 Solution

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.