Isometries Between Normed Linear Spaces

# Isometries Between Normed Linear Spaces

Definition: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces. A linear operator $T : X \to Y$ is an Isometry from $X$ to $Y$ if $\| T(x) \|_Y = \| x \|_X$ for all $x \in X$. |

The following theorem tells us that if $T$ is an isometry from $X$ to $Y$ then $T$ is continuous and $T$ is injective.

Proposition 1: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces and let $T : X \to Y$ be an isometry from $X$ to $Y$. Then:a) $T$ is continuous.b) $T$ is injective. |

**Proof of a):**Let $x_0 \in X$ and let $\epsilon > 0$ be given. Let $\delta = \epsilon$. Then if $\| x - x_0 \|_X < \delta = \epsilon$ then:

\begin{align} \quad \| T(x) - T(x_0) \|_Y = \| T(x - x_0) \|_Y = \| x - x_0 \|_X < \delta = \epsilon \end{align}

- Therefore $T$ is continuous at $x_0$. So $T$ is continuous on all of $X$.

**Proof of b)**Now suppose that $x, y \in X$ and $T(x) = T(y)$. Then $T(x) - T(y) = 0$. So:

\begin{align} \quad 0 = \| T(x) - T(y) \|_Y = \| T(x - y) \|_Y = \| x - y \|_X \end{align}

- But this happens if and only if $x - y = 0$. So $x = y$. Hence $T$ is injective. $\blacksquare$