Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 31 / 2

A function $f:\mathbb{R}\to \mathbb{R}$ satisfies Lipschitz continuity with bound $L\ge 0$ ifHere $L$ is an admissible Lipschitz bound; the least admissible value is the Lipschitz constant. The later parts use the admissible bound $1$.

The empirical spectral measures of the two real symmetric matrices arewhere ${\delta }_t$ is the Dirac measure at $t$. Thus $⟨L_N,f⟩=N^{-1}{\sum }_{i}f({\lambda }_i)$. Pair the eigenvalues in the specified increasing order and apply the triangle inequality followed by Lipschitz continuity with bound $1$:This deterministic estimate uses only the Lipschitz bound; no entry independence or distributional hypothesis is needed here.

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 $N$.

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.