Proving that a set is closed in $L^2(\mathbb{R})$ this is my first question here so i hope i don't do anything wrong. Excuse any spelling or grammar mistakes, english isn't my mother tongue.
I'm reading this paper for my bachelor thesis and have problems with corrolary 3.3 on page 8. I tried to solve this problem for some time, asked on a different forum for help but the only advice i received was to contact the authors( it was a german forum and i marked the question as answered there, because no one was able to help me, in case there are rules against cross-postings here). I did that and got a very quick response which was something of a surprise for me. The reply consisted in the promise, that he would think about this problem and get back to me. The same thing sad my professor, who gave me the paper after i asked him today.
First i will summarize everything which is needed to understand what's going on.
For $f\in L^2$ with $\lVert f\rVert =1$ and $p>1$ define $\varphi_f:\ \mathbb{R}\rightarrow \mathbb R\ ;\ a\mapsto \int|t-a|^p|f(t)|^2\mathrm dt$, this Integral and all others without specified domain of integration are taken over $\mathbb R$ or the complement of a set of measure zero. The following properties hold for $\varphi_f$:
if there is one $a_0\in\mathbb R$ with $\varphi_f(a_0)<\infty$, then 
i) $  \varphi_f(a)<\infty$ for every $a\in\mathbb R$
ii)$\varphi_f$ is stricly convex, continuous and satisfies $\lim\limits_{|a|\to\infty}\varphi_f=\infty$
iii)$\varphi_f$ has a unique minimum $\Delta_p^2(f)$ at the point $\mu_p(f)$ ( with $\Delta_p(f)\geq 0$).
Now we can define the following set for $A>0$ and $p,q>1$:
$K=\{f\in L^2:\ \|f\|=1,|\Delta_p(f)|\leq A, |\mu_p(f)|\leq A,|\Delta_q(\hat f)|\leq A,|\mu_q(\hat f)|\leq A\}$, 
where $\hat f$ is the Fourier transform of f. The authors claim without further justification that this set is closed and i wanted to proof this. Note that $\mu_p(\cdot),\ \Delta_p(\cdot)$ are not continuous, for a counterexample see page 4 of the paper just before proposition 2.2. 
That's how far i come: 
Let $f\in\overline K$, i.e. there is a sequence $(f_n)_n\in K^\mathbb N$ with $f_n\xrightarrow[n\to\infty]{L^2}f$. Since $\Delta_p(f_n),\ \mu_p(f_n)\in[-A,A]$ for all $n$, there exists a subsequence with
i)$\Delta_p(f_{n_k})\xrightarrow[k\to\infty]{}\underline\Delta_p=\liminf\Delta_p(f_n)\in[-A,A]$
ii)$\mu_p(f_{n_k})\xrightarrow[k\to\infty]{}\mu_p\in[-A,A]$.
iii)Because $f_{n_k}\xrightarrow[k\to\infty]{L^2}f$ there exists another subsequence, which i will denote by $(f_{n_k})_k$ for convenience, that converges pointwise almost everywhere to f.
That $\|f\|=1$ is easy to see and the lemma of Fatou implies
$\varphi_f(\mu_p)=\int|t-\mu_p|^p|f|^2\mathrm d t=\int\liminf|t-\mu_p(f_{n_k})|^p|f_{n_k}|^2\mathrm d t$
$\leq\liminf\int|t-\mu_p(f_{n_k})|^p|f_{n_k}|^2\mathrm d t=\liminf\Delta_p^2(f_{n_k})=\lim\Delta_p^2(f_{n_k})=\underline\Delta_p^2\leq A^2$
This means $\varphi_f(\mu_p)<\infty$, so that $\Delta_p(f)$ and $\mu_p(f)$ are well defined, and since $\Delta_p^2(f)$ is the minimum of $\varphi_f$ it holds $\Delta_p^2(f)\leq\varphi_f(\mu_p)\leq A^2\Rightarrow  |\Delta_p(f)|\leq A$. The same argument can be applied to $\hat f$ (because the Fourier transform is an unitary operator on $L^2$), so $|\Delta_q(\hat f)|\leq A$ is also satisfied.
My problem is to show that the inequality also holds for $\mu_p(f)$ and $\mu_q(\hat f)$. I tried to prove that for $a\notin [-A,A]$ it is $\varphi_f(\mu_p)\leq\varphi_f(a)$, because then $\varphi_f$ couldn't attain it's minimum outside of $[-A,A]$. 
