Find all functions with $f(x + y) + f(x - y) = 2 f(x) f(y)$ and $\lim\limits_{x\to\infty}f(x)=0$.

Determine all functions $f \colon\mathbb{R}\to\mathbb{R}$ satisfying the following two conditions:

(a) $f(x +y) + f(x - y) = 2 f(x) f(y)$ for all $x, y\in\mathbb{R}$;

(b) $\lim\limits_{x\to\infty}f(x) = 0$.

I found this problem in IMO 1985 longlist. I have been able to figure out that there is one solution of the form $f(x) = 0$ for all $x$.

Any other function satisfying the above has to have $f(0) = 1$ and $f(x) = f(-x)$.

However I don't know how to proceed.

Any thoughts?

Solution 1:

For any $x_0,y\in\mathbb{R}$, let $x=y+x_0$, then we have

$f(2y+x_0)+f(x_0)=2f(y+x_0)f(y)$

Hence,

$f(x_0) = 2f(y+x_0)f(y)-f(2y+x_0),\quad \forall y\in\mathbb{R}$

Consequently,

$\displaystyle f(x_0) = \lim_{y\to\infty}\left[f(x_0)\right] = \lim_{y\to\infty}\left[2f(y+x_0)f(y)-f(2y+x_0)\right] = 0$.

Solution 2:

Using $x=y$, we have $f(2x) = 2f(x)^2 - 1$. Pick $X>0$ large enough so that $|f(x)| < 1/2$ for all $x \ge X$. Then $|f(2X)| > 1/2$, a contradiction.

Solution 3:

This is really a comment, not a solution, but I don't have that option. If one only considers the functional equation (a), then there are lots of interesting solutions, namely $\cos ax$ and $\cosh ax$. If one "cheats" and uses imaginary arguments, one can combine these in one formula. By the standard method of inserting a power series and comparing coefficients, one can show that these are the only analytic solutions. Presumably, minimal smoothness conditions on a solution would imply analyticity. However, there are pathological non-measurable solutions. Of course, these solutions are not relevant for the original question since they do not vanish at infinity.