Characterization of Open Linear Transformation between Normed Vector Spaces

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 linear transformation.

Let $B_X^-$ and $B_Y^-$ be the closed unit balls of $X$ and $Y$ respectively.


 * $(1) \quad$ $T$ is an open map
 * $(2) \quad$ there exists $\delta > 0$ such that $\delta B_Y^- \subseteq T \sqbrk {B_X^-}$
 * $(3) \quad$ there exists $M > 0$ such that for all $y \in Y$ there exists $x \in T^{-1} \sqbrk {\set y}$ such that $\norm x_X \le M \norm y_Y$

Proof
Let $B_X^O$ and $B_Y^O$ be the open unit balls of $X$ and $Y$ respectively.

$(1)$ implies $(2)$
Suppose that $T$ is an open map.

Then $T \sqbrk {B_X^O}$ is open in $\struct {Y, \norm {\, \cdot \,}_Y}$.

Hence there exists $\delta > 0$ such that:
 * $\delta B_Y^O \subseteq T \sqbrk {B_X^O}$

We have that:
 * $\ds \frac \delta 2 B_Y^- \subseteq \delta B_Y^O$

We also have that:
 * $T \sqbrk {B_X^O} \subseteq T \sqbrk {B_X^-}$

from Image of Subset under Mapping is Subset of Image.

Hence we obtain:
 * $\ds \frac \delta 2 B_Y^- \subseteq T \sqbrk {B_X^-}$

which is the demand of $(2)$.

$(2)$ implies $(1)$
Suppose that there exists $\delta > 0$ such that $\delta B_Y^- \subseteq T \sqbrk {B_X^-}$.

We have:

From Linear Transformation between Normed Vector Spaces is Open iff Image of Open Unit Ball is Open, we have that $T$ is an open map.

Hence we have $(1)$.

$(2)$ implies $(3)$
Suppose that there exists $\delta > 0$ such that $\delta B_Y^- \subseteq T \sqbrk {B_X^-}$.

If $y = {\mathbf 0}_Y$, then $x = {\mathbf 0}_X$ has:
 * $\norm x_X \le M \norm y_Y$

for any $M > 0$.

It therefore suffices to pick an $M > 0$ that works for $y \in Y \setminus \set { {\mathbf 0}_Y}$.

Let $y \in Y \setminus \set { {\mathbf 0}_Y}$.

Then:
 * $\ds \frac {\delta y} {\norm y_Y} \in \delta B_Y^-$

so:
 * $\ds \frac {\delta y} {\norm y_Y} \in T \sqbrk {B_X^-}$

Then there exists $x \in B_X^-$ such that:
 * $\ds \frac {\delta y} {\norm y_Y} = T x$

so that:
 * $\ds \map T {\frac {\norm y_Y} \delta x} = y$

Setting:
 * $\ds x' = \frac {\norm y_Y} \delta x$

we have that $x' \in T^{-1} \sqbrk {\set y}$ with:
 * $\ds \norm {x'}_X = \frac {\norm y_Y} \delta \norm x_X \le \frac {\norm y_Y} \delta$

Hence setting:
 * $\ds M = \frac 1 \delta$

we have $(3)$.

$(3)$ implies $(2)$
Suppose that there exists $M > 0$ such that for all $y \in Y$ there exists $x \in T^{-1} \sqbrk {\set y}$ such that $\norm x_X \le M \norm y_Y$.

Let $y \in B_Y^-$.

Then there exists $x \in T^{-1} \sqbrk {\set y}$ such that:


 * $\norm x_X \le M \norm y_Y \le M$

Hence, we have:


 * $B_Y^- \subseteq T \sqbrk {M B_X^-}$

Hence, from Image of Dilation of Set under Linear Transformation is Dilation of Image:


 * $\ds \frac 1 M B_Y^- \subseteq T \sqbrk {B_X^-}$