Pointwise convergenve of mollified $f\in L^1_{loc}$

Question

Let $\Omega\subseteq\mathbb{R}^n$ open, $f\in L^1_{loc}(\Omega)$, $\eta_\epsilon(x) = \dfrac{1}{\epsilon^n}\eta(\dfrac{x}{\epsilon})$ the usual scaled mollifier, i.e. $supp (\eta_\epsilon) \subseteq B_\epsilon(0), \int_{\mathbb{R}^n} \eta_\epsilon = 1, \eta_\epsilon \ge 0$.
Then $f*\eta_\epsilon\to f$ almost everywhere as $\epsilon\to 0$.

So, starting out I noticed that by definition $|f*\eta_\epsilon(x)-f(x)|\le \int_{B_\epsilon(x)}\eta_\epsilon(x-y)|f(y)-f(x)| dy \le C\int_{B_\epsilon(x)}|f(y)-f(x)|$ where $C=\sup\eta_\epsilon$ and I assume that $\epsilon$ is small enough that $f$ does not cause any problems, that is that the integral is well defined. Now intuitively it makes sense, that the "average difference" over a ball should go to 0 if the radius goes to 0, but I need help to formally justify this.

Answer 1

Your term $C$ is dependent on $n$. It depends on the choice of mollifier but usually it has the form $C_n \epsilon^{-n}$. This gives you $$|f \ast \eta_\epsilon(x) - f(x)| \le C \epsilon^{-n} \int_{B_\epsilon(x)} |f(y) - f(x)| \, dy.$$

Now apply the Lebesgue Differentiation Theorem: if $f \in L^1_{\rm loc}(\mathbb R^n)$, then $$\lim_{\epsilon \to 0^+} \epsilon^{-n} \int_{B_\epsilon(x)} |f(y) - f(x)| \, dy \to 0$$ for almost all $x$.

Answer 2

Observe that $C$ depends on $\epsilon$ and goes to $\infty$ as $\epsilon\to0$. The proof is done in 3 steps:

1. Prove it for $f$ continuous with compact support (using uniform continuity.)
2. Using a density argument, prove it for $f\in L^1$.
3. Extend the result to $f\in L_{\text{loc}}^1$.

Answer 3

Have you heard about "Good kernels"? For more information it is better search at the net two famous theorems " Approximation of identity". It is valuable that to know about "Poisson kernels" and "Gauss-Weierstrass".