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Use the fixed common entry law from the printed setup: $Y=\sqrt{N}{X}_{ij}$ has mean zero and second moment $1$ for each upper-triangular entry, including the diagonal. LetSince $\mathbb{E}Y=0$, the removed part after centered truncation of a Wigner matrix isConsequently $\mathbb{E}{\Delta }_{ij}^2=N^{-1}\mathrm{Var}(Z_C)\le N^{-1}t(C)$. Symmetry gives $\mathrm{Tr}{\Delta }^2={\sum }_{i,j}{\Delta }_{ij}^2$. There are $N$ diagonal terms and $N(N-1)$ off-diagonal terms in this sum, soMirrored entries are counted twice, as they must be; independence of those mirrored entries is neither true nor needed for this expectation calculation.

By the preceding bound and the Markov inequality,The dominated convergence theorem applies to $Y^2{1}_{\{|Y|\ge C\}}\le Y^2$, so $t(C)\to 0$. Choose $C$ with $t(C)<{\epsilon }^3$. ThenThis second moment bound for spectral truncation depends only on $\epsilon$ and the common law of $Y$, and is uniform over functions with Lipschitz bound $1$. No fourth-moment hypothesis is required. If the scaled entry law were allowed to vary arbitrarily with $N$, a uniform choice would instead require uniform decay of its second-moment tails; the intended fixed-law setting supplies exactly that condition.