Convergence of the difference of the convolution of sequence of functions and a function

Question

Let $(g_j)_{j \in \mathbb{N}} \subseteq \mathscr{L}_1(\mathbb{R}^n)$ with $\|g_j\| = 1$ and $g_j \geq 0$ for all $j \in \mathbb{N}$. Suppose that $\lim_{j \to \infty} d_j = 0$ where $d_j := \sup\{\|x\| \mid x \in \operatorname{supp}(g_j)\}$. I want to prove that for all $f \in \mathscr{L}_1(\mathbb{R}^n)$ it is true that $$ \lim_{j \to \infty} \|f * g_j - f\|_1 = 0. $$

So far I've proved that for all $f \in \mathscr{L}_1(\mathbb{R}^n)$ and all $\epsilon > 0$ there exist some $\delta > 0$ such that $\|f - \tau_a f\|_1 < \epsilon$ for all $a \in \mathbb{R}^n$ with $\|a\| < \delta$ where $\tau_a f(x) = f(x - a)$. It is not clear to me the path to follow. I think I can use the distributivity of the convolution to product to obtain \begin{align*} \|f * g_j - f\|_1 &= \|f * g_j - h * g_j\|_1 + \|h * g_j - f\|_1\\ &\leq \|(f - h)* g_j\|_1 + \|h * g_j - f\|_1\\ &\leq \|(f - h)\|_1 \|g_j\|_1 + \|h * g_j - f\|_1 \end{align*}

for some suitable function $h \in \mathscr{L}_1(\mathbb{R}^n)$.

Answer 1

Let $\epsilon > 0$ then there exists $\delta>0$ s.t. $\Vert \tau_af-f\Vert_1<\epsilon$ for all $\Vert a \Vert < \delta$. Furthermore, there exists $N\in\mathbb{N}$ such that if $j\geq N$ then $d_j < \delta$, then supp $g_j \subset B_{\delta}(0)$. Also, $f(x) = \int_{\mathbb{R}^n}f(x)g_j(y)d\beta^n(y)$.

With this we can just write the definition of $\Vert f \ast g_j - f\Vert_1$

\begin{align*} \int_{\mathbb{R}^n}\Big|\int_{\mathbb{R}^n}(f(x-y)-f(x))g_j(y)d\beta^n(y)\Big|d\beta^n(x) &\leq \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\Big|(f(x-y)-f(x))g_j(y)\Big|d\beta^n(y)d\beta^n(x)\\ &= \cdots\\ &= \int_{B_\delta(0)}g_j(y) \int_{\mathbb{R}^n}\Big|(f(x-y)-f(x))\Big|d\beta^n(x)d\beta^n(y)\\ &< \epsilon \end{align*}