[proofplan] The argument is finite-dimensional linear algebra followed by coefficient comparison. First we use orthogonality to compute the coefficient of $f$ along each basis vector $f_j$ by pairing with $f_j$ under the Petersson inner product. Then uniqueness follows because a basis is linearly independent. Finally, since the expansion of $f$ is a finite linear combination of the $q$-expansions of the basis forms, the $n$-th Fourier coefficient is the same finite linear combination of the $n$-th Fourier coefficients. [/proofplan]

[step:Project $f$ onto each orthogonal basis vector]Define complex numbers $c_1,\dots,c_d \in \mathbb{C}$ by \begin{align*} c_j := \frac{(f,f_j)_{\mathrm{Pet}}}{(f_j,f_j)_{\mathrm{Pet}}} \end{align*} for $j \in \{1,\dots,d\}$. The denominator is nonzero because $f_j$ is a basis vector, hence $f_j \neq 0$, and the Petersson inner product is positive definite on $V$. Set \begin{align*} g := f - \sum_{j=1}^{d} c_j f_j \in V. \end{align*} For a fixed index $\ell \in \{1,\dots,d\}$, linearity of the Petersson inner product in the first variable gives \begin{align*} (g,f_\ell)_{\mathrm{Pet}} &= (f,f_\ell)_{\mathrm{Pet}} - \sum_{j=1}^{d} c_j (f_j,f_\ell)_{\mathrm{Pet}}. \end{align*} Since the basis is orthogonal, $(f_j,f_\ell)_{\mathrm{Pet}} = 0$ whenever $j \neq \ell$. Therefore \begin{align*} (g,f_\ell)_{\mathrm{Pet}} &= (f,f_\ell)_{\mathrm{Pet}} - c_\ell (f_\ell,f_\ell)_{\mathrm{Pet}} \\ &= (f,f_\ell)_{\mathrm{Pet}} - \frac{(f,f_\ell)_{\mathrm{Pet}}}{(f_\ell,f_\ell)_{\mathrm{Pet}}}(f_\ell,f_\ell)_{\mathrm{Pet}} \\ &= 0. \end{align*} Thus $g$ is orthogonal to every vector in the basis $(f_1,\dots,f_d)$.[/step]

[guided]We want to determine the coefficient of $f$ in the direction of a fixed basis vector $f_j$. Orthogonality isolates that coefficient: pairing with $f_j$ kills every other basis vector. Define \begin{align*} c_j := \frac{(f,f_j)_{\mathrm{Pet}}}{(f_j,f_j)_{\mathrm{Pet}}} \end{align*} for each $j \in \{1,\dots,d\}$. This is well-defined because $f_j \neq 0$, since $f_j$ is a basis vector, and positive definiteness of the Petersson inner product gives $(f_j,f_j)_{\mathrm{Pet}} > 0$. Now define the error term \begin{align*} g := f - \sum_{j=1}^{d} c_j f_j. \end{align*} The purpose of $g$ is to measure what remains after subtracting the proposed expansion. We show that this remainder is orthogonal to every basis vector. Fix $\ell \in \{1,\dots,d\}$. By linearity in the first variable, \begin{align*} (g,f_\ell)_{\mathrm{Pet}} &= (f,f_\ell)_{\mathrm{Pet}} - \sum_{j=1}^{d} c_j (f_j,f_\ell)_{\mathrm{Pet}}. \end{align*} Orthogonality of the basis means $(f_j,f_\ell)_{\mathrm{Pet}} = 0$ for $j \neq \ell$, so only the term $j=\ell$ remains: \begin{align*} (g,f_\ell)_{\mathrm{Pet}} &= (f,f_\ell)_{\mathrm{Pet}} - c_\ell (f_\ell,f_\ell)_{\mathrm{Pet}} \\ &= (f,f_\ell)_{\mathrm{Pet}} - \frac{(f,f_\ell)_{\mathrm{Pet}}}{(f_\ell,f_\ell)_{\mathrm{Pet}}}(f_\ell,f_\ell)_{\mathrm{Pet}} \\ &= 0. \end{align*} Thus the chosen coefficients make the proposed error term orthogonal to every basis vector.[/guided]

[step:Use the basis property to identify the expansion and prove uniqueness] Since $(f_1,\dots,f_d)$ is a basis of $V$, there exist complex numbers $b_1,\dots,b_d \in \mathbb{C}$ such that \begin{align*} g = \sum_{j=1}^{d} b_j f_j. \end{align*} Taking the Petersson inner product with $f_\ell$ for a fixed $\ell \in \{1,\dots,d\}$ gives \begin{align*} 0 = (g,f_\ell)_{\mathrm{Pet}} = \sum_{j=1}^{d} b_j (f_j,f_\ell)_{\mathrm{Pet}} = b_\ell (f_\ell,f_\ell)_{\mathrm{Pet}}. \end{align*} Since $(f_\ell,f_\ell)_{\mathrm{Pet}} \neq 0$, we get $b_\ell = 0$. This holds for every $\ell$, so $g=0$. Therefore \begin{align*} f = \sum_{j=1}^{d} c_j f_j. \end{align*} If also \begin{align*} f = \sum_{j=1}^{d} c'_j f_j \end{align*} for complex numbers $c'_1,\dots,c'_d \in \mathbb{C}$, then \begin{align*} 0 = \sum_{j=1}^{d} (c_j-c'_j) f_j. \end{align*} [Linear independence](/page/Linear%20Independence) of the basis forces $c_j-c'_j=0$ for every $j$, so $c_j=c'_j$ for every $j$. Hence the expansion is unique. [/step]

[step:Compare the Fourier coefficients term by term] For each $n \in \mathbb{N}$, let \begin{align*} A_n: V &\to \mathbb{C} \\ h &\mapsto a_n(h) \end{align*} denote the map taking a cusp form $h \in V$ to its $n$-th Fourier coefficient at the cusp $\infty$. The Fourier expansion is linear in the form: if $h,h' \in V$ and $\lambda,\mu \in \mathbb{C}$, then the $q$-expansion of $\lambda h+\mu h'$ is obtained by adding the corresponding convergent [Fourier series](/page/Fourier%20Series) term by term, so \begin{align*} A_n(\lambda h+\mu h') = \lambda A_n(h)+\mu A_n(h'). \end{align*} Thus $A_n$ is a complex-[linear map](/page/Linear%20Map). Applying $A_n$ to the already proved finite expansion \begin{align*} f = \sum_{j=1}^{d} c_j f_j \end{align*} gives \begin{align*} a_n(f) = A_n(f) = A_n\left(\sum_{j=1}^{d} c_j f_j\right) = \sum_{j=1}^{d} c_j A_n(f_j) = \sum_{j=1}^{d} c_j a_n(f_j). \end{align*} Because the sum over $j$ is finite, no limiting operation over the basis index is involved. This proves the asserted Fourier coefficient identity for every $n \in \mathbb{N}$. [/step]