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The Hoffman–Wielandt inequality says that for normal matrices $A,B$ with eigenvalues ${\alpha }_i,{\beta }_i$, there is a permutation $\pi$ such thatwhere $\Vert \cdot {\Vert }_F$ is the Frobenius norm and $\ast$ denotes the conjugate transpose. For Hermitian matrices, the increasingly ordered real eigenvalues may be paired directly: this ordering minimizes the sum of squared distances over all permutations, by removing crossed pairs. In particular, for the real symmetric matrices in this question,The square on the right is justified by symmetry. For arbitrary matrices, the appropriate expression is the conjugate-transpose product, not the ordinary square.

Put $D=|⟨L_N,f⟩-⟨{\hat{L}}_N,f⟩|$. Apply the Cauchy-Schwarz inequality to the average of the nonnegative eigenvalue differences, and then the Hoffman–Wielandt inequality:Taking nonnegative square roots gives the spectral Lipschitz bound from Frobenius distanceThe factor $N^{-1/2}$ is essential: the empirical spectral measure has total mass $1$, not $N$.

For equally sized Hermitian matrices and a real test function with Lipschitz bound one, the difference of its averages under their empirical spectral measures is at most $N^{-1/2}$ times the Frobenius norm of their difference. Pair the increasing eigenvalues, apply the triangle inequality and Cauchy-Schwarz inequality, then use the Hoffman–Wielandt inequality.