If a bounded sequence of $L^2$ functions converges in $L^1$, is the limit an $L^2$ function?

Question

I was wondering this question:

Is the $L^1$ limit of $L^2$ functions again an $L^2$ function?

Specifically, what I mean is the following: Suppose

• $f_n\to f$ in $L^1[0,1]$, in other words $\lim_n \int_0^1 |f-f_n| \, dx = 0$.
• there exists a $K>0$ such that $\int_0^1 |f_n|^2 \, dx \leq K$ for all $n$. (Note that this is stronger than just requiring $f_n\in L^2[0,1]$)

Does it follow that $\|f\|_{L_2[0,1]}^2 = \int_0^1 |f|^2 \, dx\leq K$?

I haven't been able to prove or disprove it.

Answer 1

Yes, this is true. To see this, note that since $f_n\to f$ in $L^1$ there exists a subsequence $\{n_k\}$ such that $f_{n_k}\to f$ pointwise almost everywhere. Obviously then $|f_{n_k}|^2\to|f|^2$ pointwise almost everywhere, and so by Fatou's lemma, $$\int_0^t|f|^2dx\le\liminf_{k\to\infty}\int_0^1|f_{n_k}|^2dx\le K.$$

Answer 2

Define $f_n(x) = x^{\frac{1}{n} - \frac{1}{2}}$ for every $x \in (0,1)$. Then $f_n \in L^2([0,1])$ and $\big\Vert \ f_n - \frac{1}{\sqrt{x}} \ \big\Vert_{L^1([0,1])} \to 0$ but $\frac{1}{\sqrt{x}} \not\in L^2([0,1]) $