- #1
creepypasta13
- 375
- 0
let S(R) be the schwartz space, M(R) be the set of moderately decreasing functions, F be the Fourier transform
Suppose F:S(R)->S(R) is an isometry, ie is satisfies ||F(g)|| = ||g|| for every g in S(R).
How is it possible that there exists a unique extension G: M(R)->M(R) which is an isometry, ie a function G: M(R)->M(R) so that for any g in S(R) we have G(g) = F(g), and for any g in M(R) we have ||G(g)|| = ||g||.
Suppose F:S(R)->S(R) is an isometry, ie is satisfies ||F(g)|| = ||g|| for every g in S(R).
How is it possible that there exists a unique extension G: M(R)->M(R) which is an isometry, ie a function G: M(R)->M(R) so that for any g in S(R) we have G(g) = F(g), and for any g in M(R) we have ||G(g)|| = ||g||.