Let N be a nonzero normed linear space, then N is Banach iff $S=\{x:||x||=1\}$ is Complete.
I am able to give a proof for a weaker statement than this, that being
N is Banach iff $\{x:||x|| \leqslant 1\}$
For any Cauchy sequence in N. I asscociated a Cauchy sequence in the closed unit disc. Used the completness property of S, then again returned to the original sequence to conclude its converges in N.
But how can this line of argument be refined to proof the actual statement.
Best Answer
The easy direction is, if $X$ is complete, then $S_X$ is complete, which follows immediately from $S_X$ being closed in $X$.
In brief, the proof of the other direction involves, given a Cauchy sequence $(x_n)$ in $X$,
Step 2 is the hard step. Step 1 is easy: the norm is a non-expansive map, hence uniformly continuous, and hence Cauchy continuous (i.e. preserves Cauchy sequences). Step 3 is similarly easy, and follows immediately from the continuity of the scalar multiplication operation.
So, let's suppose $(x_n)$ is Cauchy. Let $r = \lim_{n \to \infty} \|x_n\|$. If $r = 0$, then $x_n \to 0$, and we are done, so assume without loss of generality that $r > 0$. We therefore can find an $N_1$ such that $$n \ge N_1 \implies |\|x_n\| - r| < \frac{r}{2} \implies \|x_n\| > \frac{r}{2}.$$
Suppose $\varepsilon > 0$. Since $(x_n)$ is Cauchy, there must exist some $N_2$ such that $$m, n \ge N_2 \implies \|x_m - x_n\| < \frac{\varepsilon r}{4}.$$ Let $N = \max\{N_1, N_2\}$. If $m, n \ge N$, then \begin{align*} \|y_m - y_n\| &= \left\|\frac{x_m}{\|x_m\|} - \frac{x_n}{\|x_n\|}\right\| \\ &= \frac{1}{\|x_m\|}\left\|x_m - \frac{\|x_m\|}{\|x_n\|}x_n\right\| \\ &\le \frac{1}{\|x_m\|}\left(\|x_m - x_n\| + \left\|x_n - \frac{\|x_m\|}{\|x_n\|}x_n\right\|\right) \\ &= \frac{1}{\|x_m\|}\left(\|x_m - x_n\| + \Big|\|x_n\| - \|x_m\|\Big|\right) \\ &\le \frac{2}{\|x_m\|}\|x_m - x_n\| \\ &< \frac{4}{r}\cdot \frac{r\varepsilon}{4} = \varepsilon, \end{align*} proving that $y_n$ is Cauchy, and completing the final piece of the puzzle.