6. If $f(x)=2 \sin \omega x(0<\omega<1)$ has a maximum value of $\sqrt{2}$ on the interval $\left[0, \frac{\pi}{3}\right]$, then $\omega=$
6. $\frac{3}{4}$ It is known that $f\left(\frac{\pi}{3}\right)=2 \sin \left(\frac{\pi}{3} \omega\right)=\sqrt{2}$, so $\omega=\frac{3}{4}$