[Math] The Fourier series converges absolutely $\implies$ it converges uniformly.

fourier seriesfunctional-analysissequences-and-seriesuniform-convergence

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?

Best Answer

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.