The SAV step is uniquely solvable and dissipates a modified energy

Subtract two candidate solutions with the same old data. Their difference $(v,s,\nu )$ satisfies the same linear system with old data zero and the same fixed ${\alpha }_n,g_{1,h}(u^n)$. Pair $v=-M\tau A_h\nu$ with $\nu$, pair $\nu =\kappa A_{h}v+{\alpha }_{n}s{g}_{1,h}(u^n)$ with $v$, and use $s=\frac{{\alpha }_n}{2}⟨g_{1,h}(u^n),v{⟩}_h$. This gives $\kappa ⟨A_{h}v,v{⟩}_h+2s^2+M\tau ⟨A_h\nu ,\nu {⟩}_h=0$. All terms are nonnegative by [F2], so $v$ is constant. The first equation makes $v$ mean zero, hence $v=0$; the third then gives $s=0$, and the second gives $\nu =0$. Thus the square finite-dimensional linear system has trivial kernel and is invertible.

For an actual step, pairing the first SAV equation with $\mathbf{1}$ and using $A_h\mathbf{1}=0$ proves $⟨u^{n+1},\mathbf{1}{⟩}_h=⟨u^n,\mathbf{1}{⟩}_h$. Pairing that equation with ${\mu }^{n+1}$ gives $⟨{\mu }^{n+1},\delta u{⟩}_h=-M\tau ⟨A_h{\mu }^{n+1},{\mu }^{n+1}{⟩}_h$. Pairing the second equation with $\delta u$ and using the third gives $⟨{\mu }^{n+1},\delta u{⟩}_h=\frac{\kappa }{2}(⟨A_h{u}^{n+1},u^{n+1}{⟩}_h-⟨A_h{u}^n,u^n{⟩}_h+⟨A_h\delta u,\delta u{⟩}_h)+2r^{n+1}\delta r\mathrm{.}$ Since $2r^{n+1}\delta r=(r^{n+1}{)}^2-(r^n{)}^2+(\delta r{)}^2$, equating the two expressions proves the stated modified-energy identity.