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The Open Mapping Theorem Makes Surjective Operators Quantitatively Open by Vicfred

A bounded surjection between Banach spaces sends open sets to open sets. The point is to make the formal expression readable enough to audit line by line.

The mathematical object

A Banach space \(X\) is complete in its norm, and a bounded linear operator \(T:X\to Y\) has norm \(\|T\|=\sup_{\|x\|\le1}\|Tx\|\). Completeness powers the major structural theorems.

$$ T:X\to Y\ \text{bounded and surjective} $$

The typography mirrors the proof: first declare \(\mathsf D\), then state \(\mathsf C\). The symbol \(\Longrightarrow\) below is a logical dependency, not extra mathematical structure.

$$ \exists c>0,\qquad B_Y(0,c)\subseteq T(B_X(0,1)) $$

One explicit computation

Here is a concrete symbolic test. Reading it from left to right reveals which transformation is reversible and which is only an implication.

$$ T^{-1}\ \text{exists}\Longrightarrow\|T^{-1}y\|\le c^{-1}\|y\| $$

Why the identity matters

The compact alignment is a local map of the argument: assumptions on the first row, consequence on the second. Any generalisation must preserve that dependency.

$$ \begin{aligned} \mathsf{D}\;&:\quad T:X\to Y\ \text{bounded and surjective},\\[5pt] \mathsf{C}\;&:\quad \exists c>0,\qquad B_Y(0,c)\subseteq T(B_X(0,1)). \end{aligned} $$

Where it can fail

Finite-dimensional intuition can fail badly in infinite dimensions. Closed, bounded sets need not be compact, and linear maps need not be bounded automatically.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] \exists c>0,\qquad B_Y(0,c)\subseteq T(B_X(0,1)) \end{gathered}} $$

This is enough machinery for one note: an exact object, a worked case, a structural law, and a clearly marked boundary. Each layer can now be tested independently.

This article was posted on Tue 09 August 2022. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.