[proofplan] We first verify that the Petersson integral is finite by splitting the fundamental domain into a compact part and the cusp region. On the compact part, continuity gives boundedness; near the cusp, the vanishing of the constant Fourier coefficient gives exponential decay. Once the integral is known to be well-defined, linearity and conjugate symmetry follow from the corresponding algebraic properties of complex conjugation and the [Lebesgue integral](/page/Lebesgue%20Integral). Finally, positivity follows from non-negativity of $|f|^2(\operatorname{Im}z)^{k-2}$ and the identity theorem for holomorphic functions. [/proofplan]

[step:Split the fundamental domain into a compact part and a cusp region] Let \begin{align*} \mathbb{H}:=\{z=x+iy\in \mathbb{C}:y>0\} \end{align*} be the upper half-plane. We write $\mathcal{L}^2$ for two-dimensional Lebesgue measure on $\mathbb{C}\cong\mathbb{R}^2$, and we write $\mathcal{L}^1$ for one-dimensional Lebesgue measure on $\mathbb{R}$. Let $\mathcal{F}\subset \mathbb{H}$ be the [standard fundamental domain](/page/Fundamental%20Domain) \begin{align*} \mathcal{F}:=\{z=x+iy\in\mathbb{H}: |x|\leq 1/2,\ |z|\geq 1\}. \end{align*} For functions $u,v:\mathbb{H}\to\mathbb{C}$ for which the integral is absolutely convergent, define the [Petersson pairing](/page/Petersson%20Pairing) by \begin{align*} (u,v)_{\mathrm{Pet}} := \int_{\mathcal{F}} u(z)\overline{v(z)}\,y^{k-2}\,d\mathcal{L}^2(z). \end{align*} For $Y\geq 1$, define \begin{align*} \mathcal{F}_{\leq Y}&:=\{z=x+iy\in\mathcal{F}: y\leq Y\},\\ \mathcal{F}_{>Y}&:=\{z=x+iy\in\mathcal{F}: y>Y\}. \end{align*} The set $\mathcal{F}_{\leq Y}$ is compact in $\mathbb{H}$, while $\mathcal{F}_{>Y}$ is contained in the vertical strip \begin{align*} \{x+iy\in\mathbb{H}: |x|\leq 1/2,\ y>Y\}. \end{align*} [/step]

[step:Use cuspidality to obtain exponential decay at infinity]Let $f,g\in S_k(SL_2(\mathbb{Z}))$. Define the holomorphic map \begin{align*} q:\mathbb{H}&\to\{w\in\mathbb{C}:0<|w|<1\}\\ z&\mapsto e^{2\pi iz}. \end{align*} Since $f$ and $g$ are [cusp forms](/page/Cusp%20Form) for $SL_2(\mathbb{Z})$, the transformation law for the translation $z\mapsto z+1$ makes both functions $1$-periodic on $\mathbb{H}$. Hence each descends through the coordinate $q=e^{2\pi iz}$ to a [holomorphic function](/page/Holomorphic%20Function) on the punctured disc near $0$. The cusp condition at $\infty$ says precisely that these descended functions extend holomorphically across $q=0$ and vanish at $q=0$. Therefore there exist $r\in(0,1)$ and holomorphic maps \begin{align*} F,G:\{w\in\mathbb{C}:|w|<r\}\to\mathbb{C} \end{align*} such that for all $z\in\mathbb{H}$ with $|q(z)|<r$, \begin{align*} f(z)&=q(z)F(q(z)),\\ g(z)&=q(z)G(q(z)). \end{align*} Choose $Y_0\geq 1$ such that $e^{-2\pi Y_0}<r/2$. The closed disc $\{w\in\mathbb{C}:|w|\leq e^{-2\pi Y_0}\}$ is compactly contained in $\{w\in\mathbb{C}:|w|<r\}$, so continuity gives finite constants \begin{align*} M_f&:=\sup_{|w|\leq e^{-2\pi Y_0}}|F(w)|,\\ M_g&:=\sup_{|w|\leq e^{-2\pi Y_0}}|G(w)|. \end{align*} Define $C_f:=M_f$ and $C_g:=M_g$. For every $z=x+iy\in\mathcal{F}_{>Y_0}$, we have $|q(z)|=e^{-2\pi y}\leq e^{-2\pi Y_0}$, and therefore \begin{align*} |f(z)|&\leq C_f e^{-2\pi y},\\ |g(z)|&\leq C_g e^{-2\pi y}. \end{align*} Therefore, for $z=x+iy\in\mathcal{F}_{>Y_0}$, \begin{align*} |f(z)\overline{g(z)}|\,y^{k-2} \leq C_fC_g e^{-4\pi y}y^{k-2}. \end{align*} The function \begin{align*} \rho:(Y_0,\infty)\to[0,\infty),\qquad y\mapsto C_fC_g e^{-4\pi y}y^{k-2} \end{align*} is integrable on $(Y_0,\infty)$ with respect to $\mathcal{L}^1$ because exponential decay dominates polynomial growth. Since $\mathcal{F}_{>Y_0}$ is contained in the strip $[-1/2,1/2]\times(Y_0,\infty)$, we obtain \begin{align*} \int_{\mathcal{F}_{>Y_0}} |f(z)\overline{g(z)}|\,y^{k-2}\,d\mathcal{L}^2(z) \leq \int_{Y_0}^{\infty} C_fC_g e^{-4\pi y}y^{k-2}\,d\mathcal{L}^1(y) <\infty. \end{align*}[/step]

