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Let $\eta(t)$ be a smooth function such that $\eta(t)=0$ for $t \le 1, 0 \le \eta \le 1$ everywhere and $\eta(t)=1$ for $t \ge 2$. For $n\ge 3, \Gamma(x) = \dfrac{1}{|x|^2} $ and consider the function $$ \Gamma_{\epsilon}(x) = \Gamma(x) \eta(|x|/ \epsilon),$$ where $\epsilon > 0$.

  1. If $\rho\in L^{1}(\mathbb{R}^{n})\cap L^{\infty}(\mathbb{R}^{n})$ $$ w_{\epsilon}(x) = \int_{\mathbb{R}^{n}} \rho(y) \Gamma_{\epsilon}(x-y)dy \rightarrow w(x) = \int_{\mathbb{R}^{n}} \rho(y) \Gamma(x-y)dy $$ with the argument \begin{equation} w_\epsilon(x) - w(x) = \int_{|x-y| \le 2\epsilon} \rho(y) (1 - \eta(|x-y|/\epsilon))dy. \end{equation}

So \begin{equation} |w_\epsilon(x) - w(x)| \le \int_{|x-y| \le 2\epsilon} |\rho(y)| dy \le \|\rho\|_{\infty}\epsilon^n. \end{equation} But I think \begin{equation} w_\epsilon(x) - w(x) = \int_{|x-y| \le 2\epsilon} \rho(y) \Gamma(|x-y|)(1 - \eta(|x-y|/\epsilon))dy. \end{equation} This is in the page 3. In the page 4 we have 2.\begin{equation} C_n \|\rho\|_{\infty} \int_{|x-y|\le 2 \epsilon} \Bigl( \dfrac{1}{|x-y|^{n-1}} + \dfrac{1}{ \epsilon |x-y|^{n-2}} \Bigr)dy = C \|\rho\|_{\infty} \epsilon. \end{equation} But I guess \begin{eqnarray} \int_{|x-y|\le 2 \epsilon} \Bigl( \dfrac{1}{|x-y|^{n-1}} + \dfrac{1}{ \epsilon |x-y|^{n-2}} \Bigr)dy &=& \int_{|y|\le2 \epsilon} ( \dfrac{1}{|y|^{n-1}} + \dfrac{1}{ \epsilon |y|^{n-2}} \Bigr)dy \\ &=& \sigma(S^{n-1}) \int_0^{2 \epsilon}( \dfrac{1}{|r|^{n-1}} + \dfrac{1}{ \epsilon |r|^{n-2}} \Bigr)dy\\ &=& \sigma(S^{n-1})[1/n \cdot 1/(2 \epsilon)^n + 1/(n-1) \cdot 1/(2 \epsilon)^{(n-1)}] \end{eqnarray} that tends to $\infty$ when $\epsilon \rightarrow 0$.

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up vote 1 down vote accepted

Since there is no question in your post, it's not clear what we are supposed to say. Apparently, you forgot that the switch to polar coordinates brings $r^{n-1}$ into the integral. This is why you mistakenly concluded that the last integral diverges.

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