This result is due to Gelfand, Graev, Piatetski-Shapiro and has a short proof. I suggest you read Bump: Automorphic forms and representations, Prop. 3.23, pp. 285-289.

Let me switch to $G=\mathrm{PGL}_2(\mathbb{R})$ and $\Gamma=\mathrm{PGL}_2(\mathbb{Z})$ for simplicity. The idea is to consider right convolutions $\rho(\phi)$ by smooth and compactly supported functions $\phi:G\to\mathbb{C}$, which act on the space of automorphic functions $L^2(\Gamma\backslash G)$. This family of operators is sufficiently rich to distinguish automorphic functions: if $f\neq g$ then there is $\phi$ such that $\rho(\phi)f\neq\rho(\phi)g$. Thinking of $\rho(\phi)$ as an operator on $(\Gamma\cap N)\backslash G$, its kernel is given by

$$K(g,h)=\sum_{\gamma\in\Gamma\cap N}\phi(g^{-1}\gamma h).$$

As $N\cong\mathbb{R}$ and $\Gamma\cap N\cong\mathbb{Z}$, the sum can be analyzed by Poisson summation, using the characters of $N$. It turns out that by subtracting the term corresponding to the trivial character, i.e.

$$K_0(g,h)=\int_{N}\phi(g^{-1} n h) \ dn,$$

the rest of the kernel decays rapidly at infinity, hence defines a compact operator. However, on the cuspidal space the operator with kernel $K_0(g,h)$ acts by zero, hence the operator $\rho(\phi)$ is compact on the cuspidal space. In particular, $\rho(\phi)$ has an eigenvalue with finite multiplicity on any right $G$-invariant subspace of cuspidal functions, and from here the result follows by a standard linear algebra argument.

EDIT: I fixed some inaccuracies.

This result is due to Gelfand, Graev, Piatetski-Shapiro and has a short proof. I suggest you read Bump: Automorphic forms and representations, Prop. 3.2.3, pp. 285-289.

Let me switch to $G=\mathrm{PGL}_2(\mathbb{R})$ and $\Gamma=\mathrm{PGL}_2(\mathbb{Z})$ for simplicity. The idea is to consider right convolutions $\rho(\phi)$ by smooth and compactly supported functions $\phi:G\to\mathbb{C}$, which act on the space of automorphic functions $L^2(\Gamma\backslash G)$. This family of operators is sufficiently rich to distinguish automorphic functions: if $f\neq g$ then there is $\phi$ such that $\rho(\phi)f\neq\rho(\phi)g$. Thinking of $\rho(\phi)$ as an operator on $(\Gamma\cap N)\backslash G$, its kernel is given by

$$K(g,h)=\sum_{\gamma\in\Gamma\cap N}\phi(g^{-1}\gamma h).$$

As $N\cong\mathbb{R}$ and $\Gamma\cap N\cong\mathbb{Z}$, the sum can be analyzed by Poisson summation, using the characters of $N$. It turns out that by subtracting the term corresponding to the trivial character, i.e.

$$K_0(g,h)=\int_{N}\phi(g^{-1} n h) \ dn,$$

the rest of the kernel decays rapidly at infinity, hence defines a compact operator. However, on the cuspidal space the operator with kernel $K_0(g,h)$ acts by zero, hence the operator $\rho(\phi)$ is compact on the cuspidal space. In particular, $\rho(\phi)$ has an eigenvalue with finite multiplicity on any right $G$-invariant subspace of cuspidal functions, and from here the result follows by a standard linear algebra argument.

EDIT: I fixed some inaccuracies.

This result is due to Gelfand, Graev, Piatetski-Shapiro and has a short proof. I suggest you read Bump: Automorphic forms and representations, Prop. 3.23, pp. 285-289.

Let me switch to $G=\mathrm{PGL}_2(\mathbb{R})$ and $\Gamma=\mathrm{PGL}_2(\mathbb{Z})$ for simplicity. The idea is to consider right convolutions $\rho(\phi)$ by smooth and compactly supported functions $\phi:G\to\mathbb{C}$, which act on the space of automorphic functions $L^2(\Gamma\backslash G)$. This family of operators is sufficiently rich to distinguish automorphic functions: if $f\neq g$ then there is $\phi$ such that $\rho(\phi)f\neq\rho(\phi)g$. The kernel of $\rho(\phi)$ is given by a so-called automorphic kernel

$$K(g,h)=\sum_{\gamma\in\Gamma\cap N}\phi(g^{-1}\gamma h).$$

As $N\cong\mathbb{R}$ and $\Gamma\cap N\cong\mathbb{Z}$, the sum can be analyzed by Poisson summation, using the characters of $N$. It turns out that by subtracting the term corresponding to the trivial character, i.e.

$$K_0(g,h)=\int_{N}\phi(g^{-1} n h) \ dn,$$

the rest of the kernel decays rapidly at infinity, hence defines a compact operator. However, on the cuspidal space $K_0(g,h)$ acts by zero, hence the operator $\rho(\phi)$ is compact on the cuspidal space. In particular, $\rho(\phi)$ has an eigenvalue with finite multiplicity on any right $G$-invariant subspace of cuspidal functions, and from here the result follows by a standard linear algebra argument.

This result is due to Gelfand, Graev, Piatetski-Shapiro and has a short proof. I suggest you read Bump: Automorphic forms and representations, Prop. 3.23, pp. 285-289.

Let me switch to $G=\mathrm{PGL}_2(\mathbb{R})$ and $\Gamma=\mathrm{PGL}_2(\mathbb{Z})$ for simplicity. The idea is to consider right convolutions $\rho(\phi)$ by smooth and compactly supported functions $\phi:G\to\mathbb{C}$, which act on the space of automorphic functions $L^2(\Gamma\backslash G)$. This family of operators is sufficiently rich to distinguish automorphic functions: if $f\neq g$ then there is $\phi$ such that $\rho(\phi)f\neq\rho(\phi)g$. Thinking of $\rho(\phi)$ as an operator on $(\Gamma\cap N)\backslash G$, its kernel is given by

$$K(g,h)=\sum_{\gamma\in\Gamma\cap N}\phi(g^{-1}\gamma h).$$

As $N\cong\mathbb{R}$ and $\Gamma\cap N\cong\mathbb{Z}$, the sum can be analyzed by Poisson summation, using the characters of $N$. It turns out that by subtracting the term corresponding to the trivial character, i.e.

$$K_0(g,h)=\int_{N}\phi(g^{-1} n h) \ dn,$$

the rest of the kernel decays rapidly at infinity, hence defines a compact operator. However, on the cuspidal space the operator with kernel $K_0(g,h)$ acts by zero, hence the operator $\rho(\phi)$ is compact on the cuspidal space. In particular, $\rho(\phi)$ has an eigenvalue with finite multiplicity on any right $G$-invariant subspace of cuspidal functions, and from here the result follows by a standard linear algebra argument.

EDIT: I fixed some inaccuracies.