Supremum length of space curves contained in the open unit ball having always less than unity curvature I am in the process of proving that if a space curve (in $R^3$) has infinite length and the curvature tends towards $0$ as the natural parameter $s$ tends to infinity, the curve must be unbounded - i.e. not contained in any sphere of finite radius. This seems correct intuitively, but I have no guarantee it is correct, unless I am missing something obvious. One way to prove my hunch, I have deduced, is to use a lemma that any curve contained in the open unit ball with curvature always less than one must have a finite upper bound on its length (possibly $2π$, but it could be greater for all I know).
How might one go about proving such an upper bound exists, or if it exists? It might also be nice to know what the bound specifically is, too. I've thought it might be possible to pose this as a variational problem - maximizing length - and then reducing it into a simpler problem, but that appears to be hellishly complicated. Thoughts?
 A: Let ${\mathbf{x}}(s)$ be a curve in ${\mathbb{R}}^3$ with natural parameter $s$.  We will need the following lemma, the proof of which is given at the end of this answer.

Lemma: Choose a fixed point ${\mathbf{y}}$.  Then the curvature $\kappa$ satisfies
  $$
\kappa \ge \left|\dot{\theta} + \frac{1}{r}\sin{\theta}\right|,
$$
  where $r = \lVert\mathbf{x} - \mathbf{y}\rVert$ and $\theta$ is the angle between the
  velocity $\dot{\mathbf{x}}$ and $\mathbf{x} - \mathbf{y}$.

Take $\mathbf{y} = \mathbf{x}(0)$; then $r(0)=0$ and $\theta(0)=0$.  The function $r(s)$ is monotonically increasing as long as $0 \le \theta < \pi/2$ (since $\dot{r}(s) = \cos{\theta}$), so $\theta$ and $\kappa$ can be considered as single-valued functions of $r$ until that point.  The above lemma then gives
$$
\kappa(r) \ge \left|{\theta}'(r)\dot{r} + \frac{1}{r}\sin{\theta(r)}\right| = \left|{\theta}'(r)\cos{\theta(r)} + \frac{1}{r}\sin{\theta(r)}\right| = \left|\frac{1}{r}(r \sin\theta(r) )'\right|.
$$
Until the first turning point of the motion (where the velocity becomes perpendicular to the radius), we have
$$
R\sin\theta(R) \le \int_{0}^{R} r \kappa(r) dr.
$$
If the curvature is strictly below a fixed value (say, $\kappa < K$), then the integral is less than $\frac{1}{2}KR^2$ for $R>0$, and we have the result that
$$
\sin{\theta(r)} < \frac{1}{2}Kr
$$
for $r>0$.  A turning point is reached when $\theta=\pi/2$; this equation shows that the first such turning point must be at a radius greater than $2/K$, and hence the curve cannot be confined within a ball of diameter $2/K$.  Finally, the arclength before reaching a given radius $R$ is bounded by
$$
\begin{eqnarray}
s(R) &=& \int_{0}^{R} \frac{ds}{dr}dr \\ &=& \int_{0}^{R} \frac{dr}{\cos\theta(r)} \\ &<& \int_{0}^{R} \frac{dr}{\sqrt{1 - \frac{1}{4}K^2 r^2}} \\ &=& \frac{2}{K}\sin^{-1}\left(\frac{1}{2}KR\right)
\end{eqnarray}
$$
for $R \le 2/K$.  We conclude that any curve contained in the open unit ball with curvature $\kappa < 1$ must have length $s(2) < 2\sin^{-1}(1) = \pi$.  Moreover, this bound is tight, since a circular arc joining the points at $\pm (1-\epsilon^2)\hat{\mathbf{z}}$ and the point at $(1-\epsilon)\hat{\mathbf{x}}$ has length approaching $\pi$ as $\epsilon \rightarrow 0$.

