11. Given the function
$$
f(x)=a x^{2}-\frac{1}{2} x-\frac{3}{4}(a>0) \text {. }
$$

If on any closed interval of length 2, there always exist two points $x_{1} 、 x_{2}$, such that $\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \geqslant \frac{1}{4}$, then the minimum value of $a$ is $\qquad$.
II. $11 \cdot \frac{1}{4}$.
On the closed interval of length 2 $\left[\frac{1}{4 a}-1, \frac{1}{4 a}+1\right]$,
$$
\begin{array}{l}
f_{\max }(x)=f\left(\frac{1}{4 a}-1\right) \\
=f\left(\frac{1}{4 a}+1\right)=a-\frac{1}{16 a}-\frac{3}{4}, \\
f_{\text {min }}(x)=f\left(\frac{1}{4 a}\right)=-\frac{1}{16 a}-\frac{3}{4} .
\end{array}
$$

Thus, $a=f_{\max }(x)-f_{\min }(x) \geqslant \frac{1}{4}$.
When $a=\frac{1}{4}$,
$$
f(x)=\frac{1}{4} x^{2}-\frac{1}{2} x-\frac{3}{4}=\frac{1}{4}(x-1)^{2}-1 \text {. }
$$

We now prove: On any closed interval of length 2 $[t-1, t+1]$, there always exist two points $x_{1}, x_{2}$ such that
$$
\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \geqslant \frac{1}{4} \text {. }
$$

When $t \geqslant 1$, $f(x)$ is an increasing function on $[t, t+1]$.
$$
\begin{array}{l}
\text { Let } x_{1}=t, x_{2}=t+1 \text {. Then } \\
\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|=f(t+1)-f(t) \\
=\frac{1}{4}(2 t-1) \geqslant \frac{1}{4} .
\end{array}
$$

When $t<1$,
$$
\begin{array}{l}
\text { Let } x_{1}=t-1, x_{2}=t \text {. Then } \\
\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|=f(t)-f(t-1) \\
=\frac{1}{4}(2 t-1) \geqslant \frac{1}{4} .
\end{array}
$$

In summary, the minimum value of $a$ is $\frac{1}{4}$.