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 $\map D T = \paren {\map D {T_0} }^-$.

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.

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.