TITLE: Limit of convolution
QUESTION [5 upvotes]: I am having trouble calculating limit of a certain integral expression.
In some computation, I need to show that $\|f_n * g\|_{L^p} \to \|g\|_{L^1}$ as $n\to\infty$, where $f_n=(2n)^{-1/p}\chi_{(-n,n)}$ and $g$ is a positive integrable function.
Please help, thank you.

REPLY [2 votes]: It seems like Martin's answer would give a gap of 2. Choosing $n_\epsilon$ such that
$$\|g\|_1-\varepsilon\le\int_{-n_\varepsilon/2}^{n_\varepsilon/2}g\le
\int_{x-n_\varepsilon}^{x+n_\varepsilon}g$$ for $x\in[-n_{\epsilon}/2,n_{\epsilon}/2]$, we have that
$$||f_{n_\epsilon} * g||_p^p=\frac{1}{2 n_\epsilon} \int_{\mathbb{R}}\left(\int_{x-n_\epsilon}^{x+n_\epsilon} g(y) dy\right)^p dx \geq \frac{1}{2 n_\epsilon} \int_{-n_\epsilon/2}^{n_\epsilon/2}\left(\int_{x-n_\epsilon}^{x+n_\epsilon} g(y) dy\right)^p dx
 $$
$$\geq \frac{1}{2 n_\epsilon} \int_{-n_\epsilon/2}^{n_\epsilon/2}\left(||g||_1-\epsilon\right)^p dx = \frac{1}{2} (||g||_1 - \epsilon)^p$$
where the first inequality is by non-negativity, second by the choice of $n_\epsilon$ so that the inequality in Martin's answer holds.