Find a function $h:\mathbb{R}\to(0,\infty)$ with the following property. I need the proof of the existence of a bijective function $h$ with the following property, and if it is unique:
$$\forall x,a\in\mathbb{R}:a\cdot h(x)=h(x+1)$$
I know that if the assumpion $h(x)=a^x$ is true then it is a valid solution, but I don't know how to prove it without making the assumption.
 A: The comment of John Hughes suggests how to find all solutions to the functional equation.
Indeed, your assumption is equivalent to say that the function
$K(x) := h(x) / a^x$ is positive $1$-periodic, i.e. $K(x+1) = K(x)$ for every $x\in\mathbb{R}$.
Hence, any function of the form
$$
h(x) = a^x K(x),
\qquad
K\colon\mathbb{R}\to (0,\infty)\quad 1-\text{periodic}
$$
is a solution to your functional equation.
Among these solutions, you are looking for the ones that are bijective.
Clearly you must have $a\neq 1$.
In this case, if $K = $ constant you get a solution.
Unfortunately, we can also have other solutions.
Consider, for example, the case $a = e$ and let
$$
h(x) = e^x (10 + \sin(2\pi x)).
$$
Then
$$
h'(x) = e^x (10 + \sin(2\pi x) + 2\pi \cos(2\pi x)) > 0
\quad \forall x\in\mathbb{R},
$$
hence $h$ is a bijection from $\mathbb{R}$ to $(0,\infty)$.
A: If $h(x_0)=0$ for some real $x_0$ then $ah(x_0)=h(x_0+1)=0$ and the function $h$ is not bijective.It follows that $h(x)$ reaches the value $0$ in limit position when either $x\to-\infty$ or $x\to+\infty$.
If $a=1$ is fixed the $h(x)=h(x+1)$ for all real and the function is not bijective.
The only possibilities are $h(x)=a^x$ with $a\in\mathbb R_+\setminus\{1\}$ (when
 $0\lt a\lt1$ the limit value $0$ is taken for $x\to+\infty$ otherwise it is taken for $x\to-\infty$).
