Banach-Schauder Theorem/Lemma 1

Lemma
Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be Banach spaces.

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

Let $y \in Y$ and $r > 0$ be such that:


 * $\map {B_Y} {y, r} \subseteq \paren {\map T {\map {B_X} {0, m} } }^-$

where:
 * $\map {B_Y} {y, r}$ denotes the open ball in $Y$ centered at $y$ with radius $r$
 * $\map {B_X} {0, m}$ denotes the open ball in $X$ centered at $0 \in X$ with radius $m$
 * $\paren {\map T {\map {B_X} {0, m} } }^-$ denotes the topological closure of $\map T {\map {B_X} {0, m} }$.

Then:
 * $\map {B_Y} {0, r} \subseteq \paren {\map T {\map {B_X} {0, m} } }^-$

Proof
From Open Ball is Convex Set, we have:


 * $\map {B_X} {0, m}$ is convex.

Then, from Image of Convex Set under Linear Transformation is Convex, we have:


 * $\map T {\map {B_X} {0, m} }$ is convex.

From Closure of Convex Subset in Normed Vector Space is Convex, we have:


 * $\paren {\map T {\map {B_X} {0, m} } }^-$ is convex.

Let $u \in Y$ have $\norm u_Y < 1$.

Then:


 * $\norm {y + r u - y}_Y = r \norm u_Y < r$

so:


 * $y + r u \in \map {B_Y} {y, r}$

We also have:


 * $\norm {y - r u - y}_Y = r \norm u_Y < r$

so:


 * $y - r u \in \map {B_Y} {y, r}$