[guided]The only possible source of divergence in the Petersson integral is the cusp $y\to\infty$. Away from the cusp, the fundamental domain is compact, so ordinary continuity will control the integrand. Near the cusp, we use exactly the condition that $f$ and $g$ are cusp forms. Define the holomorphic map \begin{align*} q:\mathbb{H}&\to\{w\in\mathbb{C}:0<|w|<1\}\\ z&\mapsto e^{2\pi iz}. \end{align*} The cusp $y\to\infty$ corresponds to $q(z)\to 0$, because for $z=x+iy$ we have \begin{align*} |q(z)|=|e^{2\pi ix}e^{-2\pi y}|=e^{-2\pi y}. \end{align*} Because $f$ and $g$ are [cusp forms](/page/Cusp%20Form) for $SL_2(\mathbb{Z})$, the translation matrix gives $f(z+1)=f(z)$ and $g(z+1)=g(z)$. This periodicity means that, near the cusp, both functions descend through the local coordinate $q=e^{2\pi iz}$ to holomorphic functions on a punctured disc around $0$. The cusp condition says more: those descended functions extend holomorphically to $q=0$ and have value $0$ there. Therefore, after factoring out the first power of $q$, there are $r\in(0,1)$ and holomorphic maps \begin{align*} F,G:\{w\in\mathbb{C}:|w|<r\}\to\mathbb{C} \end{align*} such that whenever $|q(z)|<r$, \begin{align*} f(z)&=q(z)F(q(z)),\\ g(z)&=q(z)G(q(z)). \end{align*} This factorisation is the rigorous reason the whole infinite [Fourier series](/page/Fourier%20Series) decays uniformly like the first power of $q$: the remaining factors $F$ and $G$ are bounded on a smaller closed disc. Choose $Y_0\geq 1$ so that $e^{-2\pi Y_0}<r/2$. Since the closed disc $\{w\in\mathbb{C}:|w|\leq e^{-2\pi Y_0}\}$ is compactly contained in the domain of $F$ and $G$, continuity gives finite constants \begin{align*} M_f&:=\sup_{|w|\leq e^{-2\pi Y_0}}|F(w)|,\\ M_g&:=\sup_{|w|\leq e^{-2\pi Y_0}}|G(w)|. \end{align*} Set $C_f:=M_f$ and $C_g:=M_g$. If $z=x+iy\in\mathcal{F}_{>Y_0}$, then $|q(z)|=e^{-2\pi y}\leq e^{-2\pi Y_0}$, and hence \begin{align*} |f(z)|&=|q(z)|\,|F(q(z))|\leq C_f e^{-2\pi y},\\ |g(z)|&=|q(z)|\,|G(q(z))|\leq C_g e^{-2\pi y}. \end{align*} Multiplying the estimates gives \begin{align*} |f(z)\overline{g(z)}|\,y^{k-2} \leq C_fC_g e^{-4\pi y}y^{k-2}. \end{align*} The factor $y^{k-2}$ is only polynomial in $y$, whereas $e^{-4\pi y}$ decays exponentially. Therefore \begin{align*} \int_{Y_0}^{\infty} C_fC_g e^{-4\pi y}y^{k-2}\,d\mathcal{L}^1(y)<\infty. \end{align*} Finally, since the cusp part of the fundamental domain lies inside the strip $[-1/2,1/2]\times(Y_0,\infty)$, enlarging the domain of integration to that strip gives \begin{align*} \int_{\mathcal{F}_{>Y_0}} |f(z)\overline{g(z)}|\,y^{k-2}\,d\mathcal{L}^2(z) \leq \int_{Y_0}^{\infty} C_fC_g e^{-4\pi y}y^{k-2}\,d\mathcal{L}^1(y) <\infty. \end{align*}[/guided]

[step:Conclude absolute convergence of the Petersson integral] On the compact set $\mathcal{F}_{\leq Y_0}$, the map \begin{align*} \Phi:\mathcal{F}_{\leq Y_0}\to[0,\infty),\qquad z=x+iy\mapsto |f(z)\overline{g(z)}|\,y^{k-2} \end{align*} is continuous, hence bounded. Since $\mathcal{F}_{\leq Y_0}$ has finite $\mathcal{L}^2$-measure, \begin{align*} \int_{\mathcal{F}_{\leq Y_0}} |f(z)\overline{g(z)}|\,y^{k-2}\,d\mathcal{L}^2(z)<\infty. \end{align*} Combining this compact-part estimate with the cusp estimate gives \begin{align*} \int_{\mathcal{F}} |f(z)\overline{g(z)}|\,y^{k-2}\,d\mathcal{L}^2(z)<\infty. \end{align*} Thus $(f,g)_{\mathrm{Pet}}$ is well-defined for all $f,g\in S_k(SL_2(\mathbb{Z}))$. [/step]

