Is subset of $L^1$ functions bounded by $1 / x^2$ compact? I was given the following exercise

Given $E \subset L^1([1, +\infty))$ as follows
$$
E = \left\{ f \in L^1([1, +\infty)) \;\middle|\; \forall x \in [1, +\infty) \quad |f(x)| \leq \frac{1}{x^2} \right\}
$$

*

*is $E$ bounded?

*is $E$ closed?

*is $E$ compact?


I tried solving it with other people. First we noticed that for $E$ to be well defined in $L^1$ the condition should actually be for almost all $x \in [1, +\infty)$ otherwise it is meaningless in $L^1$.
Then I was told that $E$ was compact and that there are various (non obvious) ways to prove it: one uses the fact that given $f_n(x)$ we can use Lousin's theorem to get families of continuos functions that are then used to construct $f(x)$ over the rationals... Another one uses convolution of the $f_n$ with a smooth kernel to get a continuous function to define the limit $f(x)$... But after this we found that the sequence of functions given by
$$
f_n(x) = \left|\frac{\sin(nx)}{x^2}\right|
$$
is in $L^1$ as $\forall n \; \int_1^\infty |\sin(nx)/x^2| \mathrm dx \leq \int_1^\infty |1/x^2| \mathrm dx = \|1 / x^2\|_{L^1} < +\infty$ and $|f_n| \leq 1 /x^2$ so satisfies the condition but we could not find what this function should converge to in $E$, is this an actual counterexample and the statement is actually false or is there an explicit limit for this sequence?
Note. We got this exercise as a kind of extension of the analogous set in $\ell^1$ given by $\{ (x_n)_n \in \ell^1 \mid |x_n| \leq 1 / n^2 \}$.
 A: We have $\|f\| \le \int_1^\infty {1 \over x^2} dx = 1$ hence $E$ is bounded.
It is straightforward to see that $E$ is convex, hence $E$ is closed iff $E$ is weakly closed.
Suppose $f_n \overset{w}{\to} f$ with $f_n \in E$. Suppose $f \notin E$ and let $A_m = \{ x | |f(x)| > {1 \over x^2} + { 1\over m} \}$, we see that the Lebesgue measure $\lambda A_m >0$ for some $m$. Let $\phi(x) = 1_{A_m}(x) \operatorname{sgn} f(x)$, note that $\phi \in L^\infty$ and
$\int \phi f = \int_{A_m} |f| \ge \int_{A_m} {1 \over x^2} dx + { 1\over m} \lambda A_m$. However, note that $\int \phi f_n \le \int_{A_m} {1 \over x^2} dx $ for all $n$ and so we obtain a contradiction. Hence $ f\in E$ and so $E$ is closed.
Let $f_n$ be zero everywhere except on $[1,2)$ where $f_n(x)$ is the $n$th digit of the binary expansion of $x-1$ multiplied by ${1 \over 4}$ (so it lies in $E$). Then the distance between any two distinct $f_n$s is ${ 1\over 8} $ and so cannot contain any convergent subsequence. Hence $E$ is not compact.
A: $E$ is bounded: For all $u \in E$, $$ \| u\|_{L^1([1,\infty))}\leqslant \int_1^\infty \frac 1 {x^2} \, dx =1. $$

$E$ is closed: Let $\{u_k\} \subset E$ be a sequence of functions which converge in $L^1([1,\infty))$ to $u$. Then $$\vert u(x) \vert  \leqslant \vert u_k(x)-u(x) \vert +\vert u_k(x) \vert \leqslant \vert u_k(x)-u(x) \vert +\frac 1 {x^2} $$ for all $x\in [1,\infty)$. Since $u_k \to u$ in $L^1([1,\infty))$, $u_k \to u$ a.e. after passing to a subsequence (see here for example). Hence, after passing to this subsequence then sending $k \to \infty$ we conclude that $$ \vert u(x) \vert \leqslant \frac 1 {x^2} \qquad \text{a.e.}$$

E is not compact: Let $v_k$ be the Rademacher functions (see the first example here) except defined on $[1,2]$ instead of $[0,1]$ and set $u_k(x) = \frac 1 4 v_k \chi_{[1,2]}$. Then $$\vert u_k(x) \vert \leqslant\frac 1 4 \chi_{[1,2]} \leqslant \frac 1 {x^2}. $$ But $$\| u_k - u_\ell \|_{L^1([1,\infty))} =\frac1 4  $$ for all $k,\ell$ so $u_k$ do not have a convergent subsequence.
