14. (2007 Beijing High School Mathematics Competition for Grade 1) Given the linear function $f(x)=a x+b$ satisfies $|f(x)+f(y)-x y| \leqslant \frac{1}{4}$ for any $x, y \in$ $[0,1]$, try to determine such $f(x)$.
14. Let \( F(x, y) = a x + b + a y + b - x y \), from the given we have
\[
\begin{array}{c}
\frac{1}{4} \geqslant |F(0,0)| = 2 b \geqslant -\frac{1}{4}, \text{ i.e., } \frac{1}{8} \geqslant b \geqslant -\frac{1}{8}, \\
|F(0,1)| = |a + 2 b| \leqslant \frac{1}{4}, \\
\frac{1}{4} \geqslant F(1,1) = 2 a + 2 b - 1 \geqslant -\frac{1}{4}, \text{ i.e., } \frac{9}{8} \geqslant a + b \geqslant \frac{3}{8},
\end{array}
\]

From (2) - (3) we get \( a + 2 b - (a + b) \leqslant \frac{1}{4} - \frac{3}{8} = -\frac{1}{8} \), i.e., \( b \leqslant -\frac{1}{8} \). Combining with (1), we get \( b = -\frac{1}{8} \). Substituting \( b = -\frac{1}{8} \) into (2) and (3) respectively, we get \( a \leqslant \frac{1}{2} \) and \( a \geqslant \frac{1}{2} \). Therefore, \( a = \frac{1}{2} \).

Thus, the linear function can only be \( f(x) = \frac{1}{2} x - \frac{1}{8} \).

When \( f(x) = \frac{1}{2} x - \frac{1}{8} \), \( F(x, y) = \left| \frac{1}{2}(x + y) - \frac{1}{4} - x y \right| = \left| x - \frac{1}{2} \right| \cdot \left| \frac{1}{2} - y \right| \leqslant \frac{1}{2} \). \(\frac{1}{2} = \frac{1}{4}\) holds for any \( x, y \in [0, 1] \).

Therefore, the required linear function is \( f(x) = \frac{1}{2} x - \frac{1}{8} \).