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.
No, local integrability is not enough for pointwise convergence. Take $f=\chi_{[0,1]}$, then $f_\epsilon\to 1/2$ at $0$ and at $1$.
Continuity of $f$ is enough. Indeed, when $\epsilon$ is small, all the values of $f$ involved in the definition of $f_\epsilon(x)$ come from a small neighborhood of $x$. By virtue of continuity, they all lie close to $f(x)$. Hence $f_\epsilon(x)$, which is just a weighted average of these values, lies close to $f(x)$.
By the way, for continuous $f$ the convergence $f_\epsilon\to f$ is also locally uniform (because of uniform continuity bound on a compact set).
The pointwise convergence $f_\epsilon\to f$ pointwise also holds for Functions with every point being a Lebesgue point, which need not be continuous. (This time, convergence will not be locally uniform in general.) But this is kind of an edge case. When a function is not continuous, there is no good reason to expect this property to hold.
If you only have local integrability, that is enough for almost everywhere convergence: Why convolution of integrable function $f$ with some sequence tends to $f$ a.e.