Suppose I have a function $f(x)$ defined over $[0,\infty]$ and some condition like the following holds: $$f(y)+f(y+10)=2\cdot f(y+3) \text { for any } y\ge0$$ My intuition is that this implies that the function must take the following exponential form: $$f(x)=\alpha-\beta\cdot e^{-\gamma\cdot x}$$ However, I am not sure exactly how to prove this. Any help would be appreciated. Happy to hear intuitions, references, or a proof sketch if the answer is complicated.
Note: $\alpha$ and $\beta$ are a nuisance. To make things a little simpler, assume that $f(0)=-1$ and $f(x)\to0$ as $x\to\infty$. I think these additional conditions imply a form of: $f(x)=-e^{-\gamma\cdot x}$ (with, for the exact numbers above: $\gamma\approx0.1801$).
P.S. One thing that might help - I am fairly sure the first condition implies (additionally assuming the function is continuous and infinitely differentiable, which is fine): $$f'(y)+f'(y+10)=2\cdot f'(y+3) \text { for any } y\ge0$$ $$f''(y)+f''(y+10)=2\cdot f''(y+3) \text { for any } y\ge0$$ and so on for higher derivatives. So, if you can prove it with that information, I can do the rest...
Your intuition is correct in the sense that a solution of the type you propose exists. It is however not the only solution.
Even the added constraints of $f(0)=-1$ and $f(x) \rightarrow 0$ for $x \rightarrow \infty$ are not sufficient to uniquely specify the solution. The problem lies in the non-local relation that is imposed.
If we would ignore those constraints for the moment, we could use any function $f(x)$ on the interval $x \in [0,10)$, because the relation $$f(x+10) = 2 f(x+3) - f(x)$$ will determine the rest uniquely on the interval $[10,\infty)$. By limiting the choice we can also make it continuous, differentiable, etc.
The addition of the limit $f(x) \rightarrow 0$ for $x \rightarrow \infty$ makes it only slightly more complicated, because we could take the following function $$ f(x)=g(x) e^{-\gamma x} + c $$ for any function $g(x)$ that has period 1, i.e., $g(x+1)=g(x)$ and any constant $c$. Inserting this into the required functional relation gives $$ \left(g(x) e^{-\gamma x} + c \right) + \left(g(x+3) e^{-\gamma (x+3)} + c \right) = 2 \left(g(x+10) e^{-\gamma (x+10)} + c \right) $$ which can be rearranged to yield: $$ g(x) e^{-\gamma x} \left(1 + e^{-3\gamma} - 2 e^{-10\gamma} \right)=0 $$ From this it follows that if we take $\gamma\approx 0.180107$ so that the last factor is zero, we always have a correct function that decays to 0 in the limit $x \rightarrow \infty$. The constant $c$ and periodic function $g(x)$ can be chosen such as to fulfill various other constraints such as $f(0)=-1$.
A particular and simple family of examples would be $f(x) = -\cos(2 k \pi x) e^{-\gamma x}$ for any integer $k$. This family of functions is also continuous, differentiable, etc. and satisfies all the constraints.
In fact "all" solutions can be found by realising that the functional equation is actually a recurrence relations of 10th order and that we only need to find all the roots of the characteristic equation is given by $$ x^{10} - 2 x^3 + 1 = 0 $$ which has two real roots, $x=1$ and $x=e^{-\gamma}\approx 0.835181$, and 4 pairs of complex conjugated roots. This means that solutions need not even be continuous at any point, and can be both converging, diverging and combinations thereof.
Each pair of roots gives rise to solutions of the form $$ \cos(\alpha x + \delta) e^{-\beta x} $$ where $\delta$ is an arbitrary shift and $\alpha,\beta$ are fixed by the real and imaginary parts of the complex roots. One of the 4 sets results in $\alpha \approx 2.11591$ and $\beta \approx 0.249421$. This means that it decays faster than the real root of the characteristic equations for which we had found $\gamma \approx 0.180107$. As a results we can construct solutions $$ f(x) = \left[-1 - \epsilon \cos(\delta)\right] e^{-\gamma x} + \epsilon \cos(\alpha x + \delta) e^{-\beta x} $$ which for sufficiently small values of $\epsilon$ will form a whole family of continuous, strictly monotonic increasing functions that satisfy all the constraints.