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Let $u \in L^\infty (\mathbb R^d)$. Let $(\rho_n)$ be a sequence of mollifiers, i.e., $$ \rho_n \in C_c^{\infty} (\mathbb{R}^d), \quad \operatorname{supp} \rho_n \subset \overline{B(0,1 / n)}, \quad \int \rho_n=1, \quad \rho_n \geq 0 \text { on } \mathbb{R}^d. $$

Let $\xi:\mathbb R^d \to \mathbb R$ be measurable and $(\xi_n) \subset L^\infty (\mathbb R^d)$ such that $$ \sup_n \|\xi_n\|_\infty \le 1 \quad \text{and} \quad \xi_n \xrightarrow{n\to\infty} \xi \text{ a.e. on } \mathbb R^d. $$

Let $$ v_n := (\xi_n u) * \rho_n \quad \text{and} \quad v := \xi u. $$

I would like to prove that

  1. $v_n \xrightarrow{n\to\infty} v$ in $L^\infty (\mathbb R^d)$ w.r.t. the weak$^*$ topology $\sigma(L^\infty, L^1)$.
  2. $\int_B |v_n -v| \xrightarrow{n\to\infty} 0$ for very ball $B$.

Could you confirm if my attempt on (1.) is fine?

Clearly, both $\xi_n u$ and $\xi u$ belong to $L^\infty (\mathbb R^d)$. By Young's inequality, $v_n \in L^\infty (\mathbb R^d)$. For $f:\mathbb R^d \to \mathbb R$, we set $\check f (x) := f(-x)$.

  1. Fix $f \in L^1 (\mathbb R^d)$. It suffices to prove $\int f v_n \xrightarrow{n\to\infty} \int fv$. First, $$ \int f v_n = \int f ((\xi_n u) * \rho_n) = \int \xi_n u (f * \check \rho_n). $$

So we need to prove $\int \xi_n u (f * \check \rho_n) \xrightarrow{n\to\infty} \int \xi uf$. Clearly, $(\check \rho_n)$ is a sequence of mollifiers, so $f * \check \rho_n \xrightarrow{n\to\infty} f$ in $L^1 (\mathbb R^d)$. Let $R := \sup_n \|\xi_n u\|_\infty$. Because $\sup_n \|\xi_n\|_\infty \le 1$, we get $R<\infty$. Clearly,

  • $\| R(f * \check \rho_n) -Rf\|_1 \xrightarrow{n\to\infty} 0$.
  • each sub-sequence of $(\xi_n u (f * \check \rho_n))_n$ has a further sub-sequence that converges a.e. to $\xi u f$. This is due to $\xi_n \xrightarrow{n\to\infty} \xi$ a.e.
  • $|\xi_n u (f * \check \rho_n)| \le |R (f * \check \rho_n)|$ a.e. for all $n$.

It follows from below Lemma that $\|\xi_n u (f * \check \rho_n)-\xi u f\|_1 \xrightarrow{n\to\infty} 0$. Thus $\|fv_n - fv\|_1 \xrightarrow{n\to\infty} 0$.

Lemma Let $(X, \mathcal A, \mu)$ be a $\sigma$-finite measure space and $(E, |\cdot|)$ a Banach space. Let $p, q \in [1, \infty)$. Let $f,f_n \in \mathcal L_p (X, \mu, E)$ and $g,g_n \in \mathcal L_q (X, \mu, E)$. We assume that

  • each sub-sequence of $(f_n)$ has a further sub-sequence that converges a.e. to $f$,
  • $|f_n|^p \le |g_n|^q$ a.e. for all $n$, and $\|g_n - g\|_q \xrightarrow{n \to \infty} 0$.
  1. Let $B$ be a ball in $\mathbb R^d$ and $f := 1_B$. Then $f \in L^1 (\mathbb R^d)$. As we have proved in (1.), $\|fv_n - fv\|_1 \xrightarrow{n\to\infty} 0$. The claim then follows.

Update I have found another simpler proof of (1.). We have $$ \begin{align} \int |f v_n - fv| &= \int |\xi_n u (f * \check \rho_n - f) + \xi_n uf-\xi uf| \\ &\le \underbrace{\int |\xi_n u (f * \check \rho_n - f)|}_{A_n} + \underbrace{\int |\xi_n-\xi| |uf|}_{B_n}. \end{align} $$

We have $$ A_n \le \|u\|_\infty \|f * \check \rho_n - f\|_1 \xrightarrow{n\to\infty} 0. $$

We have $\sup_n \|\xi_n\|_\infty \le 1$ and $\xi_n \xrightarrow{n\to\infty} \xi$ a.e. on $\mathbb R^d$ imply $\|\xi\|_\infty \le 1$. So $|\xi_n-\xi| |uf| \le 2 \|u\|_\infty |f|$. By dominated convergence theorem, $B_n \xrightarrow{n\to\infty} 0$. Hence $$ \int |f v_n - fv| \xrightarrow{n\to\infty} 0. $$

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