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Suppose a sequence of absolutely continuous functions $f_{n}\to f$ in $L^{2}$ on $[0,1]$. Then does $f_{n}\to f$ almost everywhere?

I know the statement is not true for arbitrary sequences $f_{n}$ that converge in $L^{2}$ because one can choose a sequence of moving characteristic (indicator) functions where the width of the interval goes to $0$, but the sequence converges almost nowhere.

However, does imposing the absolutely continuous (or even continuous) restriction remove all such possibilities? How would one use the properties of absolutely continuous functions to prove this statement?

BaroqueFreak
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  • The answer is also negative. If you have observed that we can choose a moving characteristic function ( indicator perhaps?) to give a counter example, then "expand" that characteristic function a little bit and smooth it out, you still have a counter example.( Just my feeling, I didn't check your argument) – Paresseux Nguyen Nov 19 '20 at 00:08
  • Why don't you look for something more direct? For example $$ \sum \Vert f_n-f \Vert^2 <+\infty $$ – Paresseux Nguyen Nov 19 '20 at 00:09
  • Or $f_n$ are equicontinous ? For example Lipschitz? – Paresseux Nguyen Nov 19 '20 at 00:11

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The claim is false even if the $f_n$ are infinitely differentiable on $(0,1)$. Here is a sequence $(f_n)$ for which $f_n\to0$ in $L^2$ but $f_n$ converges nowhere to $0$.

Let $(g_n)$ be the typewriter sequence. We will construct $f_n$ to dominate $g_n$. If the $n$th member $g_n$ is a rectangle of height $1$ over the interval $[a_n,b_n]$, define $f_n$ to be $$ f_n(x):=\exp-\frac14\left[\left(\frac{x-\mu_n}{\sigma_n}\right)^2-1\right] $$ with $\mu_n:=\frac{a_n+b_n}2$ the midpoint of $[a_n,b_n]$ and $\sigma_n:=\frac{b_n-a_n}2$ the half-width. By comparison with the density for a Gaussian distribution with mean $\mu_n$ and variance $\sigma_n^2$, we have $$\int_0^1 f_n^2\le \int_{-\infty}^\infty f_n^2=\sigma_n\sqrt{2\pi e},$$ and, since $\sigma_n\to0$, we conclude $f_n\to0$ in $L^2$.

On the other hand, observe that $f_n(x)\ge1$ whenever $|x-\mu_n|\le\sigma_n$, hence $f_n\ge g_n$ for all $n$. Since $g_n(x)$ does not converge to zero for any $x$, conclude the same is true for $f_n$.

grand_chat
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