Cauchy sequence of smooth parameterized curves that does not converge

Question

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?

Answer 1

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 $X$.

Answer 2

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. $b$).

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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.