11. Let $a, b$ be real numbers, and the function $f(x)=x^{3}+a x^{2}+b x$. If there exist three real numbers $x_{1}, x_{2}, x_{3}$ satisfying $x_{1}+1 \leqslant x_{2} \leqslant x_{3}-1$, and $f\left(x_{1}\right)=f\left(x_{2}\right)=f\left(x_{3}\right)$, find the minimum value of $|a|+2|b|$.
Let $f\left(x_{1}\right)=f\left(x_{2}\right)=f\left(x_{3}\right)=c$, i.e., $x_{1}, x_{2}, x_{3}$ are the roots of the equation $x^{3} + a x^{2} + b x - c = 0$. By Vieta's formulas, we have $\left\{\begin{array}{l}x_{1} + x_{2} + x_{3} = -a, \\ x_{1} x_{2} + x_{2} x_{3} + x_{3} x_{1} = b\end{array}\right.$ $\Rightarrow a^{2} - 3 b = x_{1}^{2} + x_{2}^{2} + x_{3}^{2} - (x_{1} x_{2} + x_{2} x_{3} + x_{3} x_{1}) = \frac{1}{2}\left[\left(x_{2} - x_{1}\right)^{2} + \left(x_{3} - x_{2}\right)^{2} + \left(x_{3} - x_{1}\right)^{2}\right]$. According to the problem, we have $x_{2} - x_{1} \geqslant 1, x_{3} - x_{2} \geqslant 1, x_{3} - x_{1} \geqslant 2$, thus $a^{2} - 3 b \geqslant \frac{1}{2}(1 + 1 + 4) = 3 \Rightarrow b \leqslant \frac{a^{2}}{3} - 1$. If $|a| \geqslant \sqrt{3}$, then $|a| + 2|b| \geqslant |a| \geqslant \sqrt{3}$;
If $0 \leqslant |a| < \sqrt{3}$, then $|a| + 2|b| > |a| + 2 - \frac{2 a^{2}}{3} = -\frac{2}{3}\left(|a| - \frac{3}{4}\right)^{2} + \frac{19}{8} > \sqrt{3}$ (the minimum value is obtained at $|a| = \sqrt{3}$).
Taking $a = -\sqrt{3}, b = 0$, at this point $f(x) = x^{3} - \sqrt{3} x^{2}$ satisfies $f\left(-1 + \frac{\sqrt{3}}{3}\right) = f\left(\frac{\sqrt{3}}{3}\right) = f\left(1 + \frac{\sqrt{3}}{3}\right) = -\frac{2 \sqrt{3}}{9}$. In summary, the minimum value of $|a| + 2|b|$ is $\sqrt{3}$.