Verifying Distribution Equivalence for Fourier Series Expansion In my lecture notes, given a periodic distribution $T \in (C_{per}^\infty([-\pi,\pi]^n))'$, the Fourier coefficients are defined by
$$\hat T(m) = T({1 \over (2\pi)^n}e^{-i m \cdot x}),$$
for $m \in \Bbb Z^n$. I am then asked to verify that the formal series
$$\sum_{m \in \Bbb Z^n} \hat T(m) e^{i m \cdot x}$$
defines a distribution which agrees with $T$.
My attempt is given in an answer below.
Thanks! :) Sam
 A: First of all we use the definition given:
$$\left( \sum_{m \in \Bbb Z^n} \hat T(m) e^{i m \cdot x} \right)(\phi) = \left(\sum_{m \in \Bbb Z^n} T \left({1 \over (2\pi)^n}e^{-i m \cdot x}\right) \right) (e^{i m \cdot x}\phi).$$
Not sure what to do now. The main issue is that I don't really know what "the distribution $e^{i m \cdot x}$" means. Is it just the distribution with $e^{i m \cdot x}$ as the kernel of the integral, ie
$$S(\phi) := e^{i m \cdot x}(\phi) = \int e^{i m \cdot x}\phi(x)dx?$$
This would then sort me $S(\phi) = (2\pi)^n\hat \phi(-m)$, and so
$$\left( \sum_{m \in \Bbb Z^n} \hat T(m) e^{i m \cdot x} \right)(\phi) = \sum_{m \in \Bbb Z^n} T \left({1 \over (2\pi)^n}e^{-i m \cdot x}\right) (2\pi)^n \hat \phi(-m)$$
$$=T\left(\sum_{m \in \Bbb Z^n} \left(e^{-i m \cdot x} \hat \phi(-m)\right) \right)$$
$$=T\left(\sum_{m \in \Bbb Z^n} \left(e^{i m \cdot x} \hat \phi(m)\right) \right) = T(\phi),$$
by the Fourier inversion formula for $\phi \in \mathcal S$ (ie a Schwartz function).
Is this right? My issue in trying to work it out has always been that I don't know what form the distribution $e^{i m \cdot x}$ takes, but last night I realised that it probably means the integral form. However, I know that not all distributions arise in such a way - eg, the Dirac $\delta$-function (distribution).
[If it is right, then I've spent ages stuck on (and getting annoyed by =P!) something rather simple!]
