### Advanced Analysis, Notes 9: Banach spaces (the three big theorems)

#### by Orr Shalit

Until now we had not yet seen a theorem about *Banach spaces — *the Hahn–Banach theorems did not require the space to be complete. In this post we learn the three big theorems about operators on Banach spaces: the* principle of uniform boundedness, *the* open mapping theorem, *and the *closed graph theorem. *It is common that these three theorems are presented in texts on functional analysis under the heading “consequences of the Baire category theorem“.

#### 1. The three big theorems

Recall that a set in a metric space is said to be **nowhere dense** if the interior of is empty. A set is said to be of the** first category **(or **meager**), if it is the union of countably many nowhere dense sets. A set is of the **second category** if it is not of the first category. Finally, a set is said to be **generic** (or **residual**) if it is the complement of a set of the first category. We begin by recalling the Baire category theorem.

**Theorem 1 (Baire category theorem): ***If is a complete metric space, then is of the second category. *

**Corollary 2: ***Let be a complete metric space, and let be a sequence of closed subsets of . If , then for some , has non-empty interior. *

If is a Banach space then is a complete metric space, and the Baire category theorem has profound consequences for the operator theory on .

**Theorem 3 (the principle of uniform boundedness): ***Let be a Banach space, let be a normed space, and let be a family of operators. If for all , *

*then . *

**Remark:** The uniform boundedness theorem is often referred to as the *Banach–Steinhaus* *theorem*.

Since is of the second category (by the category theorem), Theorem 3 follows immediately from the following stronger version of the uniform boundedness principle.

**Theorem 4: ***Let be a Banach space, a normed space, and a fmaily of operators. Let be the set of all for which *

*If is of the second category, then . *

**Proof: **For all and , let . This set is closed, hence the set

is closed, too. But . As is of the second category, there is some integer such that has non-empty interior. Suppose that the open ball of radius centered at is contained in . In other words, for all , implies that for all . Rearranging this information, we find that for all ,

We conclude that for all , . That completes the proof.

We now come to the second big theorem. Recall that a map is said to be open if is open in whenever is open in .

**Theorem 5 (the open mapping theorem):** *Let and be Banach spaces, and let . If is surjective, then it is an open mapping. *

**Remark:** The open mapping theorem is sometimes referred to as the *Banach–Schauder theorem*.

**Proof: **It suffices to show that if , and is the open ball of radius around , then contains an open ball centered at . But by linearity, this reduces to showing that contains an open ball centered at .

Since is surjective, we have that . By the Baire category theorem one of the sets has non-empty interior. Thus contains an open ball, say . We want to show, as an intermediate step, that there is an open ball *centered at * that is contianed in . In fact we will show that .

Let such that . Now for all , , so . It follows that , thus . This is still not good enough, we have to show that —not its closure — contains an open ball around the origin.

Since is open if and only if is open, we may as well scale and assume that we proved that

(*) .

We will show that (*) implies that . That will complete the proof.

Now, it follows from (*) that

(**)

for all . Let . We will construct a sequence such that converges to an element in and such that . Choose such that . This is possible thanks to (**). Using (**) again, choose such that . Continuing this way, we construct a sequence such that and . It follows that , and from the continuity of is follows that .

**Remark:** Note that, unlike Theorem 4, this theorem requires both and to be complete. Where did we use the completeness of ? Of ?

**Remark:** 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 alternative proof of the open mapping theorem (perhaps we will see this, remind me).

The following is an important application of the open mapping theorem. Recall that if is a bijective linear map, then has a (unique) linear inverse. It is non-trivial that the inverse of a bounded invertible map is bounded. Show me another proof if you can.

**Corollary 6 (inverse mapping theorem):** *Let and be Banach spaces. Let be a bijective linear map which is bounded. Then the inverse of is bounded. *

**Proof:** By the open mapping theorem, the inverse of is continuous.

**Definition 7: ***Suppose that is a vector space, and let and be two norms defined on . The norms are said to be equivalent if there are constants such that for all , *

**Corollary 8: ***Let be a Banach space with a norm . Suppose that there is another norm defined on which induces a complete metric, and suppose that there exists a constant such that *

*for all . Then and are equivalent. *

**Proof:** The identity mapping from to is bounded and bijective. By the inverse mapping theorem, its inverse is bounded.

**Definition 9: ***Let . The graph of , denoted , is the set *

* is said to be closed if its graph is closed. *

Here we take the natural product norm on , namely . Other natural choices are equivalent.

It is clear that is a linear subspace of . If is continuous, then is closed. Indeed, suppose that and . Then , so , whence and .

**Theorem 10 (the closed graph theorem): **Let be a closed linear map. Then is continuous.

**Proof:** is a closed subspace of , hence a Banach space. The projections and are continuous, and is invertible. By the inverse mapping theorem, is bounded. Hence is also bounded.

#### 2. Complemented subspaces

Let be a vector space and let be a linear subspace. By linear algebra, there always exists an **alegbraic complement, **meaning a subspace such that and . This is the same as saying that every can be written as for in a unique way. The choice of is not unique. Holding a complement fixed we may define the **projection onto parallel to ** to be the map defined by . It is clear that is then the projection of onto parallel to . satisfies , and . Moreover, if is a linear operator satisfying , and are subspaces of such that and , then are are algebraic complements of one another, and is the projection onto parallel to .

Now let be a Banach space, and a closed subspace. can be complemented algebraically, but this construction is usually not useful or appropriate in the category of banach spaces. is said to be **topologocally complemented **(or simply **complemented**, when it is understood in which category we are working) if there exists a closed subspace such that and . So is topologically complemented if and only if it has a closed complement. In this situation we write .

**Theorem 11:** *A closed subspace of a Banach space is (topologically) complemented if and only if is the range of a continuous projection.*

**Proof: **Let be a continuous projection. Then , and from the discussion of the purely algebraic case. Now is closed because is continuous, and so it closed, too.

Conversely, suppose that . Then we may define algebraically the projection onto parallel to , and it holds that . We must show that is continuous. By the closed graph theorem, it suffices to show that is closed.

Let converge to . Then , and . Thus , so . Therefore . What we need to show is that . The sequences converge, so converges to . Therefore , and we conclude that .

a simple proof for the principle of uniform boundedness that does not use the Baire category theorem:

http://arxiv.org/abs/1005.1585v2

Thanks for the link! It is indeed a very nice and very simple proof.

Note remark 5 in the paper – the Baire category techniques allow to obtain a slightly stronger results, such as Theorem 4 above. One then has (slightly ?) stronger applications, such as Theorem 6 in the next post.

I just read a claim that the original proof of the Uniform Boundedness principle (by Banach and Steinhauss) did not use the Baire category theorem.

See “A Short Course on Banach Space Tmeory”, by Carothers (LMSST64), page 52:

“As the story goes, Saks, the referee of their paper, suggested an alternate proof … using the Baire category theorem…”

Carothers then says that it is a fair guess that their original proof used a “gliding hump” technique.