Isometries on Normed Linear Spaces

# Isometries on Normed Linear Spaces

 Definition: Let $X$ and $Y$ be normed linear spaces. A linear operator $T : X \to Y$ is an Isometry if for all $x \in X$ we have that $\| T(x) \| = \| x \|$.

So isometries are simply linear operators $T : X \to Y$ for which the norm of each element in the domain is equal to the norm of the corresponding image of that element in the range

The following theorem tells us that every isometry is continuous and injective.

 Theorem 1: Let $X$ and $Y$ be normed linear spaces. If $T : X \to Y$ is an isometry then $T$ is continuous and $T$ is injective.
• Proof: Let $\epsilon > 0$ be given. Let $\delta = \epsilon$. Then if $x, y \in X$ are such that $\| x - y \| < \delta = \epsilon$ then:
(1)
\begin{align} \quad \| T(x) - T(y) \| = \| T(x - y) \| = \| x - y \| < \delta = \epsilon \end{align}
• Therefore $T$ is continuous.
• Now suppose that $x, y \in X$ and $T(x) = T(y)$. Then $T(x) - T(y) = 0$. So:
(2)
\begin{align} \quad 0 = \| T(x) - T(y) \| = \| T(x - y) \| = \| x - y \| \end{align}
• But this happens if and only if $x - y = 0$. So $x = y$. Hence $T$ is injective. $\blacksquare$