One idea is to use the $\limsup$ version of Fatou's lemma, but for that i need an integrable majorant for $(|t-a|^p|f_{n_k}|)_k$ then one could argue just like before:
$\varphi_f(\mu_p)\leq\liminf\int|t-\mu_p(f_{n_k})|^p|f_{n_k}|^2\mathrm d t\leq \liminf\int|t-a|^p|f_{n_k}|^2\mathrm d t$
the last inequality holds because $\mu_p(f_{n_k})$ is the point where the minimum of $\varphi_{f_{n_k}}$ is attained and thus
$\liminf\int|t-a|^p|f_{n_k}|^2\mathrm d t\leq\limsup\int|t-a|^p|f_{n_k}|^2\mathrm d t$
$\leq\int\limsup|t-a|^p|f_{n_k}|^2\mathrm d t=\int|t-a|^p|f|^2\mathrm d t=\varphi_f(a)$.
One also knows that $K$ is relative compact, maybe this could in some way be used.
I would be very grateful for responses and remarks!
 A: The set is indeed closed. To see this, note that after all your taking
subsequences, we are reduced to showing the following: If $f_{n}\xrightarrow[n\to\infty]{L^{2}}f$
and $f_{n}\xrightarrow[n\to\infty]{\text{a.e.}}f$ and $\left|\Delta_{p}\left(f_{n}\right)\right|\leq A$
for all $n$ with $\Delta_{p}\left(f_{n}\right)\to\alpha\in\left[-A,A\right]$
and if furthermore $\left|\mu_{p}\left(f_{n}\right)\right|\leq A$
for all $n$ and $\mu_{p}\left(f_{n}\right)\xrightarrow[n\to\infty]{}\beta\in\left[-A,A\right]$,
then $\left|\mu_{p}\left(f\right)\right|\leq A$, since you already
proved yourself that $\left|\Delta_{p}\left(f\right)\right|\leq A$
under the given conditions.
To show this, we will actually show $\mu_{p}\left(f\right)=\beta$.
For this, we will use the following: If a function $g\in L^{2}$ satisfies
$\int\left|x-a\right|^{p}\cdot\left|g\right|^{2}\,{\rm d}x<\infty$
for some $a\in\mathbb{R}$, then the function
$$
\Phi_{g}:\mathbb{R}\to\mathbb{R},a\mapsto\int\left|x-a\right|^{p}\cdot\left|g\right|^{2}\,{\rm d}x
$$
is (as you noted yourself) well-defined and differentiable (as we
will now show) with derivative
$$
\Phi_{g}':\mathbb{R}\to\mathbb{R},a\mapsto p\cdot\int{\rm sign}\left(a-x\right)\cdot\left|x-a\right|^{p-1}\cdot\left|g\right|^{2}\,{\rm d}x.
$$
To see this, note that the map
$$
\mathbb{R}\to\mathbb{R},a\mapsto\left|x-a\right|^{p}\cdot\left|g\left(x\right)\right|^{2}=\left|a-x\right|^{p}\cdot\left|g\left(x\right)\right|^{2}
$$
is differentiable (because of $p>1$) with derivative
$$
\begin{cases}
p\cdot\left(a-x\right)^{p-1}\cdot\left|g\left(x\right)\right|^{2} & \text{if }a>x\\
-p\cdot\left(x-a\right)^{p-1}\cdot\left|g\left(x\right)\right|^{2} & \text{if }a<x
\end{cases}\quad=p\cdot{\rm sign}\left(a-x\right)\cdot\left|x-a\right|^{p-1}\cdot\left|g\left(x\right)\right|^{2}\text{ for }x\neq a,
$$
which extends continuously to $x=a$. Furthermore, for $\left|a\right|\leq R$
and $R\geq1$, we have the estimate
\begin{eqnarray*}
\left|p\cdot{\rm sign}\left(a-x\right)\cdot\left|x-a\right|^{p-1}\cdot\left|g\left(x\right)\right|^{2}\right| & \leq & \left(\left|x\right|+\left|a\right|\right)^{p-1}\cdot\left|g\left(x\right)\right|^{2}\\
 & \leq & \left(R+\left|x\right|\right)^{p-1}\cdot\left|g\left(x\right)\right|^{2}\\
 & \leq & \left(R+\left|x\right|\right)^{p}\cdot\left|g\left(x\right)\right|^{2}\\
 & \leq & 2^{p}\cdot\left[R^{p}\left|g\left(x\right)\right|^{2}+\left|x\right|^{p}\cdot\left|g\left(x\right)\right|^{2}\right]\in L^{1},
\end{eqnarray*}
so that the usual theorems about differentiation under the integral
(as in https://en.wikipedia.org/wiki/Leibniz_integral_rule#Measure_theory_statement)
show differentiability of $\Phi_{g}$.
