Stretching = Translation

The given functional equation is similar to $f(x) \cdot f(y) = f(x + y )$ . One solution to functional equations of this kind is $f(x) = k^{x}$ for all $x$ , where $k$ is a positive real number. If $f(y) = y = \alpha$ , then this functional equation becomes equal to the functional equation in the problem.

Suppose $f(x) = k^{x}$ satisfies our functional equation. This means that $f(y) = k^{y}$ . We want to solve the equation $f(y) = y = \alpha$ . This implies

$\begin{aligned} f(\alpha) &= \alpha \\ k^{\alpha} &= \alpha\\ \alpha \log k &= \log \alpha \\ \log k &= \frac{1}{\alpha} \log \alpha \\ k & = \alpha ^{1/\alpha} \end{aligned}$

Therefore a solution for $f(x)$ is

$\large f(x) = \left(\alpha^{1/\alpha}\right)^{x} \qquad\text{ for all } x$

The answer to this problem is Yes because we have found a non-constant function satisfying the given functional equation.

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Now try answering the bonus questions!

For any non-zero $\alpha$ , we can choose $f(x)$ freely on the interval $I=[0,|\alpha|$ ). Any real $x$ can then be written uniquely as $x=y+k\alpha$ where $y$ is on $I$ and $k$ is an integer, and we have $f(x)=f(y+k\alpha)=\alpha^kf(y)$ .

$\left(\alpha^{\frac{1}{\alpha}}\right)^x=\alpha^{\frac{x}{\alpha}}$ is a solution for any $\alpha>0$ , but take any function $p(x)$ with period $\alpha$ , e.g. $p(x)=\sin(\frac{2\pi}{\alpha}x)$ , and $p(x)·\alpha^{\frac{x}{\alpha}}$ is also a solution.

For $x=0$ , there's only the constant function $f(x)=0$ .

For $\alpha<0$ , I must say that I have no idea so far.

All group homomorphisms from the reals with addition to the reals under multiplication? Guessing