Let $\gamma$ be a curve in $\Bbb R^2$ such that it has (not necessarily strictly) increasing curvature $K$. Can this curve be a $C^2$ closed curve ? The answer I was given was no if $\gamma$ is not a circle. This was the proof :
If $\gamma : [0, a] \to \Bbb R^2$ is a $C^2$ closed curve, then $K(0) = K(a)$. So the curvature cannot be increasing except the case where it is constant which means that it is a circle.
I disagree with the "$K(0) = K(a)$" part : the curve $t \mapsto (t^2(t-1), t(t-1))$, $t \in [0,1]$ is $C^2$ and closed but calculations show that $K(0)\neq K(1)$. Putting away this counterexample, the curvature of a curve depends on its speed and acceleration and there is no reason for both of them to be equal at a same position for different time.
I agree with the initial statement however but it needs another proof in my opinion.
By closed curve, the question meant that speed and acceleration where equal at the intersection point, so $K(0)=K(1)$