Q:

Convergence in $L^2$ implies uniform convergence for the normalized functions

Good morning,
I've recently came across this question and, although it seems logical for me to be true, I still can't come up with a satisfactory answer.
The question is the following: suppose that you have a sequence $\{f_n\}_{n\in \mathbb N}\subset L^2([0,1],\mathbb{R})$ that converges in $L^2$ to some $f\in L^2([0,1],\mathbb{R})$. Define then $e_n:= \frac{f_n}{|f_n|}$ and, similarly, $e=\frac{f}{|f|}$.
Q: Is it true that $e_n\to e$ in $L^\infty([0,1],\mathbb{R})$?
My guess is yes because $|e_n-e|=2-2\cos(\theta_n)$, where $\theta_n$ is the angle between $f_n$ and $f$ and, since $f_n\to f$ a.e., then $\theta\to 0$ a.e. However, I feel that this answer is not complete, and I fear I'm missing potentially something.
Thank you in advance for your replies,
Guido

A:

Consider the sequence of characteristic functions in $L^2([0,1], \mathbb{R})$ given by
$$f_1 = \chi_{\left[0, \frac12\right]}, f_2 = \chi_{\left[\frac12, 1\right]}$$
$$f_3 = \chi_{\left[0, \frac13\right]}, f_4 = \chi_{\left[\frac13, \frac23\right]}, f_5 = \chi_{\left[\frac23, 1\right]}$$
$$f_6 = \chi_{\left[0, \frac14\right]}, f_7 = \chi_{\left[\frac14, \frac24\right]}, f_7 = \chi_{\left[\frac24, \frac34\right]}, f_9 = \chi_{\left[\frac34, 1\right]}$$
and so on. We have that $f_n \xrightarrow{n\to\infty} 0$ in $L^2$. 
However, the sequence $(e_n)_{n=1}^\infty = (f_n)_{n=1}^\infty$ does not converge in $L^\infty$. Moreover, it does not even converge almost everywhere.

