5 Weak Convergence
Let \(k \in \mathbb {N}\) and \(\epsilon {\gt} 0\). Then for any \(b {\gt} 4\),
This follows from Markov’s inequality.
We define \(\nu \) to be the random measure \(\nu (dx) = |x|^k \mu _{\mathbf{X}_n}(dx)\).
First, applying Markov’s inequality on Lemma 5.0.4 gives
Next, ‘substituting’ \(\nu (dx)\) with \(|x|^k \mu _{\mathbf{X}_n}(dx)\) gives
Substituting the term inside the expectation of Lemma 5.0.2 with the expression acquired from Lemma 5.0.5 gives
Substituting the term inside the expectation of Lemma 5.0.6 with the expression acquired in Lemma ?? gives
Let \(C_{k}\) be the Catalan number. Then, for all \(k \in \mathbb {N}\), we have:
Let \(\{ Y_{ij}\} _{1 \leq i \leq j}\) be independent random variables, with \(\{ Y_{ii}\} _{i\geq 1}\) identically distributed and \(\{ Y_{ij}\} _{1 \leq i {\lt} j}\) identically distributed. Suppose that \(r_k = \max \{ \mathbb {E}(|Y_{11}|^k),\mathbb {E}(|Y_{12}|^k)\} {\lt} \infty \) for each \(k\in \mathbb {N}\). Suppose further than \(\mathbb {E}(Y_{ij})=0\) for all \(i,j\). If \(i{\gt}j\), define \(Y_{ij} \equiv Y_{ji}\), and let \(\mathbf{Y}_n\) be the \(n\times n\) matrix with \([\mathbf{Y}_n]_{ij} = Y_{ij}\) for \(1\le i,j\le n\). Let \(\mathbf{X}_n = n^{-1/2}\mathbf{Y}_n\) be the corresponding Wigner matrix. Then, we have:
From proposition 4.1.69, we know that \(\lim _{n\to \infty } \frac{1}{n}\mathbb {E}\operatorname {Tr}(\mathbf{X}_{n}^{k}) \leq C_{k}\) (odd and even cases of \(k\)). We further know that since the limit exists, \(\lim \sup _{n\to \infty } \frac{1}{n}\mathbb {E}\operatorname {Tr}(\mathbf{X}_{n}^{k}) = \lim _{n\to \infty } \frac{1}{n}\mathbb {E}\operatorname {Tr}(\mathbf{X}_{n}^{k})\). Using lemma 5.0.8, we have:
as required.
let \(k \in \mathbb {N}\) and \(\epsilon {\gt} 0\). then for any \(b {\gt} 4\),
for \(x {\gt} b {\gt} 4 {\gt} 1 \in \mathbb {R}\), we have \(k \mapsto |x|^{k}\) is increasing with \(k\).
for \(x {\gt} b {\gt} 4 {\gt} 1 \in \mathbb {R}\), we have \(k \mapsto \mathbb {P} (\int _{|x| {\gt} b}|x|^{k}\mu \mathbf{x}_{n} (dx))\) is increasing with \(k\).
for \(x {\gt} b {\gt} 4 {\gt} 1 \in \mathbb {R}\), we have \(k \mapsto \lim \sup _{n\to \infty }\mathbb {P} (\int _{|x| {\gt} b}|x|^{k}\mu \mathbf{x}_{n} (dx))\) is increasing with \(k\).
for \(x {\gt} b {\gt} 4 {\gt} 1 \in \mathbb {R}\), we have \(k \mapsto \lim \sup _{n\to \infty }\mathbb {P} (\int _{|x| {\gt} b}|x|^{k}\mu \mathbf{x}_{n} (dx)) \geq 0\) for all \(k \in \mathbb {N}\).
for \(x {\gt} b {\gt} 4 {\gt} 1 \in \mathbb {R}\), we have \(k \mapsto \frac{1}{\epsilon }(\frac{4}{b})^{k}\) is decreasing to \(0\) as \(k\to \infty \).
