Seems to be a straight forward problem. I'm not sure what I'm missing.

Let $f_n$ be a sequence of measurable functions on $[0,1]$ bounded in $L^2$ (i.e. sup $||f_n||_{L^2}<\infty$). Suppose $f_n$ converges to $f$ in $L^1$. Show that $f\in L^2$.

Since $f_n$ converges to $f$ in $L^1$, there exists a subsequence $f_{n_k}$ which converges pointwise almost everywhere to $f$. In particular $|f-f_{n_k}|$ converges pointwise almost everywhere to zero. Hence, if there is an integrable function $g$ such that $|f-f_{n_k}|^2\le g$ almost everywhere, then we can apply the dominated convergence theorem to the sequence $|f-f_{n_k}|^2$ to get the result.

Not sure if this is the right route. Perhaps no such $g$ exists and this is the wrong way to go. Any thoughts?


Not correct; the function $g$ has no reason to exist. I think the important point here is the fact it's uniformly bounded in $L^2$.

Try applying Fatou's lemma to $(f_{n_k})^2$ - boundedness always mixes with $\lim \inf$ quite well.

  • $\begingroup$ Clearly, $f\in L^1$ so it would be nice if $L^2\subset L^1$. I don't believe that's true. I don't see that $L^1\subset L^2$ helps. $\endgroup$
    – Elbu
    Nov 12 '13 at 1:14
  • $\begingroup$ I suppose it tells us that each $f_n \in L_1$. $\endgroup$
    – Elbu
    Nov 12 '13 at 1:21
  • $\begingroup$ Edited with my idea to take advantage of the uniform boundedness in $L^2$. $\endgroup$
    – Matt Rigby
    Nov 12 '13 at 1:22
  • $\begingroup$ Thank, I'll think about that. $\endgroup$
    – Elbu
    Nov 12 '13 at 1:23
  • $\begingroup$ Fatou's Lemma makes it easy. I should have seen that. Thanks. $\endgroup$
    – Elbu
    Nov 12 '13 at 19:15

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