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 Post subject: functional equation
PostPosted: Wed, 3 Dec 2008 10:21:58 UTC 
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Suppose f: \mathbb{R} \to \mathbb{R} is a one-to-one continuous function satisfying the following conditions:

a) f\left(2x-f(x)\right)=x, for all real x.
b) There exists a real number x_0 such that f(x_0)=x_0.

Prove that f(x)=x for all real x.


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PostPosted: Thu, 4 Dec 2008 07:00:12 UTC 
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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:
-1\cong1\pmod{13}
i\cong5 \pmod{13} where
i^2=-1


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 Post subject: Re: functional equation
PostPosted: Thu, 4 Dec 2008 15:58:37 UTC 
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mathemagics wrote:
Suppose f: \mathbb{R} \to \mathbb{R} is a one-to-one continuous function satisfying the following conditions:

a) f\left(2x-f(x)\right)=x, for all real x.
b) There exists a real number x_0 such that f(x_0)=x_0.

Prove that f(x)=x for all real x.

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

By translation, we can assume that x_0=0. Write the defining relation as f(x) = 2x - f^{-1}(x) (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 f(1) and f^{-1}(1) 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 f^{-1}(y) = x and therefore f(y) = 2y - f^{-1}(y) = 2y-x. But f(y) must be positive, since f is increasing, and so 2y-x>0, or y>\tfrac12x.

Since f(y) = 2y-x it follows that f^{-1}(2y-x) = y, and so f(2y-x) = 2(2y-x) - y = 3y-2x. This must be positive, since f is increasing, and so 3y-2x>0, or y>\tfrac23x.

Now continue this argument, until a pattern becomes obvious. The next step is to say that since f(2y-x) = 3y-2x it follows that f^{-1}(3y-2x) = 2y-x, and so f(3y-2x) = 2(3y-2x) - (2y-x) = 4y-3x. This must be positive, since f is increasing, and so 4y-3x>0, or y>\tfrac34x.

So the pattern is that f(ny-(n-1)x) = (n+1)y-nx and therefore y>\tfrac{n-1}nx for all n (proof by induction). Hence y\geqslant x, contradicting the assumption that y<x.

Therefore f(x)\geqslant x for all x.

Now suppose (again for a contradiction) that there exists x > 0 with y = f(x) > x. Then f^{-1}(y) = x < y. But the defining relation f(x) + f^{-1}(x) = 2x is symmetric in f and f^{-1}, so we can apply the prevous argument with f^{-1} in place of f, and again get a contradiction.

Therefore f(x)=x for all x.


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 Post subject: reply to the question
PostPosted: Mon, 8 Dec 2008 09:10:42 UTC 
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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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