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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.

Garmekain
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    While you're thinking about that, I'll give you a little hint for just about any proof you might be considering, of any statement relate to what you were asking: look at $K(x) = \frac{h(x)}{a^x}$. What can you say about $K$? Alternatively, consider $H(x) = \ln h(x)$. – John Hughes Oct 15 '17 at 11:55

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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 $(0,\infty)$.

Rigel
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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 $x\to-\infty$).

Piquito
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