It turns out that convergence in Lp implies that the norms converge. Conversely, a.e. convergence and the fact that norms converge implies Lp convergence. Amazing!
Relationship between convergence and a.e. convergence:
Let , . If , then . Conversely, if a.e.\ and , , then . Note that the converse may fail for .
If , for all .
Proof of Lemma 1:
End Proof of Lemma 1.
Hence, using and we see that
By Minkowski’s inequality, and so that as . Done.
Assume a.e.\ and , .
For , for .
Proof of Lemma 2:
By convexity of for ,
Multiplying throughout by gives
Thus together with Lemma 1, for we have with .
Note that a.e.\ and a.e.\ which is integrable. Also, since . By Generalized Lebesgue’s DCT, we have thus
(Show that the converse may fail for ):
Consider . Then a.e.\ where , and . However .