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# If f;R to R is an odd function such that f(1+x)=1+f(x)and x^2f(1/x)=f(x) where x is not equal to zero find f(x)

mycroft holmes
272 Points
5 years ago
With the further assumption that f(x) is continuous, we can prove that f(x) = x is the only function satisfying the given conditions. (otherwise it becomes difficult to deal with transcendental reals).

Proof: since the function is odd, we have f(0) = 0 and further f(-x) = -f(x), so it suffices to prove for x>0.

By induction, we easily obtain from the 1st condition that f(n) = n for any natural number n

Now $n^2 f \left(\frac{1}{n} \right ) = f(n) \Rightarrow f\left(\frac{1}{n} \right ) = \frac{1}{n}$

Now if x = 2/n with n>2, we have if n = 2k, x = 1/k and hence f(x) =f(1/k) = 1/k = x from the above result.

Otherwise n = 2k+1, and f(n/2) = f(k+1/2) = k+f(1/2) = k+1/2 = (2k+1)/2 = n/2. Hence f(2/n) = (2/n)2 (n/2) = 2/n.

Now, let us look at x = 3/n, with n>3. If n=3k, then f(x)=x is evident.
Else n = 3k+1 and using f(n/3) = f(k+1/3) we get f(x) = x
Else n = 3k+2 and f(n/3) = f(k+2/3) and using the result we derived for f(2/n) we again have f(x) = x.

So by strong induction, we can prove that f(m/n) = m/n i.e. f(x) = x for any $x \in \mathbb{Q}$.

Now, we take any irrational number x, and consider a sequence of rational numbers {q} that converges to x (just consider the decimal expansion of x for example).

By continuity (Heine Borel Theorem) since the $q_i \rightarrow x$ we must have $f(q_i) \rightarrow f(x)$. But we have already proved that $f(q_i) = q_i$ and hence limits are unique, we must have f(x) = x.

Thus we have proved that f(x) = x for all x.