If $f(x),\forall x\in\mathbb{R}$ is continous and differentiable, and satisfies:
- $f(x_1+x_2)+f(x_1-x_2)=2f(x_1)f(x_2),\forall x_1,x_2\in\mathbb{R}$
- $f\left(1\right)=\dfrac{3}{2}$
How to prove:
$$f(x)=2^{-x-1} \left(\left(3-\sqrt{5}\right)^x+\left(3+\sqrt{5}\right)^x\right)$$
Is this the unique solution to $f(x)$?
Can we remove the continous and differentiable requirement for $f(x)$ in order to prove the uniqueness of the solution?
With the substitution
$$\begin{aligned} u &=x_1 + x_2 \\ t &= x_1 - x_2 \end{aligned}$$
we can write
$$f(u) + f(t) = 2 f\left(\frac{u+t}{2}\right) f\left(\frac{u-t}{2}\right)$$
For $u = t = 1$ we get
$$2f(1) = 2f(1)f(0)$$
which leads to
$$f(0) = 1$$
Setting either $u$ or $t$ to $1$ and the other variable to $x$ gives us:
$$\begin{aligned} f(x) + \frac{3}{2} &= 2 f\left(\frac{x+1}{2}\right) f\left(\frac{x-1}{2}\right) \\ f(x) + \frac{3}{2} &= 2 f\left(\frac{x+1}{2}\right) f\left(\frac{1-x}{2}\right) \end{aligned}$$
This can be transformed to:
$$f\left(\frac{x-1}{2}\right) = f\left(\frac{1-x}{2}\right)$$
which is equivalent to the symmetry:
$$f(x) = f(-x)$$
$f(x)$ can never be $0$. (How to prove that?) Therefore, it must be positive for all $x$.