Translation Mapping on Topological Vector Space is Homeomorphism

Theorem

Let $K$ be a topological field.

Let $X$ be a topological vector space over $K$.

Let $x \in X$.

Let $T_x$ be the translation by $x$ mapping.

Then $T_x$ is a homeomorphism.

Proof

From Translation Mapping on Topological Vector Space is Continuous, both $T_x$ and $T_{-x}$ are continuous.

It is therefore sufficient to establish that $T_{-x}$ is the inverse mapping of $T_x$.

For all $y \in X$, we have:

\(\ds \map {\paren {T_x \circ T_{-x} } } y\)

\(=\)

\(\ds \map {T_x} {y - \paren {-x} }\)

\(\ds \)

\(=\)

\(\ds \map {T_x} {y + x}\)

\(\ds \)

\(=\)

\(\ds y + x - x\)

\(\ds \)

\(=\)

\(\ds y\)

and:

\(\ds \map {\paren {T_{-x} \circ T_x} } y\)

\(=\)

\(\ds \map {T_{-x} } {y - x}\)

\(\ds \)

\(=\)

\(\ds y - x - \paren {-x}\)

\(\ds \)

\(=\)

\(\ds y - x + x\)

\(\ds \)

\(=\)

\(\ds y\)

So both $T_x \circ T_{-x}$ and $T_{-x} \circ T_x$ are the identity mapping for $X$.

So $T_{-x}$ is the inverse mapping of $T_x$, as required.

$\blacksquare$

Sources

• 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.6$: Topological vector spaces