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I'm asked to prove that for a function $f:[a,b]\subseteq \Bbb R \to \Bbb{R}^2$ such that $f(a)=f(b)$ is derivable for all $t$ on the interval then $f'(t)$ takes all possible directions. I am not sure my logic is correct, but I Was thinking about a function $f$ where $f(t)=v$. Then we would still have $f(a)=f(b)$ but we would not have f' taking any possible direction.

Brian61354270
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Wykk
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3 Answers3

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The claim is false, as the following picture shows. The curve starts and ends at the left hand side, and the initial and final tangents point in opposite directions. But the tangent never points straight down (or in the direction indicated by the red arrow).

If you add the hypotheses that

  1. $f'(u)$ is always nonzero for all $u$ in the interval and
  2. $f'(a) = f'(b)$,

then it's possible that the theorem is correct. enter image description here

John Hughes
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The most low-tech approach (under the assumption that $f^\prime$ is never $0$) seems to be the following: the $x$ and the $y$ coordinates of $f^\prime$ don't vanish simultaneously (by assumption), but they both change sign (since the integrals of both are zero).It follows that there are points where the slope of $f^\prime$ tends to $\pm \infty$ and $0.$ it follows that the slope assumes all values (assuming $f^\prime$ is continuous), which is exactly what was asked for.

Igor Rivin
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  • You might want to reflect on this proof in light of my counterexample. You're correct that the tangent line takes on all possible slopes; it's not true that the tangent vector points in all possible directions. (I was actually surprised to realize that this was the subtle explanation for the glaring difference between our answers.) – John Hughes Nov 24 '19 at 19:00
  • @JohnHughes I am aware of this issue. The question is: what is direction? It could be meant projectively (in which case what I say works) or not. – Igor Rivin Nov 24 '19 at 20:53
  • I'm sorry about the awful text formating i write from an almost obsolete iPad //I found that as a hint I'm told to consider a new function a(t) such that a(t)=v'•f(t) in which v is any vector in R^2 and v' ortogonal to v, ti would still be required for f'(u) to be nonzero o ver the interval,right – Wykk Nov 24 '19 at 23:19
  • Fair enough @IgorRivin. In the texts I've seen, the direction of a nonzero vector $v$ is often defined as $v / |v|$; I haven't seen the projective definition applied except when talking about line-fields rather than vector field; in that case, of course, the projective definition is the only one that makes sense. – John Hughes Nov 24 '19 at 23:43
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If one interprets direction as vector, as John Hughes mentioned, the result is false without some smoothness assumption. Typically one expects that the function gives a smooth map as a regular closed curve (so the tangent vectors at the endpoints agree), and so this allows one to define an analogue of the Guauss map, namely map the normalized tangent vectors of the closed curve $C$ to the unit circle, $S^1$. Then it boils down to showing that the map $C\rightarrow S^1$ is surjective. As it turns out (taking inspiration from a paper of Whitney (1937)) the endpoints assumption is not still enough, for example take the upright figure "8", with the usual orientation and assuming the middle "x" comprises of positive and negative slopes (then clearly the figure is missing the horizontal tangent pointing to the right). Hence one needs to assume the curve to be simple; in this case the problem was resolved by Hopf (1933) in his famous Umlaufsatz (see link therein). One can also find a recent literature addressing a related topic.

Pythagoras
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