Operator extension

Whenever a space $$W$$ is dense in another space $$V$$, a natural question to ask is whether the an operator defined in $$W$$ can be extended to $$V$$. This is possible for bounded linear operators as the theorem below specifies (Th. 3.3.2 ).

Let $$V_1$$ and $$V_2$$ be Banach spaces, $$W$$ a dense subspace of $$V_1$$, and $$T:W\to V_2$$ a bounded linear operator. Then, there exists a unique bounded linear operator

$$\tilde{T}:V_1\to V_2$$

for which $$\tilde{T}\mathbf{v} = T \mathbf{v}$$ for all $$\mathbf{w} \in W$$. The operator $$\tilde{T}$$ satisfies $$||\tilde{T}|| =||T||$$.

As the theorem states, the behavior of $$\tilde T$$ is obvious for $$\mathbf v \in W$$. If $$\mathbf v \in V_1\setminus W$$, then

$$\tilde T \mathbf{v} := \lim_{k\to \infty} T\mathbf{v}_k$$,

where $$\{\mathbf v_k\}_k$$ is a sequence in $$W$$ that converges to $$V_1$$. (The existence of such a sequence is guaranteed by denseness).