Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot_X}$ be a normed vector space over $\Bbb F$.
Let $\map D {T_0}$ be a linear subspace of $X$.
Let $\struct {Y, \norm \cdot_Y}$ be a Banach space over $\Bbb F$.
Let $T_0 : \map D {T_0} \to Y$ be a bounded linear transformation.
Then there exists a unique bounded linear transformation $T : \map D T \to Y$ extending $T_0$ to $\map D T = \paren {\map D {T_0} }^-$. 
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof ==
=== Existence ===
Since $T_0$ is bounded, there exists a real number $M > 0$ such that: 
:$\norm {T_0 x}_Y \le M \norm x_X$ for all $x \in \map D {T_0}$.
Let $x \in \map D T \setminus \map D {T_0}$.
From Point in Closure of Subset of Metric Space iff Limit of Sequence, there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\map D {T_0}$ with $x_n \to x$. 
We would like to set: 
:$\ds T x = \lim_{n \mathop \to \infty} T_0 x_n$
We show that this limit exists and gives a unique choice independent of the sequence $\sequence {x_n}_{n \mathop \in \N}$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {z_n}_{n \mathop \in \N}$ be sequences in $\map D {T_0}$ converging to $x$. 
We have: 
From Convergent Sequence in Normed Vector Space is Cauchy Sequence, we have that: 
:$\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence. 
So, for each $\epsilon > 0$ there exists $N \in \N$ such that: 
:$\norm {x_n - x_m}_X < \epsilon/M$
for $n, m \ge N$. 
Then, we have: 
:$\norm {T_0 x_n - T_0 x_m}_Y < \epsilon$
So $\sequence {T_0 x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $Y$. 
Since $Y$ is a Banach space, $\sequence {T_0 x_n}_{n \mathop \in \N}$ converges to $z_1$.
Swapping $\sequence {x_n}_{n \mathop \in \N}$ for $\sequence {z_n}_{n \mathop \in \N}$, we get that $T z_n \to z_2$. 
We show that $z_1 = z_2$. 
From Modulus of Limit: Normed Vector Space, we have: 
:$\norm {T_0 x_n - T_0 z_n} \to \norm {z_1 - z_2}$
On the other hand: 
From Convergent Sequence in Normed Vector Space has Unique Limit:
:$z_1 = z_2$ 
So we can extend $T_0$ to $\map D T$ by taking: 
:$\ds T x = \lim_{n \mathop \to \infty} T_0 x_n$
for any $x \in \map D T \setminus \map D {T_0}$, where $\sequence {x_n}_{n \mathop \in \N}$ is any sequence with $x_n \to x$. 
We now need to verify that the obtained $T$ is indeed linear and bounded.
Let $x, y \in \map D T$, $\lambda \in \Bbb F$ and sequences $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {y_n}_{n \mathop \in \N}$ with $x_n \to x$ and $y_n \to y$. 
Then: 
so $T$ is certainly linear. 
We finish by showing that $T$ is bounded.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence converging to $x$. 
Now, from Modulus of Limit: Normed Vector Space, we have: 
:$\norm {T_0 x_n} \to \norm {T x}$ 
and:
:$\norm {x_n} \to \norm x$ 
Since $T_0$ is bounded, there exists a real number $M > 0$ such that: 
:$\norm {T_0 x_n} \le M \norm {x_n}$ for all $x \in \map D {T_0}$. 
Taking $n \to \infty$, we have: 
:$\norm {T x} \le M \norm x$
from Limits Preserve Inequalities.
So $T$ is a bounded linear transformation extending $T_0$ to $\map D T$. 
=== Uniqueness ===
Let $T'$ be another bounded linear transformation extending $T_0$ to $\paren {\map D T}^-$. 
Let $x \in \paren {\map D T}^-$ and pick a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\map D T$ converging to $x$. 
We show that: 
:$\ds T' x = \lim_{n \mathop \to \infty} T_0 x_n$ 
We will then obtain the result from Convergent Sequence in Normed Vector Space has Unique Limit.
We have: 
So, from Limits Preserve Inequalities, we have: 
:$\norm {T' x - T_0 x_n} \to 0$
and so from Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we have: 
:$\ds T_0 x_n \to T' x$ 
From the construction of $T$ we have:
:$\ds T_0 x_n \to T' x$
From Convergent Sequence in Normed Vector Space has Unique Limit, we therefore have $T' x = T x$ for all $x \in \paren {\map D T}^-$.
So $T' = T$.
== Also see ==
* Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain gives a more general result that is admissable here, but this special case is interesting enough to justify individual treatment.
Category:Bounded Linear Transformations
Category:Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain
Category:Banach Spaces