Question: Let $f\_n: [0, 1] \to \mathbb{R}$ be a bounded sequence of functions in $L^2([0, 1])$. This means that there exists $C\_0 > 0$ so that$$\|f\_n\|\_{L^2} \le C\_0 \text{ for all }n.$$Assume that $f\_n$ converges to $f$ in $L^1([0, 1])$. I have two questions.

1. Do we have necessarily have that $f \in L^2([0, 1])$?
2. Do we necessarily have that $\|f\|\_{L^2} \le C\_0$?

Answer: 
A less functional-analytic argument is through Fatou's lemma (applied to $|f\_n|^2$), since a subsequence converges pointwise a.e.
Answer to both questions is yes. Bounded sequences in $ L^2$ have weakly converging subsequences. By uniqueness of the almost everywhere pointwise limit, these weak limits coincide with your $ f $. Furthermore, the norm of the weak limit of a subsequence is smaller than the limit inferior of the norms of the elements of the sequence.

Using a bit less abstract arguments, define a functional $\phi$ on $ L^2$ as follows. For each $ g\in C [0,1]\cap L^2[0,1] $ define $\phi(g ) $ as $\int f g $. Check that the operator norm of $\phi$, on the subspace where it ia defined equipped with the $ L^2$ norm, is smaller than $ C\_0$. Therefore we may extend $\phi $ to the full space, and conclude by observing that the $ L^2$ norm of $ f $ equals the operator norm of that extension, which is still bounded by the same constant. 

As I said, only a bit less abstract, and maybe with more details to be filled.