Now, as you noted, each map $\Phi_{g}$ is strictly convex and has
a (unique) global minimum, at $\mu_{p}\left(g\right)$. From this,
it follows that $\mu_{p}\left(g\right)$ is the unique zero of $\Phi_{g}'$.
In other words, we have $\Phi_{f_{n}}'\left(\mu_{p}\left(f_{n}\right)\right)=0$
for all $n\in\mathbb{N}$ and want to show $\Phi_{f}'\left(\beta\right)=0$.
To show this, we will use the Vitali convergence theorem in the version
found here: http://math.gmu.edu/~dwalnut/teach/Math776/Spring11/776s11lec10_notes.pdf
(Theorem 0.3). Let us verify the three prerequisites:


*

*Since $f_{n}\to f$ almost everywhere and because of $\mu_{p}\left(f_{n}\right)\to\beta$
and since the map $a\mapsto{\rm sign}\left(a-x\right)\left|x-a\right|^{p-1}$
is continuous (with value $0$ at $a=x$), we have
$$
{\rm sign}\left(\mu_{p}\left(f_{n}\right)-x\right)\left|x-\mu_{p}\left(f_{n}\right)\right|^{p-1}\cdot\left|f_{n}\left(x\right)\right|^{2}\xrightarrow[n\to\infty]{}{\rm sign}\left(\beta-x\right)\left|x-\beta\right|^{p-1}\cdot\left|f\left(x\right)\right|^{2}
$$
for almost every $x\in\mathbb{R}$.

*We first show tightness. To this end, note that we have for $R>A$
and $\left|x\right|\geq R$ that $\left|x-\mu_{p}\left(f_{n}\right)\right|\geq\left|x\right|-\left|\mu_{p}\left(f_{n}\right)\right|\geq\left|x\right|-A\geq R-A$
and thus
\begin{eqnarray*}
 &  & \int_{\mathbb{R}\setminus\left(-R,R\right)}\left|{\rm sign}\left(\mu_{p}\left(f_{n}\right)-x\right)\left|x-\mu_{p}\left(f_{n}\right)\right|^{p-1}\cdot\left|f_{n}\left(x\right)\right|^{2}\right|\,{\rm d}x\\
 & = & \int_{\mathbb{R}\setminus\left(-R,R\right)}\frac{\left|x-\mu_{p}\left(f_{n}\right)\right|^{p}}{\left|x-\mu_{p}\left(f_{n}\right)\right|}\cdot\left|f_{n}\left(x\right)\right|^{2}\,{\rm d}x\\
 & \leq & \frac{1}{R-A}\cdot\int_{\mathbb{R}\setminus\left(-R,R\right)}\left|x-\mu_{p}\left(f_{n}\right)\right|^{p}\cdot\left|f_{n}\left(x\right)\right|^{2}\,{\rm d}x\\
 & \leq & \frac{1}{R-A}\int\left|x-\mu_{p}\left(f_{n}\right)\right|^{p}\cdot\left|f_{n}\left(x\right)\right|^{2}\,{\rm d}x\\
 & \leq & \frac{\Delta_{p}\left(f_{n}\right)}{R-A}\leq\frac{C}{R-A},
\end{eqnarray*}
so that we get
$$
\sup_{n\in\mathbb{N}}\int_{\mathbb{R}\setminus\left(-R,R\right)}\left|{\rm sign}\left(\mu_{p}\left(f_{n}\right)-x\right)\left|x-\mu_{p}\left(f_{n}\right)\right|^{p-1}\cdot\left|f_{n}\left(x\right)\right|^{2}\right|\,{\rm d}x\leq\frac{C}{R-A}\xrightarrow[R\to\infty]{}0,
$$
which shows the desired tightness.

