The Big Three Pt. 4 - The Open Mapping Theorem (F-Space)

The Open Mapping Theorem

We are finally going to prove the open mapping theorem in \(F\)-space. In this version, only metric and completeness are required. Therefore it contains the Banach space version naturally.

(Theorem 0) Suppose we have the following conditions:

  1. \(X\) is a \(F\)-space,
  2. \(Y\) is a topological space,
  3. \(\Lambda: X \to Y\) is continuous and linear, and
  4. \(\Lambda(X)\) is of the second category in \(Y\).

Then \(\Lambda\) is an open mapping.

Proof. Let \(B\) be a neighborhood of \(0\) in \(X\). Let \(d\) be an invariant metric on \(X\) that is compatible with the \(F\)-topology of \(X\). Define a sequence of balls by \[ B_n=\{x:d(x,0) < \frac{r}{2^n}\} \] where \(r\) is picked in such a way that \(B_0 \subset B\). To show that \(\Lambda\) is an open mapping, we need to prove that there exists some neighborhood \(W\) of \(0\) in \(Y\) such that \[ W \subset \Lambda(B). \] To do this however, we need an auxiliary set. In fact, we will show that there exists some \(W\) such that \[ W \subset \overline{\Lambda(B_1)} \subset \Lambda(B). \] We need to prove the inclusions one by one.

The first inclusion requires BCT. Since \(B_2 -B_2 \subset B_1\), and \(Y\) is a topological space, we get \[ \overline{\Lambda(B_2)}-\overline{\Lambda(B_2)} \subset \overline{\Lambda(B_2)-\Lambda(B_2)} \subset \overline{\Lambda(B_1)} \] Since \[ \Lambda(X)=\bigcup_{k=1}^{\infty}k\Lambda(B_2), \] according to BCT, at least one \(k\Lambda(B_2)\) is of the second category in \(Y\). But scalar multiplication \(y\mapsto ky\) is a homeomorphism of \(Y\) onto \(Y\), we see \(k\Lambda(B_2)\) is of the second category for all \(k\), especially for \(k=1\). Therefore \(\overline{\Lambda(B_2)}\) has nonempty interior, which implies that there exists some open neighborhood \(W\) of \(0\) in \(Y\) such that \(W \subset \overline{\Lambda(B_1)}\). By replacing the index, it's easy to see this holds for all \(n\). That is, for \(n \geq 1\), there exists some neighborhood \(W_n\) of \(0\) in \(Y\) such that \(W_n \subset \overline{\Lambda(B_n)}\).

The second inclusion requires the completeness of \(X\). Fix \(y_1 \in \overline{\Lambda(B_1)}\), we will show that \(y_1 \in \Lambda(B)\). Pick \(y_n\) inductively. Assume \(y_n\) has been chosen in \(\overline{\Lambda(B_n)}\). As stated before, there exists some neighborhood \(W_{n+1}\) of \(0\) in \(Y\) such that \(W_{n+1} \subset \overline{\Lambda(B_{n+1})}\). Hence \[ (y_n-W_{n+1}) \cap \Lambda(B_n) \neq \varnothing \] Therefore there exists some \(x_n \in B_n\) such that \[ \Lambda x_n = y_n - W_{n+1}. \] Put \(y_{n+1}=y_n-\Lambda x_n\), we see \(y_{n+1} \in W_{n+1} \subset \overline{\Lambda(B_{n+1})}\). Therefore we are able to pick \(y_n\) naturally for all \(n \geq 1\).

Since \(d(x_n,0)<\frac{r}{2^n}\) for all \(n \geq 0\), the sums \(z_n=\sum_{k=1}^{n}x_k\) converges to some \(z \in X\) since \(X\) is a \(F\)-space. Notice we also have \[ \begin{aligned} d(z,0)& \leq d(x_1,0)+d(x_2,0)+\cdots \\ & < \frac{r}{2}+\frac{r}{4}+\cdots \\ & = r \end{aligned} \] we have \(z \in B_0 \subset B\).

