14. Given the functions
$$
\begin{array}{l}
f(x)=\left\{\begin{array}{ll}
-x^{2}+x, & x \leqslant 1 ; \\
\log _{\frac{1}{3}} x, & x>1,
\end{array}\right. \\
g(x)=|x-k|+|x-1| .
\end{array}
$$

If for any $x_{1}, x_{2} \in \mathbf{R}$, we have $f\left(x_{1}\right) \leqslant$ $g\left(x_{2}\right)$, then the range of the real number $k$ is $\qquad$
14. $k \leqslant \frac{3}{4}$ or $k \geqslant \frac{5}{4}$.

From the problem, we know that
for any $x_{1}, x_{2} \in \mathbf{R}, f\left(x_{1}\right) \leqslant g\left(x_{2}\right)$
$\Leftrightarrow$ for any $x_{1}, x_{2} \in \mathbf{R}, f_{\max } \leqslant g_{\min }$.
Notice that, when $x=\frac{1}{2}$, $f(x)$ reaches its maximum value $\frac{1}{4}$;
$$
g(x)=|x-k|+|x-1| \geqslant|k-1| .
$$

Therefore, the required condition is $|k-1| \geqslant \frac{1}{4}$.
Solving this, we get $k \geqslant \frac{5}{4}$ or $k \leqslant \frac{3}{4}$.