The nodal P1 interpolant preserves homogeneous boundary values

For each mesh triangle $K$, apply [F1] to pull $u{\mathrm{\mid }}_K$ back to the reference triangle, and then apply [F2]. This gives a continuous representative on $\bar{K}$. It is unique: if two continuous representatives differed at one point, continuity would make them differ on a nonempty open subset of $K$, contradicting [F6]. Thus each cellwise nodal value is individually well-defined.

Put $z=E_{0}u$, the zero extension from [F3]. On either side of any straight line, the half-space trace is obtained from the positive-half-space trace in [F4] by a translation and, when needed, a reflection; [F5] verifies that these changes preserve the Sobolev class and norm. The two traces of a global $H^1({\mathbb{R}}^2)$ class across such a line agree: choose $z_m\in C_c^{\mathrm{\infty }}({\mathbb{R}}^2)$ converging to $z$ by [F4]; the two traces of each $z_m$ are its same classical restriction, and boundedness of the two trace maps passes this equality to the limit. On the exterior half-space of each side of $\mathrm{\Omega }$, $z=0$, so both its exterior trace and its interior trace are zero.

If $K_{+}$ and $K_{-}$ share an interior edge, every point $x$ in the relative interior of that edge has a ball $B\subset \mathrm{\Omega }$ meeting only those two cells. By [F2], $u{\mathrm{\mid }}_B$ has a continuous representative. On each of the two open half-balls $B\cap K_{\pm }$, this representative and the corresponding representative from step 1.1 agree almost everywhere, hence everywhere by [F6]. Taking limits to $x$ shows the two cell values agree at $x$; continuity along the closed edge gives agreement at its endpoints as well. For an interior mesh vertex, take a ball about the vertex contained in $\mathrm{\Omega }$ and repeat this argument for every incident cell.

Let $x$ lie in the relative interior of a boundary mesh edge $E$. Choose a ball $B$ centered at $x$ small enough that its interior half-ball lies in the unique incident triangle $K$ and its other boundary is disjoint from all mesh edges. By [F9] choose $\chi \in C_c^{\mathrm{\infty }}(B)$ equal to one near $x$. On the interior half-space define $v=\chi u_K$ inside $B$ and zero outside, where $u_K$ is the continuous cell representative from step 1.1. The cutoff vanishes near the circular part of $\mathrm{\partial }B$, so the weak test identity shows $v\in H^1$ of that half-space; it is compactly supported and continuous on the closed half-space. Its class is $\chi z$ there. By linearity and boundedness of the trace, and by approximation with the smooth functions in step 1.2, $T(\chi z)=\chi Tz=0$ on the side. The classical-restriction clause of [F4] gives $Tv=\chi u_K{\mathrm{\mid }}_E$, so $u_K=0$ almost everywhere near $x$ on $E$. Continuity from step 1.1 makes this value zero at every point of that edge, including its endpoints.

Interior edge values agree by step 2.1. At a boundary vertex, the incident triangles form a connected fan because the square is convex; the fan is a chain under adjacency across edges containing that vertex, and at least one end cell has a boundary edge. Step 2.2 gives zero at the vertex in that end cell, and step 2.1 propagates that value through the fan. Thus all cell values at every shared vertex agree and boundary vertex values are zero. The affine traces of adjacent P1 interpolants agree because they agree at both edge endpoints, so $I_{h}u$ is continuous; on each boundary edge it is affine with both endpoint values zero, hence its trace is zero. Apply [F8] to conclude $I_{h}u\in H_0^1(\mathrm{\Omega })$.