$u_n$ is bounded in $L^1(\omega)$ implies $u_{n_{k}}$ weakly convergente in $L^2(\omega)$

Question

I saw in a book that if $u_n$ is a bounded sequence in $L^1(\omega)$ then there exists a subsequence such that $u_{n_{k}}$ converge to some $u$ in $L^2(\omega)$, how it can be possible, we have not an injection of $L^1(\omega)$ in $L^2(\omega)$, thanks.

Answer 1

Are you sure you're not missing some assumptions? I mean you are right, there is no such injections. For example pick $f_n(x)=\frac{1}{\sqrt{x}}$ as a constant sequence. Then you have $$\int_0^1|f_n(x)|\, dx =[2\sqrt{x}]_0^1=2.$$ Therefore $f_n$ is bounded in $L^1([0,1])$, but $$\int_0^1f_n^2\, dx=\int_0^1\frac{1}{x}\, dx=[\ln(x)]_0^1=\infty.$$ Hence $f_n\notin L^2([0,1])$. This means you cannot find any subsequence such that $f_n$ can converge in $L^2$.

Maybe you can give us the exact reference?