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I'm trying to prove the following statement:

Let $a$ and $b$ be points in $\mathbb R^2$ and let $X$ be the set of all continuously differentiable functions $\gamma:[0,1]\to\mathbb R^2$, with $\gamma(0)=a$ and $\gamma(1)=b$. Further let $\delta$ be the metric on $X$ defined by $$\delta(\gamma_1,\gamma_2)=\sup_{[0,1]}\Vert\gamma_1-\gamma_2\Vert.$$ Then the metric space $X$ with metric $\delta$ is not complete.

At the point I am in the analysis book I'm working through, to do this I need to find a Cauchy sequence in $X$ which does not converge (in $X$). I tried considering the sequence $\gamma_n(t)=a+t^n(b-a)$, but I cannot seem to show that it's Cauchy (I can't figure out what to use for my $N$).

I did manage to show that, for $m\gt n$, $$\delta(\gamma_n,\gamma_m)=\Vert b-a\Vert\left(\frac{n}{m}\right)^{n/(m-n)}\frac{m-n}{m},$$ but I can't figure out how to work with this to ensure that it's smaller than $\epsilon\gt0$.

Thus my question is, does this sequence work? If not, how might I come up with one that does? If so, how might I show that it's Cauchy?

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  • $\begingroup$ I've removed the tag "incompleteness" that was not relevant here $\endgroup$ – Max Nov 10 '17 at 7:59
  • $\begingroup$ Think about a sequence of curves which get more pointy with higher $n$. In the limit they are no longer differentiable. For example $\gamma_n(t) = \sqrt{x^2+1/n}$ might work I think. The limit is $|x|$. $\endgroup$ – M. Winter Nov 10 '17 at 22:24
  • $\begingroup$ @M.Winter do you mean $\sqrt{t^2+1/n}$? That wouldn't be a sequence in $X$, since the function outputs real numbers, not vectors in $\mathbb R^2$. $\endgroup$ – themathandlanguagetutor Nov 10 '17 at 22:27
  • $\begingroup$ @themathandlanguagetutor You are right about the mistake with the $x$. For the other thing, just add a constant second coordinate. $\endgroup$ – M. Winter Nov 10 '17 at 22:33
  • $\begingroup$ @M.Winter Then it still wouldn't be in $X$ because it wouldn't satisfy $\gamma_n(0)=a$ and $\gamma_n(1)=b$. $\endgroup$ – themathandlanguagetutor Nov 10 '17 at 22:40
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A rigorous way

Use

$$f_n(x)=\begin{cases}0&\text{for $x\le 0$}\\ x^{1+1/n}&\text{for $x>0$}\end{cases},\qquad f^*(x):=\begin{cases}0&\text{for $x\le 0$}\\x&\text{for $x>0$}\end{cases}$$

which gives the depicted sequence of functions and converges to $f^*$ on the interval $[-1,1]$.

$\qquad\qquad\qquad$

The important thing is: $f_n$ is differentiable for all $n$, but the limit curve is obviously not. Also, the end points are fixed, i.e. $f_n(-1)=0$ and $f_n(1)=1$ for all $n$. This means we can use this function to build our desired curve (assuming $a\not=b$)

$$\gamma_n(t)=a+f_n(2t-1)(b-a).$$

Proof.

