The shifted elliptic solution operator

Assume Countable Choice. Let $\mathrm{\Omega }\subseteq {\mathbb{R}}^n$ be open, let $L,a$ be as in Uniformly elliptic divergence-form operators and their sesquilinear forms with ellipticity constant $\theta$ and coefficient bounds $M_a,M_b,M_c$, and let $\mu \ge \beta :=\theta \mathrm{/}2+n{M}_b^2\mathrm{/}(2\theta )+M_c$, so that $a_{\mu }=a+\mu (\cdot ,\cdot {)}_{L^2}$ is bounded and coercive on $H_0^1(\mathrm{\Omega })$ with constant $\alpha :=\theta \mathrm{/}2$ (A sufficiently large shift is coercive). For $f\in L^2(\mathrm{\Omega })$ the functional $F_f(v):=(f,v{)}_{L^2(\mathrm{\Omega })}$ is conjugate-linear and bounded on $H_0^1(\mathrm{\Omega })$ with $\mathrm{\parallel }{F}_f\mathrm{\parallel }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(\mathrm{\Omega })}$: Cauchy--Schwarz gives $\mathrm{\mid }{F}_f(v)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}\mathrm{\parallel }v{\mathrm{\parallel }}_{L^2}$, and the standard $H_0^1$ norm satisfies $\mathrm{\parallel }v{\mathrm{\parallel }}_{L^2}\le \mathrm{\parallel }v{\mathrm{\parallel }}_{H_0^1}$ (Cauchy–Schwarz: $\mathrm{\mid }⟨x,y⟩\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, with equality exactly for dependent pairs, Integer-order Sobolev spaces and their norms). The shifted elliptic solution operator $K_{\mu }$ assigns to $f\in L^2(\mathrm{\Omega })$ the unique $u\in H_0^1(\mathrm{\Omega })$ with $a_{\mu }(u,v)=(f,v{)}_{L^2(\mathrm{\Omega })}\text{for every }v\in H_0^1(\mathrm{\Omega }),$ whose existence and uniqueness are The Lax--Milgram theorem; it is linear in $f$ with $\mathrm{\parallel }{K}_{\mu }f{\mathrm{\parallel }}_{H_0^1}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}\mathrm{/}\alpha$ (The Lax--Milgram solution operator has norm at most $1\mathrm{/}\alpha$). $K_{\mu }$ is first a map $L^2(\mathrm{\Omega })\to H_0^1(\mathrm{\Omega })$; it is regarded on $L^2(\mathrm{\Omega })$ through the inclusion $H_0^1(\mathrm{\Omega })\subset L^2(\mathrm{\Omega })$, and the two maps are distinguished throughout. No boundedness or boundary regularity of $\mathrm{\Omega }$ is used, and the shift is fixed and never silently changed.

Well-definedness, recorded with the definition. The form $a_{\mu }$ is bounded and coercive on the Hilbert space $H_0^1(\mathrm{\Omega })$ with the restricted Sobolev inner product and completeness supplied by The Sobolev space $H^1$ is a Hilbert space; the $L^2$ integral pairing is a Hilbert inner product by $L^2$ with the integral pairing is a Hilbert space: boundedness is the shift corollary, and coercivity holds with constant $\alpha =\theta \mathrm{/}2$ independent of $\mu$ once $\mu \ge \beta$. The datum functional $F_f$ is conjugate-linear in $v$ in the convention of Bounded, coercive and symmetric sesquilinear forms and bounded by $\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}$, so The Lax--Milgram theorem applies and produces a unique $u\in H_0^1(\mathrm{\Omega })$; for the norm estimate, testing the defining identity at $v=u$ gives $\alpha \mathrm{\parallel }u{\mathrm{\parallel }}_{H_0^1}^2\le \mathrm{Re}{a}_{\mu }(u,u)=\mathrm{Re}(f,u{)}_{L^2}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}\mathrm{\parallel }u{\mathrm{\parallel }}_{H_0^1}$. Linearity of $f\mapsto K_{\mu }f$ follows from uniqueness, and the same uniqueness makes $K_{\mu }$ independent of any choice of representative of $f$; the map $K_{\mu }$ is defined for the fixed $\mu$ and is never applied at any other shift.