Proving a certain subset is closed in $L^1$

Question

In an exam, I was asked to prove that, if $A=\{f\in L^1([0,1]):\int_0^1|f(x)|^2\mathrm{d}x\leq1\}$, then $A$ is closed in $L^1$. I tried this approach. $A$ is closed iff for all $f_n\to f$ in $L^1$ with $f_n\in A$ for all $n$, then $f\in A$. So let us take such a sequence $f_n$ and its limit $f$. We have to show $\int_0^1|f(x)|^2\mathrm{d}x\leq1$. We could try: $$\|f\|^2\leq\|f-f_n\|^2+\|f_n\|^2\leq1+\|f-f_n\|^2,$$ but then we would need to prove the first term tends to zero, which would be true were this the $L^1$ norm, but it is the $L^2$ norm, and convergence in $L^1$ does not necessarily imply convergence in $L^2$ AFAIK. I guess the condition that $\|f_n\|^2\leq1$ could tell me more, but I can't see how. Surely I can't deduce a dominated convergence, because the monotony of the integral is not invertible, so the fact $\int_0^1|f_n(x)|^2\mathrm{d}x\leq\int_0^11^2\mathrm{d}x=1$ doesn't imply the $|f_n|$s are no greater than one. Any suggestions?

Answer 1

Hint:

$(f_n)$ has a subsequence $(f_{n_k})$ that coverges to $f$ a.e.. Note $f^2_{n_k}\buildrel{k\rightarrow\infty}\over\longrightarrow f^2$ a.e., and apply Fatou's Lemma to $(f^2_{n_k})$.