Suppose that $f$ is in $L^1(\mathbb{R})$, and define $g(x) := f (ax + b)$. Compute the Fourier transform of $g$ in terms of that of $f$. Does the result hold if we merely assume that $f$ is in $L^2(\mathbb{R})$? Specify in which sense you consider the Fourier transform of f .
To solve this exercise I would be tempted to go on like this:
$$\mathcal{F}g(\xi) = \int_\mathbb{R} g(x)e^{-2\pi i x \xi} \ dx = \int_\mathbb{R} f(ax + b)e^{-2\pi i x \xi} \ dx$$
and then I would use the substitution $v := ax + b$.
However this solution seems too simple and it does not distinguish the case $L^1$ from $L^2$ and I think that more importantly we may not be able to use the first equality because that is given only for Schwartz functions (in fact we are asked to say in which way we are considering the transform)..
I saw this question on this site but noone said nothing about theese "subtleties".
Can anyone help me please?
Your proof works for functions in $L^1,$ where the Fourier transform is just given by that convergent integral. It is in $L^2$ where that formula may fail to hold. There, you define the Fourier transform via a density argument with Schwarz functions, as you mention, and so the proof cannot be done merely using this integration.
To verify the identity in $L^2,$ one must argue that the transformation $f(x) \mapsto f(ax+b)$ is a continuous transformation on $L^2$ when $a\neq 0.$ This is not so difficult to verify once you realize it is what you need to prove. Then, the fact that Schwarz functions are dense in $L^2$, combined with continuity of the Fourier transform and continuity of this mapping, implies the formula in $L^2$ from the formula for Schwarz functions.