= Solution
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. Let
$$
Z_C=Y\mathbf1_{\{|Y|\geq C\}},\qquad t(C)=\mathbb E[Y^2\mathbf1_{\{|Y|\geq C\}}].
$$
Since $\mathbb EY=0$, the removed part after <centered truncation of a Wigner matrix> is
$$
\Delta_{ij}=X_{ij}-\widehat X_{ij}=N^{-1/2}(Z_{C,ij}-\mathbb EZ_C).
$$
Consequently $\mathbb E\Delta_{ij}^2=N^{-1}\operatorname{Var}(Z_C)\leq N^{-1}t(C)$. Symmetry gives $\operatorname{Tr}\Delta^2=\sum_{i,j}\Delta_{ij}^2$. There are $N$ diagonal terms and $N(N-1)$ off-diagonal terms in this sum, so
$$
\mathbb E\left[\frac1N\operatorname{Tr}\Delta^2\right]=\operatorname{Var}(Z_C)\leq t(C).
$$
Mirrored 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>,
$$
\mathbb P\{D>\varepsilon\}\leq\mathbb P\left\{\frac1N\operatorname{Tr}\Delta^2>\varepsilon^2\right\}\leq\frac{t(C)}{\varepsilon^2}.
$$
The <dominated convergence theorem> applies to $Y^2\mathbf1_{\{|Y|\geq C\}}\leq Y^2$, so $t(C)\to0$. Choose $C$ with $t(C)<\varepsilon^3$. Then
$$
\boxed{\mathbb P\{|\langle L_N,f\rangle-\langle\widehat L_N,f\rangle|>\varepsilon\}<\varepsilon\quad\text{for every }N.}
$$
This <second moment bound for spectral truncation> depends only on $\varepsilon$ 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.
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