By the continuity of \(\Lambda\), we see \(\lim_{n \to \infty}y_n = 0\). Notice we also have \[ \sum_{k=1}^{n} \Lambda x_k = \sum_{k=1}^{n}(y_k-y_{k+1})=y_1-y_{n+1} \to y_1 \quad (n \to \infty), \] we see \(y_1 = \Lambda z \in \Lambda(B)\).

The whole theorem is now proved, that is, \(\Lambda\) is an open mapping. \(\square\)


You may think the following relation comes from nowhere: \[ (y_n - W_{n+1}) \cap \Lambda(B_{n}) \neq \varnothing. \] But it's not. We need to review some set-point topology definitions. Notice that \(y_n\) is a limit point of \(\Lambda(B_n)\), and \(y_n-W_{n+1}\) is a open neighborhood of \(y_n\). If \((y_n - W_{n+1}) \cap \Lambda(B_{n})\) is empty, then \(y_n\) cannot be a limit point.

The geometric series by \[ \frac{\varepsilon}{2}+\frac{\varepsilon}{4}+\cdots+\frac{\varepsilon}{2^n}+\cdots=\varepsilon \] is widely used when sum is taken into account. It is a good idea to keep this technique in mind.


The formal proof will not be put down here, but they are quite easy to be done.

(Corollary 0) \(\Lambda(X)=Y\).

This is an immediate consequence of the fact that \(\Lambda\) is open. Since \(Y\) is open, \(\Lambda(X)\) is an open subspace of \(Y\). But the only open subspace of \(Y\) is \(Y\) itself.

(Corollary 1) \(Y\) is a \(F\)-space as well.

If you have already see the commutative diagram by quotient space (put \(N=\ker\Lambda\)), you know that the induced map \(f\) is open and continuous. By treating topological spaces as groups, by corollary 0 and the first isomorphism theorem, we have \[ X/\ker\Lambda \simeq \Lambda(X)=Y. \] Therefore \(f\) is a isomorphism; hence one-to-one. Therefore \(f\) is a homeomorphism as well. In this post we showed that \(X/\ker{\Lambda}\) is a \(F\)-space, therefore \(Y\) has to be a \(F\)-space as well. (We are using the fact that \(\ker{\Lambda}\) is a closed set. But why closed?)

(Corollary 2) If \(\Lambda\) is a continuous linear mapping of an \(F\)-space \(X\) onto a \(F\)-space \(Y\), then \(\Lambda\) is open.

This is a direct application of BCT and open mapping theorem. Notice that \(Y\) is now of the second category.

(Corollary 3) If the linear map \(\Lambda\) in Corollary 2 is injective, then \(\Lambda^{-1}:Y \to X\) is continuous.

This comes from corollary 2 directly since \(\Lambda\) is open.

(Corollary 4) If \(X\) and \(Y\) are Banach spaces, and if \(\Lambda: X \to Y\) is a continuous linear bijective map, then there exist positive real numbers \(a\) and \(b\) such that \[ a \lVert x \rVert \leq \lVert \Lambda{x} \rVert \leq b\rVert x \rVert \] for every \(x \in X\).

This comes from corollary 3 directly since both \(\Lambda\) and \(\Lambda^{-1}\) are bounded as they are continuous.

(Corollary 5) If \(\tau_1 \subset \tau_2\) are vector topologies on a vector space \(X\) and if both \((X,\tau_1)\) and \((X,\tau_2)\) are \(F\)-spaces, then \(\tau_1 = \tau_2\).

This is obtained by applying corollary 3 to the identity mapping \(\iota:(X,\tau_2) \to (X,\tau_1)\).

(Corollary 6) If \(\lVert \cdot \rVert_1\) and \(\lVert \cdot \rVert_2\) are two norms in a vector space \(X\) such that

  • \(\lVert\cdot\rVert_1 \leq K\lVert\cdot\rVert_2\).
  • \((X,\lVert\cdot\rVert_1)\) and \((X,\lVert\cdot\rVert_2)\) are Banach

Then \(\lVert\cdot\rVert_1\) and \(\lVert\cdot\rVert_2\) are equivalent.

This is merely a more restrictive version of corollary 5.

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. 4 - The Open Mapping Theorem (F-Space)



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