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

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 $F$ in a metric space is said to be nowhere dense if the interior of $\overline{F}$ 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 $X$ is a complete metric space, then $X$ is of the second category.

Corollary 2: Let $X$ be a complete metric space, and let $F_i$ be a sequence of closed subsets of $X$. If $X = \cup_i F_i$, then for some $j$,  $F_j$ has non-empty interior.

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

Theorem 3 (the principle of uniform boundedness): Let $X$ be a Banach space, let $Y$ be a normed space, and let $\mathcal{F} \subseteq B(X,Y)$ be a family of operators. If for all $x \in X$

$\sup \{\|Tx\| : T \in \mathcal{F}\} < \infty$

then  $\sup_{T \in \mathcal{F}} \|T\| < \infty$

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

Since $X$ 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 $X$ be a Banach space, $Y$ a normed space, and $\mathcal{F} \subseteq B(X,Y)$ a fmaily of operators. Let $A \subseteq X$ be the set of all $x \in X$  for which

$\sup \{\|Tx\| : T \in \mathcal{F}\} < \infty .$

If $A$ is of the second category, then  $\sup_{T \in \mathcal{F}} \|T\| < \infty$

Proof: For all $T \in \mathcal{F}$ and $n \in \mathbb{N}$, let $A_{n,T} = \{x \in X: \|Tx\| \leq n\}$. This set is closed, hence the set

$A_n = \{x \in X : \sup_{T \in \mathcal{F}}\|Tx\| \leq n\} = \cap_{T \in \mathcal{F}} A_{n,T}$

is closed, too. But $A = \cup_{n=1}^\infty A_n$. As $A$ is of the second category, there is some integer $N$ such that $A_N$ has non-empty interior. Suppose that the open ball of radius $r$ centered at $y$ is contained in $A_N$. In other words, for all $z \in X$, $\|z-y\| implies that $\|Tz\| \leq N$ for all $T$. Rearranging this information, we find that for all $x \in B(0,r)$,

$\|Tx\| \leq \|T(x+y)\|+ \|T(y)\| \leq 2N .$

We conclude that for all $T \in \mathcal{F}$,  $\|T\| \leq 2 N r^{-1}$ . That completes the proof.

We now come to the second big theorem. Recall that a map $T:X \rightarrow Y$ is said to be open if $T(U)$ is open in $Y$ whenever $U$ is open in $X$.

Theorem 5 (the open mapping theorem): Let $X$ and $Y$ be Banach spaces, and let $T \in B(X,Y)$. If $T$ 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 $x \in X$, and $B(x,r)$ is the open ball of radius $r$ around $x$, then $T(B(x,r))$ contains an open ball centered at $Tx$. But by linearity, this reduces to showing that $T(B(0,1))$ contains an open ball centered at $0$.

Since $T$ is surjective, we have that $Y = \cup_{n=1}^\infty T(B(0,n))$. By the Baire category theorem one of the sets $\overline{T(B(0,n))} = 2 n \overline{T(B(0,1/2))}$ has non-empty interior. Thus $\overline{T(B(0,1/2)}$ contains an open ball, say $B(y,r) \subseteq \overline{T(B(0,1/2))}$. We want to show, as an intermediate step, that there is an open ball centered at $0$ that is contianed in $\overline{T(B(0,1))}$. In fact we will show that $B(0,r/2) \subseteq \overline{T(B(0,1))}$.

Let $x \in B(0,1/2)$ such that $\|Tx-y\| < r/2$. Now for all $z \in B(0,r/2)$, $\|Tx+z - y\|, so $Tx+z \in \overline{T(B(0,1/2))}$. It follows that $z = Tx + z - Tx \in \overline{T(B(0,1/2))} + T(B(0,1/2)) \subseteq \overline{T(B(0,1/2) + T(B(0,1/2))} = \overline{T(B(0,1))}$, thus $B(0,r/2) \subseteq \overline{T(B(0,1))}$. This is still not good enough, we have to show that $T(B(0,1))$ —not its closure — contains an open ball around the origin.

Since $T$ is open if and only if $2/r T$ is open, we may as well scale $T$ and assume that we proved that

(*) $B(0,1) \subseteq \overline{T(B(0,1))}$.

We will show that (*) implies that $B(0,1/2) \subseteq T(B(0,1))$. That will complete the proof.

Now, it follows from (*) that

(**) $B(0,2^{-k}) \subseteq \overline{T(B(0,2^{-k}))}$

for all $k=1,2, \ldots$. Let $y \in B(0,1/2)$. We will construct a sequence $x_1, x_2, \ldots$ such that $\sum x_k$ converges to an element in $B(0,1)$ and such that $T(\sum x_k) = y$. Choose $x_1 \in B(0,1/2)$ such that $\|y - T x_1 \| < 1/4$. This is possible thanks to (**). Using (**) again, choose $x_2 \in B(0,1/4)$ such that $\|(y - T x_1) - T x_2\|<1/8$. Continuing this way, we construct a sequence $x_1, x_2, \ldots$ such that $\|x_k\|<2^{-k}$ and $\|y - T(\sum_{k=1}^n x_k) \|< 2^{-n-1}$. It follows that $x:= \sum x_k \in B(0,1)$, and from the continuity of $T$ is follows that $Tx = y$.

