# Math Help - Why is this an isomorphism

1. ## 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!

2. ## 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, $a.e \ a$.