Characterization of Open Linear Transformation between Normed Vector Spaces

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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 linear transformation.

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


The following statements are equivalent:

$(1) \quad$ $T$ 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 $(2)$.

$\Box$

$(2)$ implies $(1)$

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

We have:

\(\ds T \sqbrk {B_X^O}\) \(\supseteq\) \(\ds T \sqbrk {\frac 1 2 B_X^-}\) Definition of Closed Unit Ball
\(\ds \) \(=\) \(\ds \frac 1 2 T \sqbrk {B_X^-}\) Image of Dilation of Set under Linear Transformation is Dilation of Image
\(\ds \) \(\supseteq\) \(\ds \frac \delta 2 B_Y^-\) by hypothesis we have $\delta B_Y^- \subseteq T \sqbrk {B_X^-}$
\(\ds \) \(\supseteq\) \(\ds \frac \delta 2 B_Y^O\)

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 $(1)$.

$\Box$

$(2)$ 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 $(3)$.

$\Box$

$(3)$ 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:

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

$\blacksquare$


Sources

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