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If $f(\tau )={\sum }_{n\ge 1}{a}_n{q}^n$ is a weight-$k$ level-one cusp form, boundedness of its invariant norm givesIntegrate $f(x+iy)e^{-2\pi inx}$ over one period and choose $y=1/n$.

The analytic input for these Mellin transforms is elementary cusp decay. For a level-one weight-$k$ cusp form, the invariant norm of a modular form $y^{k/2}|f(x+iy)|$ is invariant under the modular group. It is bounded on the standard fundamental region, because the cusp expansion decays exponentially, and hence bounded on all of $\mathbb{H}$. Fourier inversion givesTaking $y=1/n$ proves the Fourier coefficient bound for a cusp form $c_n=O(n^{k/2})$. In particular rational translates have polynomially bounded coefficients and exponential decay as $y\to \infty$, uniformly in their real parts. These facts justify the initial Mellin integrations and the later entire continuation.

Write $\tau =x+iy$ and $\sigma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. Since $ad-bc=1$,Consequently, for the invariant norm of a modular form,This computation applies to an arbitrary function; modularity is needed only to deduce $\Gamma$-invariance of $\varphi$. The exponent is $k/2$, as printed in the PDF, rather than the converted TeX's $1/2$.

Use the primitive real-analytic Eisenstein series with exactly the normalization in the PDF. Its coset expression uses ${\Gamma }_{\infty }=\{\pm \left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right):n\in \mathbb{Z}\}$, so primitive bottom rows differing by sign represent the same coset. The factor $1/2$ in the primitive-pair sum accounts for this sign identification. Unlike the sum over all nonzero integer pairs, this Eisenstein series has no additional factor $2\zeta (2w)$. Write $w$ for the Rankin–Selberg integral parameter and $s$ for the Dirichlet series parameter to keep their shift visible.

First establish an initial region where all rearrangements are absolute. The invariant norm of a modular form $y^{k/2}|f(\tau )|$ is bounded: it is $\Gamma$-invariant, continuous on a truncated fundamental domain, and tends to zero at the cusp. Integrating one period of the Fourier series givesWith $y=1/n$, this proves the Fourier coefficient bound for a cusp form $a_n=O(n^{k/2})$, and similarly $b_n=O(n^{k/2})$. Hence $D(f,g,s)$ is initially absolutely convergent for $\mathrm{Re}s>k+1$.

For $\mathrm{Re}w>1$, the primitive real-analytic Eisenstein series is locally uniformly convergent and holomorphic in $w$. On the standard fundamental domain it has at most polynomial growth in $y$, locally uniformly in this half-plane. These are the analytic properties of the Eisenstein series needed here; they can also be checked directly from its defining sum. For example, for real $\sigma >1$ and $y\ge 1$, majorize the primitive sum by the sum over all nonzero pairs and useThis bound follows by comparing a shifted one-dimensional sum with its integral and a largest summand. Summing over $m\ne 0$ gives $E(\tau ,\sigma )=O(y^{\sigma }+y^{1-\sigma })$. For complex $w$, $|E(\tau ,w)|\le E(\tau ,\mathrm{Re}w)$, and the estimates are uniform on compact parameter sets. Since $f$ and $g$ decay exponentially on the cusp end, the integral $I(f,g,w)$ is therefore holomorphic throughout $\mathrm{Re}w>1$ by dominated integration.

The product $f(\tau )\overline{g(\tau )}{y}^k$ is $\Gamma$-invariant: the two modular factors cancel the transformation $\mathrm{Im}(\gamma \tau )=y/|c\tau +d{|}^2$. With the hyperbolic area $d\mu =dxdy/y^2$, Rankin–Selberg unfolding gives, initially for $\mathrm{Re}w>2$,Absolute unfolding is justified by replacing $f\overline{g}$ by $|fg|$ and using the same finite fundamental-domain integral with $E(\tau ,\mathrm{Re}w)$. The subsequent termwise integrations can be justified particularly economically by Cauchy-Schwarz inequality: on each horizontal line,The coefficient bounds make the right side $O(y^{-k-1})$ as $y\downarrow 0$, and it decays exponentially as $y\to \infty$. Multiplication by $y^{\mathrm{Re}w+k-2}$ is integrable at zero when $\mathrm{Re}w>2$. Thus the integral of the inner Fourier series may be evaluated first. Orthogonality of the exponentials givesThe Mellin transform of each exponential isIn the same initial region the sum of the absolute values of these integrals is finite by $a_n\overline{b_n}=O(n^k)$. Consequently the exact Rankin–Selberg integral identity isIn particular the two parameters are not equal: the Dirichlet series argument is shifted by $k-1$.

