How to prove a vector-valued function is constant on some interval? Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ and $\gamma:I \rightarrow \mathbb{R}^n$ where $I \subseteq \mathbb{R}$ is an open interval.
Assume $\gamma' (t)=(\nabla f)(\gamma (t))$ for all $t \in I$, and there exist two numbers a,b with $a<b$ in I such that $\gamma(a)=\gamma(b)$, prove that there is a $\vec{p} \in \mathbb{R}^n$ such that $\gamma(t)=\vec{p}$ for all $t \in [a,b]$.
I think we the question is asking us to prove $\gamma$ is constant on$[a,b]$. I know there is a theorem in one-variable calculus saying that if $\gamma '(t)=0$ for all $t \in I$ then $\gamma$ is constant on I.
Is this theorem still true if $\gamma$ is a vector-valued function?
if the theorem is true, how can i show that it's derivative is always zero on [a,b]?
I tried Mean Value Theorem and Rolle's Theorem, but these two theorems only gives the information of the derivative at some certain point on [a,b], not for all points on [a,b], i guess the fact that $\gamma' (t)=(\nabla f)(\gamma (t))$ for all $t \in I$ would be helpful, but how can i use it?
 A: We observe that a curve $\gamma(t)$ such that
$\gamma'(t) = \nabla f(\gamma(t)) \tag{1}$
is a trajectory of the gradient flow of the function $f$; that is, an integral curve of the vector field $\nabla f$.  As such, $\gamma(t)$ is consistently, in general, crossing the level hypersurfaces of $f$ so, all else being equal, we should expect $f$ to change along $\gamma$.
However, we compute $df(\gamma(t))/dt$ as
$\dfrac{f(\gamma(t))}{dt} = \nabla f(\gamma(t)) \cdot \gamma'(t); \tag{2}$
from this and (1) we find
$\dfrac{f(\gamma(t))}{dt} = \gamma'(t) \cdot \gamma'(t) = \Vert \gamma'(t) \Vert^2 \ge 0, \tag{3}$
with equality holding precisely when 
$\gamma'(t) = 0. \tag{4}$
We thus compute
$f(\gamma(b)) - f(\gamma(a)) = \displaystyle  \int_a^b \dfrac{df(\gamma(s))}{ds}ds = \displaystyle \int_a^b \Vert \gamma'(s) \Vert^2 ds; \tag{5}$
since
$\gamma(b) = \gamma(a), \tag{5}$
the left hand side of (4) vanishes, and hence does the integral on the right.  This forces
$\Vert \gamma'(t) \Vert = 0 \tag{6}$
for $t \in [a, b]$, whence
$\gamma'(t) = 0 \tag{7}$
there.  Thus we must have
$\gamma(t) = \vec p, \tag{8}$
a constant, for $t \in [a, b]$.
Note added Tuesday 27 June 2017 10:43 AM PST:  This result should perhaps come as no surprise, since the condition $\gamma(a) = \gamma(b)$ implies the curve $\gamma(t)$ is periodic and non-trivial gradient systems don't admit periodic orbits.  Furthermore, the s (7), (8) in concert with (1) show that $\nabla f(\vec p) = 0$, i.e., $\vec p$ is a critical point of $f$.  When we realize that $\gamma(t)$ is merely sitting on a zero of the vector field $\nabla f$, the conclusions reached perhaps come as no surprise.  However, the question does provide a potentially useful alternate characterization of the situation.  Finally, that $\gamma'(t) = 0$ implies $\gamma(t)$ is constant follows from the one-dimensional case by looking at $\gamma(t)$ component-wise.  End of Note.
