# Given $f\in L^1(\mathbb{R})$ with $\|f\|_1<\infty$ and $g_n=\sqrt{n/2\pi}e^{-nx^2/2},f_n=g_n\ast f$, show that $\lim\|f_n-f\|_1=0$

Given $f$ a Lebesgue integrable function on $\mathbb{R}$ with finite $L^1$-norm, I am asked to show that $\lim_{n\to\infty} \|f_n - f\|_1 = 0$, where $f_n = f \ast g_n$ and $g_n = \sqrt{\frac{n}{2\pi}}e^{-nx^2/2}$.

A first thought was to use that $\int_\mathbb{R}\int_\mathbb{R} f\ast g = (\int_\mathbb{R} f)(\int_\mathbb{R} g)$, but am unable to manipulate the integrand into a form where that property of convolution is useful:

\begin{align} \|f_n - f\|_1 &= \int_\mathbb{R} |f_n(x) - f(x)| \, dx \\ &= \int_\mathbb{R} \bigg| \left( \int_\mathbb{R} g_n(x - y) f(y) \, dy \right) - f(x) \bigg| \, dx \end{align}

I do know that for each $n$, $\|g_n\|_1 = 1$ (it is a normalized Gaussian integral). On the other hand, $\lim_{n\to\infty} g_n(x) = 0$ for all $x \in \mathbb{R} \setminus \{0\}$; since both $f$ and $g_n$ are integrable, the inner integral on the last line above would go to zero by the Lebesgue dominated convergence theorem, wouldn't it? But that would give $\lim\|f-f_n\|_1 = \|f\|_1$.

It feels as though I am approaching this problem from the wrong direction and am unsure of how to proceed.

• When you know what your result is called, it is usually not so hard to find a proof of it. Unfortunately this result itself does not have a standard name, but it is associated to two important terms: "approximate identity" and "mollifier". Knowing that, I googled and found numerous results, including andromeda.rutgers.edu/~loftin/ra1fal10/mollifier.pdf. These notes essentially prove a generalization of your result as Theorem 3 (stated on page 5). – Ian Apr 21 '15 at 4:18
• All that is left to do after their result is to show that if $f_n$ converges uniformly on compact subsets to $f \in L^1$ then $f_n$ converges in the sense of $L^1$ to $f$. But this is not too difficult; you find compact $K$ so that $\| f_n\|_{L^1(K^c)}$ is small independent of $n$ (Royden and Fitzpatrick call the ability to do this tightness) and then you take $n$ large enough that $\| f_n - f \|_K$ is small. – Ian Apr 21 '15 at 4:23
• But $g_n$ is not compactly supported and hence it is not a mollifier, as per the definition in your link. – user233107 Apr 21 '15 at 4:47
• You're right. I think that just means you need an additional approximation: $\| f_n - f \| = \| f_n - f_{n,m} + f_{n,m} - f \| \leq \| f_n - f_{n,m} \| + \| f_{n,m} - f \|$, where $f_{n,m}$ is supported in $[-m,m]$. – Ian Apr 21 '15 at 5:08
• Then we lose that $\int_\mathbb{R} g_n(x) \, dx = 1$, i.e., $\int_{-m}^m g_n(x) \, dx \neq 1$ for all $m \in \mathbb{R}$. – user233107 Apr 21 '15 at 5:12