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For $x\in X$, choose $x_m\in D$ with $x_m\to x$ and defineThe sequence on the right is Cauchy becauseand it converges because $Y$ is complete. The same estimate shows that the limit is independent of the approximating sequence, is linear, and satisfiesIts norm is at most $\Vert T\Vert$, while restriction to $D$ gives the reverse inequality, soAny two continuous extensions agree on the dense set $D$ and hence everywhere, proving uniqueness. This is the extension of a bounded linear operator from a dense subspace.

Completeness of $X$ was never used, so the result remains true when $X$ is incomplete. Completeness of $Y$ is essential. For example, with $D=c_{00}\subset X={\ell }^2$, let the codomain be the incomplete normed space $Y=c_{00}$ with the ${\ell }^2$ norm and let $T$ be the identity. An extension would have to send every limit in ${\ell }^2$ to itself and therefore could not take values in $c_{00}$. This is the failure of dense-subspace extension into an incomplete codomain.

Given $x\in {\ell }^p$, choose a sequence $x^{(m)}\in F$ with $x^{(m)}\to x$, which is possible by part (a). The bound givesso $(T{x}^{(m)})$ is a Cauchy sequence. Since ${\ell }^p$ is a Banach space, it converges. DefineThe same estimate shows that this limit does not depend on the approximating sequence. Passing to the limit proves linearity andso $\Vert \tilde{T}\Vert \le C$. It agrees with $T$ on $F$. If two bounded extensions existed, their difference would vanish on the dense subspace $F$ and, by continuity, on its closure ${\ell }^p$. Thus the extension is unique. This is the extension of a bounded linear operator from a dense subspace.