Convergence of $f_n(x) = 2^n \cdot F(2^n (x-a_n))$ with $F(x) = e^{-x^2}$ with different notions of convergence. I had my measure theory exam this morning, and one exercise was the following: I really can't see a solution. During the semester, we talked about almost everywhere convergence, almost uniform convergence, convergence in measure an convergence in $L^p$. 
The problem is as follows:
Let be $(a_n)_n$ be a real sequence and let $F(x)=e^{-x^2}$. Define
$$f_n(x)=2^nF(2^n(x-a_n))$$
for all $x\in \Bbb{R}$ and all $n \in \Bbb{N}$.
Show if and in which meaning of convergence $F_n(x)$ converges to (some) $f$ and in that case if $$\int f=\lim_{n \to +\infty} \int f_n.$$
This is my first post, I hope I did not miss anything.
My attempt:
I noticed that $$\int f_n=\sqrt{\pi}$$ (it's quite trivial to prove this, just some substitution) and so I think that $f_n$ can converge to $F$ in $L^p$, the proof will be easy if I find a way to show that $f_n<F$ or $F<f_n$ but I can't show that (if it's possible).
 A: As you observed yourself, your sequence $(F_n)_n$ is $L^1$-normalized. But it turns out that it is quite useful to consider other $L^p$-(quasi)-norms.
We have
\begin{eqnarray*}
\int\left|F_{n}\right|^{p}\,{\rm d}x & = & 2^{np}\cdot2^{-n}\cdot\int2^{n}\cdot\left|F\left(2^{n}\left(x-a_{n}\right)\right)\right|^{p}\,{\rm d}x\\
 & \overset{y=2^{n}\left(x-a_{n}\right)}{=} & 2^{n\left(p-1\right)}\cdot\int\left|F\left(y\right)\right|^{p}\,{\rm d}y,
\end{eqnarray*}
where
\begin{eqnarray*}
\int\left|F\left(y\right)\right|^{p}\,{\rm d}y & = & \int e^{-p\cdot y^{2}}\,{\rm d}y\\
 & = & \frac{1}{\sqrt{p}}\cdot\int\sqrt{p}e^{-\left(\sqrt{p}y\right)^{2}}\,{\rm d}y\\
 & \overset{z=\sqrt{p}y}{=} & \frac{1}{\sqrt{p}}\cdot\int e^{-z^{2}}\,{\rm d}z\\
 & = & \sqrt{\frac{\pi}{p}}.
\end{eqnarray*}
Hence, we conclude
\begin{eqnarray*}
\int\sum_{n=1}^{\infty}\left|F_{n}\right|^{p}\,{\rm d}x & = & \sqrt{\frac{\pi}{p}}\cdot\sum_{n=1}^{\infty}2^{n\left(p-1\right)}\\
 & = & \begin{cases}
\infty, & \text{if }p\geq1,\\
<\infty, & \text{if }p<1.
\end{cases}
\end{eqnarray*}
In particular, this implies
$$
\sum_{n=1}^{\infty}\left|F_{n}\left(x\right)\right|^{p}<\infty
$$
for almost every $x\in\mathbb{R}$ and hence $\left|F_{n}\left(x\right)\right|^{p}\xrightarrow[n\to\infty]{}0$,
i.e. $F_{n}\left(x\right)\xrightarrow[n\to\infty]{}0$ almost everywhere. This is one of the cool instances, where one can get (a.e.) pointwise convergence by integration, although it is really hard to show $F_n \to 0$ for any $x$ "by hand".
Also, we have shown
$$
\int\left|F_{n}\left(x\right)\right|^{p}\,{\rm d}x\xrightarrow[n\to\infty]{}0\text{ for }p<1,
$$
which implies also that $F_{n}\to0$ in measure.
The above calculation also answers the question wether $\left\Vert F_{n}-f\right\Vert _{p}\to0$
(we necessarily have $f\equiv0$, because of the pointwise a.e. convergence),
namely
\begin{eqnarray*}
\left\Vert F_{n}-f\right\Vert _{p} & = & \left[2^{n\left(p-1\right)}\cdot\sqrt{\frac{\pi}{p}}\right]^{1/p}\\
 & \xrightarrow[n\to\infty]{} & \begin{cases}
\infty & \text{if }p>1,\\
\sqrt{\pi}, & \text{if }p=1,\\
0, & \text{if }p<1.
\end{cases}
\end{eqnarray*}
