# Convolution with a special approximation to the identity function

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.

-
Integrating over $\mathbb{R}^d$ is the same as integration over $(-\infty,\infty)=\mathbb{R}$ if and only if $d=1$. If the domain of integration confuses you, you should familiarize yourself with integration in $\mathbb{R}^n$ first. –  Your Ad Here Oct 16 '12 at 18:15
Ohh okay duh. Thanks for clarifying that. It was not conceptual confusion (at least not to the extent you thought) but notational confusion, since usually the normalization was done over $R^d$. So I could do $\int_{R^d}fdx(\int_{-\infty}^{\infty}\int_{R^{d-1}}K_\delta (x)dxdy)$ and use Fubini? –  ryjm Oct 16 '12 at 19:08
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 $K_\delta * f \to 0$! –  ryjm Oct 16 '12 at 22:34