## Question about the Fourier expansion of adelic Eisenstein series for $\operatorname{GL}_2$

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 $$\alpha \in k$$. Then

$$\Phi(x) = \sum\limits_{a \in k} c_{\alpha}(g,f) \psi_{\alpha}(x) \tag{1}$$

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

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).