Results 1 to 2 of 2

Thread: Why is this an isomorphism

  1. #1
    Member
    Joined
    Oct 2010
    Posts
    131

    Why is this an isomorphism

    Hello,

    We have some Polynomial P(x): $\displaystyle C^{\infty}(T^n) ->C^{\infty}(T^n)$ 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 $\displaystyle d \in \mathbb{R}, $s.t. for almost all $\displaystyle \alpha \in [0,1]^n$ we have
    $\displaystyle \left|P(k+\alpha)^{-1}\right|<=C_a * (1+\left|k\right|)^d$ for all $\displaystyle k \in \mathbb{Z}^n$.

    so it follows immediately that $\displaystyle P(D+\alpha):C^{\infty}(T^n) ->C^{\infty}(T^n)$ 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; Jul 25th 2011 at 02:00 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    My house.
    Posts
    658
    Thanks
    42

    Re: Why is this an isomorphism

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

Similar Math Help Forum Discussions

  1. Isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 10
    Last Post: Oct 27th 2010, 12:08 AM
  2. isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Sep 30th 2010, 09:52 AM
  3. isomorphism
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Mar 10th 2010, 08:50 AM
  4. Replies: 4
    Last Post: Feb 14th 2010, 03:05 AM
  5. Isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Jun 28th 2009, 11:13 PM

Search Tags


/mathhelpforum @mathhelpforum