 Projection onto a box-constrained hyperplane

LetThis closed convex set is nonempty exactly whenThese are the minimum and maximum of the linear constraint on the box. If the interval condition fails, there is no projection onto $C$.

For feasible data, minimizing $\frac{1}{2}\Vert x-y{\Vert }_2^2$ with multiplier $\nu$ for $a^{T}x=b$ separates across coordinates:The scalar function $\varphi (\nu )=a^{T}x(\nu )$ is continuous and nonincreasing. On a piece with free coordinates, $\varphi \prime (\nu )=-{\sum }_{0<x_i<1}{a}_i^2\le 0$. Its limiting values are the two box extrema, so one can choose $\nu$ with $\varphi (\nu )=b$; plateaus can make the multiplier nonunique, while the projected vector remains unique by strict convexity. For $a=0$, feasibility requires $b=0$ and coordinatewise clipping alone suffices.

For example, with $a=(1,1,1)$, $b=1$ and $y=(0.8,0.7,-0.2)$, taking $\nu =0.25$ gives the projection $(0.55,0.45,0)$. When $a$ is the all-ones vector and $b$ is an integer, the constraint set is a capped simplex. The clipped scalar equation permits efficient bisection or a breakpoint search.