Let $(g_i)$ be a sequence of functions such that $g_k: \mathbb{R}^n \to [0,\infty)$, $\text{supp}\, (g_k) \subset \overline B(0, 1/k)$ and $\int g_k (x) \, dx =1$ for all $k=1, 2, 3, \ldots$. Let $f \in L^1(\mathbb{R}^n)$. Show that $$ \lim_{k \to \infty} \|f - f*g_k\|_1 =0. $$
What I have tried:
We have the following lemma: Let $1≤p≤∞$, $f∈L_1$, $g\in L_p$. Then $f∗g\in L_p$ and $\|f∗g\|_{L_p}≤\|f\|_{L_1}\|g\|_{L_p}$. Because $f\in L_1$ and $g_k\in L_1\; \forall \,k$ we can calculate: $$\lim_{k \to \infty} \|f - f*g_k\|_1 =\lim_{k \to \infty} \int_{\mathbb{R}^n}f(x) -\int_{\mathbb{R}^n}f(y)g_k(x-y)\,dy\,dx $$ $$\leq \lim_{k \to \infty} \int_{\mathbb{R}^n}f(x)-\int_{\mathbb{R}^n}f(y)\,dy\int_{\overline B(0, 1/k)}g_k(x-y)\,dy\,dx$$ $$=\int_{\mathbb{R}^n}f(x)\,dx-\int_{\mathbb{R}^n}f(y)\,dy=0$$ Does this make any sense or is it in the right direction? Any help is appreciated.
The norm of $f$ in $L^{1}$ is not $\int f(x)dx $. It is $\int |f(x)|dx$ and absolute value is absolutely essential!.
First consider the case when $f$ is continuous with compact support. $\|f-f*g_k\|_1=\int \int |f(x)-\int f(x-y)g_k(y) dy|dx \leq \int \int |f(x)-f(x-y)|g_k(y) dydx $. The inside integral can be taken over $\|y\| <\frac 1k$. Also, the integral w.r.t. $x$ can be taken from $-T-1$ to $T+1$ if $f$ has upport in $[-T,T]$. Using uniform continuity of $f$ we see easily that $\|f-f*g_k\|_1 <\epsilon$ provided $k$ is sufficiently large.
For the general case you have to use the fact that $f$ can be approximated in $L^{1}$ norm by continuous functions with compact support. You will also need the inequality $\|f_1*f_2\|_1 \leq \|f_1\|_1 \|f_2\|_1$. This part of the proof is fairly straightforward and I will leave the details to you.