Proof of Lemma:
We will work in spherical coordinates centered at $\mathbf{y}$; then
$$
\begin{eqnarray}
{\mathbf{x}}
  &=& r\hat{\mathbf{r}}, \\
{\dot{\mathbf{x}}}
  &=& \dot{r}{\hat{\mathbf{r}}} + r\dot{\hat{\mathbf{r}}} \\
  &=& \dot{r}{\hat{\mathbf{r}}} + r v_{\perp} \hat{\mathbf{v}}_{\perp}.
\end{eqnarray}
$$
Here $\hat{\mathbf{r}}$ is the unit vector from the origin to ${\mathbf{x}}$, and $\dot{\hat{\mathbf{r}}} = v_{\perp} \hat{\mathbf{v}}_{\perp}$ is its rate of change.  Because $\hat{\mathbf{r}}$ has constant length, we have $\hat{\mathbf{v}}_{\perp}\cdot \hat{\mathbf{r}} = 0$.  Taking the time derivative of this gives
$$
0 = \dot{\hat{\mathbf{v}}}_{\perp}\cdot \hat{\mathbf{r}} + \hat{\mathbf{v}}_{\perp}\cdot \dot{\hat{\mathbf{r}}} = v_{\perp} + \dot{\hat{\mathbf{v}}}_{\perp}\cdot \hat{\mathbf{r}},
$$
which we will use later.  Now, because $s$ is a natural parameter, $$\lVert\dot{\mathbf{x}}\rVert^2 = \left(\dot{r}\right)^2 + \left(rv_{\perp}\right)^2 = 1;$$
so we can define $\theta \in [0,\pi]$ such that $\dot{r} = \cos{\theta}$ and $rv_{\perp} = \sin{\theta}$.  The velocity and acceleration become
$$
\begin{eqnarray}
{\dot{\mathbf{x}}}
  &=& \left(\cos{\theta}\right){\hat{\mathbf{r}}} + \left(\sin{\theta}\right)\hat{\mathbf{v}}_{\perp}, \\
{\ddot{\mathbf{x}}}
  &=& -\left(\sin{\theta}\dot{\theta}\right){\hat{\mathbf{r}}} + \left(\cos{\theta}\right)\dot{\hat{\mathbf{r}}} + \left(\cos{\theta}\dot{\theta}\right)\hat{\mathbf{v}}_{\perp} + \left(\sin{\theta}\right)\dot{\hat{\mathbf{v}}}_{\perp} \\
&=& -\left(\sin{\theta}\dot{\theta}\right){\hat{\mathbf{r}}} +  \left(\cos{\theta}\right)\left(\dot{\theta} + \frac{1}{r}\sin{\theta}\right)\hat{\mathbf{v}}_{\perp} + \left(\sin{\theta}\right)\dot{\hat{\mathbf{v}}}_{\perp},
\end{eqnarray}
$$
and the acceleration has (two of its three) components
$$
\begin{eqnarray}
{\ddot{\mathbf{x}}}\cdot\hat{\mathbf{r}} &=& -\left(\sin{\theta}\dot{\theta}\right) + \left(\sin{\theta}\right)\left(\dot{\hat{\mathbf{v}}}_{\perp} \cdot \hat{\mathbf{r}}\right) \\
&=& -\left(\sin{\theta}\right)\left(\dot{\theta} + \frac{1}{r}\sin{\theta}\right), \\
{\ddot{\mathbf{x}}}\cdot\hat{\mathbf{v}}_{\perp} &=& +\left(\cos{\theta}\right)\left(\dot{\theta} + \frac{1}{r}\sin{\theta}\right).
\end{eqnarray}
$$
This brings us to the result that the squared curvature
$$
\kappa^2 = \lVert\ddot{\mathbf{x}}\rVert^2 \ge \left({\ddot{\mathbf{x}}}\cdot\hat{\mathbf{r}}\right)^{2} +
\left({\ddot{\mathbf{x}}}\cdot\hat{\mathbf{v}}_{\perp}\right)^{2} = \left(\dot{\theta} + \frac{1}{r}\sin{\theta}\right)^{2},
$$
where $\theta$ is the angle between the velocity and the outward radial vector.  The lemma follows by taking the square root of both sides.
A: This is true, and follows from a theorem of Fary.  Fary's theorem says that for closed curves inside a ball, the average curvature of the curve is at least as large as the curvature of the boundary of the ball.  
Your curve isn't closed (or finite in length), so you have to think about what the theorem says in this context.  But it will work:  the average curvature is zero, and so it cannot fit in any (finite) ball.
I learned about Fary's theorem from Serge Tabachnikov; there are several proofs (of the 2-dimensional case, though most generalize to higher dimensions, as he describes at the end) in his beautiful short paper:
http://www.math.psu.edu/tabachni/prints/dna-mass2.pdf
A: Here is something along these lines, coming from estimates using elementary calculus, but it is much weaker than the lemma you want to prove.
Let $\gamma$ be a regular $C^2$ curve in $\mathbb{R}^3$ parametrized with respect to arclength $s$ (I will assume for simplicity that $s$ starts at $0$).  If the curvature $\|\gamma''(s)\|$ is always less than $K$ and $\gamma$ is contained in a ball of radius $R\leq\frac{1}{4K}$, then the length of $\gamma$ is less than $\frac{1}{K}(1-\sqrt{1-4KR})$.  
To see this, write $\gamma(s)=\gamma(0)+s\gamma'(0)+\int_0^s(s-t)\gamma''(t)dt$ (derived as in this Wikipedia article).  Moving $s\gamma'(0)$ to the left and $\gamma(s)$ to the right and applying the triangle inequality, 
$$
\begin{align}
s &\leq \|\gamma(s)-\gamma(0)\|+\|\int_0^s(s-t)\gamma''(t)dt\| \\
 &\leq 2R +\int_0^s\|(s-t)\gamma''(t)\|dt \\
 &\lt 2R +K\int_0^s(s-t)dt \\
 &=2R+\frac{K}{2}s^2.
\end{align}
$$
For this inequality to always hold, $s$ must remain smaller than the smallest root of $\frac{K}{2}x^2-x+2R$.  Thus,  $s\lt\frac{1}{K}(1-\sqrt{1-4KR})$ as claimed.  
You could still apply this to prove the original result as stated, but hopefully someone comes along with a stronger result.
