From the inequality, we have that defines a tempered distribution
(read: rapidly decreasing coefficients) on by .
It is a well known fact that the Fourier transform of is again a tempered distribution.
So, defines an isomorphism of onto itself, .
Hello,
We have some Polynomial P(x): 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 s.t. for almost all we have
for all .
so it follows immediately that 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!
From the inequality, we have that defines a tempered distribution
(read: rapidly decreasing coefficients) on by .
It is a well known fact that the Fourier transform of is again a tempered distribution.
So, defines an isomorphism of onto itself, .