Let $\Lambda \subset \mathbb{Z}^{d}$ be a finite lattice. The discrete Laplacian is defined as the linear operator $-\Delta: \mathbb{C}^{\Lambda}\to \mathbb{C}^{\Lambda}$ satisfying: $$(-\Delta \phi)(x) := \sum_{k=1}^{d}[2\phi(x)-\phi(x+e_{k})-\phi(x-e_{k})] $$ where $\{e_{1},...,e_{d}\}$ is the canonical basis of $\mathbb{R}^{d}$. I'd like to prove that this is a symmetric operator, that is, it satisfies: $$\langle \phi, -\Delta \psi\rangle = \langle -\Delta \phi, \psi\rangle$$ where $\langle \phi,\psi\rangle := \sum_{x\in \Lambda}\overline{\phi(x)}\psi(x)$ is the usual inner product on $\mathbb{C}^{\Lambda}$. Here is my proof: $$\langle \phi, -\Delta \psi \rangle = \sum_{x\in \Lambda}\overline{\phi(x)}\sum_{k=1}^{d}[2\psi(x)-\psi(x+e_{k})-\psi(x-e_{k})] = \sum_{x\in \Lambda}\sum_{k=1}^{d}2\overline{\phi(x)}\psi(x) -\sum_{x\in \Lambda}\sum_{k=1}^{d}\overline{\phi(x)}\psi(x+e_{k})-\sum_{x\in \Lambda}\sum_{k=1}^{d}\overline{\phi(x)}\psi(x-e_{k}) $$ Now, changing variables $x' = x\pm e_{k}$, we can write: \begin{eqnarray} \sum_{x\in \Lambda}\sum_{k=1}^{d}\overline{\psi(x)}\psi(x\pm e_{k}) = \sum_{x'\in \Lambda}\sum_{k=1}^{d}\overline{\phi(x\mp e_{k})}\psi(x) \tag{1}\label{1} \end{eqnarray} and putting it on our previous relation, we get: $$\langle \phi, -\Delta \psi\rangle = \sum_{x\in \Lambda}\sum_{k=1}^{d}[2\overline{\phi(x)}-\overline{\phi(x+e_{k})}-\overline{\phi(x-e_{k})}]\psi(x) = \langle -\Delta \phi, \psi\rangle$$
Is my proof correct? I'm not 100% sure about identity (\ref{1}). Any comments would be helpful!