UC知道differentiations
来源：百度知道 编辑：UC知道 时间：2024/05/24 03:46:08
suppose a function f is defined on the real numbers that for each x and y, f(x+y)=f(x)+f(y)
show that
a) if f is continuous at 0 then f is continuous at any x
b) f(n)=n*f(1); f(-n)=-f(n); f(1/n)=(1/n)*f(1), f(m/n)=(m/n)*f(1) for each positive integers m and n
c) if f is continuous for each x, then f(x)=x*f(1)
show that
a) if f is continuous at 0 then f is continuous at any x
b) f(n)=n*f(1); f(-n)=-f(n); f(1/n)=(1/n)*f(1), f(m/n)=(m/n)*f(1) for each positive integers m and n
c) if f is continuous for each x, then f(x)=x*f(1)
来来来回答你：
a)
first, f(0)=0, obvious.
because f is continuous at 0, limit of f(X)=0 when X->0.
Now, for any x, f(x)=f(x-x')+f(x'). let x'->x, we see that x-x'->0.
As we mentioned, when X->0, f(X)=0.
So, f(x')=f(x), for any x'->x.
As a result, f is continuous at any x.
b)
use induction. obvious. you can prove it.
c)
we just prf that for any rational number q, f(q)=qf(1).
now, for any real number x, we have two sequences of rational number pn and qn, s.t.
pn<x<qn
and limit of (qn-pn) is 0.
or to say that pn and qn ->x.(It's the same)
now, look at the sequence of f(x).
we have pn f(1)< f(x) < qn f(1)
take limit n->positive infinity,
xf(1)<=f(x)<=xf(1)
so, f(x)=xf(1). Done.
给我分哦 打了那么多