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:


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

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 $m$
 * $\paren {T \sqbrk {\map {B_X} {0, m} } }^-$ denotes the topological closure of $T \sqbrk {\map {B_X} {0, m} }$.

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

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:


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

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


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

From Open Ball Centred at Origin in Normed Vector Space is Symmetric:


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

From Image of Symmetric Set under Linear Transformation is Symmetric, we have:


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

From Closure of Symmetric Subset of Normed Vector Space is Symmetric:


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

Let $x \in \map {B_Y} {0, r}$.

Then $x$ can be written in the form:


 * $x = r u$

for $u$ with $\norm u_Y < 1$, in particular:


 * $\ds u = \frac x r$

Since $\map {B_Y} {y, r} \subseteq \paren {T \sqbrk {\map {B_X} {0, m} } }^-$, we have:


 * $y + r u \in \paren {T \sqbrk {\map {B_X} {0, m} } }^-$

and:


 * $y - r u \in \paren {T \sqbrk {\map {B_X} {0, m} } }^-$

Since $\paren {T \sqbrk {\map {B_X} {0, m} } }^-$ is symmetric, we have:


 * $-y + r u \in \paren {T \sqbrk {\map {B_X} {0, m} } }^-$

Writing:


 * $\ds r u = \frac 1 2 \paren {y + r u} + \frac 1 2 \paren {-y + r u}$

we have:


 * $r u \in \paren {T \sqbrk {\map {B_X} {0, m} } }^-$

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

So:


 * $x \in \paren {T \sqbrk {\map {B_X} {0, m} } }^-$

So:


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