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!

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, $\displaystyle a.e \ a$.