# Absolute convegence of Fourier series implies uniform convergence?

I've read many statements in my textbook that include something along the lines of "the Fourier series of $$f$$ converges absolutely, and hence uniformly to $$f$$". I'm not sure why this is true.

If the Fourier series converges absolutely, then it converges. That is, $$\lim_{n \to \infty} S_n(\theta)$$ converges for each $$\theta$$. Why does it follow then that this limit is uniform?

Let $$S_n(\theta)=\sum_{|k|\leq n}c_ke^{2i\pi k\theta}$$ be the partial Fourier series.
Now, suppose that for all $$\theta \in [0, 2\pi]$$, we have absolute convergence $$S_n(\theta)\rightarrow f(\theta)$$.
Let $$\varepsilon>0$$. By definition of the series being absolutely convergent, $$\sum_{k\in\mathbb Z}|c_k|<+\infty$$ Therefore, the partial sums are Cauchy: Let $$\varepsilon>0$$, there exists $$N>0$$ such that if $$n>N$$ and $$m>N$$, $$\sum_{|k|>n,m}|c_k|<\varepsilon$$ So for all $$\theta \in [0, 2\pi]$$, $$\left|S_n(\theta)-S_m(\theta)\right| \leq \sum_{|k|>n,m}|c_k| < \varepsilon$$ So $$\{S_n\}_{n\in\mathbb N}$$ is uniformly Cauchy, so it converges uniformly.
• Can you clarify how we go from the fact that the series is absolutely convergent to being able to enforce a bound of $\epsilon$? – slothropp Mar 5 at 10:02