11. (20 points) Let the function
$$
f(x)=\frac{a}{2} x^{2}+x-a
$$

have its maximum value $g(a)$ on the domain $[\sqrt{2}, 2]$.
(1) Find the analytical expression for $g(a)$;
(2) When $g(a)>g\left(\frac{1}{a}\right)$, find the range of real numbers $a$.

11. (1) When $a \geqslant 0$, $f^{\prime}(x)=a x+1>0$, so $f(x)$ is monotonically increasing on $[\sqrt{2}, 2]$. Therefore, the maximum value is obtained at the right endpoint, and
$$
g(a)=f(2)=a+2 .
$$

When $a<0$,
$$
f^{\prime}(x)=a x+1=0 \Rightarrow x=-\frac{1}{a}>0 .
$$

We discuss the following three cases.
(i) When $x=-\frac{1}{a} \in(2,+\infty)$, we have
$$
-\frac{1}{2}-\frac{1}{2} ; \\
-a-\frac{1}{2 a}, & -\frac{\sqrt{2}}{2}<a<0,
$$

Thus, $g(a)$ is a strictly increasing function for $a>-\frac{\sqrt{2}}{2}$. Therefore, $g(a)$ is non-decreasing on $\mathbf{R}$.
From $g(a)>g\left(\frac{1}{a}\right)$, we have $a>\frac{1}{a}$, which gives $a>1$ or $-1<a<0$.

However, when $-1<a \leqslant-\frac{\sqrt{2}}{2}$,
$$
-\sqrt{2} \leqslant \frac{1}{a} \leqslant-1,
$$

we have $g(a)=g\left(\frac{1}{a}\right)=\sqrt{2}$.
Therefore, the range of $a$ is
$$
\left(-\frac{\sqrt{2}}{2}, 0\right) \cup(1,+\infty) \text {. }
$$