The holomorphy of $I$ on $\mathrm{Re}w>1$ now gives the required continuation:The right side agrees with the original Dirichlet series on $\mathrm{Re}s>k+1$ and is holomorphic on the larger half-plane. No continuation of the Eisenstein series beyond $\mathrm{Re}w>1$ is needed.

It is important to supply a further argument for absolute convergence: analytic continuation alone does not imply it. For real $w>1$, both $E(\tau ,w)$ and $|f(\tau ){|}^2$ are nonnegative, and their fundamental-domain integral is finite by the cusp estimates above. Tonelli theorem therefore unfolds $I(f,f,w)$ even before knowing convergence of the Dirichlet series. On each fixed horizontal line, the Fourier series converges absolutely and uniformly, so integration in $x$ givesA second use of Tonelli theorem, now in $y$, proves the positive unfolding of a cusp-form square:Since every real $s>k$ has $s=w+k-1$ with $w>1$, we have absolute convergence of $D(f,f,s)$ for $\mathrm{Re}s>k$. Absolute convergence for complex $s$ follows by taking its real part. In fact Cauchy-Schwarz inequality also gives absolute convergence of $D(f,g,s)$ there from the two corresponding square sums.

Finally let $\sigma >(k+1)/2$ and choose $k<\beta <2\sigma -1$. The preceding square-coefficient bound and Cauchy-Schwarz inequality giveThis proves absolute convergence of cusp-form L-series from square coefficients, and henceEquivalently, the suggested elementary inequality works by choosing $k-\sigma <\alpha <\sigma -1$: after multiplication by $n^{-\sigma }$ its two bounds sum to a convergent $\sum n^{-(\sigma -\alpha )}$ and a convergent $\sum |a_n{|}^2{n}^{-(\sigma +\alpha )}$. The existence of this interval is precisely $2\sigma >k+1$.

For a cusp form, the invariant norm of a modular form $|f(\tau )|(\mathrm{Im}\tau {)}^{k/2}$ is bounded on the entire half-plane. It is invariant under the level group; on a truncated fundamental region of a modular subgroup boundedness is compactness, and near each modular cusp exponential decay of the modular cusp expansion dominates the power of height. With this bound, Fourier inversion on a horizontal interval givesChoosing $y=1/n$ proves the Fourier coefficient bound for a cusp form, $a_n=O(n^{k/2})$. Therefore $\sum |a_n|n^{-\sigma }$ converges when $\sigma >k/2+1$, and uniformly on every compact subset of that half-plane. The locally uniform limit of its holomorphic terms is holomorphic, so

For a weight-$k$ cusp form, the invariant norm of a modular form $y^{k/2}|f(z)|$, with $y=\mathrm{Im}z$, is unchanged by the modular group: $\mathrm{Im}(\gamma z)=y/|cz+d{|}^2$ and $|f(\gamma z)|=|cz+d{|}^k|f(z)|$ cancel exactly.

Every orbit meets the standard fundamental domain of the modular group $\mathcal{F}=\{z:|\mathrm{Re}z|\le 1/2,|z|\ge 1\}$, where $y\ge \sqrt{3}/2$. The Fourier expansion of a modular form at the cusp has $f(q)=qh(q)$, with $h$ holomorphic near zero, since $f$ is a cusp form. Thus $|f(x+iy)|=O(e^{-2\pi y})$ uniformly in $x$ as $y\to \infty$, and $y^{k/2}|f|$ tends to zero there. On the remaining compact part of $\mathcal{F}$ it is bounded by continuity. Invariance therefore gives

For every $y>0$, Fourier coefficient extraction givesFor $n\ge 1$, minimize the last factor by taking $y=k/(4\pi n)$. This proves the Fourier coefficient bound for a cusp formThe assertion concerns positive indices; the constant coefficient vanishes because $f$ is a cusp form.

For a cusp form $f$, the invariant norm of a modular formis modular invariant, bounded on $\mathcal{F}$, and tends to zero at its cusp. Part a therefore makes it bounded throughout $\mathfrak{h}$: $|f(x+iy)|\le C{y}^{-k/2}$.

The correct PDF expansion is $f(\tau )={\sum }_{n\ge 1}{a}_n(f)q^n$. Fourier inversion givesand henceChoosing $y=1/n$ proves the Fourier coefficient bound for a cusp form $|a_n(f)|\le C^{\prime }{n}^{k/2}$.

Put $\tau =x+iy$. The modular transformation law andshow that the invariant norm of a modular formis invariant under $\Gamma (1)$. On the region from part (a), it is bounded: it is continuous on every truncated region, while the cusp form condition makes it tend to zero as $y\to \infty$. Thusfor all $x\in \mathbb{R}$ and $y>0$.

The Fourier coefficient formula on one period givesand henceChoosing $y=1/n$ yieldsThis proves the Fourier coefficient bound for a cusp form.