0
$\begingroup$

Question is to prove that

$X$ is complete if and only if $S(X)=\{x: \|x\|=1\}$ is complete.

The condition $S(X)=\{x :\|x\|=1\}$ being complete seems to be very weak to conclude that X is complete.

I was trying to solve this but could not succeed.

Let $(x_n)$ be a Cauchy sequence in $X$, we want to find a Cauchy sequence in $S(X)$.

Obvious corresponding sequence in $S(X)$ is $y_n=\dfrac{x_n}{\|x_n\|}$.

We then see if this $(y_n)$ is Cauchy and then quote the condition that $S(X)$ is complete to conclude that $(y_n)$ converges and thus $(x_n)$ converges.

But then, it seems to be not natural why should $(y_n)$ is Cauchy if $(x_n)$ is Cauchy. $$\|y_n-y_m\|=\left|\left|\dfrac{x_n}{\|x_n\|}-\dfrac{x_m}{\|x_m\|}\right|\right|=\frac{1}{\|x_n\| \|x_m\|}\left|\left|x_m||x_n||-x_n||x_m||\right|\right|$$

As $(x_n)$ is Cauchy, we see that it is bounded and so we have $M>0$ such that $\|x_n\|\leq M$ for all $n$.

This says $\|y_n-y_m\|\leq \frac{M}{\|x_n\| \|x_m\|}\|x_n-x_m\|$ which does not say anything about $(y_n)$ being Cauchy.

Assuming something more, as $S'(X)=\{x: \|x\|\leq 1\}$ is complete we can then consider the sequence $(y_n)$ with $y_n=x_n/M$ we then have $||y_n||\leq 1$ and $\|y_n-y_m\|\leq 1/M \|x_n-x_m\|$ which says that $(y_n)$ is Cauchy and for the same reason as above we can say $x_n$ is Cauchy.

Consider the case when $X= \mathbb{Q}$ the collection of rational numbers, then $S(X)=\{-1,1\}$ which is complete but $\mathbb{Q}$ is not complete.

So, I guess we need to have the condition the unit disk is complete.

Clearly $\Bbb Q$ is not a $\Bbb R$-normed space, it is a $\Bbb Q$-normed space. So, $\Bbb Q$ is not a complete as $\Bbb Q$-normed space.

After all it turns out that question is correct.

Please give only hints.

$\endgroup$
1
$\begingroup$
  1. Show that the sequence $\lVert x_n \rVert$ is Cauchy (reverse triangle inequality).
  2. Assume $x_n$ does not converge to $0$. Then $\lVert x_n \rVert$ has a strictly positive lower bound for $n$ big enough. Use this to conclude that $y_n$ is cauchy.
$\endgroup$
  • $\begingroup$ Can you tell me what is wrong with the example of rational numbers? $\endgroup$ – user87543 Sep 14 '16 at 12:02
  • $\begingroup$ The rational numbers are no $\mathbb{R}$-normed space. That's the problem. From $\lVert x_n \rVert$ cauchy you can't conclude that it converges in $\mathbb{Q}$. $\endgroup$ – Paul K Sep 14 '16 at 12:03
  • $\begingroup$ I have already said that it is not a $\mathbb{R}$ normed space. Even as $\mathbb{Q}$ space it is not complete. $\endgroup$ – user87543 Sep 14 '16 at 12:08
  • $\begingroup$ And that's why in functional analysis you work in $\mathbb{R}$ or $\mathbb{C}$ vectorspaces. You really can't expect a normed space to be complete if the underlying field isn't complete. $\endgroup$ – Paul K Sep 14 '16 at 12:17
  • $\begingroup$ Ok.. I got it.. As $\mathbb{Q}$ is not complete field, my example is not valid.. Thanks $\endgroup$ – user87543 Sep 14 '16 at 12:50
1
$\begingroup$

Suppose $X$ is not complete. Then, you can isometrically embed it in a dense subspace of its completion, say $X\stackrel\iota\hookrightarrow \overline X$.

Now, the subspace spanned by a vector $\overline v\in \overline X\setminus\iota(X)$ kinda makes "a hole" in $S(\overline X)$, which can be brought back to $S(X)$ using the fact that $\iota$ is an isometry and $\iota(X)$ is dense.

$\endgroup$
  • $\begingroup$ Can you tell me what is wrong with the example of rational numbers? $\endgroup$ – user87543 Sep 14 '16 at 12:02
  • $\begingroup$ I think it is implicitly assumed that $X$ is a $\Bbb R$ or $\Bbb C$ vector space. For instance, at some point in the proof I suggest, you'll want to use the fact that, if $v\in \overline X\setminus\{0\}$, then $$v\in \iota(X)\iff \frac{v}{\lVert v\rVert}\in\iota(X)$$ this is not the case for the metric completion of a $\Bbb Q$-vector space, because $\Bbb Q$ itself is not complete. $\endgroup$ – user228113 Sep 14 '16 at 12:50
  • $\begingroup$ I don't know of a notion of "completeness as $\Bbb Q$-normed space", and I doubt you can sensibly maneuver around the fact that $$1,\ 1.4,\ 1.41,\ 1.414,\ 1.4142,\ 1.41421,\ldots$$ is Cauchy but without a limit. $\endgroup$ – user228113 Sep 14 '16 at 12:55
  • $\begingroup$ Yes.. Got it :) $\endgroup$ – user87543 Sep 15 '16 at 7:09
1
$\begingroup$

Perhaps a simpler method: Suppose $\| x_n \| \not \to 0$. If $(x_n)$ is Cauchy, then $( \| x_n \| )$ is a Cauchy sequence in $\mathbb{R}$, and thus convergent. Let $h = \lim_n \| x_n \| > 0$. Set $z_n = \frac{1}{h} x_n$. Then \begin{align*}\|y_n - y_m\| & \leq \| y_n - z_n \| + \|z_n - z_m\| + \|z_m - y_m\| \\ & = \| (1 - \|x _n \| / h) y_n \| + \| (1/h) (x_n - x_m) \| + \| (1 - \| x_m \| / h) y_m \| \\ & = | 1 - \|x _n \| / h | + \frac{1}{h} \| x_n - x_m \| + |1 - \| x_m \| / h| \\ & = \frac{1}{h} \left( |h - \|x _n \| | + \| x_n - x_m \| + |h - \| x_m \| | \right) . \end{align*}.

Now we show $(y_n)$ is Cauchy. Fix $\epsilon > 0$. Then since $(x_n)$ is Cauchy, there exists $N_1$ such that if $m, n \geq N_1$, then $\| x_n - x_m \| < \epsilon h / 3$. Moreover, since $h = \lim_n \| x_n \|$, there exists $N_2$ such that for $n \geq N_2$, we have $| h - \| x_n \| | < \epsilon h / 3$. Set $N = \max (N_1, N_2)$. Then for $m, n \geq N$, we get \begin{align*} \| y_n - y_m \| & \leq \frac{1}{h} \left( |h - \|x _n \| | + \| x_n - x_m \| + |h - \| x_m \| | \right) \\ & < \frac1h ( \epsilon h / 3 + \epsilon h / 3 + \epsilon h / 3 ) & = \epsilon . \end{align*}

But also, since $\| z_n - y_n \|$ goes to $0$, this suggests a candidate for the limit of $x_n$ in terms of that of $y_n$.

$\endgroup$
  • 1
    $\begingroup$ This is helpful. Thanks :) $\endgroup$ – user87543 Sep 15 '16 at 7:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy