Weak convergence in $L^1$ $f_n \to f$ and bounded $L^2$ norm implies the limit is in $L^2$

Question

The Entire problem

Let $f_n \to f$ weakly in $L^1$ and assume $\|f_n\|_2\leq M$ then prove $f \in L^2$. I am not sure how to proceed. Part b) is easy as the span of characteristic functions is dense in $L^2$

Any hints would be appriciated.

Answer 1

The following fact can be found on MSE:

If $f$is measurable and $\int |fh| <\infty$ for every $h \in L^{2}$ then $f \in L^{2}$.

So let $h \in L^{2}$ and consider any simple function $s$ such that $0 \leq s \leq |h|$. Then $\int fs =\lim \int f_ns \leq \sup (\int |f_n|^{2})^{1/2} (\int |h|^{2})^{1/2} $. Taking sup over all such $s$ we get $\int |fh| <\infty$.

There are several proofs for the result I have used. Here is a Functional Analytic proof: Define a continuous linear functional $T_n$ on $L^{2}$ by $T_nh=fhI_{A_n}$ where $A_n= \{x: |f(x)| \leq n, |x| \leq n\}$. Then Holder's / C-S inequality shows that $\|T_n\| \leq \sqrt {\int_{A_n} |f|^{2}}.$ By choosing $h$ to be $sgn (f) f I_{A_n}$ it is easy to see that $\|T_n\| = \sqrt {\int_{A_n} |f|^{2}}$ . By hypothesis $\sup_n |T_n(h)| \leq \int |fh| <\infty$ for each $h \in L^{2}$. By Uniform Boundedness Principle this implies $\sup_n \|T_n\| <\infty$. But $\sup_n \|T_n\|=\sup_n \sqrt {\int_{A_n} |f|^{2}}=\int |f|^{2}$. Hnce $f \in L^{2}$.