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

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 $\map D T = \paren {\map D T}^-$.
Let $\map \BB {\map D {T_0}, Y}$ be the space of bounded linear transformations on $X$.
Let $\norm {\, \cdot \,}_{\map \BB {\map D {T_0}, Y} }$ be the norm on $X$.
Let $\struct {Y, \norm \cdot_Y}$ be a Banach space over $\Bbb F$.
Let $\map \BB {\map D T, Y}$ be the space of bounded linear transformations $\map D T \to Y$.
Let $\norm {\, \cdot \,}_{\map \BB {\map D T, Y} }$ be the norm on $X \to Y$.
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 $\map D T = \paren {\map D {T_0} }^-$. 
Further, $\norm {T_0}_{\map \BB {\map D {T_0}, Y} } = \norm T_{\map \BB {\map D T, Y} }$.

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 $\sequence {y_n}_{n \mathop \in \N}$ be sequences in $\map D {T_0}$ converging to $x$. 
We have: 
{{begin-eqn}}
{{eqn	| l = \norm {T_0 x_n - T_0 x_m}_Y
		| r = \norm {\map {T_0} {x_n - x_m} }_Y
}}
{{eqn	| o = \le
		| r = M \norm {x_n - x_m}_X
}}
{{end-eqn}}
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 $\sequence {y_n}_{n \mathop \in \N}$, we get that $T y_n \to z_2$. 
We show that $z_1 = z_2$. 
From Modulus of Limit: Normed Vector Space, we have: 
:$\norm {T_0 x_n - T_0 y_n}_Y \to \norm {z_1 - z_2}_Y$
On the other hand: 
{{begin-eqn}}
{{eqn	| l = \norm {T_0 x_n - T_0 y_n}_Y
		| r = \norm {\map {T_0} {x_n - y_n} }_Y
}}
{{eqn	| o = \le
		| r = M \norm {x_n - y_n}_X
}}
{{eqn	| o = \le
		| r = M \norm {x_n - x}_X + M \norm {y_n - x}_X
		| c = using the triangle inequality for the norm
}}
{{eqn	| o = \to
		| r = 0
		| c = Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence
}}
{{end-eqn}}
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: 
{{begin-eqn}}
{{eqn	| l = \map T {\lambda x + y}
		| r = \lim_{n \mathop \to \infty} \map {T_0} {\lambda x_n + y_n}
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \paren {\lambda T_0 x_n + T_0 y_n}
}}
{{eqn	| r = \lambda \lim_{n \mathop \to \infty} T_0 x_n + \lim_{n \mathop \to \infty} T_0 y_n
		| c = Combined Sum Rule for Sequences in Normed Vector Space
}}
{{eqn	| r = \lambda T x + T y
		| c = recalling the definition of $T$
}}
{{end-eqn}}
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: 
:$\norm {T_0 x_n}_Y \to \norm {T x}_Y$ 
and:
:$\norm {x_n}_X \to \norm x_X$ 
Since $T_0$ is bounded, there exists a real number $M > 0$ such that: 
:$\norm {T_0 x_n}_Y \le M \norm {x_n}_X$ for all $x \in \map D {T_0}$. 
Taking $n \to \infty$, we have: 
:$\norm {T x}_Y \le M \norm x_X$
from Limits Preserve Inequalities.
In general if:
:$\norm {T_0 x}_Y \le M \norm x_X$ for all $x \in \map D {T_0}$
then we obtain that:
:$\norm {T x}_Y \le M \norm x_X$ for all $x \in \map D T$
Hence $\norm T_{\map \BB {\map D T, Y} } \le \norm {T_0}_{\map \BB {\map D {T_0}, Y} }$. 
Since $T$ extends $T_0$, we have:
:$\norm {T_0}_{\map \BB {\map D {T_0}, Y} } \le \norm T_{\map \BB {\map D T, Y} }$
We conclude that $T$ is a bounded linear transformation extending $T_0$ to $\map D T$ with:
:$\norm {T_0}_{\map \BB {\map D {T_0}, Y} } \le \norm T_{\map \BB {\map D T, Y} }$
{{qed|lemma}}