4
$\begingroup$

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.

  • 2
    Could you give a link to google book's or write down the cited theorems?2012-06-29
  • 0
    It's Exercise (4.28), on page 127, of the Springer edition. Just found it.2012-06-29
  • 0
    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 $p=\infty$.2012-06-29
  • 0
    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.2012-06-29
  • 1
    You can prove that $$ \text{support}(\rho_n)=\frac{1}{n}\text{support}(\rho). $$2012-06-29
  • 1
    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 $x$.2012-06-29

1 Answers 1