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
Let $\map D {f_0}$ be a linear subspace of $X$.
Let $f_0 : \map D {f_0} \to \Bbb F$ be a bounded linear functional.
Then there exists a unique bounded linear functional $f : \map D f \to \Bbb F$ extending $f_0$ to $\map D f = \paren {\map D {f_0} }^-$.
Corollary 2
Let $D$ be an everywhere dense linear subspace of $X$.
Let $T_1 : X \to Y$ and $T_2 : X \to Y$ be bounded linear transformations with:
- $T_1 x = T_2 x$ for all $x \in D$.
Then:
- $T_1 = T_2$
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:
\(\ds \norm {T_0 x_n - T_0 x_m}_Y\) | \(=\) | \(\ds \norm {\map {T_0} {x_n - x_m} }_Y\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds M \norm {x_n - x_m}_X\) |
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:
\(\ds \norm {T_0 x_n - T_0 z_n}\) | \(=\) | \(\ds \norm {\map {T_0} {x_n - z_n} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds M \norm {x_n - z_n}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds M \norm {x_n - x} + M \norm {z_n - x}\) | using the triangle inequality for the norm | |||||||||||
\(\ds \) | \(\to\) | \(\ds 0\) | Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence |
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:
\(\ds \map T {\lambda x + y}\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \map {T_0} {\lambda x_n + y_n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\lambda T_0 x_n + T_0 y_n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lambda \lim_{n \mathop \to \infty} T_0 x_n + \lim_{n \mathop \to \infty} T_0 y_n\) | Combined Sum Rule for Sequences in Normed Vector Space | |||||||||||
\(\ds \) | \(=\) | \(\ds \lambda T x + T y\) | recalling the definition of $T$ |
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 $\map D T$.
$\Box$
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:
\(\ds \norm {T' x - T_0 x_n}\) | \(=\) | \(\ds \norm {T' x - T' x_n}\) | $T$ extends $T_0$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds C \norm {x - x_n}\) | for some real number $C > 0$, since $T'$ is bounded | |||||||||||
\(\ds \) | \(\to\) | \(\ds 0\) | Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence |
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 $T' = T$.
$\blacksquare$
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.