Chapter 10: Problem 24

Show that a linear map \(T: V \rightarrow W\) between topological vector spaces is continuous everywhere on \(V\) if and only if it is continuous at the origin \(0 \in V\).

Short Answer

Expert verified

The theorem has been proved. A linear map between topological vector spaces is continuous everywhere if and only if it is continuous at the origin.

Step by step solution

Understand Definitions

Before starting the proof, it is vital to understand the definitions: (1) A topological vector space is a vector space which also has a topology such that vector addition and scalar multiplication are continuous functions. (2) A linear map between two vector spaces is a function that preserves the operations of addition and scalar multiplication. (3) A function is continuous at a point if the limit of the function at that point is equal to the value of the function at that point. (4) In topological spaces, a function is continuous if the preimage of every open set is open.

Assume T is continuous everywhere

Under the assumption that the map \(T: V \rightarrow W\) is continuous everywhere, it is necessary to show that it is also continuous at the origin. Because \(T\) is continuous everywhere, \(T^{-1}(W)\) is open in \(V\) for any open subset \(W\) of \(W\). Hence, \(T^{-1}(W) \cap 0\) is open in \(V\) for any open subset \(W\) of \(W\). This shows that \(T\) is continuous at \(0\), since the preimage of every open set in \(W\) containing \(0_W\) is an open set in \(V\) containing \(0_V\). This completes one part of the proof.

Assume T is continuous at the origin

Under the assumption that the map \(T: V \rightarrow W\) is continuous at the origin, it is necessary to show that it is also continuous everywhere. Given any point \(v \in V\), by applying the linear property of \(T\) and its continuity at \(0_V\), one can show that \(T\) is continuous at \(v\). By the topological property of vector spaces, the map \(T: V \rightarrow W\) is continuous at each point \(v\) of \(V\), hence \(T\) is continuous everywhere in \(V\). This completes the other part of the proof.