Followed by this entry where the following questions have been discussed
- Boundedness implies weakly convergence: If is a reflexive Banach space and is a bounded sequence. Then up to a subsequence converges weakly to some in .
- Weakly convergence becomes strongly convergence via compact operator: A compact operator between Banach spaces maps every weakly convergent sequence in into one that converges strongly in .
Now I shall discuss more results which appear frequently in the calculus of variation.
Let us recall over a minifold , the Sobolev norm (or ) is defined by
It is immediately to see that if strongly in , i.e. then strongly in and strongly in .
It turns out to discuss what happen to weakly convergence. Actually, we shall prove the following important result, called the third fundamental result.
Weakly convergence in Sobolev spaces implies weakly convergence in spaces. We assume in . We shall prove both and converge weakly to and in , respectively.
By the principle of uniform boundedness, any weakly convergence sequence is bounded in the norm. Consequently, and are bounded in . By the weak compactness of balls in , there is a subsequence such that
in (i.e., both are in ). Since the weak convergence in implies the convergence in the dual space of -the space of test functions. It follows that and, hence, . It follows that and thus
in as desired.
Now we consider the reverse case. We shall prove the following
Weakly convergence in spaces plus the boundedness implies weakly convergence in Sobolev spaces. We assume in and is bounded. We shall prove that and in .
Since is bounded in , by the first fundamental result, in for some . By the third fundamental result above, in . It follows from the uniqueness of weak limit that which implies .
In order to prove in , we shall use the following result whose proof is based on the simple contradiction argument.
Let be a topological space. A sequence converges to (in the topological of ) if and only if any subsequence of contains a sub-subsequence that converges to .
Let us pick a particular subsequence of and rename it back to for simplicity. It suffices to prove that contains a subsequence that converges to weakly in . Followed the proof of the third fundamental result, there is a subsequence of and a function such that
both in . It follows from the definition of weakly convergence in that in fact we get
in . The reason is the following:
for any . Having this and the fact that weak limit is unique we deduce that . The latter now implies
in . The proof follows.