The associated elliptic operator is densely defined, symmetric and lower bounded

Range inclusion. By [F1] every element of $\mathrm{ran}{K}_{\mu }$ lies in $D(L)$, so it suffices to show that $\mathrm{ran}{K}_{\mu }$ is dense in $L^2(\mathrm{\Omega })$. Let $y\in L^2(\mathrm{\Omega })$ be orthogonal to $\mathrm{ran}{K}_{\mu }$. Then $(y,K_{\mu }y{)}_{L^2}=0$, while the defining equation of $K_{\mu }$ at $v=K_{\mu }y$ gives $a_{\mu }(K_{\mu }y,K_{\mu }y)=(y,K_{\mu }y{)}_{L^2}=0$; coercivity [F2] yields $\mathrm{\parallel }{K}_{\mu }y{\mathrm{\parallel }}_{H_0^1}^2\le \frac{2}{\theta }{a}_{\mu }(K_{\mu }y,K_{\mu }y)=0$, so $K_{\mu }y=0$. For every $v\in H_0^1(\mathrm{\Omega })$ the defining equation then gives $(y,v{)}_{L^2}=a_{\mu }(K_{\mu }y,v)=0$, and density of $H_0^1(\mathrm{\Omega })$ in $L^2(\mathrm{\Omega })$ ([F3]) forces $y=0$. Since $K_{\mu }$ is linear, its range is a linear subspace; by [F3] its orthogonal complement is trivial, so $\mathrm{ran}{K}_{\mu }$ is dense; hence its superset $D(L)$ is dense in $L^2(\mathrm{\Omega })$.

Symmetry. Let $u,v\in D(L)$ and write $f=Lu$, $g=Lv$. By [F5], $(Lu,v{)}_{L^2}=(f,v{)}_{L^2}=a(u,v)$ and $(u,Lv{)}_{L^2}=(u,g{)}_{L^2}=\overline{(g,u{)}_{L^2}}=\overline{a(v,u)}$. Since the coefficients are Hermitian and $b=0$ with real $c$, the form $a$ is symmetric, $a(u,v)=\overline{a(v,u)}$ (The $L^2$ operator associated with a symmetric elliptic form, Bounded, coercive and symmetric sesquilinear forms), so $(Lu,v{)}_{L^2}=(u,Lv{)}_{L^2}$.

Lower bound. For $u\in D(L)$, [F5] with $v=u$ gives $(Lu,u{)}_{L^2}=a(u,u)$, which is real by symmetry; Garding's inequality [F4] yields $a(u,u)\ge -\beta \mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}^2$, and a second application with the positive shift gives $a(u,u)=a_{\mu }(u,u)-\mu \mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}^2\ge -\mu \mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}^2$ because $a_{\mu }(u,u)\ge 0$ by [F2]. No boundary regularity of $\mathrm{\Omega }$ was used.