TITLE: Convergence in $L^1$-norm Involving Convolution
QUESTION [2 upvotes]: Let $g_k(x) = \frac{1}{k^n}g(x/k)$, where $g$ is Lebesgue integrable on $\mathbb{R}^n$, zero outside of some ball with finite radius, and $\int g = 1$. I am trying to show that
$$\lim_{k \to 0} ||f - f \ast g_k||_1 = 0,$$
for all $f \in L^1(\mathbb{R}^n)$ and $k > 0$.

Thus far: We have the convolution
$$ (f \ast g_k)(x) = \frac{1}{k^d}\int_{\mathbb{R}^n}f(x - t)g(t/k)dt.$$
So
$$
\begin{align}
||f - f \ast g_k||_1 &= \int_{\mathbb{R}^n}\left|f(x) - (f\ast g_k)(x)\right|dx \\
&= \int_{\mathbb{R}^n}\left|f(x) - \frac{1}{k^n}\int_{\mathbb{R}^n}f(x-t)g(t/k)dt\right|dx \\
&= \int_{\mathbb{R}^n}\left|\frac{f(x)}{k^n}\int_{\mathbb{R}^n}g(t/k) - \frac{1}{k^n}\int_{\mathbb{R}^n}f(x-t)g(t/k)\right|dx \tag{$\int g = 1$} \\
&\leq \frac{1}{k^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\left|f(x) - f(x-t)\right|g(t/k)dtdx
\end{align}
$$
I'm not sure exactly how to proceed in detail, but I have a rough sketch: We know $g$ is zero outside of some ball, so introduce characteristic/indicator function which uses this fact? Apply Tonelli's to rearrange to get something that looks like:
$$\int ||f(x) - f(x-t)||_1. \tag{1}$$ Then we can use the fact that  a function in $L^1$ space has the property, $(1) \to 0$ as $t \to 0$. Is this line of thinking correct?

REPLY [1 votes]: Hints: First make the substituion $s=\frac  1 k t$. Note that the integral w.r.t $s$ can be taken over $kC$ where $C $ is the support of $g$. Interchange the order of integration and use the following well known fact:
If $f$ is integrable then $\int_{\mathbb R^{n}} |f(x)-f(x-y)|dx \to 0$ as $ y \to 0$. 
[Rudin's RCA has a proof of this result].
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