Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain/Corollary 2

Corollary to 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.

Let $\struct {Y, \norm \cdot_Y}$ be a Banach space.

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

Define $T_0 : D \to Y$ by:

$T_0 x = T_1 x = T_2 x$ for all $x \in D$.

Then $T_0$ is a bounded linear transformation, and so from Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain there exists a unique bounded linear transformation extending $T_0$ to $X$.

Since $T_1$ and $T_2$ are both bounded linear transformation extending $T_0$ to $X$, so we must have $T_1 = T_2$.

$\blacksquare$