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I'm working my way through Stein and Shakarchi's Real Analysis, and I'm having some trouble figuring this exercise out.

Given the function $K_\delta$ that satisfies the normal approximation to the identity properties

$|K_\delta(x)| \leq A\delta^{−d}$ for all $\delta > 0$

$|K_\delta(x)| \leq A\delta/|x|^{d+1}$ for all $\delta > 0$

but

$\int_{-\infty}^\infty{K_\delta(x)dx}= 0$ for all $\delta > 0$

prove that if $f$ is integrable on $R^d$ then $(f*K_\delta)(x) \to 0$ as $\delta \to 0$ for a.e $x$.

Is it enough to use $\int_{R^d}(f*K_\delta)(x)=\int_{R^d}f\int_{R^d}K_\delta$? The domain of integration is confusing me - integrating over $R^d$ the same as integrating over $-\infty$ to $\infty$ in this case, correct? Then $\int_{R^d}(f*K_\delta)(x)=0$. Am I missing something?

Thanks for the help.

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    Okay I thought about it a bit, and I think I found an easy way. All I have to do is take any other approximation to the identity function $J_\delta$ and define a new function $G_\delta=J_\delta + K_\delta$. Then it's clear that $G_\delta$ satisfies all the conditions for a normal approximation to the identity function, so $G_\delta * f \to f$. But $J_\delta * f \to f$ by construction, and since $K_\delta * f = (G_\delta * f) - (J_\delta * f)$ we must have that $K_\delta * f \to 0$!2012-10-16

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