[proofplan] We define the old subspace as the span of degeneracy images from proper divisors of $N$, and define the new subspace as its Petersson-orthogonal complement. The Atkin-Lehner-Li old-new theorem gives that the old subspace is Hecke-stable and that this orthogonal complement is also Hecke-stable, yielding the direct [orthogonal decomposition](/theorems/436). Finally, the commuting normality of the Hecke operators on the new subspace gives a simultaneous orthogonal eigenbasis; normalising the first nonzero Fourier coefficient gives the newforms of level $N$. [/proofplan]

[step:Define the old subspace using degeneracy maps from lower levels] For each divisor $M$ of $N$ with $M < N$ and each positive divisor $d$ of $N/M$, define the degeneracy map \begin{align*} \alpha_{M,d}: S_k(\Gamma_0(M)) &\to S_k(\Gamma_0(N)) \\ f &\mapsto \alpha_{M,d}(f), \qquad \alpha_{M,d}(f)(z) = d^{k/2} f(dz). \end{align*} The factor $d^{k/2}$ is the standard Petersson-normalised convention; replacing it by a nonzero scalar gives the same image subspace. Since $dM$ divides $N$, the usual transformation law for cusp forms shows that $\alpha_{M,d}(f)$ is a cusp form of weight $k$ on $\Gamma_0(N)$. Define \begin{align*} S_k(\Gamma_0(N))^{\mathrm{old}} := \sum_{\substack{M \mid N \\ M < N}} \sum_{d \mid N/M} \alpha_{M,d}\bigl(S_k(\Gamma_0(M))\bigr) \subset S_k(\Gamma_0(N)). \end{align*} This is a finite-dimensional subspace because $N$ has finitely many divisors and each $S_k(\Gamma_0(M))$ is finite-dimensional for $k \ge 2$. [/step]

[step:Take the Petersson orthogonal complement as the new subspace]Equip $S_k(\Gamma_0(N))$ with the Petersson inner product \begin{align*} (f,g)_N := \int_{\Gamma_0(N)\backslash \mathbb{H}} f(z)\overline{g(z)} y^k\, d\mu_{\mathrm{hyp}}(z), \end{align*} where $\mathbb{H}:=\{x+iy \in \mathbb{C}: y>0\}$ is the complex upper half-plane, $\mathcal{L}^2$ is two-dimensional Lebesgue measure on $\mathbb{R}^2$, and $d\mu_{\mathrm{hyp}}(z)=y^{-2}\,d\mathcal{L}^2(x,y)$ is hyperbolic area measure. The integral is finite because $f$ and $g$ are cusp forms. Define \begin{align*} S_k(\Gamma_0(N))^{\mathrm{new}} := \bigl(S_k(\Gamma_0(N))^{\mathrm{old}}\bigr)^{\perp_N}, \end{align*} where $\perp_N$ denotes orthogonality with respect to $(\cdot,\cdot)_N$. Since $S_k(\Gamma_0(N))$ is finite-dimensional and $(\cdot,\cdot)_N$ is positive definite, elementary finite-dimensional [Hilbert space](/page/Hilbert%20Space) theory gives \begin{align*} S_k(\Gamma_0(N)) = S_k(\Gamma_0(N))^{\mathrm{old}} \oplus S_k(\Gamma_0(N))^{\mathrm{new}}. \end{align*} The sum is direct because a subspace intersects its orthogonal complement only in $\{0\}$.[/step]

[guided]The purpose of the Petersson inner product is to turn the finite-dimensional [vector space](/page/Vector%20Space) of cusp forms into a Hilbert space, so that the phrase "orthogonal complement" has an exact meaning. For $f,g \in S_k(\Gamma_0(N))$, set \begin{align*} (f,g)_N := \int_{\Gamma_0(N)\backslash \mathbb{H}} f(z)\overline{g(z)} y^k\, d\mu_{\mathrm{hyp}}(z), \end{align*} where $\mathbb{H}:=\{x+iy \in \mathbb{C}: y>0\}$ is the complex upper half-plane, $\mathcal{L}^2$ is two-dimensional Lebesgue measure on $\mathbb{R}^2$, and $d\mu_{\mathrm{hyp}}(z)=y^{-2}\,d\mathcal{L}^2(x,y)$ is hyperbolic area measure. The measure has been specified explicitly, and the integral converges because cusp forms decay at the cusps. Define the new subspace by \begin{align*} S_k(\Gamma_0(N))^{\mathrm{new}} := \bigl(S_k(\Gamma_0(N))^{\mathrm{old}}\bigr)^{\perp_N}. \end{align*} Now use the finite-dimensional [orthogonal decomposition theorem](/theorems/241) for inner product spaces. Its hypotheses are satisfied because $S_k(\Gamma_0(N))$ is finite-dimensional for $k \ge 2$, and the Petersson pairing is a positive definite Hermitian inner product on cusp forms. Therefore every form $h \in S_k(\Gamma_0(N))$ decomposes uniquely as \begin{align*} h = h_{\mathrm{old}} + h_{\mathrm{new}}, \end{align*} with $h_{\mathrm{old}} \in S_k(\Gamma_0(N))^{\mathrm{old}}$ and $h_{\mathrm{new}} \in S_k(\Gamma_0(N))^{\mathrm{new}}$. Equivalently, \begin{align*} S_k(\Gamma_0(N)) = S_k(\Gamma_0(N))^{\mathrm{old}} \oplus S_k(\Gamma_0(N))^{\mathrm{new}}. \end{align*} The directness follows because if a vector lies in both a subspace and its orthogonal complement, then its inner product with itself is zero, hence the vector is zero.[/guided]