  • $f_n'(x)$ equals $(1+1/n)x^{1/n}$ for $x>0$ and hence both defining cases agree on $f_n'(0)=0$. Therefore $f_n$ is indeed differentiable. Obviousoly $f^*$ is not.
  • We have $\sup_{x\in[-1,1]}|f_n(x)-f^*(x)|=\sup_{x\in[0,1]}|x^{1+1/n}-x|$. It is basic anaylsis to find the maximum at $$\hat x=\left(1+\frac1n\right)^{-n},\qquad |\hat x^{1+1/n}-\hat x|=\left|\frac1{n-1}\cdot \left(1+\frac1n\right)^{-n}\right|$$ which vanishes for $n\to\infty$. Hence $f_n\to f^*$ in the given metric.
  • Now we have \begin{align} d(\gamma_n,\gamma^*)&=\sup_{t\in[0,1]}\|\gamma_n(t)-\gamma^*(t)\|\\ &=\|b-a\|\cdot\sup_{t\in[0,1]}|f_n(2t-1)-f^*(2t-1)|\\ &=\|b-a\|\cdot\sup_{t\in[-1,1]}|f_n(t)-f^*(t)|\\&\to0, \end{align} as $n\to\infty$. Hence $\gamma_n$ converges (and therefore is Cauchy), but the limit is not in $X$.
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  • $\begingroup$ I'm not quite satisfied with saying that it's Cauchy just because it converges (in some set), simply because that's a result we've really only shown for sequences that converge in $\mathbb R^d$ with the usual metric. That said, I'll look more at this example and see if I either can't show this result or show it's Cauchy directly. $\endgroup$ – themathandlanguagetutor Nov 13 '17 at 17:42
  • $\begingroup$ @themathandlanguagetutor well its a classical fact (and not too hard to see) that if a sequence is Cauchy in some metric space, then also in any metric space which contains the sequence. This is because Cauchy only needs the elements of the sequence and does not refer to the containing space otherwise. $\endgroup$ – M. Winter Nov 13 '17 at 21:32
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Just a hint

Choose some sequence of curves which gets more and more pointy with increasing $n$. For example

$$\gamma_n(t):=\begin{pmatrix}\!\sqrt{t^2+\frac 1n}\; \\ t\end{pmatrix}.$$

The limit of this sequence is $\gamma^*(t)=(|t|,t)^\top$, which is not differentiable in $t=0$. We can see this via

$$\|\gamma_n(t)-\gamma^*(t)\|\le \frac 1 {\sqrt n},$$

which is not hard to show. Since the sequence converges, it must be Cauchy. But the limit is not in $X$. It should not be too hard to find a way to make the endpoint match $a$ and $b$, e.g. by adding segments on both ends of $\gamma_n$ which start (resp. end) in $a$ (resp. $b$).


Another idea, for which I will not give an explicit formula (but the proofs are obvious) is motivated by the picture below:

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$

You again take a wedge ($\wedge$) like curve as the limit curve (which is not in $X$). By rotating and scaling you can make the endpoints fit $a$ and $b$ (if $a\not=b$). The sequence curves $\gamma_n$ share initial- and end segment with $\wedge$ but the pointy end is replaced by a circle arc which fits in such a way that the curve stays differentiable.

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  • $\begingroup$ What's wrong with my hint? $\endgroup$ – M. Winter Nov 10 '17 at 22:51
  • $\begingroup$ Even if I can modify $\gamma_n$ to have the right endpoints (which I'm still not seeing how to do) $\gamma^*$ is indistinguishable from $(t,0)$ on $[0,1]$, so it should still be in $X$. $\endgroup$ – themathandlanguagetutor Nov 10 '17 at 22:59
  • $\begingroup$ @themathandlanguagetutor Please take a minute to think about the hint which I gave you. This is no solution. Think about this curve as defined on $[-1,1]$ or so. You should be able to work out how to reparametrize $\gamma_n$ nd $\gamma^*$ to make the relevant part fit into $[0,1]$. This is no "give me a final answer" site, but a "help me to understand how to work out an answer myself" site. $\endgroup$ – M. Winter Nov 10 '17 at 23:03
  • $\begingroup$ I'm sorry, I was just very fixed on the interval $[0,1]$. Does this look like it should work: $\gamma_n(t)=a+\left(\left(\sqrt{(2t-1)^2+1/n}+(t-1)\sqrt{1+1/n}\right)(b_1-a_1),t(b_2-a_2)\right)$? $\endgroup$ – themathandlanguagetutor Nov 10 '17 at 23:19
  • $\begingroup$ @themathandlanguagetutor Also sorry if I sounded rude. You are definitely on the right track. I am not sure if your formula works for $t=1$. Of course there might be better solutions to this problem (especially ones with an easier final proof that they are Cauchy), I just wanted to point ot the "pointy end" intuition. And your initial idea might still work. $\endgroup$ – M. Winter Nov 10 '17 at 23:27

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