Let $\theta(x)$ equal $k\exp(-\frac{1}{1-||x||} )$ if $||x||<1$, and equal 0 if $||x||\geq1.$ Here $||.||$ designates the Euclidian norm in $\mathbb{R}^{^{n}}$, and the constant $k$ is chosen such that $\int_{\mathbb{R}^{n}}\theta(t)d\lambda(t)=1$. If we set $$\theta_{\varepsilon}(x)=\frac{1}{\varepsilon^{N}}\theta(\frac{x}{\varepsilon}),$$ we obtain a $\mathcal{C}^{\infty}$ function whose support is the Euclidian ball $B(x,\varepsilon)$, for all $\varepsilon>0$. Then we can define by convolution $$f_{\varepsilon}(x):= (f\ast \theta_{\varepsilon})(x).$$
We know that if $f$ is locally integrable in $\mathbb{R}^{n}$, then $f_{\varepsilon}$ is infinitely differentiable. My question is: under which condition(s), do we have $f_{\varepsilon}(x)\rightarrow f(x)$ (pointwise), as $\varepsilon$ $\rightarrow0$? Is (localy) integrability enough? Do you know a readable book on this specific subject? Thanks for your reply.