Consider smooth closed space curve (parametrised by its arc length) in 3 dimensional space.Does there exist atleast one point at which both the curvature & torsion attain extremum values?
Imagine a 2D curve defined as follows: $x(t) := \kappa(s), y(t) := \tau(s)$;
if it were true that a space curve always contains a point where curvature and torsion are simultaneously extremal, then the associated 2D curve (as defined above), would always contain a corner of its bounding box.
Even if infinite curvatures and torsions are allowed, counterexamples can easily be constructed: take a space curve, whose associated 2D curve as defined previously, is an archimedian spiral.

1$\begingroup$ You appear to be ignoring the OP's condition that the curve be closed, which, of course, rules out curvaturetorsion profiles such as Archimedean spirals, etc. Also, without producing an example where it doesn't, how do you know that the curvaturetorsion profile curve of a closed space curve won't always touch at least one corner of its bounding box? I agree that it looks very unlikely on the face of it, but that, in and of itself, is not evidence for belief. $\endgroup$ – Robert Bryant Dec 27 '13 at 13:44

$\begingroup$ @Robert: Ok, the Archimedean spiral is not a closed curve, but that can be easily fixed by using instead a curve in the plane, which is the hull curve of a convolution of an Archimedean spiral with "very small" circles, whose center is on the spiral, letting the parameter $t$ of the planar curve tend to infinity. Your other question is more delicate but it can be answered, that the integration of periodic functions for curvature and torsion essentially results in periodic functions for the three coordinates of the space curve as a consequence of the FrenetSerret formulas. $\endgroup$ – Manfred Weis Dec 27 '13 at 15:48

1$\begingroup$ Actually, your last statement is not true at all. For example, if you take $\kappa$ and $\tau$ to be constant (and nonzero), which is about as periodic as you can get, the resulting space curve (a helix) will certainly not be periodic. In fact, determining when a pair of periodic functions is the curvature and torsion of a periodic space curve is highly nontrivial. $\endgroup$ – Robert Bryant Dec 27 '13 at 17:59
No. Consider a curve on the unit $2$sphere, parametrized by arc length $ds$ with geodesic curvature $\rho(s)$. One easily computes that, as a curve in $3$space, one has $$ \kappa(s)^2 = 1 + \rho(s)^2 \qquad\text{and}\qquad \tau(s) = \frac{\rho'(s)}{1+\rho(s)^2}. $$ Thus, if what you were asking were true, then every closed curve on the $2$sphere would have a point where both $\rho$ and $\rho'$ had critical points. However, considering simple examples shows that this is not so.
You can convince yourself that this isn't true in the plane by considering an ellipse, where $\rho$ has critical points only where $\rho'$ vanishes (and, at those $4$ points, i.e., the vertices, $\rho''$ is not zero). Now consider the analog of the ellipse on the $2$sphere, i.e., the curve cut out by $x^2+y^2+z^2=1$ and $x^2/a^2+y^2/b^2z^2 = 0$ where $a>0$ and $b>0$ are very small. This will have geodesic curvature on the $2$sphere closely approximating that of the corresponding ellipse in the plane and will furnish an explicit counterexample.