In section 8.4 of Folland's Real Analysis it's stated that
if $f \in L^1(\mathbb{T}^n)$ has a Fourier series $\widehat f \in l^1(\mathbb{Z}^n)$, then the Fourier series $\sum_{\kappa} \widehat f(\kappa)e^{2\pi i \kappa \cdot x} $ converges absolutely and uniformily to a function $g$. Since $l^1 \subset l^2$, it follows from theorem 8.20 that $f \in L^2$ and that the series converges to $f$ in the $L^2$ norm. Hence $f = g$ almost everywhere [...].
Theorem 8.20 states that the set of functions $ E_\kappa(x) = e^{2\pi i \kappa \cdot x}$ is an orthonormal basis for $L^2(\mathbb{T}^n)$.
I don't understand why if follows from this theorem that $f \in L^2(\mathbb{T}^n)$. Can anyone help me with this?
I've searched for this result in other sources and found a different proof of the assertion "$f = g$ almost everywhere" in Loukas Grafakos' Classical Fourier Analysis. It's proof uses arguments involving Fejér Kernels that are a lot more complicated than "Theorem 8.20 implies $f \in L^2$".
Thanks in advance for any help.
Start with a function $f_{a,b,\epsilon}$ that is $0$ on $[0,a], [b,2\pi]$ is $1$ on $[a+\epsilon,b-\epsilon]$ and is linear on $[a,a+\epsilon]$ as well as on $[b-\epsilon,b]$. This function $f_{a,b\epsilon}$ has a uniformly convergent Fourier series $\sum_{n=-\infty}^{\infty}c_{a,b,\epsilon,n}e^{inx}$ that converges to the given functions $f_{a,b,\epsilon}$, at least for $\epsilon$ small enough. Therefore, $$ \int_{0}^{2\pi} f_{a,b,\epsilon}(x)f(x)dx=\int_0^{2\pi}\sum_{n=-\infty}^{\infty}c_{n,a,b,\epsilon}e^{inx}f(x)dx \\ = \sum_{n=-\infty}^{\infty}c_{n,a,b,\epsilon}\int_{0}^{2\pi}f(x)e^{inx}dx = 0. $$ Now, letting $\epsilon\downarrow 0$ gives $$ \int_{a}^{b}f(x)dx =0. $$ This is true for all $0 < a < b < 2\pi$. By the Lebesgue differentiation theorem, $f=0$ a.e.. So $f=0$ as a function in $L^1$.