講課提綱第五次1 4極限與連續(xù)函數(shù)的性質(zhì)

§1.4P60Ex1.4若函數(shù)f(x)在[a,b] ，則對(duì)于正整數(shù)n，存在xn[a,b]，使f(xn)nkBolzano定理，存在收斂子列k

xkxk

A[abf(xA（

k

f(xnk)

fA)

f(xnk) f(x在[a,bx[a,b，總存在

0x(xxxx時(shí)，有f(x)Mx(xkx,xk

(xkx,xk

)k11

,

1k ,Mxn，則f(x)M(x[abf(x在[ab上連續(xù)，所以其在[abMsupf(x)，則根據(jù)上確界的定義，對(duì)任意的正整數(shù)nxn[ab]axf(xn)M1nkBolzano定理，存在收斂子列k

xkxk

A[abf(xA

k

f(xnk)

fAf(xn)M1nfA)MMf(x在[abf(x在[a,b上取不到上確界MMf(x)0(x[a,b])g(x)

Mf

gx在[a,bg(x)0M10gx在[a,bg(x)

Mf

M1f(x)M

M。這與M是f(x)在[a,b]上的上確界1（

令c1a1b1（f(c1)0f(c1)0，則取[a2,b2a1c12取[a2,b2c1,b1令c2a2b2（f(c20，f(c20，則取[a3,b3a2c2]2則，取[a3,b3]c2,b2]

f(bn0存在

an

bn[abf(x在點(diǎn)f()

n

f(an)0，f()

n

f(bn)0f()02（

f(x)A有上界，從而有上確界，記csupA（下面證明acbf(x在點(diǎn)af(a)0，所以存在10f(a1)0，即a1A，因此aa1cf(x在點(diǎn)bf(b)0，所以存在20x[b2,b]f(x0。這說明[b2,b中沒有A中的元素，故b2A的上界，所以cb2bacbf(x)0(x[cbf(c)0不能成立（Why？f(c)0f(x在[ab[cdf1在[cdf(xfy0(cdx0

f1y00

f1(y)f1(y0)

x0

f1(y0)

f1(y)

f1y0x0f(xf(x0)yf(x0f(x0y0yy0f(x0y0取miny0f(x0),f(x0y0y

f1yf1y0)f1yy0f1yf1yy0f1(y00)

f1(y0)

f1y00f1yy0(cdf1y00)f1y00),f1y0ab]yy0f1y)

f1y0f1y00yy0f1(y)

f1y0xf1y00),f1y0y[cd]，使得f1(y)x。這與f1(y)的值域是[a, （1）f(x1在(0,x1x11x1x

x00x1x02x2x0x

2xx0

，知

0。這說明不僅與02x0（2）f(x)sinx在(,)

x2x0(,x2

0sinxsin0

2cosxx02

x

，知。這說明f(xI上有定義，若對(duì)任意的0，總存在0x1Ix2Ix1I

f(x1f(x2)f(xNotef(xI上不一致連續(xù)的嚴(yán)格敘述：存在00，對(duì)任意的0，存在x1Ix2Ix1

f(x1f(x2)033（1） 3333

xx3x23x03xf(x)x3在[1,0“x30

(x2xx0x2)x

x2x

01x0100xx0xx0x3x3x2x00

11

“k1f(x)k2g(x)k1f(x0)k2g(x0)

f(x)f(x0)

g(x)g(x0)Notef(x)x在(0,）x2xx1卻在(0,）x定理（一致連續(xù)與連續(xù)f(xII定理（一致連續(xù)與有界性f(x)在區(qū)間(ab上一致連續(xù)，則其在(ab上Note2f(x在(abf(x在(abf(xsin1在(0,1x

1

1112k2證明：取1f(x在區(qū)間(a112k2xab),xabxb

ba

f(xf(x)1

a

k0,1,2,n0x(xk1xkf(x1)kf(x)f(x1)kxabf(x1n0

f(x)

f(x1n0f(x在(abf(x),g(x在區(qū)間(abf(x)g(x)在區(qū)間(ab上也f(xg(x在區(qū)間(ab上一致連續(xù)，所以有