What is the Fourier transform of $1$? 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 ? 

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