Let $f,f\_n \in L^1(\mathbb{R},\mathbb{R}\_+)$ with $\int\_{\mathbb{R}} f = \int\_{\mathbb{R}} f\_n = 1$, $(\sqrt{f\_n})'$ bounded in $L^2$, $\nabla \sqrt{f}\in L^2$ and such that $$\int\_{ p+[0,1/n]} f\_n = \int\_{ p+[0,1/n]} f$$ for any $p \in \mathbb{Z}/n$ and any $n \in \mathbb{N}$. How to prove that $f\_n$ converges to $f$ ae ?



Convergence of local means implies converge ae?
$\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb N}
\newcommand{\Z}{\mathbb{Z}}$
We should, and will, assume that 
\begin{equation}
\int\_p^{p+1/n}f\_n=\int\_p^{p+1/n}f \tag{0}
\end{equation}
for all $n\in\N$ and all $p\in\Z/n$. (If (0) is assumed only for $p\in\N/n$, then the conclusion will obviously be false in general.) 


The key here is 



> 
> **Lemma 1:** If a function $g$ is such that $\int\_\R g^2=1$ and $\int\_\R g'^2=C^2$ for some real constant $C\ge0$, then $|g|\le2\sqrt C$. 
> 
> 
> 


Lemma 1 will be proved at the end of this answer. 


Let $f\_0:=f$, $g\_0:=g:=\sqrt f$, and $g\_n:=\sqrt f\_n$, so that for $n=0,1,\dots$
\begin{equation\*}
 \int\_\R g\_n^2=1 \quad \text{and}\quad\int\_\R {g'\_n}^2\le C
\end{equation\*}
for some real $C>0$, whence, by Lemma 1,
\begin{equation\*}
 |g\_n|\le2\sqrt C. \tag{1}
\end{equation\*}


Next, for all $n=0,1,\dots$ and all real $x$ and $y$ such that $x\le y$, 
\begin{equation\*}
 |g\_n(y)-g\_n(x)|\le\int\_x^y|g'\_n(u)|\,du\le\sqrt{y-x}\,\sqrt{\int\_\R g'\_n(u)^2\,du}
 \le C\sqrt{y-x},
\end{equation\*}
whence, in view of (1), 
\begin{equation\*}
 |f\_n(y)-f\_n(x)|=|g\_n(y)-g\_n(x)||g\_n(y)+g\_n(x)|\le 4C\sqrt C\sqrt{y-x}. \tag{2}
\end{equation\*}


Take now any $x\in\R$. Then for each $n\in\N$ and some $p=p\_{n,x}\in\Z/n$, we have $x\in[p,p+1/n]$ and hence, by (2), 
\begin{equation\*}
 |f\_n(x)-\bar f\_{n,p}|\le 4C\sqrt C\sqrt{1/n} 
\end{equation\*} 
and 
\begin{equation\*}
 |f(x)-\bar f\_{n,p}|=|f(x)-\bar f\_{n,0,p}|\le 4C\sqrt C\sqrt{1/n}, 
\end{equation\*} 
where 
\begin{equation}
 \bar f\_{n,p}:=\frac1{1/n}\int\_p^{p+1/n}f\_n=\bar f\_{n,0,p}:=\frac1{1/n}\int\_p^{p+1/n}f, 
\end{equation}
which yields 
\begin{equation\*}
 |f\_n(x)-f(x)|\le 8C\sqrt C\sqrt{1/n}, 
\end{equation\*}
so that $f\_n\to f$ uniformly. 




---


It remains to present


*Proof of Lemma 1:* Without loss of generality, $g(0)\ge0$ and $C\in(0,\infty)$. For any $x\in[0,x\_\*]$, where $x\_\*:=\frac14\,g(0)^2/C^2$, we have 
\begin{equation\*}
 g(x)\ge g(0)-\int\_0^{x\_\*}|g'(u)|\,du\ge g(0)-\sqrt{\int\_0^{x\_\*}\,du}\;\sqrt{\int\_\R g'(u)^2\,du}
= g(0)/2, 
\end{equation\*}
whence 
\begin{equation\*}
 1=\int\_\R g(x)^2\,dx\ge\int\_0^{x\_\*}g(x)^2\,dx\ge \tfrac14\,g(0)^2 x\_\*=\tfrac1{16}\,g(0)^4/C^2, 
\end{equation\*}
whence $|g(0)|\le2\sqrt C$. Similarly, $|g(t)|\le2\sqrt C$ for all real $t$. $\Box$


