The Big Three Pt. 3 - The Open Mapping Theorem (Banach Space)

What is open mapping

An open map is a function between two topological spaces that maps open sets to open sets. Precisely speaking, a function \(f: X \to Y\) is open if for any open set \(U \subset X\), \(f(U)\) is open in \(Y\). Likewise, a closed map is a function mapping closed sets to closed sets.

You may think open/closed map is an alternative name of continuous function. But it's not. The definition of open/closed mapping is totally different from continuity. Here are some simple examples.

  1. \(f(x)=\sin{x}\) defined on \(\mathbb{R}\) is not open, though it's continuous. It can be verified by considering \((0,2\pi)\), since we have \(f((0,2\pi))=[-1,1]\).
  2. The projection \(\pi: \mathbb{R}^2 \to \mathbb{R}\) defined by \((x,y) \mapsto x\) is open. Indeed, it maps an open ball onto an open interval on \(x\) axis.
  3. The inclusion map \(\varphi: \mathbb{R} \to \mathbb{R}^2\) by \(x \mapsto (x,0)\) however, is not open. An open interval on the plane is locally closed but not open or closed.

Under what condition will a continuous linear function between two TVS be an open mapping? We'll give the answer in this blog post. Open mapping theorem is a sufficient condition on whether a continuous linear function is open.

Open Mapping Theorem

Let \(X,Y\) be Banach spaces and \(T: X \to Y\) a surjective bounded linear map. Then \(T\) is an open mapping.

The open balls in \(X\) and \(Y\) are defined respectively by \[ B_r^X=\{x \in X:\lVert x \rVert<r\}\quad\text{and}\quad B_r^Y=\{y \in Y:\lVert y \rVert<r\} \] All we need to do is show that there exists some \(r>0\) such that \[ B_r^Y \subset T(B_1^X) \] Since every open set in \(X\) or \(Y\) can be expressed as a union of open balls. For a ball in \(X\) centered at \(x \in X\) with radius \(r\), we can express it as \(x+B_r^X\). After that, it becomes obvious that \(T\) maps open set to open set.

First we have \[ X=\bigcup_{n=1}^{\infty}B_n^{X}. \] The surjectivity of \(T\) ensures that \[ Y=\bigcup_{n=1}^{\infty}T(B_n^X). \] Since \(Y\) is Banach, or simply a complete metric space, by Baire category theorem, there must be some \(n_0 \in \mathbb{N}\) such that \(\overline{T(B_{n_0}^{X})}\) has nonempty interior. If not, which means \(T(B_n^{X})\) is nowhere dense for all \(n \in \mathbb{N}\), we have \(Y\) is of the first category. A contradiction.

Since \(x \to nx\) is a homeomorphism of \(X\) onto \(X\), we see in fact \(T(B_n^X)\) is not nowhere dense for all \(n \in \mathbb{N}\). Therefore, there exists some \(y_0 \in \overline{T(B_1^{X})}\) and some \(\varepsilon>0\) such that \[ y_0+B_\varepsilon^Y \subset \overline{T(B_1^X)} \] the open set on the left hand is a neighborhood of \(y_0\), which should be in the interior of \(\overline{T(B_1^X)}\).

On the other hand, we claim \[ \overline{T(B_1^X)} - y_0 \subset \overline{T(B_2^X)}. \] We shall prove it as follows. Pick any \(y \in \overline{T(B_1^X)}\), we shall show that \(y-y_0 \in \overline{T(B_2^X)}\). For \(y_0\), there exists a sequence of \(y_n\) where \(\lVert y_n \rVert <1\) for all \(n\) such that \(Ty_n \to y_0\). Also we are able to find a sequence of \(x_n\) where \(\lVert x_n \rVert <1\) for all \(n\) such that \(Tx_n \to y\). Notice that we also have \[ y-y_0=\lim_{n \to \infty}T(x_n-y_n), \] since \[ \lVert x_n -y_n \rVert \leq \lVert x_n \rVert+\lVert y_n \rVert <2, \] we see \(T(x_n-y_n) \in T(B_2^X)\) for all \(n\), it follows that \[ y-y_0 \in \overline{T(B_2^X)}. \] Combining all these relations, we get \[ B_\varepsilon^Y \subset \overline{T(B_2^X)}. \] Since \(T\) is linear, we see \[ 2B_{\varepsilon/2}^{Y} \subset \overline{T(2B_1^X)}=2\overline{T(B_1^X)}. \] By induction we get \[ B_{\varepsilon/2^n}^Y \subset \overline{T(B_{1/2^{n-1}}^X)} \] for all \(n \geq 1\).