[step:Use Atkin-Lehner-Li theory to obtain Hecke stability of the new subspace] For each prime $p$ dividing $N$, let \begin{align*} U_p: S_k(\Gamma_0(N)) &\to S_k(\Gamma_0(N)) \end{align*} denote the usual bad-prime Hecke operator, characterised on Fourier expansions by \begin{align*} U_p\left(\sum_{m=1}^{\infty} a_m e^{2\pi i m z}\right) = \sum_{m=1}^{\infty} a_{pm} e^{2\pi i m z}. \end{align*} The Atkin-Lehner-Li old-new theorem, applied with weight $k \ge 2$ and level $N$, gives the following precise stability statement for the old subspace defined by the degeneracy images above: $S_k(\Gamma_0(N))^{\mathrm{old}}$ is stable under every $T_\ell$ with $\ell \nmid N$ and every $U_p$ with $p \mid N$, and the Petersson adjoint of each such operator also preserves $S_k(\Gamma_0(N))^{\mathrm{old}}$. Therefore, if $A$ is one of the operators $T_\ell$ with $\ell \nmid N$ or $U_p$ with $p \mid N$, then for $v \in S_k(\Gamma_0(N))^{\mathrm{new}}$ and $w \in S_k(\Gamma_0(N))^{\mathrm{old}}$, \begin{align*} (Av,w)_N = (v,A^*w)_N = 0, \end{align*} because $A^*w \in S_k(\Gamma_0(N))^{\mathrm{old}}$ and $v$ is orthogonal to the old subspace. Thus $Av \in S_k(\Gamma_0(N))^{\mathrm{new}}$, so the new subspace is stable under every operator appearing in the theorem. [/step]

[step:Invoke the Atkin-Lehner-Li newform theorem to obtain a normalised eigenbasis]Let $\mathbb{T}_N^{\mathrm{new}}$ denote the algebra of endomorphisms of $S_k(\Gamma_0(N))^{\mathrm{new}}$ generated by the restrictions of all $T_\ell$ with $\ell \nmid N$ and all $U_p$ with $p \mid N$. The Atkin-Lehner-Li newform theorem, applied to the Petersson-orthogonal complement of the old subspace just constructed, states that $S_k(\Gamma_0(N))^{\mathrm{new}}$ has a basis consisting of cusp forms $f$ whose Fourier expansion \begin{align*} f(z)=\sum_{m=1}^{\infty} a_m(f)e^{2\pi i m z} \end{align*} satisfies $a_1(f)\ne 0$ and such that $f$ is an eigenvector for every $T_\ell$ with $\ell \nmid N$ and every $U_p$ with $p \mid N$. For each basis vector $f$, replace $f$ by $a_1(f)^{-1}f$. Since every operator in $\mathbb{T}_N^{\mathrm{new}}$ is linear, this scalar rescaling preserves all eigenvalue equations and gives $a_1(f)=1$. These $a_1$-normalised simultaneous eigenforms are, by definition in the same theorem, the newforms of level $N$.[/step]

[guided]The previous step showed that each operator $T_\ell$ with $\ell \nmid N$ and each operator $U_p$ with $p \mid N$ restricts to an endomorphism of $S_k(\Gamma_0(N))^{\mathrm{new}}$. We now use the eigenbasis part of the Atkin-Lehner-Li newform theorem, rather than trying to prove it from an unverified normality assertion for the bad-prime operators. The theorem applies because $k \ge 2$, $N \ge 1$, the old subspace has been defined using all degeneracy maps from proper divisors of $N$, and the new subspace is its Petersson-orthogonal complement. The conclusion of the theorem is exactly the structural statement needed here: $S_k(\Gamma_0(N))^{\mathrm{new}}$ has a basis of cusp forms $f$ that are simultaneous eigenvectors for every $T_\ell$ with $\ell \nmid N$ and every $U_p$ with $p \mid N$. Moreover, for each such basis element the first Fourier coefficient is nonzero. Writing the Fourier expansion as \begin{align*} f(z)=\sum_{m=1}^{\infty} a_m(f)e^{2\pi i m z}, \end{align*} the theorem gives $a_1(f)\ne 0$. Therefore the scalar multiple $a_1(f)^{-1}f$ is defined. Because each $T_\ell$ and $U_p$ is linear, if $Af=\lambda_A f$ for one of these operators $A$, then \begin{align*} A\bigl(a_1(f)^{-1}f\bigr)=a_1(f)^{-1}Af=\lambda_A a_1(f)^{-1}f. \end{align*} Thus the rescaled form remains a simultaneous eigenform for all the stated operators, and its first Fourier coefficient is $1$. These $a_1$-normalised simultaneous eigenforms are precisely the newforms of level $N$.[/guided]

[step:Combine the orthogonal decomposition and eigenbasis conclusions] The Petersson orthogonal construction gives \begin{align*} S_k(\Gamma_0(N)) = S_k(\Gamma_0(N))^{\mathrm{old}} \oplus S_k(\Gamma_0(N))^{\mathrm{new}}. \end{align*} The Atkin-Lehner-Li stability and newform theorem show that $S_k(\Gamma_0(N))^{\mathrm{new}}$ has a basis of $a_1$-normalised simultaneous eigenforms for every $T_\ell$ with $\ell \nmid N$ and every $U_p$ with $p \mid N$. Hence the old-new direct-sum decomposition and the existence of the newform basis both follow. [/step]