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

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

• hint : you should write out $\|f * g_j - f\|_1$ as an integral and use what you wrote about the operator $\tau_a$. Commented Apr 25 at 11:27

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*}