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
.
Proof:
Assume .
(Case: ).
Lemma 1:
If ,
for all
.
Proof of Lemma 1:
Hence .
End Proof of Lemma 1.
Hence, using and
we see that
Thus
Hence .
(Case: .)
By Minkowski’s inequality, and
so that
as
. Done.
Converse:
Assume a.e.\ and
,
.
Lemma 2:
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
.