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.