# The Fourier series converges absolutely $\implies$ it converges uniformly.

Let $S_N(f)$ be the $N$th partial sum of the Fourier series for $f$. I.e.

$$S_N(f) = \sum_{n = -N}^{N} \hat{f}(n) e^{2\pi i n x / L}$$

Suppose that the Fourier series converges absolutely, i.e.

$$\sum_{n = -\infty}^{\infty} \left ( |\hat{f}(n)e^{2\pi i n x/L}| = |\hat{f}(n)|\right ) \lt \infty.$$

Then sequence of functions that are the partial sums converges uniformly.

I think this can be generalized to:

Let $f_n$ be a sequence of functions $f_n:S \rightarrow T$, let $S_N = \sum_{n = -N}^{N} f_n$, Then if the series $\sum_{N = -\infty}^{\infty} f_n$ converges absolutely, then the sequence of functions $S_N$ converges uniformly.

Attempted Proof . Let $L = \sum_{N=-\infty}^{\infty} |f_n|$. Then for all $x\in S$, $\epsilon \gt 0$, there exists $M$ such that $N\gt M \implies d_T(\sum_{n = -N}^{N} |f_n(x)|, L(x)) \lt \epsilon$. We want to show that there exists a $f:S\rightarrow T$, such that for all $\epsilon \gt 0$, there exists $M$ such that for all $x\in S$, $N \gt M$, $d_T(S_N(x), f(x)) \lt \epsilon$. What next?

It is not true. Let $S=[0,1)$ and $f_n(x)=x^n$ if $n>0$, $f_n(x)=0$ if $n\le0$. The series converges absolutely but not uniformly.

• Please see my approach below. I can't find the trick yet. It' might be something about $|e^i{\dots}| = 1$ always, idk. – BananaCats Category Theory App Oct 3 '13 at 23:23

Let $f_n : S \rightarrow T$ be a sequence of continuous functions $f_n : S \rightarrow T$. Suppose $\lim_{N\rightarrow \infty} \sum_{k = -N}^{N} |f_n| = L(x)$ for each $x \in S$. Then for all $\epsilon \gt 0$ there's $M$ such that $N\gt M \implies |\sum|f_n(x)| - L(x)| \lt \epsilon$. But absolute convergence of the sequence of numbers $|f_n(x)|$ implies convergence when $T$ is Cauchy-complete. So assume $T$ is and $\sum f_n \rightarrow f$ for some map $f$. We want to show that $\sum f_n \rightarrow f$ uniformly.

According to Julián Aguirre, this is not true generally. So there's something about the fourier series ($g_n(x) = \hat{g}(n)e^{-i 2 \pi n x / L}$) that makes this so.

For all $\epsilon \gt 0$ does there exist $N$ such that for all $M \gt N$, $x\in S$, $\ |\sum_{n = -M}^{M} g_n(x) - g(x)| \lt \epsilon$?

I want to say that $g(x) - \sum^M g_n(x) = \sum_{n = -\infty}^{-(M+1)} g_n(x) + \sum_{n=M+1}^{\infty} g_n(x)$. Since the series is absolutely convergent then any sub-sum is too.
$$|\sum_{n = -M}^{M} g_n(x) - g(x)| \leq |\sum_{n = -\infty}^{-(M+1)} g_n(x)| + |\sum_{n=M+1}^{\infty} g_n(x)| \leq \epsilon$$

since the two sums on the right go to zero. But this proves it in general, so something's wrong.

• The only thing special about Fourier series in this respect is that $|e^{ix}|=1$ for $x\in\mathbb{R}$. This is a general result in the line you are looking for. Let $f_n\colon X\to\mathbb{C}$ be a sequence of functions such that $0<C_1\le|f_n(x)|\le C_2$ for all $n\in\mathbb{N}$ and all $x\in X$, and let ${a_n}$ be a sequence of complex numbers. If $\sum_na_nf_n(x)$ converges absolutely for some $x\in X$, then it converges uniformly in $X$. – Julián Aguirre Oct 4 '13 at 9:53
• @ Julián Aguirre : Can you provide reference to this result? – user149418 May 20 at 17:54