a. Prove that for any bounded sequence of complex numbers $\{a_n\}_{n=1}$ the cosine series $$ \sum_{n=1}^\infty a_n\cos nx$$ is convergent in the sense of distributions on $\mathbb{R}$.
b. What else can be said if $\{a_n\} \in l^2$?
c. What else can be said if $\{a_n\} \in l^1$?
This is a problem from an old qualifying exam...Right now I'm quite puzzled about b and c--I'm guessing if a is solved, it might make those questions clearer...
I started by letting $f_k(x) = \sum_{n=1}^k \cos nx$ and $\phi \in \mathcal{D}(\mathbb{R})$. then $$f_k\phi = \int_{\mathbb{R}} f_k(x) \phi(x) dx = \sum_{n=1}^k \int_{\mathbb{R}}a_n\cos nx \phi(x)dx $$ we could bound the $a_n$ but I don'e see how that will help as of now...
A quick way to prove (a) is to consider the series $$\sum_{n=1}^\infty \frac{a_n}{n^2}\cos nx$$ It converges uniformly, therefore also converges in the sense of distributions.
Furthermore, taking the derivative preserves convergence of distributions, since the derivative passes to the test function. Differentiating the above series twice gives the original one.
(This argument generalizes to any sequence of coefficients that are bounded by some power of $n$.)
For b) I presume you are expected to say that the series converges in $L^2(0,2\pi)$. Indeed, the orthogonality of cosines yields $$\int_0^{2\pi} \left(\sum_{n=k}^m a_n \cos nx\right)^2\,dx = \sum_{n=k}^m \int_0^{2\pi} a_n^2 \cos^2 nx\,dx = \pi \sum_{n=k}^m a_n^2 $$ so the partial sums of the series form a Cauchy sequence in $L^2(0,2\pi)$.
By periodicity, the same happens on other intervals of length $2\pi$, and therefore on all bounded intervals.
To summarize, in case b) the sum is a function of period $2\pi$ that is square integrable on its period.
In case c) the series converges uniformly (Weierstrass M-test). Hence the sum is a continuous periodic function.