= Solution
Put $A_N=\begin{pmatrix}0&-1\\N&0\end{pmatrix}$. All calculations here use the <Hecke-normalized rational slash operator> of the PDF. Its <automorphy factor> identity gives $(f[\alpha]_k)[\beta]_k=f[\alpha\beta]_k$. For $\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma_1(N)$,
$$
A_N\gamma A_N^{-1}=\begin{pmatrix}d&-c/N\\-Nb&a\end{pmatrix}\in\Gamma_1(N).
$$
This proves that $A_N$ normalizes the <Gamma 1 congruence subgroup>. Right-action associativity shows $f[A_N]_k$ has the same transformation law as $f$. It is holomorphic in the <complex upper half-plane>; at each <modular cusp>, the rational change of variable preserves cusp decay by the upper-triangular factorization used in 2(i). Consequently $W_N$ maps the <cusp form> space to itself.
For a scalar matrix $-NI$, the printed slash normalization gives
$$
f[-NI]_k=(-1)^kN^{k-2}f.
$$
Since $A_N^2=-NI$, the two phase and determinant factors cancel:
$$
\boxed{W_N^2f=i^{2k}N^{2-k}f[-NI]_k=f.}
$$
Explicitly, the <Fricke involution> in the question is
$$
W_Nf(\tau)=i^kN^{-k/2}\tau^{-k}f\left(-\frac1{N\tau}\right).
$$
Its phase is essential for the involution and the functional-equation sign.
Suppose $W_Nf=\varepsilon f$. Define $F(y)=f(iy/\sqrt N)$ for $y>0$. Substitution at $\tau=iy/\sqrt N$ gives
$$
\boxed{F(1/y)=\varepsilon y^kF(y).}
$$
At infinity $F$ decays exponentially because $f$ is a <cusp form>. This identity gives $F(y)=O(y^{-k}e^{-c/y})$ near zero for some $c>0$. Thus its <Mellin transform> converges for every complex $s$, locally uniformly with all $s$ derivatives.
Initially, in a right half-plane where the <Dirichlet series> converges absolutely, termwise integration and the <gamma function> integral give
$$
\int_0^\infty F(y)y^{s-1}dy
=N^{s/2}(2\pi)^{-s}\Gamma(s)\sum_{n\geq1}\frac{a_n}{n^s}
=\Lambda_N(f,s).
$$
Such a half-plane exists without any eigenform hypothesis. Indeed the invariant quantity $(\operatorname{Im}\tau)^{k/2}|f(\tau)|$ is bounded on the modular quotient, by cusp decay and compactness away from the cusps. Fourier coefficient extraction on the line $\operatorname{Im}\tau=1/n$ gives $a_n=O(n^{k/2})$, so $\operatorname{Re}s>1+k/2$ suffices for the interchange.
Split the integral at one and change $y$ to $1/y$ in its lower part. The <phase-normalized Fricke functional equation> becomes transparent:
$$
\boxed{\Lambda_N(f,s)=\int_1^\infty F(y)\left(y^{s-1}+\varepsilon y^{k-s-1}\right)dy.}
$$
The integral defines an <entire function>: exponential decay dominates every power and every logarithmic derivative uniformly on compact $s$ sets. Replacing $s$ by $k-s$ and using $\varepsilon^2=1$ now proves
$$
\boxed{\Lambda_N(f,k-s)=\varepsilon\Lambda_N(f,s).}
$$
This gives the full analytic continuation of the completed <L-function of a cusp form>, not merely a formal identity in its initial convergence region.
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