A sequence of mollifiers $(\rho_n)_{n \geq 1}$ is any sequence of functions on $\mathbb{R}^d$ such that $$ \rho_n \in C_c^{\infty} (\mathbb{R}^d), \quad \operatorname{supp} \rho_n \subset \overline{B(0,1 / n)}, \quad \int \rho_n=1, \quad \rho_n \geq 0 \text { on } \mathbb{R}^d. $$
Then we have a result from Brezis's Functional Analysis, i.e.,
Theorem 4.22. Assume $f \in L^p(\mathbb{R}^d)$ with $1 \leq p<\infty$. Then $(f*\rho_n) \xrightarrow[]{n \to \infty} f$ in $L^p(\mathbb{R}^d)$.
Now I would like to prove below result, i.e.,
Theorem If $f \in L^1 (\mathbb R^d)$, then $$ \lim_{h \to 0} \int |f(x+h)-f(x)| d x = 0. $$
- Could you confirm if my below attempt is fine?
- Is there a shorter approach without using mollifiers?
Proof Let $(\rho_n)$ be a sequence of mollifiers. We have $$ \begin{align} &\int |f(x+h)-f(x)| d x \\ \le &\int |f(x+h)- (f * \rho_n) (x+h)| d x + \int |(f * \rho_n) (x+h) - f(x)| d x \\ = &\int |f(x)- (f * \rho_n) (x)| d x + \int |(f * \rho_n) (x+h) - f(x)| d x. \end{align} $$
Let $K := \operatorname{supp} \rho_n$. Let $\lambda$ be the Lebesgue measure on $\mathbb R^d$. Then $$ \begin{align} &|(f * \rho_n) (x+h)| \\ \le{} &\int_K |f(x+h-y)| \rho_n (y) d y \\ \le{} &\|\rho_n\|_{L^\infty} \int_K |f(x+h-y)| d y \quad \text{by Hölder's inequality} \\ \le{} &\|\rho_n\|_{L^\infty} \int_K |f(x+y)| d y \quad \text{by translation-invariance of } \lambda. \end{align} $$
We define $g:\mathbb R^d \to \mathbb R$ by $g(x) := \|\rho_n\|_{L^\infty} \int_K |f(x+y)| d y$. Then $$ \begin{align} \|g\|_{L^1} &= \|\rho_n\|_{L^\infty} \int_{\mathbb R^d} \int_K |f(x+y)| d y dx \\ &= \|\rho_n\|_{L^\infty} \int_K \int_{\mathbb R^d} |f(x+y)| d x dy \quad \text{by Tonelli's theorem}\\ &= \|\rho_n\|_{L^\infty} \int_K \|f\|_{L^1} dy \\ &= \lambda(K) \|\rho_n\|_{L^\infty} \|f\|_{L^1} <\infty. \end{align} $$
Hence $g \in L^1(\mathbb R^d)$. Notice that $f * \rho_n$ is smooth, so $$ (f * \rho_n) (x+h) - f(x) \xrightarrow[]{h \to 0} (f * \rho_n) (x) - f(x) \quad \forall x \in \mathbb R^d. $$
On the other hand $$ |(f * \rho_n) (x+h) - f(x)| \le |g(x)| + |f(x)|. $$
Clearly, $g+f \in L^1(\mathbb R^d)$. By dominated convergence theorem, $$ \lim_{h\to 0} \int |(f * \rho_n) (x+h) - f(x)| d x = \int |(f * \rho_n) (x) - f(x)| d x. $$
Hence $$ \lim_{h\to 0} \int |f(x+h)-f(x)| d x \le 2\int |(f * \rho_n) (x) - f(x)| d x. $$
The claim then follows by Theorem 4.22. and by taking the limit $n \to \infty$.
Update I have found a simpler approach. Still we need a result from Brezis's Functional Analysis, i.e.,
Corollary 4.23. Let $\Omega$ be an open subset of $\mathbb R^d$. Then $\mathcal C_c^\infty (\Omega)$ is dense in $L^p(\Omega)$ for $1 \leq p<\infty$.
Fix $\varepsilon >0$. By Corollary 4.23., there is $g \in \mathcal C_c^\infty (\mathbb R^d)$ such that $\|f-g\|_{L^1} < \varepsilon$. By dominated convergence theorem, $$ \lim_{h \to 0} \int |g(x+h)-g(x)| d x = 0. $$
We have $$ \begin{align} & \int |f(x+h)-f(x)| d x \\ \le{} & \int |f(x+h)-g(x+h)| d x + \int |g(x+h)-g(x)| d x + \int |g(x)-f(x)| d x \\ ={} & 2 \|f-g\|_{L^1} + \int |g(x+h)-g(x)| d x \quad \text{by translation-invariance of } \lambda. \end{align} $$
Hence $$ \lim_{h \to 0} \int |f(x+h)-f(x)| d x \le 2 \varepsilon. $$
The claim then follows by taking the limit $\varepsilon \downarrow 0$.