7. The minimum distance from a point on the parabola $y^{2}=2 x$ to the line $x+y+$ $1=0$ is $\qquad$ .
$$
=7 . \frac{\sqrt{2}}{4} \text {. }
$$

Let point $P\left(\frac{b^{2}}{2}, b\right)$ be the point where the distance to the line $x+y+1=0$ is minimized. Then
$$
d=\frac{\left|\frac{b^{2}}{2}+b+1\right|}{\sqrt{2}} \geqslant \frac{\sqrt{2}}{4} .
$$