Let $f_n$ be a sequence in $L^1(\Bbb R)\cap L^2(\Bbb R)$ ;$f\in L^{1}(\Bbb R)$ and $g\in L^2(\Bbb R)$.

If $f_n\xrightarrow {L_1(\Bbb R)} f, f_n\xrightarrow {L_2(\Bbb R)} g$,then show that $f=g$(almost everywhere).


We have $\int_\Bbb R |f_n-f|<\infty $ and $\int_\Bbb R |f_n-g|^2<\infty .$

Now since $f_n\xrightarrow {L_1(\Bbb R)}f\implies $ there exists a subsequence $f_{n_k}$ of $f_n$ converging pointwise to $f$ a.e. $\implies ||f_{n_k}-f||_{L_1(\Bbb R)}\to 0\implies \int _\Bbb R|f_{n_k}-f|\to 0$.

Similarly there exists a subsequence $f_{n_k}$ of $f_n$ converging pointwise to $g$ a.e. $\implies ||f_{n_k}-g||_{L_2(\Bbb R)}\to 0\implies \int _\Bbb R|f_{n_k}-g|^2\to 0$.

Two questions:

  • Is it true that $\int _\Bbb R|f_{n_k}-g|^2\to 0\implies \int _\Bbb R|f_{n_k}-g|\to 0$.

  • If I can show that $\int |f-g|=0$ then that will imply that $f=g $ a.e.

But I am stuck on these questions.Please help.

  • 2
    $\begingroup$ Do you know that when à sequence of functions converges in $L_1(A)$ (or $L_2(A)$) it is possible to find a subsequence that converges to same limit on A a.e. ? $\endgroup$ – user322559 Mar 13 '17 at 7:32
  • $\begingroup$ The only reference I have is a PDF of the course I followed but it's in French. $\endgroup$ – user322559 Mar 13 '17 at 8:10
  • $\begingroup$ @Ben: See the "properties" section here en.m.wikipedia.org/wiki/Convergence_in_measure $\endgroup$ – PhoemueX Mar 13 '17 at 8:43
  • $\begingroup$ almost any book on measure theory has the a.e. subsequence result, for example Rudin RCA. $\endgroup$ – zhw. Mar 13 '17 at 19:56
  • $\begingroup$ @Errol.Y;I just read about the subsequence criterion;How to proceed now $\endgroup$ – Learnmore Mar 16 '17 at 5:05

Because $f_n\to f$ in $L^1$, there is a subsequence converging to $f$ pointwise a.e.. Then that subsequence still converges in $L^2$ to $g$, so it has a further subsequence converging pointwise a.e. to $g.$ This subsubsequence now converges pointwise to both $f$ and to $g$ a.e. implying that $f=g$ a.e.

This is likely the method Errol. Y hinted at in a comment.

Regarding your attempts and related questions:

  • Taking two different subsequences with no relation between them would make it harder to show they have the same pointwise limit.
  • No you cannot conclude from $\int|f_n-g|^2\to 0$ that $\int|f_n-g|\to 0$, and you cannot even conclude that $\int|f_n-g|$ is ever finite, because you do not no a priori that $g$ is in $L^1$. (Of course we know in this problem that it ends up being true because $g=f$.)
  • If you could show that $\int|f-g|=0$, then it would follow that $f=g$ a.e., but I don't see a way to go this way, again especially because you do not know at first even that $g$ is in $L^1$. An $L^2$ limit of $L^1$ functions need not be in $L^1$. The conclusion of this problem does show that $g$ is in $L^1$, by showing that it equals $f$.
  • $\begingroup$ Really thankful to you for such a detailed answer! $\endgroup$ – Learnmore Mar 16 '17 at 6:25

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