Does Strong Convergence in $L^1$ Imply Weak Convergence in $L^2$?

Question

If I have $f_n \to f$ in $L^1(D)$, where $D \subset \mathbb{R}$ is compact, is it accurate to say $f_n \rightharpoonup f$ in $L^2(D)$?

The argument is as follows: consider a simple function $\phi = \sum_i a_i \chi_{D_i}$. Then \begin{align*} \lim_{n\to \infty} \int_D \phi f_n & = \lim_{n\to\infty} \sum_i a_i \int_{D_i} f_n \\ & = \sum_i a_i \int_{D_i} f \\ & = \sum_i \int_{D_i} a_i f \\ & = \int_D \phi f \end{align*}

Then weak convergence would follow from density of simple functions in $L^2$. This seems to make sense, but I couldn't find this result anywhere else - seems surprising for a result that appears so elementary.

Answer 1

This is false. Consider, for example, $f_n=n^{3/4}\chi_{(0,1/n)}$ on $D=[0,1]$. Then $f_n\to 0$ in $L^1$, but the sequence does not converge weakly in $L^2$ (it would be bounded in norm if it did).

The problem with your argument is that you are implicitly assuming that $\|f_n\|_2$ is bounded; otherwise, there's no obvious way to carry out the approximation argument going from simple functions to general functions.