# 高微continuous&archimedes的問題~

(a)for each of the following statements, determine whether it is true or false and justify the answers

1.A continuous function f:(a,b)→R defined on an open interval (a,b) is bounded.

2.A continuous function f:[a,b]→R defined on an closed interval [a,b] is bounded.

(b)

Suppose that function f:[a,b]→R is Lipschitz; that there is a constant c>=0 such that

│f(u)-f(v)│<= c│u-v│ for all points u,v in [a,b]

For a partition P of [a,b] , prove that

0<=U(f,P)-L(f.P)<= c[b-a]‧gapP

(hint: use the extreme value thm. on each partition interval and {[Xi-Xi-1]^2，i從1~n的總和}<=[b-a]‧gapP )

Update:

Rating

(1) False: Define f: (0,1)->|R by f(x)=1/x then f is continuous on (0,1)

but f is not bounded

(2)True:proof

Fix x€[a,b], choose δ_x>0 such that |f(y)-f(x)|<1 whenever y€[a,b] and y€B(x,δ_x)

then {B(x,δ_x):x€[a,b]} form an open cover of [a,b],By compacness of [a,b], there is N such that {B(x_1,δ_1),....,B(x_N,δ_N)} also covers [a,b],

Now for each y€[a,b], y€B(x_i,δ_i) for some i,1<=i<=N

then |f(y)-f(x_i)|<1

=>|f(y)|<|f(x_i)|+1 for 1<=i<=N

=>|f(y)|<M M=max||f(x_1),.......|f(x_N)|}+1

=> f is bounded

(b) Take a partition P={a=x_o<x_1<...<x_n=b}

Let M_i=sup{f(x):x€[x_(i-1),x_i]} mi=inf{f(x):x€[x_(i-1),x_i]}

i=1,2,...,n

then 0<=U(f,p)-L(f,p)=Σ(M_i-m_i)(x_i-x_(i-1))

=Σ sup{f(f)-f(y): x,y€[x_(i-1),x_i]}[x_i-x_(i-1)]

=Σsup{|f(x)-f(y)|:x,y€[x_i,x_(i-1)]}[x_i-x_(i-1)]

<=Σ(i=1~n)c*|x-y|[x_i-x_(i-1)]

<=c*(b-a)^2

2009-03-13 00:12:35 補充：

你的gapP 是什麼

可以再寫清楚一點

2009-03-18 17:40:56 補充：

δ_x 是δ下標x

Fix 是固定的

• (a) 1.No. f(x) = 1/x, x in (0,1) unbounded

(a) 2. Yes. [a,b] is a compact set and f is continuous on [a, b] ==> f([a,b]) is compact set ==> f([a,b]) is bounded.