*Finally, we show uniform integrability. Let $\varepsilon>0$. We
want to show that there is some $\delta>0$ with
$$
\int_{E}\left|x-\mu_{p}\left(f_{n}\right)\right|^{p-1}\cdot\left|f_{n}\left(x\right)\right|^{2}\,{\rm d}x=\int_{E}\left|{\rm sign}\left(\mu_{p}\left(f_{n}\right)-x\right)\left|x-\mu_{p}\left(f_{n}\right)\right|^{p-1}\cdot\left|f_{n}\left(x\right)\right|^{2}\right|\,{\rm d}x\overset{!}{<}\varepsilon
$$
for all $n\in\mathbb{N}$ and all measurable $E\subset\mathbb{R}$
with $\lambda\left(E\right)<\delta$, where $\lambda$ is the Lebesgue
measure.
By the previous step, there is $R>0$ with
$$
\int_{E\setminus\left(-R,R\right)}\left|x-\mu_{p}\left(f_{n}\right)\right|^{p-1}\left|f_{n}\left(x\right)\right|^{2}\,{\rm d}x\leq\int_{\mathbb{R}\setminus\left(-R,R\right)}\left|x-\mu_{p}\left(f_{n}\right)\right|^{p-1}\cdot\left|f_{n}\left(x\right)\right|^{2}\,{\rm d}x<\frac{\varepsilon}{2}
$$
for all $n\in\mathbb{N}$. Thus, it suffices to show to show
$$
\int_{E\cap\left(-R,R\right)}\left|x-\mu_{p}\left(f_{n}\right)\right|^{p-1}\left|f_{n}\left(x\right)\right|^{2}\,{\rm d}x<\frac{\varepsilon}{2}
$$
for all $E\subset\mathbb{R}$ with $\lambda\left(E\right)<\delta$.
But for $x\in\left(-R,R\right)$, we have
$$
\left|x-\mu_{p}\left(f_{n}\right)\right|^{p-1}\leq\left(\left|x\right|+\left|\mu_{p}\left(f_{n}\right)\right|\right)^{p-1}\leq\left(A+R\right)^{p-1}
$$
and thus
$$
\int_{E\cap\left(-R,R\right)}\left|x-\mu_{p}\left(f_{n}\right)\right|^{p-1}\left|f_{n}\left(x\right)\right|^{2}\,{\rm d}x\leq\left(A+R\right)^{p-1}\cdot\int_{E\cap\left(-R,R\right)}\left|f_{n}\left(x\right)\right|^{2}\,{\rm d}x\leq\left(A+R\right)^{p-1}\cdot\int_{E}\left|f_{n}\left(x\right)\right|^{2}\,{\rm d}x.
$$
But since $f_{n}\to f$ in $L^{2}$, we have $\left|f_{n}\right|^{2}\to\left|f\right|^{2}$in
$L^{1}$, so that the converse direction of Vitali's convergence theorem
(Theorem 0.4 in the pdf linked above) shows that $\left(\left|f_{n}\right|^{2}\right)_{n}$
is uniformly integrable. Thus, there is $\delta>0$ with $\int_{E}\left|f_{n}\right|^{2}\,{\rm d}x<\frac{\varepsilon}{2\left(A+R\right)^{p-1}}$
for all $E\subset\mathbb{R}$ with $\lambda\left(E\right)<\delta$,
which shows tightness.
All in all, Vitali's theorem shows
\begin{eqnarray*}
0=\Phi_{f_{n}}'\left(\mu_{p}\left(f_{n}\right)\right) & = & p\cdot\int{\rm sign}\left(\mu_{p}\left(f_{n}\right)-x\right)\left|x-\mu_{p}\left(f_{n}\right)\right|^{p-1}\left|f_{n}\left(x\right)\right|^{2}\,{\rm d}x\\
 & \xrightarrow[n\to\infty]{} & p\int{\rm sign}\left(\beta-x\right)\left|x-\beta\right|^{p-1}\left|f\left(x\right)\right|^{2}\,{\rm d}x\\
 & = & \Phi_{f}'\left(\beta\right).
\end{eqnarray*}
As seen above, this implies $\beta=\mu_{p}\left(f\right)$ and thus
$\left|\mu_{p}\left(f\right)\right|=\left|\beta\right|\leq A$ as
desired.
