Find continuous $f$ such that $f(x+1)=f(x)+f(\frac{1}{x})$

Question

Find all real continuous functions that verifies : $$f(x+1)=f(x)+f\left(\frac{1}{x}\right) \ \ \ \ \ \ (x\neq 0) $$

I found this result $\forall x\neq 1 \ \ f(x)=f\left(\frac{x}{x-1} \right)$ and I tried to study the behaviour of the function $g$ defined as $g(x)=\frac{x}{x-1}$ and compare it with $x$ in order to use fixed point theorem but it won't work.

I need a hint and thanks.

Answer 1

Some partial results:

• Letting $x+1= {1\over x} $ we get $$x_{1,2}= {-1\pm \sqrt{5}\over 2}$$ are zeroes for $f$.
• If $x\to {1\over x}$ we get $$f(1+{1\over x}) = f({1\over x})+f(x)=f(1+x)$$ so $$\boxed{f(1+{1\over x}) = f(1+x);\;\;\;\forall x}$$

So taking $x\to 0$ in this equation we get $$f(1) = \lim _{x\to \pm \infty}f(x)$$ and on the other hand we have, from starting equation $$f(1)-f(0) =\lim _{x\to \pm \infty}f(x)$$ so $f(0)=0$ and we have $x_3=0$.

• From boxed equation we get $x\to x-1$ we get, for $\forall x\ne 1$: $$f(x) = f\Big({x\over x-1}\Big)$$

Since last formula is not defined for $x=1$ it is tempting to check whether $$\boxed{g(x) = {x^2+x-1\over x-1}}$$ satisfies the condition, which does, but unfortunately, it is no defined on whole $\mathbb{R}$.

• Finally, if we set $V$ as a set of all solution to this equation, we see that $V$ is a vector space over $\mathbb{R}$.

Answer 2

Some results

Assume that $f$ is defined and continuous on all of $\mathbb{R}$ even if the equation $f(x+1) = f(x) + f(\frac{1}{x})$ is not defined for $x=0.$

From the given identity $ f(x+1) = f(x) + f(\frac1x) $ we get $ f(\frac1x) = f(x+1) - f(x) \to f(1) - f(0) $ as $x\to0$. Thus $\lim_{R\to\infty}f(R) = \lim_{R\to\infty}f(-R) = f(1) - f(0).$

Furthermore, we have $$\begin{align} f(2) &= f(1+1) = f(1) + f(\frac{1}{1}) = f(1) + f(1) \\ f(3) &= f(2+1) = f(2) + f(\frac{1}{2}) = f(1) + f(1) + f(\frac{1}{2}) \\ f(4) &= f(3+1) = f(3) + f(\frac{1}{3}) = f(1) + f(1) + f(\frac{1}{2}) + f(\frac{1}{3}) \\ \vdots \\ f(n) &= f(1) + \sum_{k=1}^{n-1} f(\frac{1}{k}) \\ \end{align}$$ Taking limits as $n\to\infty$ gives $ \lim_{n\to\infty} f(n) = f(1) + \sum_{k=1}^{\infty} f(\frac{1}{k}) , $ but since $\lim_{n\to\infty} f(n) = f(1) - f(0)$ this implies $$ f(0) = -\sum_{k=1}^{\infty} f(\frac{1}{k}) . $$ However, the convergence of the series implies that $|f(\frac1k)|\to 0$ as $k\to\infty.$ By continuity of $f$ this means that $f(0)=\lim_{k\to\infty}f(\frac1k)=0.$ And from that follows that $ \sum_{k=1}^{\infty} f(\frac{1}{k}) = 0 $ and that $ \lim_{R\to\infty}f(\pm R) = f(1) . $

We also get that $$ f(0) = f((-1)+1) = f(-1) + f(\frac1{-1}) = 2 f(-1) , $$ i.e. $f(-1) = 0.$