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While doing my research I came across an interesting question. Let $\{\bf{x}_n \}\subset R^m$ be any sequence of points with $\bf{x}_n\rightarrow \bf{0}$, as $n \rightarrow \infty$. Is there a smooth path $c: [0,a) \rightarrow \mathbf{R}^m$, $a>0$, $c(0)=\bf{0}$, a (decreasing) sequence $\{t_k\}\subset (0,a)$, an (infinite) subsequence $\{\bf{x}_{n_k} \}\subset \{\bf{x}_n \}$ so that $c(t_k)=\bf{x}_{n_k}$? Can we make the path analytic at zero?

Any help, or suggestion would be appreciated.

Mohammadreza Bidar
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    Using $C^\infty$ bump functions you can connect the dots with a smooth path, but it will probably not be analytic at $0$. You can probably place the points near 0 to prevent any analytic behavior. Maybe something like $x_n = (-1)^n x_1/n$ is too wiggly to be on an analytic path. – kimchi lover Sep 11 '17 at 02:16

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I'll just tackle the analytic case: Assume that you have a sequence $y_k = \phi(t_k)$ where $\phi$ analytic at $0$, $\phi$ has $0$ of order $d$ at $0$. Then we have $\frac{\log |y_k|}{\log |t_k|}\to d$. Now if you have a sequence $(x'_n, x''_n)\in \mathbb{R}^2$ approaching $(0,0)$ such that $\frac{\log|x'_n|}{\log|x''_n|}\to \sqrt{2}$, then no subsequence can be interpolated by an analytic function.

$\bf{Added:}$ Consider the $C^{\infty}$ case now. We'll show the following: if the sequence $a_n$ decreases fast enough to $0$ then there exists a smooth function $f$ defined around $0$ so that $f(\frac{1}{n}) = a_n$. Indeed, consider $\phi_n$ smooth bump function such that $\phi_n(\frac{1}{m}) =\delta_{mn}$. Now the take $a_n$ so that $|a_n| \le \frac{1}{2^n \|\phi_n\|_n}$, where the norm $\|\psi\|_n = \sup_{t\in [-1,1], 1\le k \le n} |\psi^{(k)}(t)|$. Then the function $f$ equals the series $\sum a_n \phi_n$, which converges to a smooth function.

Now it's enough to choose a subsequence of $x_n$ that converges to $0$ fast enough and thus can be interpolated at $\frac{1}{n}$. I think it can be arranged that the whole sequence can be interpolated by a smooth function at a sequence of values ( we have to retard enough the convergence to $0$ of this).

orangeskid
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  • That was wonderful! Thanks. – Mohammadreza Bidar Sep 11 '17 at 04:47
  • If $(x'_n,x''_n)=\phi(t_k)=(\phi_1(t_k),\phi_2(t_k))$, $\dfrac{\log|x'_n|}{\log|x''_n|} \rightarrow \sqrt{2}$ means $\dfrac{\log|\phi_1(t_k)|}{\log|\phi_2(t_k)|} \rightarrow \sqrt{2}$, I don't see a contradiction directly here. – Mohammadreza Bidar Sep 11 '17 at 04:56
  • Thanks! Great question too! – orangeskid Sep 11 '17 at 05:00
  • @Mohammadreza Bidar: For analytic functions the order is non-zero for each component, so we get $\frac{\log|\phi_j(t_k)|}{\log |t_k|} \to d_j$, for $j=1,2$, hence the ratio is rational, so cannot be $\sqrt{2}$. – orangeskid Sep 11 '17 at 05:09
  • Yeah, got it, thanks. – Mohammadreza Bidar Sep 11 '17 at 05:10
  • I have one question about this: does it work for non-zero derivative? There are multiple definitions of smooth path and smooth curve and op didn't specify which one they are using. – Stefan Octavian Feb 13 '21 at 18:28
  • @Stefan Octavian: It may not work for non-zero derivative, with the same argument like in first part. If you have a smooth map $\phi(t) = (\phi_1(t), \phi_2(t))$, and assume that not all of the derivatives of $\phi$ at $0$ are $(0,0)$. Say $\phi_1^{(k)}(0) \ne 0$, with $k$ minimum. Then $\lim_{t\to 0} \frac {\log |\phi_2(t)|}{\log|\phi_1(t)|}$ is rational ( with $k$ a denominator) or $+\infty$. It has to do with the order of the zero ... infinitely small of a certain order... – orangeskid Feb 13 '21 at 22:28
  • For any two points, we can construct a cubic bezier curve between them with any lateral derivatives at the endpoints. Take any sequence converging to $0$ and build such bezier curves between successive points of the sequence that the lateral derivatives at the joint points are equal and the sequence of derivatives converges to a non-zero value. The resulting curve is $C^1$ and there is a sequence of points converging to $0$ where the derivative has non-zero limit, so the derivative of the curve there cannot be $0$. – Stefan Octavian Feb 14 '21 at 13:47
  • @Stefan Octavian: I see, your definition of smooth is $C^1$, in general it is $C^{\infty}$, Your construction might work. You can just write it up in an answer. – orangeskid Feb 14 '21 at 20:48
  • @Stefan Octavian: I see, a path like $(t, t^{\sqrt{2}})$ is $C^1$. I think your idea is quite OK. – orangeskid Feb 15 '21 at 04:36