Three, (15 points) Given the linear function $f(x)=a x+b$
for any $x, y \in[0,1]$ it satisfies
$|f(x)+f(y)-x y| \leqslant \frac{1}{4}$.
Determine such $f(x)$.


For a linear function $f(x)=a x+b$, let
$$
F(x, y)=a x+b+a y+b-x y .
$$

According to the given, $F(0,0)=2 b \geqslant-\frac{1}{4}$, which means
$$
b \geqslant-\frac{1}{8} \text {. }
$$
$$
F(0,1)=a+2 b \leqslant \frac{1}{4} \text {. }
$$

Also, $F(1,1)=2 a+2 b-1 \geqslant-\frac{1}{4}$, which means $a+b \geqslant \frac{3}{8}$.
Subtracting (3) from (2) gives
$$
\begin{array}{l}
a+2 b-(a+b) \leqslant \frac{1}{4}-\frac{3}{8}=-\frac{1}{8} \\
\Rightarrow b \leqslant-\frac{1}{8} .
\end{array}
$$

Combining with equation (1), we get $b=-\frac{1}{8}$.
Substituting $b=-\frac{1}{8}$ into equations (2) and (3), we get
$$
a \leqslant \frac{1}{2}, a \geqslant \frac{1}{2} \Rightarrow a=\frac{1}{2} \text {. }
$$

Therefore, the required linear function is $f(x)=\frac{1}{2} x-\frac{1}{8}$. When $f(x)=\frac{1}{2} x-\frac{1}{8}$,
$$
\begin{array}{l}
F(x, y)=\left|\frac{1}{2}(x+y)-\frac{1}{4}-x y\right| \\
=\left|x-\frac{1}{2}\right| \cdot\left|\frac{1}{2}-y\right| \leqslant \frac{1}{2} \times \frac{1}{2}=\frac{1}{4}
\end{array}
$$

This holds for any $x, y \in[0,1]$.