My reference is Daniel Bump's book, *Automorphic Forms and Representations*, Chapter 3.7.  Let $k$ be a number field,  $G = \operatorname{GL}_2$, $B$ and $T$ the usual Borel subgroup and maximal torus of $G$.  For $\chi$ an unramified character of $T(\mathbb A)/T(k)$, let $V$ be the space of "smooth" functions $f: G(\mathbb A) \rightarrow \mathbb C$ satisfying $f(bg) = \chi(b) \delta_B(b)^{\frac{1}{2}}f(g)$ which are right $K$-finite.  For $f \in V$ and $g \in G(\mathbb A)$, define the Eisenstein series

$$E(g,f) = \sum\limits_{\gamma \in B(k) \backslash G(k)} f(\gamma g)$$

For suitable $\chi$, the series converges absolutely for all $g \in G(\mathbb A)$.  Now $E(g,f)$ has a "Fourier expansion," which as explained by Bump is gotten as follows: the function $$\Phi: \mathbb A/k \rightarrow \mathbb C$$ $$\Phi(x) = E( \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix}g,f)$$
is continuous, hence is in $L^2(\mathbb A/k)$, and therefore has a "Fourier expansion" over the characters of $\mathbb A/k$.  If $\psi$ is a fixed character of $\mathbb A/k$, then $\psi_{\alpha}: x \mapsto \psi(\alpha a)$ comprise the rest of them, for $\alpha \in k$.  The characters $\psi_{\alpha}$ form an orthonormal basis of $L^2(\mathbb A/k)$, so by orthogonality,
$$\Phi(x) = \sum\limits_{a \in k} c_{\alpha}(g,f) \psi_{\alpha}(x) \tag{1}$$

$$c_{\alpha}(g,f) = \int\limits_{\mathbb A/k} E( \begin{pmatrix} 1 & y \\ & 1 \end{pmatrix} g,f) \psi(-\alpha y) dy$$

According to Bump, we may simply set $x = 0$, giving us the Fourier expansion for the Eisenstein series

$$E(g,f) = \sum\limits_{\alpha \in k} c_{\alpha}(g,f)$$

**My question**: Why is this last step valid?  The right hand side of equation (1) converges to $\Phi(x)$ in the $L^2$-norm.  As far as I know, this is not an equation of pointwise convergence.  In general, the Fourier series of a continuous function need not converge pointwise to that function everywhere (in the classical case $\mathbb R/\mathbb Z$, the Fourier series of a continuous function converges pointwise to that function *almost* everywhere).