Let $(V,\left\lVert\cdot\right\rVert)$ be a normed vector space whose unit sphere $\left\{v \in V: \left\lVert v \right\rVert = 1\right\}$ is sequentially compact. Show that any closed ball $\left\{v \in V: \left\lVert v \right\rVert \leq R\right\}$ must be sequentially compact. Show that $V$ is complete.
My only idea was to note that if $(v_n)$ is a sequence in the closed ball, then $\left(\frac{v_n}{\left\lVert v_n \right\rVert}\right)$ is a sequence in the unit sphere, so has a convergent subsequence, but that doesn't really seem to help.
Best Answer
You have started the proof correctly. Now consider two cases:
Case 1) $0$ is a limit point of $(\|v_n\|)$. In this case show that $(v_n)$ has a subsequence converging to $0$.
Case 2. There is a subsequence of $(\|v_n\|)$ converging to some positive number $r$. In this case use $v_n=\frac {v_n} {\|v_n\|} \|v_n\|$ to get a subsequence of $(v_n)$ which converges.