I have some problem in proving the following question.
Let $\mathcal{H}$ be some second-order Hamilton operator on $l^2(\mathbb{Z}^d)$, which is the $l^2$-space on the discrete lattice. Given H is also self-adjoint and $\lambda_2=\sup\sigma(\mathcal{H})$ as the largest spectral value. Let $\tilde{\mathcal{H}}$ denote the restriction of $\mathcal{H}$ in the subspace consisting of component-wise symmetric functions $\mathfrak{G}^2$ as following:
$$\mathfrak{G}^2:=\lbrace f\in l^2(\mathbb{Z}^{2d}):\:f(x,y)=f(y,x)\:\: \forall(x,y)\in\mathbb{Z}^d \times\mathbb{Z}^d \rbrace.$$
The claim is: The supremum of the spectrum of $\tilde{\mathcal{H}}^2$ coincides $\lambda_2$.
could be helpful：I proved that $\tilde{\mathcal{H}}$ is an endomorphism on $\mathfrak{G}^2$. And $\mathfrak{G}^2$ is a closed subspace.
Somehow I can't prove the claim even in an elementary way. But I do believe the claim is true. Could someone gives me a little bit advice how to prove this?
I do appreciate a lot for your help in advance.