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Math Help - Why is this an isomorphism

  1. #1
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    Why is this an isomorphism

    Hello,

    We have some Polynomial P(x): C^{\infty}(T^n) ->C^{\infty}(T^n) between the set of all smooth maps on the torus.
    Now my book says following:
    Since the smooth functions on the torus are characterised by having rapidly decreasing Fourier coefficients, and we have the following equation: there is a d \in \mathbb{R}, s.t. for almost all \alpha \in [0,1]^n we have
    \left|P(k+\alpha)^{-1}\right|<=C_a * (1+\left|k\right|)^d for all k \in \mathbb{Z}^n.

    so it follows immediately that P(D+\alpha):C^{\infty}(T^n) ->C^{\infty}(T^n) is an isomorphism for almost every \alpha.

    The equation above are no problems to me. I know them. But i really don't see, why this is an immediate consequence from the assumptions above??
    Do you have an idea, why the F-Transformation and the fact, that it is decreasing rapidly can help me to show that P is an isom.??

    if someone wants to see the book, Partial differential equations ... - Google Bücher
    it is Prop.10.1 on page 248

    Regards!
    Last edited by Sogan; July 25th 2011 at 02:00 PM.
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  2. #2
    Super Member Rebesques's Avatar
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    Re: Why is this an isomorphism

    From the inequality, we have that T_a=P(D+a)\in \mathcal{D}'(\mathbb{T}^n) defines a tempered distribution
    (read: rapidly decreasing coefficients) on C^{\infty}(\mathbb{T}^n) by T_a(u)=\int P(D+a)u .
    It is a well known fact that the Fourier transform \mathcal{F}(T_a) of T_a is again a tempered distribution.
    So, T_a=P(D+a)\in \mathcal{D}'(\mathbb{T}^n) defines an isomorphism of C^{\infty}(\mathbb{T}^n) onto itself, a.e \ a.
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