convergence of integrals (almost unity for convolution)

Question

Suppose that $\lambda^{d}$ is the Lebesgue-Borel measure on $\mathbb{R}^d$ and, if $r>0$, $H_r$ is the map on $\mathbb{R}^d$ defined by $H_r(\overline{x})=r\overline{x}$. Furthermore, $K$ is a non-negative real function over $\mathbb{R}^d$ such that $\int_{\mathbb{R}^d}Kd\lambda^{d}=1$. If $f$ is a continuous real function over $\mathbb{R}^d$ with compact support, define $I_r(f)=\int_{\mathbb{R}^d}f\circ H_r^{-1}.Kd\lambda^{d}=\int_{\mathbb{R}^d}f(\dfrac{\overline{x}}{r})Kd\lambda^{d}$. Prove that $\lim_{r\rightarrow \infty}I_r(f)=f(\overline{0})$. I can prove (using dominated convergence) that the sequence $\left ( I_n(f) \right )_{n \in \mathbb{N}}$ converges to $f(\overline{0})$, but how go from there?

Answer 1

Quick Answer: It is impossible to prove and the conclusion is wrong without extra assumption imposed on $f$.

Reason: For each $r>0$, if $f=g$, $\lambda^{d}-a.e.$, and with compact support and suitable boundedness condition, then $I_{r}(f)=I_{r}(g)$. In another word, modifying $f$ on a measure zero set will not alter the value $I_{r}(f)$ and hence the limiting value $\lim_{r\rightarrow\infty}I_{r}(f)$ (provided that the limit exists). However, $f(\bar{0})$ relies on the value of $f$ at $\bar{0}$.

Answer 2

Suppose that $\lim_{r\rightarrow \infty}I_r(f)\neq f(\overline{0})$. Then there exists $\epsilon > 0$ with the property that for each $R>0$, one can find an $r>R$ with $\left | I_r(f) - f(\overline{0}) \right |\geqslant \epsilon$. This leads to $1<t_1<t_2<...<t_n<...$ with $\left | I_{t_j}(f) - f(\overline{0}) \right |\geqslant \epsilon$. Applying dominated convergence to $g_n = f(\dfrac{\overline{x}}{t_n})K$, justified by $g_n \rightarrow f(\overline{0})K$ and $\left | g_n(\overline{x})\right |\leqslant ||f||K(\overline{x})$ leads to $I_{t_n}(f)\rightarrow f(\overline{0})$, a contradiction.