Functional Equation $f(x+y)-f(x-y)=2f'(x)f'(y)$

Solution 1:

1. OP has already derived

   • $f(2x)-f(0)~=~2f^{\prime}(x)^2~\geq~0$.

   • $f^{\prime}(0)~=~0$.

   • $f(y)-f(-y)~=~0$ even function.

   • $x\mapsto f(x)$ solution $\quad\Rightarrow\quad$ $x\mapsto f(x)+c$ solution.

2. Assume that $f$ is not a constant function. Then there exists a point $y_0\in\mathbb{R}$ such that $f^{\prime}(y_0)\neq 0$. From the bootstrap/recurrence relation $$ f^{(n+1)}(x)~=~\frac{f^{(n)}(x+y_0)-f^{(n)}(x-y_0)}{2f^{\prime}(y_0)}, \qquad n~\in~\mathbb{N}_0,$$ we can deduce inductively that $f\in C^{\infty}(\mathbb{R})$ is infinitely many times differentiable.

3. Under the assumption that $f$ is 4 times differentiable, one can derive the ODE $$f^{(3)}(x)~=~f^{\prime}(x)f^{(4)}(0).$$

4. Combined with what OP already has derived, it follows that the full solution is

   $$ f(x)~=~\lambda^{-2}\cosh(\lambda x)+ c\qquad\vee\qquad f(x)~=~\frac{x^2}{2}+c, \qquad\vee\qquad f(x)~=~c, $$ $$ \lambda~\in~\mathbb{C}\backslash\{0\}, \qquad c ~\in~\mathbb{C}.$$

Solution 2:

Assuming function is twice differentiable,

Differentiating w.r.t $y$ and evaluating at $y=0$, I got $ f''(0)=1 $.

Differentiating w.r.t. $y$ and then w.r.t. $x$ gives $$ f''(x+y) + f''(x-y) = 2f''(x)f''(y) $$ which at $y=x$ gives $ g(2x) + 1 = 2(g(x))^2 $ where $g(x) = f''(x)$.

Does this help?