Injective Linear Transformation between Normed Vector Spaces sends Closed Unit Ball to Closed Unit Ball iff Isometric Isomorphism

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Theorem

Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$.

Let $T : X \to Y$ be a injective linear transformation.

Let $B_X^-$ be the closed unit ball of $X$.

Let $B_Y^-$ be the closed unit ball of $Y$.


Then $T$ is an isometric isomorphism if and only if:

$T \sqbrk {B_X^-} = B_Y^-$


Proof

Necessary Condition

Suppose that $T$ is an isometric isomorphism.

Then:

$\norm {T x}_Y = \norm x_X$ for each $x \in X$.

Hence if $\norm x_X \le 1$, we have $\norm {T x}_Y \le 1$.

So $T \sqbrk {B_X^-} \subseteq B_Y^-$.

Conversely, let $y \in B_Y^-$.

Since $T$ is an isometric isomorphism:

$T$ is bijective and there exists $x \in X$ such that $y = T x$.

Since $T$ is a linear isometry, we have:

$\norm x_X = \norm {T x}_Y \le 1$

So we have $x \in B_X^-$.

So we have $T \sqbrk {B_X^-} = B_Y^-$.

$\Box$

Sufficient Condition

Suppose that:

$T \sqbrk {B_X^-} = B_Y^-$

We first show that $T$ is surjective.

We have:

\(\ds T \sqbrk X\) \(=\) \(\ds T \sqbrk {\bigcup_{n \mathop = 1}^\infty n B_X^-}\)
\(\ds \) \(=\) \(\ds \bigcup_{n \mathop = 1}^\infty T \sqbrk {n B_X^-}\) Image of Union under Mapping: General Result
\(\ds \) \(=\) \(\ds \bigcup_{n \mathop = 1}^\infty n T \sqbrk {B_X^-}\) Image of Dilation of Set under Linear Transformation is Dilation of Image
\(\ds \) \(=\) \(\ds \bigcup_{n \mathop = 1}^\infty n B_Y^-\)
\(\ds \) \(=\) \(\ds Y\)

We now show that $T$ is a bounded linear transformation.

We have:

\(\ds \sup_{x \in B_X^-} \norm {T x}_Y\) \(=\) \(\ds \sup_{y \in T \sqbrk {B_X^-} } \norm y_Y\)
\(\ds \) \(=\) \(\ds \sup_{y \in B_Y^-} \norm y_Y\)
\(\ds \) \(=\) \(\ds 1\)

So $T$ is a bounded linear transformation with $\norm T_{\map B {X, Y} } = 1$.


We move to proving that $T^{-1}$ is a bounded linear transformation.

Since $T$ is bijective, we have:

$T^{-1} \sqbrk {B_Y^-} = B_X^-$

from Preimage of Image of Subset under Injection equals Subset.

From Inverse of Linear Transformation is Linear Transformation, $T^{-1}$ is a linear transformation.

Repeating the previous calculation with $T$ swapped for $T^{-1}$, we obtain that:

$\ds \sup_{y \in B_Y^-} \norm {T^{-1} y}_X = \sup_{x \in B_X^-} \norm x_X = 1$

So $T^{-1}$ is a bounded linear transformation with $\norm {T^{-1} }_{\map B {Y, X} } = 1$.


Now, we have:

$\norm {T x}_Y \le \norm x_X$ for all $x \in X$

and:

$\norm {T^{-1} y}_X \le \norm y_Y$ for all $y \in Y$.

In particular, taking $x \in X$ and $y = T x$ in the latter inequality, we obtain:

$\norm x_X \le \norm {T x}_Y$ for all $x \in X$.

Hence we obtain:

$\norm x_X \le \norm {T x}_Y \le \norm x_X$

Hence:

$\norm {T x}_Y = \norm x_X$

So $T$ is an surjective linear isometry.

From Linear Isometry is Injective: Corollary, we have that $T$ is an isometric isomorphism.

$\blacksquare$

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