Does the sequence of Stieltjes measures build concentrations?

Question

Let $\mu$ be a finite non-negative regular Borel measure on $\mathbb{R}^2$ with no point masses. I am trying to understand the behaviour of certain Stieltjes measures when singularities appear: Assume for the product measures $\mu\otimes \mu$: $$\int_{\mathbb{R}^2\times \mathbb{R}^2} \log^{+}\left(\frac{1}{|x-y|}\right)d(\mu\otimes \mu)(x,y)<\infty$$ and $$\int_{|x-y|<\rho} \log^{+}\left(\frac{1}{|x-y|}\right)d(\mu\otimes \mu)(x,y)\to 0 \text{ when } \rho\to 0$$ which means that $\mu$ cannot concentrate in $\{x=y\}$.

Question We consider the sequence of Stieltjes measures $\mu_n(x)=\int_{\mathbb{R}^2} \eta_n(x-y) d\mu(y)$ where $\eta_n$ is the sequence of standard mollifiers, more precisely $\eta_n(x)=\frac{1}{\frac{1}{n^2}}\eta\left(\frac{x}{\frac{1}{n}}\right)=n^2\eta(nx)$ where $$\eta(x)=\begin{cases}\exp\left(\frac{1}{|x|^2-1}\right) &\text{ if } |x|<1\\ 0 &\text{ if} |x|\geq 1\end{cases} $$ I want to understand whether $$\int_{|x-y|<\rho} \log^{+}\left(\frac{1}{|x-y|}\right)d(\mu_n\otimes \mu_n)(x,y)\to 0 \text{ when } \rho\to 0$$ uniformly in $n$ in the sense that: for every $\varepsilon$ there exists some $\rho_0=\rho_0(\varepsilon)$ and $N_0=N_0(\varepsilon,\rho_0)$ so that for all $\rho<\rho_0$ and $n\geq N_0$ $$\int_{|x-y|<\rho} \log^{+}\left(\frac{1}{|x-y|}\right)d(\mu_n\otimes \mu_n)(x,y)<\varepsilon?$$ This would mean that this sequence cannot build concentrations at $\{x=y\}$.

Direct attempt $$\int_{|x-y|<\rho} \log^{+}\left(\frac{1}{|x-y|}\right)d(\mu_n\otimes \mu_n)(x,y)=\int_{|x-y|<\rho} \int_{\mathbb{R}^2}\int_{\mathbb{R}^2}\log^{+}\left(\frac{1}{|x-y|}\right)\eta_n(x-u)\eta_n(y-v) d\mu(u)d\mu(v)dy dx$$ Rewrite the integral over $\{|x-y|<\rho\}$ to and integral over $\mathbb{R}^2\times \mathbb{R}^2$ by introducing a cutoff function $\chi_{\rho}(x-y)$ which is 1 for $|x-y|<\rho$ and $0$ if $|x-y|>2\rho$. Now we can rewrite the integral and change variables: $\tilde{x}=x-u$, $\tilde{y}=y-v$. Then $$\int_{\mathbb{R}^2}\int_{\mathbb{R}^2} \int_{\mathbb{R}^2}\int_{\mathbb{R}^2}\log^{+}\left(\frac{1}{|x-y|}\right)\chi_{\rho}(x-y)\eta_n(x-u)\eta_n(y-v) d\mu(u)d\mu(v)dy dx\\= \int_{\mathbb{R}^2}\int_{\mathbb{R}^2} \int_{\mathbb{R}^2}\int_{\mathbb{R}^2}\log^{+}\left(\frac{1}{|\tilde{x}+u-y-\tilde{y}|}\right)\chi_{\rho}(\tilde{x}+u-y-\tilde{y})\eta_n(\tilde{x})\eta_n(\tilde{y}) d\mu(u)d\mu(v)d\tilde{y}d\tilde{x}.\\ $$ Because of the support of the standard mollifier we can restrict to $|\tilde{x}|,|\tilde{y}|<\frac{1}{n}$. In order to make use of the assumption I estimated the logarithm by the triangle inequality $$\log^{+}\left(\frac{1}{|\tilde{x}+u-y-\tilde{y}|}\right)\leq \log^{+}\left(\frac{1}{||u-v|-|\tilde{x}-\tilde{y}||}\right)\leq \log^{+}\left(\frac{1}{||u-v|-\frac{2}{n}|}\right)$$ and similar for the cutoff function but I guess thats just too strong. However I don't see how I can compare this expression just to $\log^{+}\left(\frac{1}{|u-v|}\right)$... my first intuition was that I only have to choose a bigger surround of $\{x=y\}$, more precisely I thought that I only have to choose $\rho+\frac{1}{n}$ instead of $\rho$ to get this estimate, but I don't see how I can get there.

Does anybody have an idea how I can make use of the assumption?

