17. Let $0 \leqslant x \leqslant \pi, 0 \leqslant y \leqslant 1$. Find the minimum value of the function $f(x, y)=(2 y-1) \sin x+(1-y) \sin (1-y) x$.
17. It is known that for all $0 \leqslant x \leqslant \pi$, we have $\sin x \geqslant 0, \sin (1-y) x \geqslant 0$.

Thus, when $\frac{1}{2} \leqslant y \leqslant 1$, $f(x, y) \geqslant 0$, and the equality holds when $x=0$.
When $0 \leqslant yx>\sin x$,
and. $\sin (x+\delta)=\sin x \cdot \cos \delta+\cos x \cdot \sin \delta \leqslant \sin x+\delta \cos x$.
Therefore, $\frac{\sin x}{x}-\frac{\sin (x+\delta)}{x+\delta}=\frac{(x+\delta) \sin x-x \sin (x+\delta)}{x(x+\delta)}$
$$
\geqslant \frac{(x+\delta) \sin x-x(\sin x+\delta \cos x)}{x(x+\delta)}=\frac{\delta \sin x-\delta x \cos x}{x(x+\delta)}=\frac{\delta \cos x(\tan x-x)}{x(x+\delta)}>0 .
$$

Thus, the function $f(x)=\frac{\sin x}{x}$ is decreasing on $\left(0, \frac{\pi}{2}\right)$.
Since $g(x)=\sin x$ is decreasing on $\left[\frac{\pi}{2}, \pi\right]$, it follows that $f(x)=\frac{\sin x}{x}$ is also decreasing on $\left[\frac{\pi}{2}, \pi\right]$.
Moreover, $00$,
then $\frac{\sin (1-y) x}{1-y} \leqslant \frac{1-y}{1-2 y} \sin (1-y) x, \sin x \leqslant \frac{1-y}{1-2 y} \sin (1-y) x$.
Thus, $(2 y-1) \sin x+(1-y) \sin (1-y) x \geqslant 0$, which means $f(x, y) \geqslant 0$.
When $y=0$, the equality holds.
In summary, when $x=0$ or $y=0$, $f(x, y)_{\min }=0$.