14. A function $f(x)$ defined on $\mathbf{R}$ satisfies: If for any $x_{1}, x_{2} \in \mathbf{R}$, we have $f\left(\frac{x_{1}+x_{2}}{2}\right) \leqslant \frac{1}{2}\left[f\left(x_{1}\right)+\right.$ $f\left(x_{2}\right)$ ], then the function $f(x)$ is called a convex function on $\mathbf{R}$.
Given the quadratic function $f(x)=a x^{2}+x(a \in \mathbf{R}, a \neq 0)$.
(1) Prove that when $a>0$, the function $f(x)$ is a convex function;
(2) If $x \in[0,1]$ and $|f(x)| \leqslant 1$, try to find the range of the real number $a$.
14. (1) For any $x_{1}, x_{2} \in \mathbf{R}, a>0$,

since $f\left(x_{1}\right)+f\left(x_{2}\right)-2 f\left(\frac{x_{1}+x_{2}}{2}\right)$
$$
\begin{array}{l}
=a x_{1}^{2}+x_{1}+a x_{2}^{2}+x_{2}-2\left[a\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\frac{x_{1}+x_{2}}{2}\right] \\
=a x_{1}^{2}+a x_{2}^{2}-\frac{1}{2} a\left(x_{1}^{2}+x_{2}^{2}+2 x_{1} x_{2}\right)=\frac{1}{2} a\left(x_{1}-x_{2}\right)^{2} \geqslant 0,
\end{array}
$$

thus $f\left(\frac{x_{1}+x_{2}}{2}\right) \leqslant \frac{1}{2}\left[f\left(x_{1}\right)+f\left(x_{2}\right)\right]$.
Therefore, the function $f(x)$ is a concave function.
(2) From $|f(x)| \leqslant 1 \Leftrightarrow-1 \leqslant f(x) \leqslant 1 \Leftrightarrow-1 \leqslant a x^{2}+x \leqslant 1$.

When $x=0$, $a \in \mathbf{R}$;
$$
\left\{\begin{array}{l}
a \geqslant-\frac{1}{x^{2}}-\frac{1}{x}=-\left(\begin{array}{cc}
1 & 1 \\
2
\end{array}\right)^{2}+\frac{1}{4}, \\
a \leqslant \frac{1}{x^{2}}-\frac{1}{x}=\left(\frac{1}{x}-\frac{1}{2}\right)^{2}-\frac{1}{4}
\end{array}\right. \text { always holds. }
$$

Since $x \in(0,1]$, thus $\frac{1}{x} \geqslant 1$.
When $\frac{1}{x}=1$, $-\left(\frac{1}{x}+\frac{1}{2}\right)^{2}+\frac{1}{4}$ achieves the maximum value $-2$, and $\left(\frac{1}{x}-\frac{1}{2}\right)^{2}-\frac{1}{4}$ achieves the minimum value 0.
Therefore, $-2 \leqslant a \leqslant 0$, combining with $a \neq 0$, we get $-2 \leqslant a<0$.
In summary, the range of $a$ is $[-2,0)$.