Solve functional equation $f(x+y) + f(x-y) = 2f(x)f(y)$

Question

The statement of the problem : Let $f: \mathbb R \rightarrow \mathbb R $ with the 2 properties :

1. $f(x+y) + f(x-y) = 2f(x)f(y) , \forall x , y \in \mathbb R$

2. There exists a "the smallest" strictly positive number a, with the property that f(a) is the maximum of the function and f(a)>0 .

Prove that f is a periodic function with period a.

My approach :

For $x = y = 0 \implies 2f(0) = 2f(0)^2$ so $f(0)\in\{0,1\}$;

If $f(0) = 0 \implies 2f(x)=0 \implies f(x)=0 , \forall x \in \mathbb R$ , so it's clearly periodic.(EDIT actually it is false because if the function was 0 then it would not correspond with 2) so f(0) is necessarily 1).

Now if $f(0) = 1$ , for $x = 0$ we get $f(y) + f(-y) = 2f(y) \implies f(y)=f(-y) , \forall x \in \mathbb R $ . So it is enough to determine $f$ on the interval $[0,\infty)$.

EDIT : so I managed to prove that f(a)=1 and that the final answer might be something related to a trigonometric function , probably cosine , since $cos(-x)=cos(x)$ as above.I don't really have any ideas about how I should continue, but I'm waiting for suggestions or solutions.

Any and all proofs will be helpful. Thanks a lot!

Answer 1

$2f(x) = f(x+0) + f(x-0) = 2f(x)f(0)$ for all $x$. If there exist any $w$ so that $f(w) \ne 0$ then $2f(w) =2f(w)f(0)$ and $f(0)=1$. And if $f(x)=0$ for all $x$ that violates property 2. So $f(0) = 1$.

$f(a+a) +f(a-a) = 2f^2(a)$ so $f(2a)+1 = 2f^2(a)$. But $f(a) \ge 1$. But if $f(a) > 1$ we'd have $f(2a) + 1 \le f(a) + 1< f(a)+f(a) \le f^2(a) + f^2(a) =2f^2(a) = f(2a)+1$. That's a contradiction so $f(a) = f(0) = 1$.

It follows that $f(2a) + 1=2f^2(a)=2$ so $f(2a) =1$ and by strong induction that $f(na) = f(a) =1$. [suppose $f(ka) =f(a)=1$ for all $k \le n$ then $f(na + a) + f(na-a)= 2f(na)f(a)=2$. But $f(na+a) + f(na-a)=f((n+1)a) + f((n-1)a) = f((n+1)a) + 1$ so $f((n+1)a) = 1 =f(a)$.]

Now consider $f(x+a) + f(x-a) = 2f(x)f(a) = 2f(x)$. If $f(x+a) \ne f(x-a)$, wolog $f(x-a) < f(x+a)$, then there is a positive $d$ so that $f(x-a) = f(x)-d; f(x+a)=f(x) + d$.

This leads to a bit of problem $f(x + 2a)+f(x) = f((x+a)+a)+f((x+a)-a)=2f(x+a)f(a)=2f(x) + 2d$ so $f(x+2a) = f(x)+2d$ and by induction for any natural $n$ we get $f(x + na) = f(x) + nd$. But by archimedian principal we can find an $n$ are $f(x+na)=f(x) + nd > f(a)$. A contradiction.

So $f(x+a) =f(x-a)$ and $f(x+a)+f(x-a) = 2f(x)$ so $f(x+a)= f(x-a) =f(x)$ for all $x$. (Note if $f(x-a) > f(x+a)$ our argument would have let to $f(x-na) =f(x)+nd>f(a)$ for some positive $d$ and natural $n$)

Answer 2

(Too big for a comment), If we additionally assume that $f$ is $C^2(\mathbb{R})$, then this regularity condition uniquely determines the type of the function. Subtracting both sides by $2f(x)$ and then taking the limit as $y$ goes to zero. We obtain, $$ \frac{1}{y^2}(f(x+y)+f(x-y)-2f(x))=\frac{2}{y^2}f(x)(f(y)-1)$$ or $f''(x) = 2f(x)\left(\lim_{y \to 0}\frac{f(y)-1}{y^2}\right) = Lf(x)$, where $L = \lim_{y\to 0} f''(y)$ (Obtained via LH rule). Since the function $f$ is bounded, $L < 0$ and $f(x) = \cos{(\sqrt{-L}x)}$. I have used the boundary conditions that $f(0) =1$ and by taking partial derivative with respect to $y$ and plugging in $x=y=0$, $f'(0) = 0$. The function achieves maximum value at $\frac{2\pi}{\sqrt{-L}}$, therefore from property 2, $a = \frac{2\pi}{\sqrt{-L}}$ and $f$ is periodic with period $a$, thus $$ \boxed{f(x) = \cos{\left(\frac{2\pi}{a}x\right)} }$$

Answer 3

Partial answer

Plugging in $y = 0$ gives us $2f(x) = 2f(x)f(0)$, so $f(0) = 1$. ($f(x) = 0$ would also be a solution, but would violate property #2.)

Plugging in $x = 0$ gives us $f(y) + f(-y) = 2f(0)f(y)$. From this, and the previous result that $f(0) = 1$, we get $f(-y) = f(y)$. In other words, $f$ is an even function.

Plugging in $y = x$ gives us $f(2x) + f(0) = 2f(x)^2$, or $f(2x) = 2f(x)^2 - 1$. The same result can be obtained with $y = -x$, since $f$ is an even function.

Let $M = f(a)$ be the maximum value of the function. By the recurrence relation $f(2x) = 2f(x)^2 - 1$, we have $f(2a) = 2M^2 - 1$. Because $M$ is the maximum value of $f$, we have $2M^2 - 1 \le M$, or $2M^2 - M - 1 \le 0$. Equivalently, $-\frac{1}{2} \le M \le 1$. But since we know that $f(0) = 1$, them $M \ge 1$. Thus, the only possible value of $M$ is $M = 1$. So, $f(2a) = 2\times 1^2 - 1 = 1$. Similarly, $f(4a) = 2f(2a)^2 - 1 = 2\times 1^2 - 1 = 1$. And continuing this reasoning, we get $\forall n \in\mathbb{N} : f(2^na) = 1$.

Now, plug $y = a$ into the original equation, getting $f(x+a) + f(x-a) = 2f(x)f(a)$. Since $f(a) = 1$, this means $f(x+a) + f(x-a) = 2f(x)$. And similarly, $f(x+2^na) + f(x-2^na) = 2f(x)$.

The recurrence $f(x+a) + f(x-a) = 2f(x)$ can easily be satisfied by any function that's $a$-periodic (so $f(x - a) = f(x) = f(x + a)$). However, right now I'm struggling to prove that this is a necessary condition on $f$, so for now I'm marking my answer as "partial".