Limit of convolution 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.
 A: By a well known inequality (Young), 
$$
\|f_n*g\|_p\le\|f_n\|_p\|g\|_1,
$$
and
$$
\|f_n\|_p^p=\int_{\Bbb R}((2n)^{-1/p}\chi_{(-n,n)})^p=1.
$$
So, is enough to prove that
$$
\|f_n*g\|_p\ge\|g\|_1-\epsilon
$$
for $n$ large enough.
Now,
$$
\|f_n*g\|_p^p=
{1\over 2n}\int_{\Bbb R}\left(\int_{x-n}^{x+n} g\right)^p\, dx
$$
and the idea is: the interior integral 
$$\int_{x-n}^{x+n} g\approx\|g\|_1$$
when $n$ is large and $x$ is not very far form zero. Namely, if $\varepsilon>0$,
then $\exists n_\varepsilon\in{\Bbb N}$ s.t. for $x\in[-n_\varepsilon,n_\varepsilon]$:
$$
\|g\|_1-\epsilon/2\le\int_{-n_\epsilon}^{n_\epsilon}g
$$
Now, let be $n\ge n_\epsilon$. Then,
$$
n-n_\epsilon\le x\le n-n_\epsilon \implies [-n_\epsilon,n_\epsilon]\subset[x-n,x+n]
$$
and
$$||f_n * g||_p^p=
\frac{1}{2n}\int_{\mathbb{R}}\left(\int_{x-n}^{x+n}g\right)^p dx\ge
\frac{1}{2n}\int_{n_\epsilon-n}^{n-n_\epsilon}\left(\int_{x-n}^{x+n}g\right)^p dx\ge
$$
$$
\frac{1}{2n}\int_{n_\epsilon-n}^{n-n_\epsilon}\left(\int_{-n_\epsilon}^{n_\epsilon}g\right)^p dx\ge
\frac{1}{2n}\int_{n_\epsilon-n}^{n-n_\epsilon}(||g||_1-\epsilon)^p dx =
{n-n_\epsilon\over n}(||g||_1-\epsilon/2)^p.
$$
Now, make $n$ large enough.
EDIT: Interesting consequence: this proves the sharpness of Young's inequality in this case: Equality in Young's inequality for convolution.
A: 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.
