= 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 gives
$$
\lVert Tx^{(m)}-Tx^{(l)}\rVert_p
\leq C\lVert x^{(m)}-x^{(l)}\rVert_p,
$$
so $(Tx^{(m)})$ is a <Cauchy sequence>. Since $\ell^p$ is a <Banach space>, it converges. Define
$$
\widetilde Tx=\lim_{m\to\infty}Tx^{(m)}.
$$
The same estimate shows that this limit does not depend on the approximating sequence. Passing to the limit proves linearity and
$$
\lVert\widetilde Tx\rVert_p\leq C\lVert x\rVert_p,
$$
so $\lVert\widetilde T\rVert\leq C$. 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>.
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