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Any ideas on how to get started with this?

Let $\rho \in L^1(\mathbb{R}^N)$ with $\int \rho=1$. Set $\rho_n(x)=n^N \rho(nx)$. Let $f \in L^p(\mathbb{R}^N)$. Show that $\rho_n \star f \to f$ in $L^p(\mathbb{R}^N)$.

The proof of Theorem 4.22 would almost go through for this case. The problem is that I don't know anything about the support of the $\rho_n$'s.This seems to be crucial for the proof of Proposition 4.21 which is used to prove the theorem.

chango
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    Could you give a link to google book's or write down the cited theorems? – martini Jun 29 '12 at 14:34
  • It's Exercise (4.28), on page 127, of the Springer edition. Just found it. – Siminore Jun 29 '12 at 14:39
  • You know that $\int_{\mathbb{R}^n\backslash B_R(0)} \rho $ becomes arbitrily small if $R$ is sufficiently large, that should suffice. I don't think this is true if $p=\infty$. –  Jun 29 '12 at 14:42
  • Yes @Thomas, it is probably a matter of uniform integrability at infinity. Probably an approximation of $\rho$ with a compactly-supported function may also work. – Siminore Jun 29 '12 at 14:54
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    You can prove that $$ \text{support}(\rho_n)=\frac{1}{n}\text{support}(\rho). $$ – HorizonsMaths Jun 29 '12 at 14:53
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    I don't know your approach of proof, but you should note that, unlike in the case of smooth $\rho$ with compact support, you will only get convergence in $L^p$, $1 \le p < \infty$. In particular, as pointed out by an example in Stein's book on Harmonic analysis (Chapter II, § 5.16), $\lim_{n\rightarrow\infty} \rho_n \star f(x)$ may fail to exist for almost every $x$. –  Jun 29 '12 at 15:45

2 Answers2

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Ok, so from what I gather I should proceed in the following way:

Define $$ \bar \rho = \frac{I_{B(R)} \: \rho}{\|I_{B(R)} \: \rho\|_1}, $$ where $R>0$ is such that $\int_{\mathbb{R}^N \setminus B(R)} \rho < \epsilon$. Then I write

$$ \| \rho_n \star f - \bar \rho_n \star f + \bar \rho_n \star f -f \|_p \leq \| \rho_n \star f - \bar \rho_n \star f \|_p + \| \bar \rho_n \star f -f \|_p $$ $$ \leq \| \rho_n - \bar \rho_n\|_1 \|f\|_p + \| \bar \rho_n \star f -f \|_p $$

et cetera

chango
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We can use the following trick Brezis uses in his proof of the Kolmogorov-Riesz-Fréchet Theorem (Theorem 4.26 in the textbook). For a.e. $x$, using Hölder's inequality we have \begin{align}|\rho_{n}\star f(x)-f(x)| &\leq \int |f(x-y)-f(x)||\rho_{n}(y)|\:\text{d}y\\ &\leq \Big(\int |f(x-y)-f(x)|^p|\rho_{n}(y)|\:\text{d}y\Big)^{1/p}\|\rho\|_{1}^{1/p'}.\end{align} Thus, \begin{align}\int |\rho_{n}\star f(x)-f(x)|^p\:\text{d}x &\leq \|\rho\|_{1}^{p/p'}\int\int |f(x-y)-f(x)|^p|\rho_{n}(y)|\:\text{d}y\:\text{d}x \\ &=\|\rho\|_{1}^{p/p'}\int |\rho_{n}(y)|\int |f(x-y)-f(x)|^p\:\text{d}x\:\text{d}y \\ &= \|\rho\|_{1}^{p/p'}\int |\rho(z)|\int |f(x-z/n)-f(x)|^p\:\text{d}x\:\text{d}z \\ &= \|\rho\|_{1}^{p/p'}\int |\rho(z)| \|\tau_{z/n}f-f\|_{p}^{p}\:\text{d}z.\end{align} The sequence of integrands converges pointwise a.e. to $0$ as $n\rightarrow\infty$ and is dominated by $2^{p+1}\|f\|_{p}^{p}|\rho(z)|\in L^{1}(\mathbb{R}^{n})$. Thus, we may use the dominated convergence theorem to conclude.

Karthik Kannan
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