Fourier transforms of smooth, compactly supported functions

[chub]

Homework Statement

$$f, \hat{f} \in C_c^\infty(\mathbb{R}^n)$$

Homework Equations

$$\hat{f} = \int_{\mathbb{R}^n} f(x) e^{-2\pi i \xi \cdot x} \,dx$$
$$\check{f} = \int_{\mathbb{R}^n} \hat{f}(x) e^{2\pi i \xi \cdot x} \,d\xi$$

The Attempt at a Solution

As $$C_c^\infty \subset \mathcal{S}$$ (the Schwarz space) we know that the Fourier transformation is invertible, and that
$$f = (\hat{f})\check{} = (\check{f})\hat{}$$
in other words
$$f = \int_{\mathbb{R}^n} \hat{f} e^{2 \pi i \xi \cdot x} \,d\xi = \int_{\mathbb{R}^n} \check{f} e^{-2 \pi i \xi \cdot x} \,dx$$

Somehow these must be zero. I am familiar with the idea that the smoother f is, the faster its transform must decay at infinity; and vice versa. Since $$C^\infty, C_c$$ are the ultimate of both this must somehow indicate that the situation is "too good to be true" in a nontrivial way. But I do not know how to implement this. Advice?

 

[mighty2000]

Thank you for your post. It is great to see that you are working with the Fourier transformation and its properties.

Firstly, let me clarify that the Fourier transform is not necessarily zero for smooth functions. It is only guaranteed to decay at infinity for functions in the Schwartz space, as you mentioned. However, the decay rate at infinity can vary depending on the smoothness of the function. For example, a C^\infty function will have a faster decay than a C^2 function.

To understand why this is the case, we need to look at the definition of the Schwartz space. The Schwartz space is defined as the space of all infinitely differentiable functions whose derivatives decay faster than any polynomial at infinity. This means that the functions in this space have very fast decaying derivatives, which in turn leads to a fast decay of the Fourier transform.

On the other hand, functions in C_c^\infty have compact support, meaning that they are zero outside of a bounded set. This implies that their derivatives will also have compact support, and hence, their Fourier transform will not decay as fast at infinity. In fact, the faster the decay of the derivatives at infinity, the slower the decay of the Fourier transform.

In summary, the smoothness of a function affects the decay rate of its Fourier transform, but it is not necessarily zero. I hope this helps clarify your doubts.