- #1

- 713

- 5

let's suppose I have two functions [itex]f,g\in L^2(\mathbb{R})[/itex] and I consider the inner product [tex]\left\langle f,g \right\rangle = \int_\mathbb{R} f(x)g(x)dx[/tex]

If I transform the function

*f*in the following way [itex]f(x) \mapsto f(\phi(u))[/itex], where [itex]\phi:\mathbb{R}\rightarrow \mathbb{R}[/itex] is smooth and bijective, I can still calculate the inner product [tex]\left\langle f \circ \phi,g \right\rangle = \int_\mathbb{R} f(\phi(u))g(u)du[/tex] Instead, if [itex]\phi:U\rightarrow \mathbb{R}[/itex] is smooth and bijective but

*U*is not necessarily ℝ, I can't calculuate the inner product [itex]\left\langle f \circ \phi,g \right\rangle[/itex] anymore.

Does this happen because in the first case [itex]\phi[/itex] acted as a mapping [itex]L^2(\mathbb{R}) \rightarrow L^2(\mathbb{R})[/itex] to the same vector space, while in the second case we had a mapping [itex]L^2(\mathbb{R}) \rightarrow L^2(U;\mathbb{R})[/itex] which is a different vector space. Am I right?