Extending an Isometry in Schwartz Space to Moderately Decreasing Functions

[creepypasta13]

Homework Statement

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). Show 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||.

hint: You may use that for any g in M(R) there exists a sequence {g_n} subset in S(R) such that ||g_n - g|| converges to 0

note: make sure you prove that both that G exists, and that it is unique

Homework Equations

The Attempt at a Solution

i was thinking of showing that F is 1-1, onto, and linear, just to expand my options. also using the hint to show that there exists F(h_k) converging to F(f), and then ||h_k - f|| converges to 0, and then defining h_k(t) as equal to f(t) for all |t| < k, and 0 otherwise

 

[creepypasta13]

i heard that to do it for the L2, you extend it by density, but i don't know what that means