What is the Fourier transform of $1$?

Question

For sure, $g(x)=1$ has no fourier transform, but it has a Fourier transform in distribution sense. We have for $\varphi \in \mathcal C_0^\infty (\mathbb R)$, $$\left<\hat 1,\varphi \right>:=\left<1,\hat \varphi \right>=\int_{\mathbb R}\hat \varphi (x)dx=\int_{\mathbb R}\hat \varphi (x)e^{2i\pi0x}dx=\varphi (0)=\left<\delta ,\varphi \right>.$$ Do at the end $\hat 1=\delta $. Could someone explain what it mean ? Because in somehow, $\hat 1$ doesn't make sense strongly, but in distribution it is $\delta $, and I'm not sure how to interpret this. Could someone help to try to understand ?

Answer 1

$g$ is a locally integrable function, hence a tempered distribution. Fourier transform of a tempered distribution $u$ is defined by $\hat {u} (\phi)=u(\hat {\phi})$.

Answer 2

The distribution $u = 1$ is the linear map on test functions $\left<u,\phi\right> = \int_\Bbb{R} \phi(x)dx$ its Fourier transform $\hat{u}$ is the linear map on test functions such that $\left<\hat{u},\phi\right>=\left<u,\hat{\phi}\right>$.

ie. $$\left<\hat{u},\phi\right> = \int_\Bbb{R} \hat{\phi}(x)dx= \phi(0) = \left<\delta,\phi\right>$$

So $\hat{u}$ and $\delta$ are the same linear map on test functions, they are the same distribution.

If we define those things like that it is because from the beginning we are considering $u$ as the limit of a sequence of (test) functions $u_n$, sequence which maybe doesn't converge to a function but whose corresponding sequence of linear map $\left<u_n,.\right>$ on test functions converges.