Show that $\displaystyle\int_{\mathbb{R}} \lim \inf _{\epsilon \to 0} |f_\epsilon| \leq \displaystyle\int_{\mathbb{R}} |f|$

Question

Let $f \in L^1 (\mathbb{R})$ and $\varphi_\epsilon$ be a mollifier, that is $\varphi_\epsilon = \epsilon^{-1} \varphi (\frac{x}{\epsilon})$, $\varphi: \mathbb{R} \to \mathbb{R}$ is a function satisfying $\varphi \geq 0$, and $\varphi$ has a compact support and $\int \varphi =1$. Let $f_\epsilon = f \ast\varphi_\epsilon$ be the convolution. Show that $\displaystyle\int_{\mathbb{R}} \displaystyle\lim \inf_{\epsilon \to 0} |f_\epsilon| \leq \int_{\mathbb{R}} |f|$

My solution

Let's consider $f_n = |f * \varphi_{1/n}|$. Since the mollifier $\varphi_\epsilon$ is such that $\varphi_\epsilon \geq 0$, we have $f_n \geq 0$ for all $n$. Then we can apply Fatou's lemma:

$$\int_{\mathbb{R}} \lim \inf_{n\to \infty} f_n \leq \lim \inf_{n\to \infty} \int_{\mathbb{R}} f_n$$

We know that for $f \in L^1 (\mathbb{R})$, the mollifier has the properties that $\lim_{\epsilon \to 0} f * \varphi_\epsilon = f$ a.e. and $\lim_{\epsilon \to 0} \int_{\mathbb{R}} |f * \varphi_\epsilon - f| = 0$.

Therefore, $\lim_{n \to \infty} \int_{\mathbb{R}} |f * \varphi_{1/n}| = \int_{\mathbb{R}} |f|$. Substituting this into the inequality gives the desired result. Did I miss somthing in the proof.

Answer 1

If you already know that $\lim_{\epsilon\rightarrow 0}f_\epsilon=f$ a.e., then you have $\liminf_{\epsilon\rightarrow 0}\vert f_\epsilon\vert=\vert f\vert $ a.e. so that $\int_\mathbb{R} \liminf_{\epsilon\rightarrow 0}\vert f_\epsilon\vert \, dx=\int_\mathbb{R} \vert f\vert \, dx$ and the question is obvious.

If you don't want to use this result, then you actually use Fatou's lemma and write $\int_\mathbb{R} \liminf_{\epsilon\rightarrow 0}\vert f_\epsilon\vert \, dx \leq \liminf_{\epsilon\rightarrow 0} \int_\mathbb{R}\vert f_\epsilon\vert \, dx$. As $\vert f_\epsilon\vert\leq \vert f\vert*\varphi_\epsilon$, we have $\int_\mathbb{R}\vert f_\epsilon\vert \, dx\leq \int_\mathbb{R}\vert f\vert*\varphi_\epsilon\vert \, dx=\int_\mathbb{R}\vert f\vert \, dx$, so that $\liminf_{\epsilon\rightarrow 0}\int_\mathbb{R}\vert f_\epsilon\vert \, dx\leq \int_\mathbb{R}\vert f\vert \, dx$.