Banach-Schauder Theorem/Lemma 2

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 $r > 0$ be such that:


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

where:
 * $\map {B_Y} {0, r}$ denotes the open ball in $Y$ centered at $0 \in Y$ 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:
 * $\map {B_Y} {0, r} \subseteq \map T {\map {B_X} {0, 2 m} }$

Proof
We first show that:


 * $\map {B_Y} {0, 2^{-n} r} \subseteq \paren {\map T {\map {B_X} {0, 2^{-n} m} } }^-$ for each $n \in \N$.

Let $y \in \map {B_Y} {0, 2^{-n} r}$.

Then:


 * $\norm y_Y < 2^{-n} r$

so:


 * $\norm {2^n y}_Y < r$

So:


 * $2^n y \in \map {B_Y} {0, r}$

Then:


 * $2^n y \in \paren {\map T {\map {B_X} {0, m} } }^-$

From Point in Closure of Subset of Metric Space iff Limit of Sequence, we have:


 * there exists a sequence $\sequence {y_k}_{k \mathop \in \N}$ in $\map T {\map {B_X} {0, m} }$ such that $y_k \to 2^n y$.

For each $k \in \N$, there exists $x_k \in \map {B_X} {0, m}$ such that $y_k = T x_k$.

Then we have, from linearity and Multiple Rule for Sequence in Normed Vector Space:


 * $\map T {2^{-n} x_k} \to y$

with:


 * $\norm {2^{-n} x_k}_X < 2^{-n} m$ for each $k \in \N$

So:


 * $2^{-n} x_k \in \map {B_X} {0, 2^{-n} m}$ for each $k \in \N$.

Then:


 * $\map T {2^{-n} x_k} \in \map T {\map {B_X} {0, 2^{-n} m} }$ for each $k \in \N$.

so:


 * $y \in \paren {\map T {\map {B_X} {0, 2^{-n} m} } }^-$

So:


 * $\map {B_Y} {0, 2^{-n} r} \subseteq \paren {\map T {\map {B_X} {0, 2^{-n} m} } }^-$ for each $n \in \N$.

from the definition of set inclusion.

Now we show that:


 * $\map {B_Y} {0, r} \subseteq \map T {\map {B_X} {0, 2 m} }$

Let $z \in \map {B_Y} {0, r}$. Then we have:


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

From Condition for Point being in Closure, there exists $y_1 \in \map T {\map {B_X} {0, m} }$ such that:


 * $\norm {z - y_1}_Y < 2^{-1} r$

Since $y_1 \in \map T {\map {B_X} {0, m} }$ there exists $x_1 \in \map {B_X} {0, m}$ such that:


 * $y_1 = T x_1$

so that:


 * $\norm {z - T x_1}_Y < 2^{-1} r$

Since:


 * $z - T x_1 \in \map {B_Y} {0, 2^{-1} r}$

so:


 * $z - T x_1 \in \paren {\map T {\map {B_X} {0, 2^{-1} m} } }^-$

Then from Condition for Point being in Closure, there exists $y_2 \in \map T {\map {B_X} {0, 2^{-1} m} }$ such that:


 * $\norm {\paren {z - T x_1} - y_2}_Y < 2^{-2} r$

Since $y_2 \in \map T {\map {B_X} {0, 2^{-1} m} }$, there exists $x_2 \in \map {B_X} {0, 2^{-1} m}$ such that:


 * $y_2 = T x_2$

so:


 * $\norm {z - T x_1 - T x_2}_Y < 2^{-2} r$

Repeating this process, for each $n \in \N$ we can find $x_n \in \map {B_X} {0, 2^{-n + 1} m}$ such that:


 * $\ds \norm {z - \sum_{k \mathop = 1}^n T x_k}_Y < 2^{-n} r$

From the linearity of $T$ we then have:


 * $\ds \norm {z - \map T {\sum_{k \mathop = 1}^n x_k} }_Y < 2^{-n} r$

We have:

So the sequence $\sequence {\sum_{k \mathop = 1}^n \norm {x_k}_X}_{n \mathop \in \N}$ is bounded and increasing.

From the Monotone Convergence Theorem and Infinite Series preserves Strict Inequality, we therefore have:


 * $\ds \sum_{k \mathop = 1}^\infty \norm {x_k}_X < 2 m < \infty$

Since $X$ is Banach, from Absolutely Convergent Series is Convergent iff Normed Vector Space is Banach we have:


 * $\ds \sum_{k \mathop = 1}^\infty x_k$ converges to a limit $x \in X$.

Note that we have by Norm of Summation:


 * $\ds \norm {\sum_{k \mathop = 1}^n x_k}_X \le \sum_{k \mathop = 1}^n \norm {x_k}_X $ for each $n \in \N$.

So, taking $n \to \infty$ and using Norm is Continuous, we have:


 * $\ds \norm {\sum_{k \mathop = 1}^\infty x_k}_X < 2 m$

So we have:


 * $x \in \map {B_X} {0, 2 m}$

Further, since $2^{-n} r$, we have:


 * $\ds \norm {z - \map T {\sum_{k \mathop = 1}^n x_k} }_Y \to 0$

from the Squeeze Theorem for Real Sequences.

From Modulus of Limit: Normed Vector Space, we have:


 * $\ds \lim_{n \mathop \to \infty} \map T {\sum_{k \mathop = 1}^n x_k} = z$

From Continuity of Linear Transformations, we have that:


 * $T$ is continuous

since $T$ is bounded.

So, from Sequential Continuity is Equivalent to Continuity in Metric Space:


 * $\ds z = \map T {\lim_{n \mathop \to \infty} \sum_{k \mathop = 1}^n x_k} = T x$

So:


 * $z \in \map T {\map {B_X} {0, 2 m} }$

So by the definition of set inclusion:


 * $\map {B_Y} {0, r} \subseteq \map T {\map {B_X} {0, 2 m} }$

as desired.