Why is the wiener algebra a subset of continuous functions

Question

I am studying Fourier analysis and I am currently reading about the Wiener algebra. The Wikipedia page claims that $A(\mathbb{T}) \subset C(\mathbb{T})$ (see: https://en.wikipedia.org/wiki/Wiener_algebra ) however other sources define it to be in $C(\mathbb{T})$. This confused me and I have been trying to prove that $A(\mathbb{T}) \subset C(\mathbb{T})$ but I do not even really know where to start. Any help would be appreciated.

TLDR: I am struggling to show $A(\mathbb{T}) \subset C(\mathbb{T})$

Thank you!

Edit 1:

I have managed to prove that $S_N(f)$ converges in the $C(\mathbb{T})$ norm, but I do not think that this allows me to conclude that $f \in C(\mathbb{T})$ since I do not think I have that $S_N(f) \to f$ or rather I don't know why $S_N(f)$ must converge to the "right" thing

Answer 1

I have figured it out (literally a few mins after giving a bounty). Put $g = \lim_{N \to \infty} ||S_Nf||_{C(\Pi)}$ then as mentioned in the post this converges. Note that because $ \{ \hat{f}(n) \} \in \ell_1 $ we have that $\{ \hat{f}(n) \} \in \ell_2 $ hence $g \in L_2$ as these are isometrically isomorphic spaces given by an $S_N$ correspondence. Then since the Lebesgue measure on the torus is finite it follows that $g \in L^2 \implies g \in L^1$ therefore because all the Fourier coefficients of $g-f$ are $0$ and $g-f \in L^1$ we have that $g-f=0$ a.e and thus $g=f$ a.e

Answer 2

To prove that $A(\mathbb{T})\subset C(\mathbb{T})$, we need to show that every function $f$ in $A(\mathbb{T})$ is continuous on $\mathbb{T}$.

$A(\mathbb{T})$ is a set of complex-valued functions that are continuous on $\mathbb{T}$ (the unit circle in the complex plane) and have a Fourier series that converges uniformly to $f$ on $\mathbb{T}$.

So, let's take an arbitrary function $f$ in $A(\mathbb{T})$, and let $F_n$ be the $n$th partial sum of the Fourier series of $f$. Because the Fourier series converges uniformly to $f$, we have $F_n(x)\to f(x)$ uniformly on $\mathbb{T}$ as $n\to\infty$. Since each $F_n$ is a continuous function on $\mathbb{T}$, it follows that $f$ is also continuous on $\mathbb{T}$, since the uniform limit of continuous functions is continuous.

In other words, every function in $A(\mathbb{T})$ is continuous on $\mathbb{T}$, so $A(\mathbb{T})\subset C(\mathbb{T})$.

Answer 3

Let $f\in A(\mathbb{T}),$ i.e. $f\in L^1(\mathbb{T})$ and the Fourier coefficients of $f$ are absolutely summable. The function $$g(x)=\sum_{n=-\infty }^\infty \hat{f}(n)e^{inx}$$ belongs to $C(\mathbb{T})$ as the series is uniformly convergent due to the Weierstrass majorization theorem. Moreover $$\hat{g}(n)={1\over 2\pi }\int\limits_0^{2\pi}g(x)e^{-inx}\,dx \\=\sum_{n=-\infty}^\infty \hat{f}(n){ 1\over 2\pi}\int\limits_0^{2\pi}e^{ikx}e^{-inx}\,dx=\hat{f}(n)$$ The latter implies $f=g,$ hence $f\in C(\mathbb{T}).$

Remark It was possible to change the order of infinite sum and the integral because the series was uniformly convergent.