A function with Fourier coefficients in $\ell^1$ is continuous

Question

If a function $f:\mathbb{T}\to\mathbb{C}$, given by $f(t)=\sum_{k}a_ke^{ikt}$, verifies that its Fourier coefficients belong to $\ell^1$, that is, $$\sum_{k}|a_k|<\infty$$

Can we say that $f$ is a continuous function?

Answer 1

Let

$s = \displaystyle \sum_{-\infty}^\infty \vert a_k \vert, \tag 1$

and for integer $N > 0$ set

$s_N = \displaystyle \sum_{-N}^N \vert a_k \vert; \tag 2$

then $s_N$ is a monotonically increasing sequence of non-negative real numbers, and

$\displaystyle \lim_{N \to \infty} s_N = \sum_{-\infty}^\infty \vert a_k \vert = s < \infty, \tag 3$

and thus we have, for any $\epsilon > 0$, a sufficiently large $M \in \Bbb Z$ such that $N > M$ implies

$\displaystyle \sum_{\vert k \vert > N} \vert a_k \vert = s - s_N < \epsilon. \tag 3$

Now consider the sequence of functions $f_N: \Bbb T \to \Bbb C$

$f_N(t) = \sum_{\vert k \vert \le N} a_k e^{ikt}; \tag 4$

if $N_2 \ge N_1 > M$ we have

$\vert f_{N_2}(t) - f_{N_1}(t) \vert = \vert \displaystyle \sum_{N_1 < k \le N_2} a_k e^{ikt} \vert \le \sum_{N_1 < k \le N_2} \vert a_k e^{ikt} \vert \le \sum_{k > N_1} \vert a_k \vert < \epsilon, \tag 5$

independently of $t$. Thus he sequence of continuous functions $f_N(t)$ converges uniformly on $\Bbb T$; hence the limit function

$f(t) = \displaystyle \sum_k a_k e^{ikt} \tag 6$

is also continuous.

Answer 2

Yes, because each term is a continuous function of $t$ and the series converges uniformly by the Weierstrass M-test.