### Advanced Analysis, Notes 7: Banach spaces (dual spaces and duality, Lp spaces, the double dual, quotient spaces)

#### by Orr Shalit

Today we continue our treatment of the dual space of a normed space (usually Banach) . We start by considering a wide class of Banach spaces and their duals.

#### 1. spaces

Perhaps the best studied class of examples of infinite dimensional Banach spaces which are not Hilbert spaces are the spaces. In this section we collect some important facts about spaces which we shall use to illustrate some results or in applications to analysis.

In this section, we will assume some experience with measure theory.

Let be a measure space. Recall that this means that is some set, is a -algebra of subsets of (this just means that is a collection of subsets of that contains and is closed under taking complements and countable unions) and is a function that satisfies:

- , and
- If is a countable sequence of disjoint sets, then .

The function is called a measure, and elements of are called measurable sets. All our measure spaces will be assumed finite, meaning that there is a sequence of measurable sets such that and for all .

For every measurable function and every , we define

We also define

For we define to be the set of all measurable functions for which . Of course, we will identify functions which are equal almost everywhere. The spaces arise as spaces when and is the counting measure.

**Definition:** *A function is said to be simple if it as the form *

*where , all have finite measure, and denotes the characteristic function of . *

We shall denote by the vector space of all simple functions. clearly, is contained in for all . It is a standard and important fact, that we shall use below, that if , then is dense in

**Theorem 1: ***For , is a Banach space. For , the space of simple functions of finite measure support are dense in . The set of simple functions is dense in .*

**Proof:** We prove the theorem for ; is left as an excercise. By Minkowski’s inequality is a normed space, and the issue we delve on is completeness.

We will use the fact that a normed space is complete if and only if every absolutely convergent sequence in the space is convergent (see the homework exercises). Thus, it suffices to show that if are elements of such that , then converges in the norm of to an element of this space.

Let be such a sequence. For every , define

The sequence converges everywhere to a to a function that has values in . For all ,

By the monotone convergence theorem,

thus $F \in L^p$. In particular, the function is finite (meaning that its value is not ) almost everywhere. It follows that the sequence converges absolutely for almost every , hence it defines a measurable function. Since , we have that .

Now we will show that in the norm of . For this we have to show that

(*).

But $latex |f – \sum_{k=1}^N f_k |^p\rightarrow 0$ pointwise almost everywhere. On the other hand,

for all . Thus (*) follows from the dominated convergence theorem. That completes the proof.

Given , it is customary to denote , so that . is called the conjugate exponent of .

**Theorem 2 (Holder’s inequality):** .

For a proof, see this.

It is immediate from Holder’s inequality that every gives rise to a functional by way of

By Holder’s inequality, . A little more effort shows that (this will be shown below). It is a deeper fact (to be shown further down below) that is actually an isometric isomorphism from **onto** .

**Theorem 3:** *Let . Every continuous linear functional on is of the form for some . *

We may summarize the above discussion as for all . The rest of this section is devoted to proving this fact. The proof will follow pretty much the proof from Folland’s book “Real Analysis”. We fix a -finite measure space , and let be related by .

**Lemma:** *For all , . *

**Proof:** We already know that , so it remains to show the reverse. We may assume that .

Assume first that . Define . Here is the function that returns for every complex number the phase . One checks readily that and that . Thus

so .

Now we treat the case . Fix . The set has nonzero measure. Let be a subset of such that . Let . Then , and . Moreover,

Since this holds for all , we find . The other inequality has already been established, thus .

**Lemma:** Let be measurable and assume that

satisfies . Then and .

**Proof:** By Holder’s inequality. , so we will only prove (and this will also give us ). We treat the cases and separately.

The case :

There is a sequence of simple functions satisfying and pointwise (such a sequence, which is easy to construct by hand, is constructed in any course on measure theory). Define . Then , . Then

where the first inequality follows from Fatou’s Lemma.

The case :

Let , and put . We shall show that , from which it would follow that .

Indeed, if , then (from -finiteness) there is such that . Define . Then , and ; this is a contradiction to the very definition of . Thus , and the proof is complete.

**Lemma:** With the notation of the previous lemma,

.

**Proof:** It is clear that the right hand side is less than or equal to . For the reverse inequality, let , and let have -norm such that

Let . Then . Note, however, that is not necessarily simple. But there is a sequence of simple functions such that . By the dominated convergence theorem, . This shows established the required inequality.

**Proof of Theorem 3:** We already know that the map given by is isometric (here as above is the functional ). It remains to show that this map is surjective. Recall that we are assuming that ; it is your job to find where we use this assumption.

Assume first that is a finite measure. Let . We seek a function such that . To obtain this function, define a function by

for . By finiteness of , the function is in so is well defined. In fact, it turns out that is a complex valued measure. Ineed, if is the dsijoint union , then pointwise and hence (by monotone convergence) in norm, whence

Now if has -measure zero, then in , hence . Thus , so by the Radon-Nikodym Theorem, it follows that there is some such that for all . From this we have that

(*)

for all . The lemma above implies that and that . Examining (*) again, we obtain by density of that .

