Let $f : U \to \mathbb{R}$ be continuously differentiable, where $U \subset \mathbb{R}^n$ is open. I would like to show that $\nabla f$ has no non-constant closed integral curve.
Q1: Let the curve be denoted $\gamma : [0,1] \to \mathbb{R}^n$. Does constant mean $\gamma(t)= c$ for all $t \in [0,1]$? Does this just mean the curve is a point in $\mathbb{R}^n$?
Here's my proposed solution. Let $F(t) = f(\gamma(t))$. Assume for a contradiction that $\nabla f$ contains $\gamma$, i.e. $\gamma$ is an integral curve, i.e. we have $\gamma'(t) = \nabla f(\gamma(t))$. Now let's take the derivative of $F$ with respect to $t$. By the chain rule and the assumption above we have $$ F'(t) = \nabla f(\gamma(t)) \cdot \gamma'(t) = \nabla f(\gamma(t)) \cdot \nabla(f(\gamma(t)) = \Vert \nabla f(\gamma(t))\Vert^2 \geq 0 . $$ Let's integrate $F'(t)$. By the Fundamental Theorem of Calculus and the fact that $\gamma$ is closed, we have $\int_0^1 F'(t) dt = F(1) - F(0) = 0$. But then, since we know that $F'(t) \geq 0$, the integral being zero implies that $F'(t)=0$, i.e. $F(t) = f(\gamma(t))$ is constant. But this is a contradiction to $\gamma$ being non-constant, hence no such $\gamma$ exists.
Is this a correct proof? I'm wondering in my last line why can't $f$ be constant? I guess then $\nabla f$ would obviously have no non-constant closed curves, so we exclude this case from our considerations to begin with? Thanks.