What Are the Solutions for These Prove Function Questions?

[guox1560]

Realli need help on those two questions! any1 can help with that, or post the solution. Thank you guys!

Homework Statement

1. Suppose f : R -> R is a function such that
f(2x - f(x)) = x
for all x and let r be a fixed real number.
(a) Prove that if there exists y such that f(y) = y + r, then f(y - nr) =
(y - nr) + r for all positive integers n.
(b) Prove that if, in addition to the assumptions in part a), f is also oneto-
one, then f(y - nr) = (y - nr) + r is true for all integers n.

2. Suppose that f: R -> R is a one-to-one function such that the following
are true
f is continuous at all points a 2 R; that is to say that
lim f(x) = f(a) for all a 2 R.
x->a

* f(0) = 0, and
* f(2x - f(x)) = x for all x 2 R.

Prove that f(x) = x for all x 2 R. (Hint: use the precise definition of limits
and the result of question 1.)

 

[SammyS]

For 1 (a):

Look at f(2y-f(y)).

Then look at f(2(y-r)-f(y-r)).

These suggest using induction.​

For 1 (b):

If f is one-to-one, then f^-1 exists.

It looks like f^-1 has properties similar to f.​

 

[guox1560]

For 1 (a):

Look at f(2y-f(y)).

Then look at f(2(y-r)-f(y-r)).

These suggest using induction.​

For 1 (b):

If f is one-to-one, then f^-1 exists.

It looks like f^-1 has properties similar to f.​

Can you write down the solution for me??

 

[SammyS]

Can you write down the solution for me??

The following is a quote from the Rules for posting on this site:

"On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made."​

If I post the solution I will be reprimanded, and you will retain very little about how to solve this problem.

A little more help:

In order to "Look at f(2y-f(y))." as I suggested:

You are given that, f(2x-f(x))=x .
So, obviously, f(2y-f(y)) = y. You might say, "So what?"
You are also told to suppose there is a y, for fixed real number, r, such that f(y) = y+r. Plug this into 2y-f(y) and simplify, then use the result as the argument (in red) of f((2x-f(x)).​

Once you get the hang of that, do a similar thing to simplify, Then look at f(2(y-r)-f(y-r)).