1.4.4 * Given the function $f(x)=\log \frac{1}{2}\left(\frac{\sqrt{a^{2}+1}-a}{a}\right)^{x}$ is a decreasing function on $\mathbf{R}$. Find the range of real number $a$.
Parse $f(x)$ as a decreasing function on $\mathbf{R}$, so, $\frac{\sqrt{a^{2}+1}-a}{a}>1$. Since $\sqrt{a^{2}+1}-a>$ $|a|-a \geqslant 0$, it follows that $a>0$. Thus, $\sqrt{a^{2}+1}-a>a$, which implies $a^{2}+1>4 a^{2}$, leading to $a^{2}<\frac{1}{3}$, and hence $0<a<\frac{\sqrt{3}}{3}$.