Answer 1

Reformulating the Problem. By normalizing both $\mu$ and $\eta$, we may assume both $\mu(\mathrm{d}x)$ and $\eta(x)\,\mathrm{d}x$ are probability measures. Then by introducing independent RVs $X_1, X_2 \sim \mu$ and $\xi_1, \xi_2 \sim \eta$, the assumption boils down to

$$ \lim_{\rho \to 0^+} \mathbb{E}\left[ \log^+\left(\frac{1}{\| Z \|}\right) \mathbf{1}_{\{\|Z\| < \rho\}} \right] = 0, \qquad Z = X_1 - X_2, $$

and the mollified integral becomes

$$ \mathbb{E}\biggl[ \log^+\biggl(\frac{1}{\| Z - \frac{1}{n}\xi \|}\biggr) \mathbf{1}_{\{\|Z - \frac{1}{n}\xi\| < \rho\}} \biggr], \qquad \xi = \xi_1 - \xi_2. \tag{*} $$

(Here, we utilized the fact that the density of $\frac{1}{n}\xi_k$'s is $\eta_n$ and the density of $X_k - \frac{1}{n}\xi_k$'s is $\mu_n$.) This leads us to consider the function $L = L_{\rho, \delta}$ on $\mathbb{R}^2$ defined by

$$ L(z) = L_{\rho, \delta}(z) := \mathbb{E}\biggl[ \log^+\biggl(\frac{1}{\| z - \frac{1}{2} \delta \xi \|}\biggr) \mathbf{1}_{\{\|z - \frac{1}{2}\delta \xi\| < \rho\}} \biggr] $$

where $\delta > 0$. Note that the mollified integral, $\text{(*)}$, becomes

$$ \text{(*)} = \mathbb{E}[L_{\rho,\frac{2}{n}}(Z)]. \tag{2} $$

In light of this, it suffices to bound $L(z)$ by a constant multiple of $\log^+(1/\|z\|)$ truncated at a level depending only on $\delta$ and $\rho$.

Analysis of $L(z)$. Note that the density of $\xi$ is a continuous function supported on $\mathsf{B}_2(0)$, where $\mathsf{B}_r(a)$ stands for the disk of radius $r$ centered at $a$. From this, the density of $\frac{1}{2}\delta \xi$ can be bounded by a constant multiple of $\delta^{-2} \mathbf{1}_{B_\delta(0)}$, leading to:

$$\begin{align*} L(z) &\lesssim \delta^{-2} \int_{\mathsf{B}_{\delta}(0)} \log^+\biggl(\frac{1}{\| z - y \|}\biggr) \mathbf{1}_{\{\|z - y \| < \rho\}} \, \mathrm{d}y \\ &= \delta^{-2} \int_{\mathsf{B}_{\delta}(z) \cap \mathsf{B}_\rho(0)} \log^+\biggl(\frac{1}{\| y \|}\biggr) \, \mathrm{d}y \end{align*}$$

• Case 1. When $\|z\| \leq \delta$, we have $\mathsf{B}_{\delta}(z) \cap \mathsf{B}_\rho(0) \subseteq \mathsf{B}_{\rho \wedge ( \|z\|+\delta)}(0)$. Hence, by invoking the polar coordinates, the integral can be bounded by

  $$ L(z) \lesssim \delta^{-2} f(\rho \wedge ( \|z\|+\delta)), \qquad \text{where} \quad f(r) := \int_{0}^{r} s \log^+(1/s) \, \mathrm{d}s. $$

  To make use of this bound, note that $f'(r)$ is increasing for $r < e^{-1}$ and hence $f(r) \leq rf'(r)$ in this range. So, when $\delta < (2e)^{-1}$, the above inequality implies

  $$L(z) \lesssim \delta^{-2} f(2\delta) \lesssim \log^+\biggl(\frac{1}{\delta}\biggr) \leq \log^+\biggl(\frac{1}{\|z\|}\biggr). $$

• Case 2. When $\|z\| > \delta$, the disk $\mathsf{B}_{\delta}(z)$ is confined in a circular sector with the central angle of $2\arcsin(\frac{\delta}{\|z\|})$. Using this, it is not hard to check that

  $$ L(z) \lesssim \delta^{-2} \frac{\delta}{\|z\|} [ f(\rho \wedge ( \|z\|+\delta)) - f(\rho \wedge ( \|z\|-\delta)) ] $$

  Since $f'(r) = r \log^+(1/r)$ is concave on $(0, 1)$, for $0 < a < b < 1$ we have

  $$ f(b) - f(a) \leq (b - a) f'\left( \frac{a+b}{2} \right). $$

  (This can be argued either by geometry or using Jensen's inequality.) Hence, if $\|z\| + \delta < e^{-1}$, then by the above inequality and the monotonicity of $f'$ on $(0, e^{-1})$,

  $$ L(z) \lesssim \frac{1}{\|z\|} f' \biggl( \frac{ (\rho \wedge ( \|z\|+\delta)) + (\rho \wedge ( \|z\|-\delta)) }{2} \biggr) \leq \frac{f'(\|z\|)}{\|z\|} = \log^+\biggl(\frac{1}{\|z\|}\biggr). $$

  We also note that, when $\|z\| > \rho + \delta$, we have $L(z) = 0$.

Combining altogether, it is straightforward to establish:

Theorem. There exists a constant $C > 0$ such that $$ L_{\rho,\delta}(z) \leq C \log^+\biggl(\frac{1}{\|z\|}\biggr)\mathbf{1}_{\mathsf{B}_{\delta+\rho}(0)}(z) $$ for any $\rho, \delta \in (0, \frac{1}{3e})$.

In light of the above theorem and equation $\text{(2)}$, we can deduce the desired uniform convergence.