Show that $X$ has not periodic orbits. Let $X=\nabla f$, where $f$ is a function of class $C^{r},r\geq 2$ defined on open set $\triangle\subset\mathbb{R}^{n}$. Show that $X$ has not periodic orbits.
My approach: If $X$ has periodic orbit, $\gamma=\{\varphi(t,p);0\leq t\leq T\}$ of periodic $T$, where $\varphi(t,p)$ is trajectory on $X$. Then $X(f(\varphi(t)))=\nabla f(\varphi(t))=\nabla f(\varphi(t))*\varphi'(t)$. In particular, $$X(f(\varphi(T)))-X(f(\varphi(0)))=0$$
I want find to contradiction, any hint will be appreciated.
 A: $f(\phi(T))-f(\phi(0))=$
$\int_0^T{{df(\phi(t))}\over{dt}}dt =$
$\int_0^Tdf(\phi(t)).\phi'(t))=$
$\int_0^T\nabla 
f(\phi(t)).X(\phi(t))=-\int_0^T\|X(\phi(t))\|^2dt<0$. Contradiction.
A: I'm using the notation $\gamma(t)$ for $\phi(t; p)$.
Observe that
$\dfrac{df(\gamma(t))}{dt} = \nabla f(\gamma(t)) \cdot \dot \gamma(t)  \tag{1}$
by the chain rule; but
$\dot \gamma(t) = \nabla (\gamma(t)), \tag{2}$
whence (1) becomes
$\dfrac{df(\gamma(t))}{dt} = \dot \gamma(t) \cdot \dot \gamma(t) = \Vert \dot \gamma(t) \Vert^2  \ge 0,  \tag{3}$
with equality holding wherever $\dot \gamma(t) = 0$.  It then follows that
$f(\gamma(T)) - f(\gamma(0)) = \displaystyle \int_0^T\dfrac{df((\gamma(s))}{ds}ds = \displaystyle \int_0^T \Vert \dot \gamma(s)) \Vert^2 ds > 0 \tag{4}$
unless $\dot \gamma(t) = 0$ for $0 \le t \le T$.  If $\dot \gamma(t) = 0$ on this interval, then $\gamma(t) = \gamma(0)$ is constant, and 
$\nabla f(\gamma(t)) = 0, \tag{5}$
which implies $f(\gamma(t))$ remains at a zero or equilibrium point of $\nabla f$; NOT a periodic orbit in the general sense of the term.  When $\dot \gamma(t) \ne 0$ somewhere on $[0, T]$, the continuity of $ \dot \gamma(t)$ forces the integral on the right of (4) to be strictly positive; then if
$\gamma(T) = \gamma(0), \tag{6}$
the left hand side vanishes; this contradiction shows that (6) cannot hold, so there can be no periodic trajectory of $\nabla f$.
Nota Bene: There is an error in our OP's main equation.  It should read (using my notation $\gamma(t))$:
$X(f(\gamma(t))) = \nabla f(\gamma(t))(f(\gamma(t))) = \nabla f(\gamma(t)) \cdot \nabla f(\gamma(t)) = \nabla f(\gamma(t)) \cdot \dot \gamma(t); \tag{7}$
perhaps this explains his/her difficulty with this problem.
