Banach-Schauder Theorem/Lemma 1

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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:

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

$\blacksquare$