# Convergence of Fourier series for a sum which is not uniform convergent

Given $$\sum_{n=1}^\infty\frac{\cos nt}{n}$$is it a fourier series in

a. $L^2(\mathbb T)$?

b. $C(\mathbb{T})$?

Usually when we get a series we use Weierstrass M test in order to find the sum is uniformly convergent, hence the function to which the series converges is continuous. Here, this test won't work since $$\sum_{n=1}^\infty\bigg|\frac{\cos nt}{n}\bigg|\le \sum_n\frac 1 n=\infty$$ We notice that the fourier coefficients (if it's a fourier series) are defined by $c_k=\frac{1}{2k}$.

How can we prove that in this case the series converges to a function ? Suppose we did so and not by uniform convergence: how would we disprove the continuity of the function?

1. $\mathcal F: L^2(\mathbb T) \to \ell^2(\mathbb Z)$ is an isometry if defined via the complex fourier transform. Can you see that the complex fourier coefficients are in $\ell^2(\mathbb Z)$?
2. Evaluate the series at $t=0$. Can this be continuous?
• But if we can't? (For example take $\mathbb{R}$ with $\tau_{disc}$, which doesn't contain any compact subspace)? – user65985 Feb 11 '15 at 10:20