P0 Galerkin BEM for the single layer is well posed and quasioptimal

Define $a(\lambda ,\mu )=⟨\mu ,V\lambda ⟩$. Boundedness follows from [F1] and the dual norm. For $u=S\lambda$ and $v=S\mu$, the weak equations [F2] give $a(\lambda ,\mu )={\int }_{{\mathbb{R}}^3}\mathrm{\nabla }u\cdot \mathrm{\nabla }v=a(\mu ,\lambda )$, so $a$ is symmetric; with [F1] it is coercive. The load $F_f(\mu )=⟨\mu ,f⟩$ is bounded by duality.

The fixed square-sum chart norm makes $H^{1\mathrm{/}2}$ a real Hilbert space, hence its continuous dual $H^{-1\mathrm{/}2}$ is Hilbert under the fixed dual norm by the trace-space convention in [F3]. Apply Lax–Milgram using [A1] to obtain the unique exact solution $\lambda$. By [F3], the restriction of $a$ to $X_h$ remains bounded and coercive and the restricted load is bounded, so Lax–Milgram on the finite-dimensional real Hilbert space $X_h$ gives the unique ${\lambda }_h$.

Apply [F4] with $V=H^{-1\mathrm{/}2}(\mathrm{\Gamma })$, $V_h=X_h$, common norm $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{H^{-1\mathrm{/}2}}$, extension ${\tilde{a}}_h=a$, and $F_h=F_f{\mathrm{\mid }}_{X_h}$. The bounded-form constant is at most $C$ and the coercivity constant is at least $c$ by [F1]; the consistency defect is zero because both discrete objects are restrictions of the exact form and load. Its estimate is therefore $\mathrm{\parallel }\lambda -{\lambda }_h\mathrm{\parallel }\le (1+C\mathrm{/}c){\inf }_{{\mu }_h\in X_h}\mathrm{\parallel }\lambda -{\mu }_h\mathrm{\parallel }$, as stated.