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.

See also: Two fundamental results in the calculus of variation

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