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From asymptotic distribution and vague convergence to uniform convergence, with numerical applications

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arxiv 2309.03662 v1 pith:SE5JK42N submitted 2023-09-07 math.NA cs.NAmath.PR

classification math.NAcs.NAmath.PR
keywords lambdaomegainftyldotsasymptoticdistributionqquadbounded
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abstract

Let $\{\Lambda_n=\{\lambda_{1,n},\ldots,\lambda_{d_n,n}\}\}_n$ be a sequence of finite multisets of real numbers such that $d_n\to\infty$ as $n\to\infty$, and let $f:\Omega\subset\mathbb R^d\to\mathbb R$ be a Lebesgue measurable function defined on a domain $\Omega$ with $0<\mu_d(\Omega)<\infty$, where $\mu_d$ is the Lebesgue measure in $\mathbb R^d$. We say that $\{\Lambda_n\}_n$ has an asymptotic distribution described by $f$, and we write $\{\Lambda_n\}_n\sim f$, if \[ \lim_{n\to\infty}\frac1{d_n}\sum_{i=1}^{d_n}F(\lambda_{i,n})=\frac1{\mu_d(\Omega)}\int_\Omega F(f({\boldsymbol x})){\rm d}{\boldsymbol x}\qquad\qquad(*) \] for every continuous function $F$ with bounded support. If $\Lambda_n$ is the spectrum of a matrix $A_n$, we say that $\{A_n\}_n$ has an asymptotic spectral distribution described by $f$ and we write $\{A_n\}_n\sim_\lambda f$. In the case where $d=1$, $\Omega$~is a bounded interval, $\Lambda_n\subseteq f(\Omega)$ for all $n$, and $f$ satisfies suitable conditions, Bogoya, B\"ottcher, Grudsky, and Maximenko proved that the asymptotic distribution (*) implies the uniform convergence to $0$ of the difference between the properly sorted vector $[\lambda_{1,n},\ldots,\lambda_{d_n,n}]$ and the vector of samples $[f(x_{1,n}),\ldots,f(x_{d_n,n})]$, i.e., \[ \lim_{n\to\infty}\,\max_{i=1,\ldots,d_n}|f(x_{i,n})-\lambda_{\tau_n(i),n}|=0, \qquad\qquad(**) \] where $x_{1,n},\ldots,x_{d_n,n}$ is a uniform grid in $\Omega$ and $\tau_n$ is the sorting permutation. We extend this result to the case where $d\ge1$ and $\Omega$ is a Peano--Jordan measurable set (i.e., a bounded set with $\mu_d(\partial\Omega)=0$). See the rest of the abstract in the manuscript.

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