**1. Weak convergence and weakly compact**

We introduce some notations used in this article.

Let be a normed linear space and denote the dual space of . For and , we write instead of . A sequence in converges *weakly* to in if for any . We write in for simplicity.

Let us give some basic examples on this concept.

**Example 1.** When , note that where . Then a sequence in converges weakly to if for any . By the Riesz representation theorem on spaces, there exists such that

So weakly in if and only if

for any .

**Proposition 2.** If is compact Hausdorff space and is bounded in . Then for each

if and only if weakly in .

The proof of this fact uses the following Riesz representation theorem on . For the proof, see Rudin, RCA for example.

**Theorem 3** (Riesz representation theorem)**.** Let be a compact Hausdorff space and let be the set of signed Radon measures on such that

is finite. Define a norm on by setting

Then the dual space of is . So

for any bounded linear functional on , there

exists a such that

*Proof of Proposition 2.* Suppose . Then for any bounded linear functional on , there exists a signed Radon measure with such that

for all . Then

Since is bounded in and , by the dominated convergence theorem, we conclude that

This shows converges weakly to in .

Conversely, suppose converges weakly to in . Then for each , one can easily check that is a Radon measure on . So

This shows for each

.

>We list one of the well-known properties of weak convergence.

**Proposition 4.** Let be a sequence in . Then

- If strongly in , then weakly in .
- If weakly in , then is bounded and .
- If weakly and if strongly in , then .

*Proof.* (i) Let be a bounded linear functional on . Then we have

Letting , we see that . So converges weakly to .

(ii) We prove this by using the Banach-Steinhaus theorem. For each , define by

and by

Then

So

Since is a Banach space, by the Banach-Steinhaus theorem,

Since

this shows that .

Note that there exists such that and . So from this, we have

Now taking liminf to above inequality to get

(iii) This follows from

From (ii), is bounded. So letting , we get the desired result.

Let be a sequence in . We say

converges to weakly-star (or weakly{*}) if

for any . One can easily deduce similar results like Proposition

??.

In arbitrary normed linear space, usually the unit ball is not compact in norm topology. Also, it may not be compact in weak topology. However, in weak star topology, the unit ball is always compact. This is due to Banach and Alalogu.

**Theorem 5.** Every bounded sequence in has a weakly-star convergent sequence.

*Proof. *For simplicity, we assume is separable. Then there exists a countable dense subset of . We may assume that satisfies for all . Now consider

Then for the first line, by the Bolzano-Weierstrass theorem, there exists a convergent subsequence of . For such , there exists a convergent subsequence of of . Continuing this process, then we see that

exists. Now define

Then exists for all . Since is bounded in , is a Cauchy sequence for each in . So from this, is Cauchy for all in . Hence,

exists. Then is linear and

and so

So is a bounded linear functional and . So weakly star in .

For the general case, one can find the proof, e.g., Rudin or Yosida. We omit the proof of general case.

We say is \emph{reflexive }if for every , there

is such that

for all . When is reflexive, we identify and

.

**Theorem 6.** If is reflexive, then every bounded sequence in has a weakly convergent sequence.

*Proof.* Let be a bounded sequence in . Since is reflexive, is a bounded linear functional on and for all . Hence by Theorem 5, there exists a convergent subsequence of in which converges to . So for any ,

i.e.,

This shows converges weakly to in .

### 2. Compact operators and completely continuous

Recall that a mapping is said to be *continuous* if in for any sequence in . A mapping is *weakly continuous* if weakly in whenever weakly in . A mapping is said to be *completely continuous* if strongly in whenever weakly in . Finally, a mapping is *compact* if is continuous and for every bounded sequence in , there exists a subsequence such that converges strongly in .

*Remark.* Let be a bounded set in and be linear. Suppose is compact in . Then is bounded. When is linear, to check the compactness, it suffices to check is compact in whenever is a bounded set in .

By definition and Proposition 4, one can check that completely continuity implies continuity. Also, it implies weak continuity.

**Proposition 7.** If is reflexive and is completely continuous, then is compact.

* Proof. * Let be a bounded sequence in . Then by Theorem ??, there exists a subsequence for which converges weakly to in . Since is completely continuous, converges strongly to . Hence is compact.

**Example 8.** Define by . Then is compact. Let be a bounded sequence in . Then is bounded. So by the Bolzano-Weierstrass theorem, there exists a subsequence

for which converges. So is compact. Define . Then forms an orthonormal basis for . Then from Parseval’s inequality, we see that for any . So converges weakly to . But

So is not weakly continuous. Hence is not completely continuous.

**Example 9.** The identity map

is completely continuous but not compact. Note that

is non-reflexive.

**Theorem 10.** Let be linear. Then

- if is continuous, then is weakly continuous.
- if is compact, then it is completely continuous.
- if are Banach spaces, then is weakly continuous if and only if is continuous.

Proof. (i) Let and let be a bounded linear functional on . Then is

linear and

So is a bounded linear functional on . Hence

i.e.,

So weakly in .

(ii) Let weakly in . Then by Proposition 4, is bounded in . Since is compact, there is a subsequence

for which converges strongly to in . Then by (i), and so weakly in . It remains to show .

Note that weakly in . So for any bounded linear functional on ,

So for any . Now by the consequence of Hahn-Banach theorem, there exists a bounded linear functional such that

So in . Hence, strongly in . Now we are left to show that strongly in Suppose not. Then there exists and a subsequence such that

(1)

Note that is a bounded sequence. So using the compactness of again, there exists a subsequence such that in . This contradicts (1). Hence, strongly in .

(iii) Suppose is weakly continuous. Consider

Then is weakly closed subspace of . Hence is strongly closed in . Thus, is continuous by the closed graph theorem.

A normed linear space is *continuously embedded* in if and there exists a constant such that for all . We write . We say is *compactly embedded* in if is continuously embedded in and if is bounded in , then there exists a subsequence which converges strongly in .

**Example 11. **By the Sobolev embedding theorem, the inclusion map , is linear and continuous. But the embedding is not compact. Indeed, take

Then for every and is bounded in .

Note

and similarly,

But

which cannot be bounded. So it cannot have a convergent subsequence in . Also, is linear and continuous but not compact. Hence it is not completely continuous.

**Corollary 12.** Let . If is compactly embedded in and converges weakly to in , then converges strongly to in . In addition, if is reflexive, then the converse holds.

*Proof. *Since is compactly embedded in , by (i) is completely continuous. Then the corollary is proved by the definition of completely continuous. If is reflexive, the inclusion map is compact by Proposition 7.