Re: [問題] IMO 2011 in Netherlands Day 1

作者: LimSinE (r=e^theta)   2011-07-20 01:19:02
※ 引述《FAlin (FA(ハガレン))》之銘言:
: 3. Let f : R → R be a real-valued function defined on the set of real numbers
: that satisfies
: f(x+y) ≦ yf(x) + f(f(x))
: for all real numbers x and y. Prove that f(x) = 0 for all x ≦ 0.
又到了函方時間了:::
f=0 是一解,假設非此解
1."f(x)/x →-inf(無限大), as x→inf"
若不然,存在A ,對任意x,存在序列yn→inf,使得
A(x+yn) <= f(x+yn) <= yn f(x) + f(f(x))
x固定yn→inf可得 f(x)>=A,代回得 A<=f(x+y)<=yf(x)+f(f(x)) for all x,y
再讓x固定,y→+inf、-inf 得 f(x)=0,矛盾
2.
代入y=0,得 f(x)<=f(f(x)),和1. 得f有上界M,故f(x+y)<=yf(x)+M
3."f<=0"
若存在 f(z)>0,則as x→-inf,f(x)→-inf 從而 f(f(x))→-inf
[此因 f(z+y)<=yf(z)+M,對y取→-inf]
故 f(0) <= x(f(-x)) + M → -inf, as x→inf,矛盾
4.
代入y=f(x)-x,得 f(f(x))<=(f(x)-x)f(x)+f(f(x))
故 (f(x)-x)f(x)>=0。
因此,若f(a)<0,則f(a)<=a;否則 f(a)=0
特別的a=f(x)時,由2.知 f(a)>=a,故 f(a)=a,或f(a)=0
即對所有x, f(f(x))=f(x) 或 0
5."存在z使得 f(f(z))=0"
若不然,則所有x,f(f(x))=f(x)。
又對y>=0 f(x+y)<=yf(x)+f(f(x))<=f(f(x))=f(x)
知f為遞減,因此f。f遞增,兩者又相等,f只好是常數。和1.矛盾
6. f(0)=f(f(f(z))<=0,但 f(f(f(z))>=f(f(z))=0,故 f(0)=0
7. 若x<0,0=f(0)=f(x+(-x))<=(-x) f(x) <=0,故 f(x)=0

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