7. For any real numbers $a, b$, the minimum value of $\max \{|a+b|,|a-b|,|1-b|\}$ is $\qquad$ .
7. $\frac{1}{2}$.

From the problem, we know
$$
\begin{array}{l}
\max \{|a+b|,|a-b|,|1-b|\} \\
\geqslant \frac{|a+b|+|a-b|+2|1-b|}{4} \\
\geqslant \frac{|a+b-(a-b)|+|2-2 b|}{4} \\
\geqslant \frac{|2 b|+|2-2 b|}{4} \geqslant \frac{|2 b+(2-2 b)|}{4} \\
=\frac{1}{2},
\end{array}
$$

When $a=0, b=\frac{1}{2}$, the above expression achieves the minimum value $\frac{1}{2}$.