I'm trying to prove that a sequence of functions $(k_N)_{N\in\mathbb{N}}$


where $r>0$, $k>2$ and

Edit: I have forgot to say, that $k\in\mathbb{N}$ is the size of the matrix $Q$ i.e. $Q$ is $k\times k$. This means that $\det Q=0$ and $Q$ is positive semidefinte.

I have also changed $k$ from beeing $>0$ to $k>2$.

End of the Edit

$$Q:=\left(\begin{array}{cccccccc} (1-\frac{1}{k})& -\frac{1}{k} & -\frac{1}{k} & \cdots & -\frac{1}{k} \\ -\frac{1}{k} & (1-\frac{1}{k}) & -\frac{1}{k} & \cdots & -\frac{1}{k} \\ \vdots & \ddots & & &\vdots \\ -\frac{1}{k} & \dots & &-\frac{1}{k} &(1-\frac{1}{k}) \end{array}\right)$$

goes to a delta function when $N\to\infty$. What I'm missing ist showing, that given $\varepsilon>0$ and $\eta>0$ there exists $n_0$ such that $\forall n>n_0$


I was looking for some expression for the error function of multivariate gaussians, but could find nothing of the sort. Is there a nice way of representing the error functions so that I don't have to diagonalize $Q$ and treat the function as several independent gaussians?


  • $\begingroup$ What is $r(\beta')$ and what is $\beta'$? $\endgroup$ Dec 27, 2020 at 19:35
  • $\begingroup$ $r(\beta')$ is a constant of my system. I'm working with a fixed $\beta'$. $\endgroup$
    – Rafael
    Dec 27, 2020 at 19:38

1 Answer 1


$\newcommand\R{\mathbb R}$Let us write $x$ for $\vec x$ and $t$ for $\eta$. Then $$\int_{x\in\R^k\colon|x|>t}k_n(x)\,dx=\infty$$ for all real $t>0$.

Indeed, we can write $$Q=I_k-\frac1k\,1_k1_k^T,$$ where the matrix $I_k$ is the $k\times k$ identity matrix and $1_k$ is the $1\times k$ column matrix with unit entries, that is, $1_K=e_1+\dots+e_k$, where $(e_1,\dots,e_k)$ is the standard basis of $\R^k$. So, $Q$ is the matrix of the orthogonal projector onto the orthogonal complement -- say $V$ -- of (the span of) the vector $1_k$; in particular, $Q1_k=0$.

Fix any orthonormal basis $(b_1,\dots,b_k)$ of $\R^d$ such that $b_1=1_k/\sqrt k$ (so that $(b_2,\dots,b_k)$ is an orthonormal basis of $V$). Let $(x_1,\dots,x_k)$ denote the coordinates of an arbitrary vector $x\in\R^d$ in the orthonormal basis $(b_1,\dots,b_k)$. Then $$(x,Qx)=x_2^2+\dots+x_k^2.$$ Therefore and because the Lebesgue measure is spherically invariant, for all real $t>0$ we have $$ \begin{aligned} &\int_{x\in\R^k dx\,\colon|x|>t}dx\, k_n(x) \\ &\ge\int_{x\in\R^k\colon x_1>t}dx\, k_n(x) \\ &=\int_t^\infty dx_1\int_\R dx_2\cdots\,\int_\R dx_k \exp\{-n^2r(x_2^2+\dots+x_k^2)\}\Big(\sqrt{\frac{r}{\pi}}\Big)^k n^{2k} =\infty, \end{aligned}$$ because the inner $(k-1)$-fold integral is a strictly positive real number not depending on $x_1$.

  • $\begingroup$ Thank you for your answer! I have realized that the definition of $k_N$ is wrong. The function should be normalized. I'm going to edit the definition. $\endgroup$
    – Rafael
    Dec 28, 2020 at 2:10
  • $\begingroup$ Clearly, no strictly positive normalizing factor can possibly help here. $\endgroup$ Dec 28, 2020 at 19:10
  • $\begingroup$ After reading this comment I was really confused about what went wrong with my idea. I have just seen, that I forgot to mention an important detail: $k\in\mathbb{N}$ is also the size of the matrix, so that $Q$ is positive semi-definite. I'm terribly sorry for the misunderstanding. I have been working on the problem leading to this question for so long, that it didn't even occurred to me to mention. $\endgroup$
    – Rafael
    Dec 29, 2020 at 0:39
  • $\begingroup$ I'm editing the question again, but your clever decomposition of the matrix has already helped me a great deal! $\endgroup$
    – Rafael
    Dec 29, 2020 at 0:40
  • $\begingroup$ I did not have a problem with $k$ denoting the dimension and $k_n$ denoting the kernel. The problem is with the fact that $Q$ is a singular (positive-semidefinite) matrix, rather than a positive-definite one. $\endgroup$ Dec 29, 2020 at 3:57

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