A*******r
发帖数: 217 | 1 Find all functions f : R≥0 → R≥0 such that, for all x, y ∈ R≥0,
f（（x + f(x)）/2+ y)= 2x − f(x) + f(f(y))
and
(f(x) − f(y))(x − y) ≥ 0. | f***r
发帖数: 1126 | 2 suppose f is monotonic.
Suppose f(0)=c.
(1) choose x=0. f(f(y))-c=f(*)>=f(0)=c
therefore f(f(y))>=2c for all y
(2)choose x=c
f(c/2 +f(c)/2+y)=2c-f(c)+f(f(y))=2c-f(f(0))+f(f(y))<=f(f(y))
f is monotonic, so
c/2+f(c)/2+y<=f(y) for all y.
(3) choose y=0
f(x/2+f(x)/2)+f(x)=2x+f(c). Using the inequality of (2) twice and obtain
2x+f(c)>=f(x+c/4+f(c)/4)+x+c/2+f(c)/2
>=x+c/4+f(c)/4+c/2+f(c)/2+x+c/2+f(c)/2
>=2x+f(c)+5c/4+f(c)/4
>=2x+f(c)
Therefore c=f(c)=0 and all the inequalities should be equalities. f(x)=x for
all x. | f***r
发帖数: 1126 | 3 The following is a proof that f should be monotonic.
Suppose we have f(a)=f(b)=d for some a<=b. Then for all x in [a,b] we should
have f(x)=d.
Take x=(a+b)/4+b/2 and y=(x+a)/2 both of which are in [a,b].
We have
f(3(a+b)/4+d/2)=(a+b)/2+b-d+f(d).
Similarly take x=(a+b)/4+a/2 and y=(x+b)/2 both of which are in [a,b]. We
have
f(3(a+b)/4+d/2)=(a+b)/2+a-d+f(d).
The above two equations imply that a=b. Therefore f is monotonic. |
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