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Fourier transform isometry

  1. Jun 3, 2009 #1
    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||.
     
  2. jcsd
  3. Jun 4, 2009 #2
    What is the space of moderately decreasing functions?

    The way you do it for L2 is to extend F:S->S by density of S in L2.
     
  4. Jun 4, 2009 #3
    extend it by density? what does that mean?
     
  5. Jun 4, 2009 #4
    Suppose [itex]X[/itex] and [itex]Y[/itex] are Banach spaces, [itex]S\subset X[/itex] is some subspace, and that [itex]T:S\to Y[/itex] is a bounded linear mapping. There exists extensions of [itex]T[/itex], which are bounded linear mappings [itex]\overline{T}:X\to Y[/itex]. The claim is that if [itex]S[/itex] is dense in [itex]X[/itex], then there exists only one extension [itex]\overline{T}[/itex], and it is the unique extension of [itex]T[/itex] which is unique by the density of [itex]S[/itex].

    The proof starts with a remark that the inequality

    [tex]
    \| Tx_n - Tx_k\| \leq \|T\| \|x_n - x_k\|
    [/tex]

    implies that when [itex]x_1,x_2,\ldots\in S[/itex] is an Cauchy sequence, then so is [itex]Tx_1, Tx_2,\ldots\in Y[/itex].
     
  6. Jun 5, 2009 #5
    is there any other way to prove it without banach spaces? we've never covered that in my class and its not covered in my textbook
     
  7. Jun 5, 2009 #6
    I was merely trying to mention the fact in a general context. Hilbert spaces are always Banach spaces, and [itex]L^2(\mathbb{R})[/itex] space is a Hilbert space, so what I said does not force anyone away from [itex]L^2[/itex]-stuff.
     
  8. Jun 5, 2009 #7
    you're tying to prove :"The claim is that if is dense in X, then there exists only one extension , and it is the unique extension of T which is unique by the density of S.", right?

    i dont understand how the inequality you mentioned implies that the extension is unique
     
  9. Jun 5, 2009 #8
    The idea is to define a new "extended" map F2 such that F2 agrees with F on S, and the way it is extended to L2 is via limits of sequences in S.

    In other words, F2(s) = F(s) for all s in S, and F2(x) = limn->infty F(sn) for x not in S, where sn is some sequence in S converging to x.

    There are a few things to check to make sure this is legit:
    1) If 2 sequences converge to the same point x, F2(x) is the same regardless of which sequence is chosen.
    2) F2 is bounded, linear.
    3) F2 is the only bounded linear extension of F (uniqueness)

    This entire process is known as "extending by density", and shows up all the time. It shows up so much that authors rarely go through the whole process but rather will say things like "the result follows by density".
     
    Last edited: Jun 5, 2009
  10. Jun 6, 2009 #9
    but how do you show the extension is unique? by using the inequality mentioned a few posts above?
     
  11. Jun 6, 2009 #10
    That inequality is important and highly tied up in the whole affair, but generally you would use it on the previous parts. Uniqueness is actually the easy part. Once you know that F2 is a bounded linear extension of F from a dense subset to the whole space, then you can argue as follows:

    Suppose there were 2 bounded linear extensions F2 and F3. Since F2 and F3 are continuous, they are sequentially continuous. On the other hand, they must agree on S. Therefore, for any convergent sequence in S, sn->x,
    F2(x) <- F2(sn) = F3(sn) -> F3(x).
    The only way this is possible is if F2(x)=F3(x), since a sequence can only converge to one thing (in a metric space).

    (why? If the points F2(x) and F3(x) are a nonzero distance d away from each other, then just take large enough n such that ||F2(sn) - F2(x)|| < d/2 and ||F2(sn) - F3(x)|| < d/2 and apply the triangle inequality )
     
  12. Jun 6, 2009 #11
    i think i'm able to prove the existence part, but i dont see how showing F2 is linear is relevant
     
  13. Jun 6, 2009 #12
    It would kind of suck if the Fourier transform wasn't linear. Half the stuff you do with Fourier transforms would go out the widow. )-:

    Anyways, it just so happens that it is linear, so you might as well prove it.

    EDIt: Oh also, just remembered, you need linearity for boundedness and continuity to be equivalent, so it is important for the proofs.
     
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