We shall show however \[ B_{\varepsilon/4}^Y \subset T(B_1^X). \] For any \(u \in B_{\varepsilon/4}^Y\), we have \(u \in \overline{T(B_{1/2}^X)}\). There exists some \(x_1 \in B_{1/2}^{X}\) such that \[ \lVert u-Tx_1 \rVert < \frac{\varepsilon}{8}. \] This implies that \(u-Tx_1 \in B_{\varepsilon/8}^Y\). Under the same fashion, we are able to pick \(x_n\) in such a way that \[ \lVert u-Tx_1-Tx_2-\cdots-Tx_n \rVert < \frac{\varepsilon}{2^{n+2}} \] where \(\lVert x_n \rVert<2^{-n}\). Now let \(z_n=\sum_{k=1}^{n}x_k\), we shall show that \((z_n)\) is Cauchy. For \(m<n\), we have \[ \lVert z_n - z_m \rVert =\left\Vert\sum_{k=m+1}^nx_k \right\Vert \leq \sum_{k=m+1}^{n}\lVert x_k\rVert < \frac{1}{2^{m+1}} \] Since \(X\) is Banach, there exists some \(z \in X\) such that \(z_n \to z\). Further we have \[ \lVert z\rVert = \lim_{n \to \infty}\lVert z_n \rVert \leq \sum_{k=1}^{\infty}\lVert x_n \rVert < 1 \] therefore \(z \in B_1^X\). Since \(T\) is bounded, therefore continuous, we get \(T(z)=u\). To summarize, for \(u \in B_{\varepsilon/4}^Y\), we have some \(z \in B_{1}^X\) such that \(T(z)=y\), which implies \(T(B_1^X) \supset B_{\varepsilon/4}^Y\).

Let \(U \subset X\) be open, we want to show that \(T(U)\) is also open. Take \(y \in T(U)\), then \(y=T(x)\) with \(x \in U\). Since \(U\) is open, there exists some \(\varepsilon>0\) such that \(B_{\varepsilon}^{X}+x \subset U\). By the linearity of \(T\), we obtain \(B_{r\varepsilon}^Y \subset T(B_{\varepsilon}^X)\) for some small \(r\). Using the linearity of \(T\) again, we obtain \[ B_{r\varepsilon}^Y + y \subset T(B_{\varepsilon}^X+x) \subset T(U) \] which shows that \(T(U)\) is open, therefore \(T\) is an open mapping.

Remarks

One have to notice that the completeness of \(X\) and \(Y\) has been used more than one time. For example, the existence of \(z\) depends on the fact that Cauchy sequence converges in \(X\). Also, the surjectivity of \(T\) cannot be omitted, can you see why?

There are some different ways to state this theorem.

  • To every \(y\) with \(\lVert y \rVert < \delta\), there corresponds an \(x\) with \(\lVert x \rVert<1\) such that \(T(x)=y\).
  • Let \(U\) and \(V\) be the open unit balls of the Banach spaces \(X\) and \(Y\). To every surjective bounded linear map, there corresponds a \(\delta>0\) such that

\[ T(U) \supset \delta{V}. \]

You may also realize that we have used a lot of basic definitions of topology. For example, we checked the openness of \(T(U)\) by using neighborhood. The set \(\overline{T(B_1^X)}\) should also remind you of limit point.

The difference of open mapping and continuous mapping can be viewed via the topologies of two topological vector spaces. Suppose \(f: X \to Y\). If for any \(U \in \tau_X\), we have \(f(U) \in \tau_Y\), where \(\tau_X\) and \(\tau_Y\) are the topologies of \(X\) and \(Y\), respectively. But this has nothing to do with continuity. By continuity we mean, for any \(V \in \tau_Y\), we have \(f^{-1}(V) \in \tau_U\).

Fortunately, this theorem can be generalized to \(F\)-spaces, which will be demonstrated in the following blog post of the series. A space \(X\) is an \(F\)-space if its topology \(\tau\) is induced by a complete invariant metric \(d\). Still, completeness plays a critical rule.

The series

Since there is no strong reason to write more posts on this topic, i.e. the three fundamental theorems of linear functional analysis, I think it's time to make a list of the series. It's been around half a year.

The Big Three Pt. 3 - The Open Mapping Theorem (Banach Space)

https://desvl.xyz/2020/08/12/big-3-pt-3/

Author

Desvl

Posted on

2020-08-12

Updated on

2021-10-06

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