微積分講義chap3.3

1、3. Continu ous functions3.19. Defi ni tionER, E and f: E R.1. f is continuous ata E,if given >0,>0 s.t 0< x a , x E f(x) f(a).2. f is continuous onEif f is continuous ata , a E.3.20. TheoremLet I be an open interval such that contains a poina and f : I R .Then f is continuous at a I iff lim

2、 f (x) f (a).x a3.21 TheoremSuppose thatE is a non empty subset oR, a E and f : E R. Then follow ing stateme nts are equivale nt:(1) f is continu ous ata.(2) If xn con verges toa and xn E , the n f(xn)f(a) as n3.22 TheoremSuppose thatE is a non empty subset oR, a E and f, g : E R . If f and g are co

3、n ti nu ous at a, the n(1) f+g and f-g are continuous ata.(2) f gg is continuous ata.(3) f is continuous at a for each real number(4) f is continuous at a provided f is well-defined.gg3.23. Defi nitionf : A R B R.g : B R C R.Define g o f : A Rby (g。f)(x)=g(f(x).g。f is called the composition off and

4、g.3.24. TheoremA,B Rf : A Rg : B Rf(A) B, where f(x)=f(x) x A1. If A=I is an open interval,a I, L= limx -f(x)B and g is con ti nu ous atLlim (go f)(x)=g(lim f (x)=g (L).x ax a2. f is continuous ata, g is continuous atf(a) Proof.g。f is con ti nu ous ata.1. Given>0, to find >0 s.t0< x,xg(f(x)

5、 g(L)2g is con ti nu ous atLgive n>0,1 >0 s.t yg(y)g(L)lim f (x) =Lx afor 1 >0,>0 s.tOv x,xf(x) L0< x a,x Af(x)g(f(x)g(f(x)g(L)Hma(go f)(x)=g(L)2.The Proof is similar to (1).3.25. Defini tio n.ER, Ean d f : E R.f is bounded if M> 0 s.t f (x)E.3.26. Theorem.9 (Extreme Value Theorem)

6、 I: closed, boun ded in terval ofR.f: I R is continuous onl(1). f is boun ded.(2). If M= sup f (x) andm= inf f (x)x IX IXm andxms.t f(xM )=M andf(Xm)=m.Proof.1. We show that f(x)|x a,b is bounded.Suppose not Given n>0, there is anxn a,b s.tf (xn)>n. Xn is a sequenee of a,b xn is a bounded sequ

7、eneeBy Bolanzo -Weierslrasssubsequenee «nk of xn s.t Xnke, asnkfor a real nu mberc bel ongs toa,b.f is continuouslim f (xnk) =f(c)n k(f(Xnk)>nknlim f(Xnk)=)nkf is boun ded2.S=f(x) |x a,b is boundedsupS and infS .We find xM & xm s.t f(x M )= supS, andf(xm)= infS.Then f(xM ) f(y), for each

8、 y a,b f(Xm) f(y), for each y a,bSuppose that no such<Mf(x)<supS for all x a,b.Defi ne g : a,b R byg(x)=1supSf(x)g is con ti nu ous on a,b.By (1). g is boun dedc>0,s.t g(x) c , x a,b.1supS f (x)c f(x)supS-一, x a,b c1(supS supS-一 <M) cXm s.t f(XM)=SupSsimilarly Xm s.t f(Xm)=infS.3.28. The

9、orem>0 s.t f(x)>0f is a real valued fun ctio n.f is continuous atx0 and f(x 0 )> 0 for 0<|x- x 0|< and x Domf.Proof.Given =週>0f is con ti nu ous at x°>0 s.t 0<|x- x o|<, x Domf f (x) f(x°)2 -炷 Vf(X)-f(Xo)V 空2 2 字 <f(x)< 警 ovfg3.29. TheoremLet a,b be a close

10、d interval of R and f : a,b R be continuous.If y0 R lies betweenf(a) and f(b) then x0 (a,b) s.t f(x0)= y0. Proof.We assume thaf(a)< y0<f (b).Let E=x a,b:f(x)< y0. a E E .E a,b E is boun dedBy completeness axiom ofR supE and a supE bLet x0=supE.By Theorem 2.11 xnE s.t lim xn=x 0 a,bnf is con

11、tinu ousf(X0)=lnim f(xj y0supposef (X0)y°f(x°)v y。y。- f(x°)>0Let h(x)= y0-f(x),x a,b h is continu ous ona,b h(x°)>0By Theorem 3.28>0 s.t h(x)>0 for 0< xand x a,b. So,y0 -f(x)>0.y°>f(x) for X0Moreover, x0 bwe can find an x s.t x>x 0f(x°)= y°X

12、。and x a,b.and y0>f(x)(X0=supx|f(x)vy。)3.30 Example力x 0f(x)=1,x0lim f(x) 1 and lim f (x)1x 0x 0 xm0f(x) does not exist. f is note。ntinuou at 03.31. Examplef(x)=sin 1x，x 0,1,x0.1Let g(x)= ,x 0xg is continu ous on R01By Theorem 3.24 f(x)=si n( ) is contin uous on R0 x1lim sin() do not existx 0 xf i

13、s not continu ous at 03.32 Example.1,x Q,f(x)=Show that f is no where continu ous.0,x Q.ProofLet x RBy density of rational and irration numberan Q s.t an x and bn Q s.t bn xlim f (an)1, lim f (bn)0nnlim f (y) does not exist.y xf is not continuous atx, x R3.33 Exampley,x pq q,f(x)= q qThen f is disc on ti nu ous atx Q (0,1) but continu ous0,x Q.at x Qc(0,1).Proof.Let a Qc (0,1).Let xn (0,1), xn a.We show ljm f (xn) =f(a).SupposeXn QLet xn = - be a irrduced form, it is sufficient to show qn as n qnS