Barycentric interpolation and nearest-site queries are correct

The matrix with columns $p_1-p_0,p_2-p_0$ is invertible by [A1], so the two coordinate equations uniquely determine ${\lambda }_1,{\lambda }_2$ and then ${\lambda }_0=1-{\lambda }_1-{\lambda }_2$. Thus the coordinates in [F1] are unique, and their nonnegativity is exactly membership in the convex combinations defining the triangle.

Parameterize an edge by $x(t)=(1-t)p_0+t{p}_1$. Its barycentric interpolant is $(1-t)f_0+t{f}_1$. Apply [F3] with nodes $0,1$ and data $f_0,f_1$: the unique cardinal polynomial is $f_0+(f_1-f_0)t$, which equals the edge interpolant, including its separate endpoint values.

Each ${\lambda }_i(x)$ is affine in $x$, so $I_{T}f={\sum }_i{\lambda }_i{f}_i$ is affine and has value $f_i$ at vertex $p_i$. An affine function agreeing at three noncollinear vertices is unique because its three coefficients are determined by an invertible $3\times 3$ augmented vertex matrix. On a shared edge, both triangle restrictions are affine in the edge parameter and agree at its two endpoints, so they coincide throughout the edge; the same endpoint argument handles a shared vertex.

A nonempty finite set of real distances $\{\mathrm{\parallel }x-p{\mathrm{\parallel }}^2:p\in P\}$ has a minimum. The finite set of identifiers attaining it has a least element, so the specified query returns exactly one minimizing site. Its inside/outside status is decided by the defining convex-hull membership test, equivalently by triangle barycentric membership when $x$ lies in the triangulated hull; outside the hull no triangle is returned but the same finite minimum still exists.