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Assume that $h_\lambda(x)=\frac{1}{\pi} \frac{\lambda}{\lambda^2+x^2}$, for $\lambda>0$, $x \in \mathbb{R}$.

I know that if $f\in L^p$ then $\lim_{\lambda \rightarrow 0} \|f*h_\lambda -f\|_p =0$, for $1\leq p< \infty$ ( Rudin, Real and complex analysis, Thr.9.10).

How to prove that if $f\in L^1$ then $$\lim_{\lambda \rightarrow 0} f*h_\lambda (x)=f(x) \textrm{ a.e. ?}$$

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Doesn't $\lim_{\lambda \rightarrow 0} \|f*h_\lambda -f\|_p =0$ imply $\lim_{\lambda \rightarrow 0} \|f*h_\lambda(x) -f(x)\| =0$ a.e.? –  Alex Becker Mar 18 '12 at 16:14
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Are you sure that $h_\lambda$ is this function? As presented, $\int h_\lambda = \sqrt{2\pi}$, and it's easy to check on an example that $f \star h_\lambda(x) \not\rightarrow f(x)$ ae. (take $f(x) = \mathbf{1}_{[0,1]}$ and plug the limit in a calculator for example). –  Najib Idrissi Mar 18 '12 at 17:03
    
If you rescale $h_\lambda$ st. its integral is 1 however, you can rewrite $f * h_\lambda(x) - f(x) = \int (f(x-y) - f(x))h_\lambda(y) \mathrm{d}y$. It's an useful starting point. –  Najib Idrissi Mar 18 '12 at 17:09
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It is somewhat lengthy but not difficult to prove that the Poisson kernel is majorized by the Hardy-Littlewood maximal function and therefore that (since it concentrates around zero) it differentiates $L^p$. You can either do this by dyadic decomposition or by finding $\varphi$ so that $h_\lambda (x) = \int_0^\infty \varphi(r)\frac{1}{m(B_r (0))} \chi_{B_r (0))}(x) dr$ and then by using Fubini's theorem. Once you have proven majorization you need to show that the result holds for continuous functions. You can find a full writeup of the style of the argument in any text on harmonic analysis. –  Chris Janjigian Mar 18 '12 at 19:38
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@AlexBecker no it does not. You can get convergence in measure along any particular sequence from that condition, but not a.e. unless you know something else about the form of the kernel. In this case, since it satisfies a weak-type maximal inequality you can show that it differentiates $L^p$. –  Chris Janjigian Mar 18 '12 at 19:42

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