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 Post subject: functional equationPosted: Wed, 3 Dec 2008 10:21:58 UTC
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Joined: Sun, 23 Mar 2008 10:55:47 UTC
Posts: 29
Suppose is a one-to-one continuous function satisfying the following conditions:

a) , for all real .
b) There exists a real number such that .

Prove that for all real .

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 Post subject: Posted: Thu, 4 Dec 2008 07:00:12 UTC
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Joined: Sun, 4 May 2003 16:04:19 UTC
Posts: 2906
since f is one-to-one and continuous, it has an inverse g.

Hence g(f(2x-f(x)) = g(x) = 2x-f(x)

f(x) + g(x) = 2x
not sure if it helps but it looks simpler than what you had before. and i didn't even use the part b)

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Has anyone noticed that the below is WRONG? Otherwise this statement would be true:

where

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 Post subject: Re: functional equationPosted: Thu, 4 Dec 2008 15:58:37 UTC
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Joined: Sat, 7 Jan 2006 18:29:24 UTC
Posts: 1401
Location: Leeds, UK
mathemagics wrote:
Suppose is a one-to-one continuous function satisfying the following conditions:

a) , for all real .
b) There exists a real number such that .

Prove that for all real .

Here's a somewhat cumbersome proof. Maybe someone can improve on it.

By translation, we can assume that . Write the defining relation as (as suggested by bugzpodder).

Since f is invertible and continuous, it is monotone. In fact, it must be increasing (because if it is decreasing, with f(0)=0, then the inverse function would also be decreasing, so that both and would be negative, and their sum could not be equal to 2).

Suppose (for a contradiction) that there exists x > 0 with 0 < f(x) = y < x. Then and therefore . But f(y) must be positive, since f is increasing, and so 2y-x>0, or .

Since it follows that , and so . This must be positive, since f is increasing, and so 3y-2x>0, or .

Now continue this argument, until a pattern becomes obvious. The next step is to say that since it follows that , and so . This must be positive, since f is increasing, and so 4y-3x>0, or .

So the pattern is that and therefore for all n (proof by induction). Hence , contradicting the assumption that y<x.

Therefore for all x.

Now suppose (again for a contradiction) that there exists x > 0 with y = f(x) > x. Then . But the defining relation is symmetric in and , so we can apply the prevous argument with in place of , and again get a contradiction.

Therefore f(x)=x for all x.

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 Post subject: reply to the questionPosted: Mon, 8 Dec 2008 09:10:42 UTC
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Joined: Mon, 8 Dec 2008 08:44:12 UTC
Posts: 3
Location: Chandigarh
as the function is continuous and differentiable therefore we can differentiate it.
f'(2x-f(x))*(2-f'(x))=1
putting x=x0,
as f(xo)=xo,
f'(xo)*(2-f'(xo))=1 we get f'(xo)=1 as the only solution.now applying f(xo+h)=f(xo)+h*(f'(xo) where h tends to zero.
this implies f(xo+h)=f(xo)+h.now extending it to any real value x we get f(x)=x.hence this relation is proved.my soln. is based upon the continuity and differentiability of the function. had it not been diferentiable i would not have been able to differentiate it and lastly by using the basic definition of f'(x) i came to the desired conclusion. if u have any queries.please post and i will be more than willing to join the discussion.

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