for \(x {\gt} b {\gt} 4 {\gt} 1 \in \mathbb {R}\), the sequence \(k \mapsto \lim \sup _{n\to \infty }\mathbb {P} (\int _{|x| {\gt} b}|x|^{k}\mu \mathbf{x}_{n} (dx))\) has:
We know that for all \(k \in \mathbb {N}\), \(\lim \sup _{n\to \infty }\mathbb {P}(\int _{|x| {\gt} b}|x|^{k}\mu \mathbf{x}_{n}(dx)) \geq 0\) (lemma 5.0.14). We also see that the sequence is bounded by \(\frac{1}{\epsilon } (\frac{4}{b})^{k}\) (lemma 5.0.10), which is decreasing to \(0\) as \(k\to \infty \) (lemma 5.0.15). Via squeeze theorem (where we use the zero sequence as another bound), we have:
as required.
for any sequence \((a_{k})_{k=1}^{\infty } \in \mathbb {R}\), if \(a_{k}\) has:
for all \(k \in \mathbb {N}\), \(a_{k} \geq 0\) (nonnegative sequence)
for all \(k \in \mathbb {N}\), we have \(a_{k+1} \geq a_{k}\) (strictly increasing sequence)
\(\lim _{k\to \infty } a_{k} = 0\)
then for all \(k \in \mathbb {N}\), we have \(a_{k} = 0\)
let \(k \in \mathbb {N}\) and \(\epsilon {\gt} 0\). Then for any \(b {\gt} 4\),
Consider sequence \((a_{k})_{k =1}^{\infty }\) defined as \(a_{k} = \lim \sup _{n\to \infty }\mathbb {P}(\int _{|x| {\gt} b}|x|^{k} \mu \mathbf{x}_{n}(dx))\). Then, we use the conditions from lemma 5.0.17:
From lemma 5.0.14, we know that \(a_{k} \geq 0\) for all \(k \in \mathbb {N}\), so the sequence is nonnegative.
From lemma 5.0.13, we know that \(a_{k+1} \geq a_{k}\) for all \(k \in \mathbb {N}\), so the sequence is strictly increasing.
From lemma 5.0.16, we know that \(\lim _{k\to \infty } a_{k} = 0\), so the sequence converges to zero.
Therefore the sequence \(a_{k} = \lim \sup _{n\to \infty }\mathbb {P}(\int _{|x| {\gt} b}|x|^{k} \mu \mathbf{x}_{n}(dx)) = 0\) for all \(k \in \mathbb {N}\), as required.
Fix a bounded, continuous function \(f \in C_{b}(\mathbb {R})\), fix \(\epsilon {\gt} 0\), fix \(b {\gt} 4\). Then, there exists a polynomial \(P_{\epsilon }\) such that:
Let \(f \in C_{b}(\mathbb {R})\), fix \(\epsilon {\gt} 0\), and \(b {\gt} 4\), and let \(P_{\epsilon }\) be the polynomial from lemma 5.0.19. Then, we have:
If \(\left|\int f\, d\mu _{X_n} - \int f\, d\sigma _1\right|{\gt}\epsilon \), then \(\left|\int f\, d\mu _{X_n} - \int P_\epsilon \, d\mu _{X_n}\right|{\gt}\epsilon /3, \left|\int P_\epsilon \, d\mu _{X_n} - \int P_\epsilon \, d\sigma _1\right| {\gt}\epsilon /3,\) or \(\left|\int P_\epsilon \, d\sigma _1 - \int f\, d\sigma _1\right|{\gt}\epsilon /3\).
Pigeonhole.
