Definition of "Extension" of Bounded Linear Transformation

Question

I have been given the problem of proving the B.L.T. Theorem for my homework which states,

Every bounded linear transformation $\mathsf{T}$ from a normed vector space X to a complete, normed vector space Y can be uniquely extended to a bounded linear transformation $\tilde{\mathsf{T}}$ from the completion of X to Y. In addition, the operator norm of $\mathsf{T}$ is $c$ iff the norm of $\tilde{\mathsf{T}}$ is $c$.

What exactly does this theorem mean by, 'extension from $\mathsf{T}$ to $\tilde{\mathsf{T}}$ ?'

Answer 1

If $T: X \to Y$ is a linear continuous map and $X \subset X'$ where $X'$ is the completion of $X$, a (linear continuous) extension $\tilde{T}$ is another linear continuous map $\tilde{T}: X' \to Y$ such that $\left. \tilde{T} \right|_X = T$.

Answer 2

The space $X$ is dense in its completion $\tilde{X}$. So every $\tilde{x}\in \tilde{X}$ is the limit of a sequence $\{ x_n \}\subset X$. This sequence is a Cauchy sequence in $X$. Because $T$ is bounded, then $\{ Tx_n \}$ is a Cauchy sequence in $Y$ (which is complete) and, hence, converges to some $y\in Y$. If $\{ x_n ' \}$ is another such sequence in $X$ converging to $\tilde{x}$, then $\{ Tx_n - Tx_n' \}$ converges to $0$ in $Y$. So, there is no ambiguity in defining $\tilde{T}\tilde{x}=y$. You can check that this extension is linear and bounded by the same bound for $T$. It is proper to say that $T$ has a unique bounded linear extension $\tilde{T} : \tilde{X}\rightarrow Y$.