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?
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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 $\bar{0}$.
Danny Pak-Keung Chan
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Yes, of course, $f$ is supposed to be continuous, I forgot to mention that. 'edited'. – Romanda de Gore May 17 '18 at 23:13
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But that does not help, does it. Okay, got the point. Must have misinterpreted something. – Romanda de Gore May 17 '18 at 23:16
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But of course, I would like to know why dominated convergence does not apply. If $g_n = f(\dfrac{\overline{x}}{n})K$, then, $g_n \rightarrow f(\overline{0})K$ and $|g_n(\overline{x})|\leqslant ||f||K(\overline{x})$, which I thought sufficient. – Romanda de Gore May 17 '18 at 23:40
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But of course, the continuity requirement saves the argument, because if two continuous functions differ at one point, then they differ in an open region of that point, hence cannot be equal $\lambda^d$- a.e. – Romanda de Gore May 18 '18 at 10:08
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That is correct. Therefore, I stated at the beginning that some extra conditions must be imposted on $f$, or otherwise the statement is false. – Danny Pak-Keung Chan May 18 '18 at 14:24
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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.