\(\mathbb {P}\left( \left|\int f\, d\mu _{X_n} - \int f\, d\sigma _1\right|{\gt}\epsilon \right) \le \mathbb {P}\left(\left|\int f\, d\mu _{X_n} - \int P_\epsilon \, d\mu _{X_n}\right|{\gt}\epsilon /3\right) + \mathbb {P}\left(\left|\int P_\epsilon \, d\mu _{X_n} - \int P_\epsilon \, d\sigma _1\right| {\gt}\epsilon /3\right) + \mathbb {P}\left(\left|\int P_\epsilon \, d\sigma _1 - \int f\, d\sigma _1\right|{\gt}\epsilon /3\right)\)
Triangle equality: use MeasureTheory.lintegral_edist_triangle. (Not sure if this is needed)
\(\mathbb {P}\left(\left|\int P_\epsilon \, d\sigma _1 - \int f\, d\sigma _1\right|{\gt}\epsilon /3\right)\) is identically zero.
By construction, \(|P_\epsilon -f|{\lt}\epsilon /6\) on \([-b,b]\), which includes the support \([-2,2]\) of \(\sigma _1\).
Break up the integral.
If \(\left|\int f\, d\mu _{X_n} - \int P_\epsilon \, d\mu _{X_n}\right|{\gt}\epsilon /3\), then \(\int |f-P_\epsilon |\mathbb {1}_{|x|\le b}\, d\mu _{X_n}{\gt}\epsilon /6\) or \(\int |f-P_\epsilon |\mathbb {1}_{|x|{\gt} b}\, d\mu _{X_n}{\gt}\epsilon /6\).
Pigeonhole.
Use previous two lemmas.
\( \limsup _{n\to \infty } \left(\mathbb {P}\left(\int |f-P_\epsilon |\mathbb {1}_{|x|\le b}\, d\mu _{\mathrm{X}_n}{\gt}\epsilon /6\right) \right) = 0\)
\( \limsup _{n \to \infty } \left(\mathbb {P}\left(\left|\int P_\epsilon \, d\mu _{\mathrm{X}_n} - \int P_\epsilon \, d\sigma _1\right| {\gt}\epsilon /3\right) \right) = 0 \)
\(\mathbb {P}\left( \left|\int f\, d\mu _{\mathrm{X}_n} - \int f\, d\sigma _1\right|{\gt}\epsilon \right) \leq \mathbb {P}\left(\int |f-P_\epsilon |\mathbb {1}_{|x|{\gt} b}\, d\mu _{\mathrm{X}_n}{\gt}\epsilon /6\right)\)
\( \mathbb {P}\left(\int |f-P_\epsilon | \mathbb {1}_{|x|\ge b}\, d\mu _{\mathrm{X}_n} {\gt} \epsilon /6\right) \le \mathbb {P}\left(\int c|x|^k\mathbb {1}_{|x|\ge b}\, \mu _{\mathrm{X}_n}(dx) {\gt} \epsilon /6\right) \)
Let \(k =\)deg\(P_\epsilon \), and since \(f\) is bounded, \(|f(x) - P_{\epsilon }(x)| \leq \| f\| _{\infty } + |P_\epsilon (x)|\). Also note it is on interval \(|x| {\gt} b\), which completes the proof.
\(\limsup _{n \to \infty }\mathbb {P}\left(\int c|x|^k\mathbb {1}_{|x|\ge b}\, \mu _{\mathrm{X}_n}(dx) {\gt} \epsilon /6\right) = 0 \)
\(\limsup _{n \to \infty } \mathbb {P}\left(\int |f-P_\epsilon | \mathbb {1}_{|x|\ge b}\, d\mu _{\mathrm{X}_n} {\gt} \epsilon /6\right) = 0\)
Let \(\mathrm{X}_n=n^{-1/2}\mathrm{Y}_n\) be a sequence of Wigner matrices, with entries satisfying \(\mathbb {E}(Y_{ij})=0\) for all \(i,j\) and \(\mathbb {E}(Y_{12}^2)=t\). Then the empirical law of eigenvalues \(\mu _{\mathrm{X}_n}\) converges in probability to \(\sigma _t\) as \(n\to \infty \). Precisely: for any \(f\in C_b(\mathbb {R})\) (continuous bounded functions) and each \(\epsilon {\gt}0\),