[step:Prove linearity in the first variable] Let $f,g,h\in S_k(SL_2(\mathbb{Z}))$ and let $\lambda\in\mathbb{C}$. By absolute convergence, all integrals below are finite. Using pointwise distributivity in $\mathbb{C}$ and linearity of the Lebesgue integral, we compute \begin{align*} (f+\lambda g,h)_{\mathrm{Pet}} &= \int_{\mathcal{F}} (f(z)+\lambda g(z))\overline{h(z)}\,y^{k-2}\,d\mathcal{L}^2(z)\\ &= \int_{\mathcal{F}} f(z)\overline{h(z)}\,y^{k-2}\,d\mathcal{L}^2(z) + \lambda\int_{\mathcal{F}} g(z)\overline{h(z)}\,y^{k-2}\,d\mathcal{L}^2(z)\\ &= (f,h)_{\mathrm{Pet}}+\lambda(g,h)_{\mathrm{Pet}}. \end{align*} [/step]

[step:Prove conjugate symmetry] Let $f,g\in S_k(SL_2(\mathbb{Z}))$. Since $y^{k-2}$ is real-valued and nonnegative, and since complex conjugation commutes with absolutely convergent Lebesgue integrals, we have \begin{align*} \overline{(f,g)_{\mathrm{Pet}}} &= \overline{\int_{\mathcal{F}} f(z)\overline{g(z)}\,y^{k-2}\,d\mathcal{L}^2(z)}\\ &= \int_{\mathcal{F}} \overline{f(z)\overline{g(z)}\,y^{k-2}}\,d\mathcal{L}^2(z)\\ &= \int_{\mathcal{F}} g(z)\overline{f(z)}\,y^{k-2}\,d\mathcal{L}^2(z)\\ &= (g,f)_{\mathrm{Pet}}. \end{align*} [/step]

[step:Show that zero Petersson norm forces vanishing on an open set] Let $f\in S_k(SL_2(\mathbb{Z}))$ and suppose \begin{align*} (f,f)_{\mathrm{Pet}}=0. \end{align*} Then \begin{align*} 0 = \int_{\mathcal{F}} |f(z)|^2y^{k-2}\,d\mathcal{L}^2(z). \end{align*} The integrand \begin{align*} \Psi:\mathcal{F}\to[0,\infty),\qquad z=x+iy\mapsto |f(z)|^2y^{k-2} \end{align*} is nonnegative and continuous on the interior of $\mathcal{F}$. If there were a point $z_0\in\operatorname{int}(\mathcal{F})$ with $f(z_0)\ne 0$, continuity of $f$ would give a radius $r>0$ and a constant $c>0$ such that the open Euclidean ball \begin{align*} B(z_0,r):=\{z\in\mathbb{C}:|z-z_0|<r\} \end{align*} satisfies $B(z_0,r)\subset\operatorname{int}(\mathcal{F})$ and \begin{align*} |f(z)|^2y^{k-2}\geq c \end{align*} for every $z=x+iy\in B(z_0,r)$. Hence \begin{align*} 0 = \int_{\mathcal{F}} |f(z)|^2y^{k-2}\,d\mathcal{L}^2(z) \geq \int_{B(z_0,r)} c\,d\mathcal{L}^2(z) = c\,\mathcal{L}^2(B(z_0,r)) >0, \end{align*} a contradiction. Therefore $f$ vanishes on $\operatorname{int}(\mathcal{F})$. [/step]

[step:Apply the identity theorem to obtain positive definiteness] The function $f:\mathbb{H}\to\mathbb{C}$ is holomorphic because $f\in S_k(SL_2(\mathbb{Z}))$. The upper half-plane $\mathbb{H}$ is connected, and $\operatorname{int}(\mathcal{F})$ is a nonempty open subset of $\mathbb{H}$. We use the identity theorem for holomorphic functions in the following form: if a holomorphic function on a connected open subset of $\mathbb{C}$ vanishes on a nonempty open subset, then it vanishes identically. Since these hypotheses hold for $f$ on $\mathbb{H}$ and since $f$ vanishes on $\operatorname{int}(\mathcal{F})$, the theorem implies that $f=0$ on all of $\mathbb{H}$. Thus, if $f\ne 0$, then $(f,f)_{\mathrm{Pet}}\ne 0$. Since \begin{align*} (f,f)_{\mathrm{Pet}} = \int_{\mathcal{F}} |f(z)|^2y^{k-2}\,d\mathcal{L}^2(z) \end{align*} is a real nonnegative number, it follows that \begin{align*} (f,f)_{\mathrm{Pet}}>0. \end{align*} This proves positive definiteness and completes the proof. [/step]