I am working on the following problem:
Let $S^1 := \mathbb{R}/2\pi \mathbb{Z}$ and suppose $p(x) = a_0 + a_1x + \cdots + a_kx^k$ is a polynomial such that for all $n \in \mathbb{Z}$ we have $p(in) \not= 0$. Define the operator $P : C^{\infty}(S^1) \rightarrow C^{\infty}(S^1)$ by $Pu := p(\partial /\partial x)u$. Show that for all $f \in C^{\infty}(S^1)$ there exists a function $u \in C^{\infty}(S^1)$ such that $Pu = f$.
What I have done is the following. Suppose such a $u$ exists. Then $(Pu)^{\widehat{}} = \widehat{f}$. And we can calculate, using the Fourier transform, that $(Pu)^{\widehat{}}(n) = p(in)\widehat{u}(n)$. So $\widehat{u}(n) = \frac{\widehat{f}(n)}{p(in)}$ (recall $p(in) \not= 0$). So assuming there exists $u \in C^{\infty}(S^1)$ with $Pu = f$ we have, by Fourier inversion, that $$ u(x) = \sum_{n \in \mathbb{Z}}\frac{\widehat{f}(n)}{p(in)}e^{inx} $$ If I now define $u$ by this formula, how can I show that $u \in C^{\infty}(S^1)$? My thought is this: since $p$ is a polynomial, we have that for every $N > 0$ there exists a positive integer $M_N$ such that for all $n \geq M_N$ we have $\lvert p(in) \rvert > N$ since the highest power term in $p$ will dominate. This means that $\left\lvert \frac{\widehat{f}(n)}{p(in)}e^{inx}\right\rvert < \frac{\lvert\widehat{f}(n)\rvert}{N}$ for all $n \geq M_N$. Since $f$ is smooth, its Fourier series is absolutely convergent and, as seen by this calculation, the absolute value of its Fourier coefficients dominate the corresponding coefficients in the series definition of $u$. Hence $u$ converges absolutely to a smooth function (where smoothness can be verified by term-by-term differentiation, which is allowed because of the absolute - and therefore uniform - convergence). How does this sound?
In general, I am confused about convergence of Fourier-related series with regard to which type of convergence I should be concerned with demonstrating. Should the properties I want the limiting function to have (e.g. well-defined everywhere, continuous, integrable, etc) be the exact motivation for which type of convergence I want (e.g. point-wise, uniform, $L^1$, etc)?
Your argument is solid, here are the ingredients, which for the most part you already have.
From smoothness to decay
The Fourier coefficients of a $C^\infty$ function decay faster than any power of $n$. Indeed, fix an integer $k$; the Fourier coefficients of the $k$th derivative of $f$ are bounded, hence $|\hat f(n)| = O(n^{-k})$. And this is valid for every $k$.
Fourier multiplier
Dividing by $p(in)$ preserves the property (1), since $p(in)$ is large for large $n$, as you observed.
From decay to smoothness
The decay of coefficients implies that for every $k$, the series differentiated term-by-term $k$ times converges uniformly. This implies that the sum is $C^k$. And this is valid for every $k$.