Let $u: \Bbb R^2\to\Bbb R$. We define $f(t)=u(\cos(t),\sin(t))$.
Given that for every $x,y$ such that $x^2+y^2=1$ the following is true: $$y\frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}\geq0$$
Prove that $f'(t)\leq0$ for every $t$ in $[0,2\pi]$.
What I did until now?
I have calculated $$ f'(t) = \cos(t) \frac{\partial u}{\partial y}-\sin(t) \frac{\partial u}{\partial x} $$ According to what is given, I get that $f'(t)\leq0$ for every $t$ and not only for $t\in [0,2\pi]$.
Where did I make the mistake?
$f'(t) = \frac {\partial u}{\partial x} \frac{d(\cos t)}{dt} + \frac {\partial u}{\partial y}\frac {d(\sin t)}{dt} = \frac {\partial u}{\partial x} (-\sin t) + \frac {\partial u}{\partial y}\cos t = -y\frac {\partial u}{\partial x} + x\frac {\partial u}{\partial y}$
and by our given condition $-y\frac {\partial u}{\partial x} + x\frac {\partial u}{\partial y} \le 0$
$f'(t)\le 0$ for all $t.$
Since $f(t)$ is periodic. It seems to me that this condition would imply that $f$ is either constant or discontinuous.