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

Remark: A corollary of the open mapping theorem is point 4 of Theorem 6 in Notes 7: that the quotient map $\pi : X \rightarrow X/M$ 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 $T$ is a bijective linear map, then $T$ 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 $X$ and $Y$ be Banach spaces. Let $T : X \rightarrow Y$ be a bijective linear map which is bounded. Then the inverse of $T$ is bounded.

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

Definition 7: Suppose that $X$ is a vector space, and let $\|\cdot\|$ and $\| \cdot\|'$ be two norms defined on $X$. The norms are said to be equivalent if there are constants $c, C$ such that for all $x \in X$

$c\|x\| \leq \|x\|' \leq C\|x\| .$

Corollary 8: Let $X$ be a Banach space with a norm $\|\cdot\|$. Suppose that there is another norm $\| \cdot \|'$ defined on $X$ which induces a complete metric, and suppose that there exists a constant $C$ such that

$\|x\|' \leq C\|x\|$

for all $x \in X$. Then $\|\cdot\|$ and $\| \cdot \|'$ are equivalent.

Proof: The identity mapping from $(X,\|\cdot\|)$ to $(X,\| \cdot\|')$ is bounded and bijective. By the inverse mapping theorem, its inverse is bounded.

Definition 9: Let $T:X \rightarrow Y$. The graph of  $T$denoted $G(T)$, is the set

$G(T) = \{(x,y) \in X \oplus Y : y = Tx\} .$

$T$ is said to be closed if its graph is closed.

Here we take the natural product norm on $X \oplus Y$, namely $\|(x,y)\| = \|x\|_X + \|y\|_Y$. Other natural choices are equivalent.

It is clear that $G(T)$ is a linear subspace of $X \oplus Y$.  If $T$ is continuous, then $G(T)$ is closed. Indeed, suppose that $(x_n, T x_n) \in G(T)$ and $(x_n,T x_n) \rightarrow (x,y)$. Then $x_n \rightarrow x$, so $T x_n \rightarrow Tx$, whence $y = Tx$ and $(x,y) \in G(T)$.

Theorem 10 (the closed graph theorem): Let $T : X \rightarrow Y$ be a closed linear map. Then $T$ is continuous.

Proof: $G(T)$ is a closed subspace of $X \oplus Y$, hence a Banach space. The projections $\pi_X : G(T) \rightarrow X$ and $\pi_Y : G(T) \rightarrow Y$ are continuous, and $\pi_X$ is invertible. By the inverse mapping theorem, $\pi_X^{-1}$ is bounded. Hence $T = \pi_Y \circ \pi_X^{-1}$ is also bounded.

#### 2. Complemented subspaces

Let $X$ be a vector space and let $M$ be a linear subspace. By linear algebra, there always exists an alegbraic complement, meaning a subspace $N$ such that $M \cap N = \{0\}$ and $M + N = X$. This is the same as saying that every $x \in X$ can be written as $m+n$ for $m \in M, n \in N$ in a unique way. The choice of $N$ is not unique. Holding a complement fixed we may define the projection onto $M$ parallel to $N$ to be the map $P : X \rightarrow M$ defined by $P(m+n) = m$. It is clear that $I - P$ is then the projection of $X$ onto $N$ parallel to $M$. $P$ satisfies $P^2 = P$, $R(P) = N(I-P) = M$ and $N(P) = N$. Moreover, if $Q$ is a linear operator satisfying $Q^2 = Q$, and $M,N$ are subspaces of $X$ such that $R(Q) = M$ and $N(Q) = N$, then $M$ are $N$ are algebraic complements of one another, and $Q$ is the projection onto $M$ parallel to $N$.

Now let $X$ be a Banach space, and $M$ a closed subspace. $M$ can be complemented algebraically, but this construction is usually not useful or appropriate in the category of banach spaces. $M$ 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 $N \subseteq X$ such that $X = M + N$ and $M \cap N = \{0\}$. So $M$ is topologically complemented if and only if it has a closed complement. In this situation we write $X = M \oplus N$.

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

Proof: Let $P$ be a continuous projection. Then $X = R(P) + N(P)$, and $R(P) \cap N(P) = \{0\}$ from the discussion of the purely algebraic case. Now $N(P)$ is closed because $P$ is continuous, and $R(P) = N(I-P)$ so it closed, too.

Conversely, suppose that $X = M \oplus N$. Then we may define algebraically the projection $P$ onto $M$ parallel to $N$, and it holds that $R(P) = M$. We must show that $P$ is continuous. By the closed graph theorem, it suffices to show that $G(P)$ is closed.

Let $(x_k, P x_k) \in G(P)$ converge to $(x,y)$. Then $x_k = m_k + n_k$, and $P x_k = m_k$. Thus $m_k \rightarrow y$, so $y \in M$. Therefore $y = Py$. What we need to show is that $y = Px$. The sequences $x_k, m_k$ converge, so $n_k = x_k - m_k$ converges to $x - y$. Therefore $x - y \in N = N(P)$, and we conclude that $Px = Py = y$.