cn e esily checked y considering the sequence of functions f n (t) = t =p (;n] (t):

14 A more generl result (lso known s Hrdy s inequlity): If f is nonnegtive loclly integrle function on (; ) nd p; r > ; then x p r F p (x)dx p p t p r f p (t) dt: r (5) Moreover, if the right hnd side is nite, so is the left hnd side. This cn e deduced (vi rescling) from the following lemm (pplied to u = x p ; p > ; nd h = f): Lemm. Suppose tht u : (; )! R is convex nd incresing nd h is nonnegtive loclly integrle function. Then u x x dx h(t) dt x u(h(x)) dx x :

15 Proof. In fct, y Jensen s inequlity, u = x dx x dx h(t)dt x u(h(t)) dt x x u(h(t)) [;x] (t) dt dx x = x 2 u(h(t)) t = x 2 dx dt u(h(t)) dt t :

16 Exercises. Prove the inequlities :43 < ex2 dx < + e 2 ; 2e < ex2 dx + e2 x2 dx < + e 2 ; e 2 x < e 2 (e ) e ln x dx < e 2 : 2. Let f : [; ]! R e di erentile function hving ounded derivtive. Prove tht V r(f) ( )2 2 sup x f (x) where V r(f) represents the vrince of f: Hint: Put M = sup x f (x) : Then pply the Cheyshev inequlity for the pir of functions f(x)+ Mx nd f(x) Mx (hving opposite monotony). 2

17 3. If f is integrle, then jf(x)j dx Hint: Consider the identity 3 f(x)dx sup x f (x) : ( ) f(x) = x f(t)dt + (t x ( ) f (t)dt t)f (t)dt: 4. Suppose tht f is continuously di erentile on [; ]: Prove tht nd sup jf(x)j x jf(=2)j jf(t)j + jf (t)j dt jf(t)j + 2 jf (t)j dt:

18 3. Suppose tht f is continuously di erentile on [; ] nd f() = f() = : Then sup jf(t)j t 2 f (t) dt: 5. For t > rel numer, consider the function f : (; )! R; f(x) = x t : i) Use the Lgrnge Men Vlue Theorem to compre f(7) f(6) with f(9) f(8); ii) Prove the inequlity 7 t + 8 t < 6 t + 9 t ; iii) Compute R 2 7 t dt: iv) Conclude tht 67 ln ln 8 < 56 ln ln 9 : 6. Consider the sequence ( n ) n de ned y the formul n = r 2 + dx q p 2x {z } n sqr :

19 Prove tht 2 n r 2 + q p 2 {z } n sqr nd nd the limit of the sequence ( n ) n : for ll n 7. Infer from the Cuchy-Schwrz inequlity tht ln(n + ) ln n < qn(n + ) for n nturl nd =2 sin 3=2 xdx < r 3 : 8. Prove the inequlities: 2x2 dx 3=2; esin x dx e sin x dx 2 : 9. Compute lim n! R n+ n x sin x dx nd lim x! R 3x 2x t2 e t2dx:

20 . (The Bernoulli inequlity). i) Prove tht for ll x > we hve nd ( + x) + x if 2 ( ; ) [ (; ) ( + x) + x if 2 [; ]; equlity occurs only for x = : ii) The sustitution + x! x=y followed y multipliction y y leds us to Young s inequlity (for full rnge of prmeters).. (The integrl nlogue of the AM-GM inequlity). Suppose tht f : [; ]! (; ) is continuous function. Prove tht e R ln f(x)dx f(x)dx:

21 2. (Ostrowski s inequlity). Suppose tht f : [; ]! R is di erentile function. Prove tht f(x) f(t) dt 4 + x + 2 ( ) 2 2 C A ( ) su 3. (. Opil). Let f : [; ]! R e continuously di erentile function such tht f() = : Prove tht f(x)dx = ( x)f (x)dx nd infer from this formul the inequlities: f(x)dx f jf(x)j (x) dx 2 2 sup 2 x f (x) f (x) 2 dx:

22 References [] Constntin P. Niculescu, An Introduction to Mthemticl Anlysis, Universitri Press, Criov, 25. [2] C. P. Niculescu nd L.-E. Persson, Convex Functions nd their Applictions. A Contemporry Approch. CMS Books in Mthemtics 23, Springer Verlg, 26.