Continuous $f \ge 0$ s.t. $f(x)\sin(x) = f(\frac{\pi}{2} - x)$?

Question

Does there exists a continuous function $f: \left[0,\frac{\pi}{2}\right] \to [0,\infty)$ with $f(0) = f\left(\frac{\pi}{2}\right) = 0$ s.t. $$f(x)\sin(x) = f\left(\frac{\pi}{2} - x\right)$$ or is there no such a function? Maybe someone sees an easy proof for one of it.

I tried to show existence using Banach's fix-point theorem by defining $$I(f) = \frac{f \circ g}{\sin}$$ with $$g(x) = \frac{\pi}{2} - x$$

If $I$ has a fix-point it would solve our equation. $I$ is a self-map by add the additional restriction $f'(0) = f'\left(\frac{\pi}{2}\right) = 0$ the the function space but unfortunately it's not a contraction w.r.t to the sup-norm because of $$||I(f) - I(g)||_\infty \ge ||f-g||_\infty$$

Obviously $f$ is a linear operator but I don't know if this will help to find a solution…

Thx in advance.

Answer 1

Change $x$ to $\frac {\pi} 2 -x$ to get $f(\frac {\pi} 2 -x)\cos x=f(x)$. Multiply this by $\sin x$ to get $f(\frac {\pi} 2 -x)\cos x \sin x=f(x)\sin x =f(\frac {\pi} 2 -x)$. Since $|\sin x \cos x| \leq \frac 1 2$ we get $f(\frac {\pi} 2 -x)=0$ so $f \equiv 0$,

Answer 2

Hint. We have that $$f(\frac{\pi}{2} - x)\sin(\frac{\pi}{2} - x) = f(x)$$ and therefore, after letting $f(\frac{\pi}{2} - x)=f(x)\sin(x)$ and $\sin(\frac{\pi}{2} - x)=\cos(x)$, we find $$f(x)\sin(x)\cos(x)=f(x)$$ or $$f(x)\underbrace{(2-\sin(2x))}_{>0}=0.$$ What may we conclude?

Answer 3

If there exists $x_0$ for which $f(x_0)\ne 0$, then we have $$f(x_0)\sin(x_0)=f(\pi/2-x_0)$$ But also, putting $x=\pi/2-x_0$, $$f(\pi/2-x_0)\sin(\pi/2-x_0)=f(x_0)$$ Together, using $x_0\ne 0$, these give $$\sin(x_0)=f(\pi/2-x_0)/f(x_0)=1/\sin(\pi/2-x_0)$$ But $\sin(x_0)<1$ and $1/\sin(\pi/2-x_0)>1$, a contradiction. Therefore the only solution is $$f(x)=0\text{ for all }x$$