Exponential as solution of a functional equation.

Question

the function $f(t)=2^t$ is a solution of the functional equation: $$f(t+1)=2\cdot f(t),\ f(0)=1$$ Is this unique? Is this unique as continuous funtion? Is this unique as differential funtion?

Thank you.

Answer 1

Choose any continuous function $f(t)$ such that $f(t)=1$ whenever $t$ is an integer, then $$2^t\cdot f(t)$$ will also work. For instance, $f(t)=\cos(2\pi t)$.

Answer 2

No, no, and no.

You may define your function to be whatever you like on $[0,1)$ (with very weak restrictions if you want it to be continuous or differentiable) and then extend it both ways: to $[1,2)$, and to $[-1,0)$, and so on.

Consider this, for one: $$f(t) = 2^{t+\sin(2\pi t)}$$ It is infinitely differentiable and analytic, yet still fits the equation.

Note that I could have put an arbitrary multiplier before the $\sin(2\pi t)$, which would produce an infinite family of such functions.