Coercivity of the adjoint makes the form-operator range dense

Given: Countable Choice; a real or complex Hilbert space $H$; a bounded coercive sesquilinear form $a$ with constants $M,\alpha$; its operator $A\in \mathcal{B}(H)$ with $a(u,v)=(Au,v)$; and the adjoint form $a^{\ast }$ with operator $A^{\ast }$.

The adjoint form $a^{\ast }(u,v)=\overline{a(v,u)}$ is bounded with bound $M$ and coercive with the same constant $\alpha$, and its operator is the Hilbert adjoint $A^{\ast }$; also $a$ coercive with constant $\alpha$ makes $A$ bounded below with constant $\alpha$ (The adjoint of a coercive form is coercive with the same constants, A bounded form is represented by a unique bounded operator, A coercive form operator is bounded below, The Hilbert-space adjoint of a bounded operator).

Orthogonal complements: $(\mathrm{ran}A{)}^{\perp }=\ker A^{\ast }$, and for every linear subspace $M$ of $H$ one has $M^{\perp \perp }=\bar{M}$, with $\{0{\}}^{\perp }=H$ (Kernel–range orthogonality for Hilbert adjoints, The double orthogonal complement of a subspace is its closure, Orthogonality and the orthogonal complement).

A bounded-below operator on a Banach space has closed range: applied to $A:H\to H$, whose domain $H$ is complete, this gives that $\mathrm{ran}A$ is closed (A bounded-below operator has closed range).

Coercivity of $a$ with constant $\alpha$ means $\mathrm{Re}a(u,u)\ge \alpha \mathrm{\parallel }u{\mathrm{\parallel }}^2$ for all $u$ (Bounded, coercive and symmetric sesquilinear forms).