※ 引述《cherry770519 (＊桃子＊)》之銘言：
: 2 1
: 1. If f: R -> R is of class C show that f is not 1-1
Pick a point (x0,y0) such that df(x0,y0) does not vanish.
If df vanishes everywhere then f is identically constant, therefore not 1-1.
By the implicit function theorem the equation
f(x(t),y(t)) = c = f(x0,y0)
1
locally has a C solution γ(t) = (x(t),y(t)) with γ'(t) != 0.
Hence {γ(t) | -δ<t<δ} consists of more than a single point, and they all
belong to the preimage of c under f.
: 2 1 2
: 2. If f: R -> R is of class C show that f does not carry R onto R
: 2
: in fact, show that f(R) contains no open subset of R
It suffices to show that f(|R) has 2-dimenstional measure zero.
Since coutable unions of measure zero sets are of measure zero,
proving f([-K,K]) has measure zero, for each positive K, is enough.
Put M = max |f'|. Divide the interval [-K,K] into n pieces of the same length
[-K,K]
2K/n. By the mean value theorem, each subinterval is mapped into a closed
disk of radius not greater than M(K/n). Hence f([-K,K]) is contained in
n closed disks of radius M(K/n) and the total area will be at most
2
nπ(M(K/n)),
which tends to zero as n goes to infinity.
: 3. Given a sequence (Xn) in R,define
: ---
: lim Xn := lim sup{Xk , k >= n} lim Xn := lim inf{Xk , k >= n}
: n->無限大 --- n->無限大
: (a) find the lim Xn and lim Xn of the sequence Xn= sin(n*(pi)/4)
: (b) For any bounded sequence (Xn) , show that
: ---
: x= lim Xn if and only if for any @ > 0 ,
: there is only finitely many Xn's satisfying Xn > = x + @
: but there is infinitely many Xn's satisfying Xn > = x - @
(a) 1;-1
(b) Put Yn = sup{Xn, Xn+1,...}. {Yn} decreases to limsup Xn = L.
To each ε>0, there correspond N such that Yn < L+ε for each n>N.
Hence Y_N+1 < L+ε and therefore Xn < L+ε for each n>N.
If there exists ε>0 and N such that Xn < L-ε for all n>N,
then Yn ≦ L-ε for all n>N. Since {Yn} decreases L, this is impossible.
Conversely, aussume J is a number with such properties.
Given ε>0, there exists N with Xn < J+ε for all n>N.
Hence Yn≦J+ε for all n>N and therefore L ≦ J.
Suppose L < J, then there would be only finitely many Xn's
with Xn > L+(J-L)/2, but the second property of J asserts that
there are infinitely many Xn's with Xn > J-(J-L)/2.
Thus L=J.
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