7. Let $f(x)=\frac{\sin \pi x}{x^{2}}(x \in(0,1))$. Then
$$
g(x)=f(x)+f(1-x)
$$

the minimum value of $g(x)$ is . $\qquad$
7.8.

From the given, we have
$$
\begin{aligned}
f^{\prime}(x) & =\frac{\pi x \cos \pi x-2 \sin \pi x}{x^{3}}, \\
f^{\prime \prime}(x) & =\frac{\left(6-\pi^{2} x^{2}\right) \sin \pi x-2 \pi x \cos \pi x}{x^{3}} .
\end{aligned}
$$

Next, we need to prove
$$
\left(6-\pi^{2} x^{2}\right) \sin \pi x-2 \pi x \cos \pi x>0.
$$

Let $t=\pi x \in(0, \pi)$. Then
$\left(6-\pi^{2} x^{2}\right) \sin \pi x-2 \pi x \cos \pi x$
$$
=\left(6-t^{2}\right) \sin t-2 t \cos t=p(t).
$$

Thus, $p^{\prime}(t)=\left(4-t^{2}\right) \cos t$.
It is easy to see that $p^{\prime}(t)$ is increasing on the interval $\left(0, \frac{\pi}{2}\right)$, decreasing on the interval $\left(\frac{\pi}{2}, 2\right)$, and increasing on the interval $(2, \pi)$.

Also, $p(0)=0, p^{\prime}(0)>0$, so for $t \in(0, \pi)$, $p(t)>0$, which means that for $x \in(0,1)$, $f^{\prime \prime}(x)>0$.
Therefore, $f(x)$ is a convex function on the interval $(0,1)$. By Jensen's inequality, we know that for $x \in(0,1)$,
$$
\begin{array}{l}
\frac{f(x)+f(1-x)}{2} \\
\geqslant f\left(\frac{x+(1-x)}{2}\right)=f\left(\frac{1}{2}\right)=4.
\end{array}
$$

Thus, $g(x) \geqslant 8$, with equality holding if and only if $x=\frac{1}{2}$.
Therefore, the minimum value of $g(x)$ is 8.