A conforming P1 Dirichlet discretization gives a first Ritz upper bound

Adjacent triangles share the same endpoint values of $\varphi$, so their linear traces agree on each interior edge. The boundary edges have zero values at both endpoints, so their linear traces vanish identically. Thus [F1] places $\varphi$ in $H_0^1(\mathrm{\Omega })$. It is nonzero because $\varphi (c)=1$. Each triangle has base length $1\mathrm{/}2$, height $1\mathrm{/}2$ and area $1\mathrm{/}8$; the hat changes from $1$ at $c$ to $0$ across height $1\mathrm{/}2$, so $\mathrm{\mid }\mathrm{\nabla }\varphi \mathrm{\mid }=2$ on each cell. Therefore $a(\varphi ,\varphi )=8\cdot \frac{1}{8}\cdot 2^2=4.$

On a triangle, $\varphi$ is the barycentric coordinate of $c$. On the reference triangle $\{(x,y):x>0,y>0,x+y<1\}$, direct integration gives ${\int }_0^1{\int }_0^{1-x}{x}^{2}dydx=\frac{1}{12}=\frac{1}{6}\left(\frac{1}{2}\right)\mathrm{.}$ An affine change of variables preserves the ratio of this integral to the triangle area, so ${\int }_{K_j}{\varphi }^2=\mathrm{\mid }{K}_j\mathrm{\mid }\mathrm{/}6=1\mathrm{/}48$. Summing the eight cells gives $b(\varphi ,\varphi )=1\mathrm{/}6$.

Because $V_h$ is one-dimensional, its sole Ritz value is its Rayleigh quotient, ${\lambda }_{h,1}=a(\varphi ,\varphi )\mathrm{/}b(\varphi ,\varphi )=4\mathrm{/}(1\mathrm{/}6)=24$. By [F2], the continuous first eigenvalue minimizes the same quotient over all nonzero $H_0^1$ functions. Since $\varphi$ is an admissible nonzero function by step 1.1, ${\lambda }_1\le a(\varphi ,\varphi )\mathrm{/}b(\varphi ,\varphi )=24={\lambda }_{h,1}$.