Finally, it remains to prove the theorem for a -finite measure. As the heart of the proof is behind us, this is left as an exercise for the reader.

#### 2. Duality and the double dual

Corollary 14 in the previous lecture is a generalization of the fact that in an inner product space, if a vector is orthogonal to all other vectors in the space then . This suggests the following notation. In some texts, when considering the action of on , the notation is used instead of . This invites one to think of the relationship between and as something more symmetrical. Sometimes, elements of the dual space are denoted as . Let’s use this notation for a little while to make it familiar.

One can learn many things about a Banach space from its dual . For example:

**Exercise A:** If is separable, then is separable, too.

**Example:** The example of shows that the converse is false. In addition, the exercise shows that .

Let . We denote . If then we denote . The spaces and are called the annihilator of and the pre-annihilator of , respectively. With this notation we can state Corollary 16 from the previous lecture as follows: *A subspace contains in its closure if and only if . In other words, . *

**Exercise B: **True or false: ?

Every induces a function by way of

For every normed space , we denote by the (norm) closed unit ball of .

**Proposition 4:** *The map is an isometry from into . *

**Proof:** Linearity is trivial. To see that the map is norm preserving,

where for the second equality we used Corollary 13 from the last lecture.

**Definition 5: ***A Banach space is said to be reflexive if the map is an isometry of onto . *

**Example:** is reflexive for all . and are examples of non-reflexive spaces (as Exercise A and the example following it show). It follows non of these spaces is isomorphic to an space for .

#### 3. Quotient spaces

Let be a normed space, and let be a closed subspace. In linear algebra we learn how to form the **quotient space** , which is defined to be the set of all cosets , , with the operations

It is known and also easy to show that these operations are well defined, and give the structure of a vector space. Our goal now is to make into a normed space, and to prove that when is complete, so is .

To abbreviate, let us write for . We define

Let us denote by the quotient map .

**Theorem 6:** *With the above defined norm, the following hold:*

*is a normed space.**is a contraction: for all .**For every such that , there is an with such that .*- is open in if and only if is open in .
*If is a closed subspace containing , then is closed.**If is complete, then so is .*

**Proof: **1. is trivial. follows from the fact that is closed. For every , let be such that and are very close to and , respectively. Then

and the right hand side is very close to , so can only be very slightly bigger than . That proves the triangle inequality.

**Exercise C:** Complete the proof of the theorem.

**Remark:** Note that by 4 of the above theorem, the topology induced by the quotient norm is precisely the *quotient topology* which one defines in topology.

**Theorem 7: ***Let be a normed space and a closed subspace. Then*

*is isometrically isomorphic to .**is isometrically isomorphic to .*

**Proof: **1. Define be

where is any extension of to . This is well defined, because if is another extension of then . Well defined-ness implies linearity (think about it). is also surjective, sine for all . It remains to prove that is isometric.

Let and let be an extension. Then . But each functional extends , so , whence . On the other hand, as ranges over all , ranges over all extensions of . By Hahn-Banach, there is some extension, say , such that . Thus the infimum is attained and .

2. Define a linear map by

It is obvious that , since vanishes on . We need to show that is a surjective isometry.

By 2 of Theorem 6, . On the other hand,

Using 3 of Theorem 6, maps the unit ball of onto the unit ball of , so the right hand side is equal to

It remains to show that the range of is equal to . Let . Then . It follows that if , then , so we may define a functional on by

By definition . The kernel of is equal to , and is closed by Theorem 6. By Exercise F in Notes 6, is continuous. That completes the proof.

#### 4. The adjoint of an operator

The idea of the proof in 2 of Theorem 7 — the construction of the map — is one of general applicability.

Let . Then we define by way of

**Theorem 8:** *Let and be normed spaces, and let . Then the above defined satisfies:*

*, so is linear and bounded.**is satisfies for all . Moreover, is the unique function from to with this property.*- .

**Exercise D: **Prove Theorem 8.

**Theorem 9:** *Let and be normed spaces, and let . Then and . *

**Proof:** Thanks to the notation, the proof is *exactly* the same as in the Hilbert space case (note, though, that both assertions require proof, and do not follow one from the other by conjugation).

**Corollary 10:** *For , . *

**Proof:** This follows from Corollary 16 in the previous lecture.

**Exercise E:** Give an example of an operator for which .

[…] A corollary of the open mapping theorem is point 4 of Theorem 6 in Notes 7: that the quotient map is open. However this fact is elementary, and can also be used to give an […]

[…] the end of Notes 7 we obtained the relations and . We also noted that in general […]

The definition of the set of functionals which vanish on S: “=0” is missing

Thanks Alon, I corrected it.

Thanks for all your corrections!