Unconditional weighted L2 stability for θ at least one half

By [F3] the step has a unique solution. Write $\delta U=U^{n+1}-U^n$. The identity $U_{\theta }=U^n+\theta \delta U$ gives $⟨\delta U,U_{\theta }{⟩}_h=\frac{1}{2}(\mathrm{\parallel }{U}^{n+1}{\mathrm{\parallel }}_h^2-\mathrm{\parallel }{U}^n{\mathrm{\parallel }}_h^2)+(\theta -\frac{1}{2})\mathrm{\parallel }\delta U{\mathrm{\parallel }}_h^2$ by expanding the two squared norms.

Taking the inner product of [F1] with $U_{\theta }$ and using step 1.1 gives $\frac{1}{2}(\mathrm{\parallel }{U}^{n+1}{\mathrm{\parallel }}_h^2-\mathrm{\parallel }{U}^n{\mathrm{\parallel }}_h^2)+(\theta -\frac{1}{2})\mathrm{\parallel }\delta U{\mathrm{\parallel }}_h^2+\mathrm{\Delta }t⟨A_h{U}_{\theta },U_{\theta }{⟩}_h=\mathrm{\Delta }t⟨q^n,U_{\theta }{⟩}_h$. The two terms after the norm difference on the left are nonnegative by $\theta \ge 1\mathrm{/}2$ and [F2]. Cauchy–Schwarz and the triangle inequality give $\mathrm{\mid }⟨q^n,U_{\theta }{⟩}_h\mathrm{\mid }\le \mathrm{\parallel }{q}^n{\mathrm{\parallel }}_h\mathrm{\parallel }{U}_{\theta }{\mathrm{\parallel }}_h\le \mathrm{\parallel }{q}^n{\mathrm{\parallel }}_h\max (\mathrm{\parallel }{U}^n{\mathrm{\parallel }}_h,\mathrm{\parallel }{U}^{n+1}{\mathrm{\parallel }}_h)$.

Put $a=\mathrm{\parallel }{U}^n{\mathrm{\parallel }}_h$, $b=\mathrm{\parallel }{U}^{n+1}{\mathrm{\parallel }}_h$, and $r=\mathrm{\parallel }{q}^n{\mathrm{\parallel }}_h$. If $b\le a$, the claimed bound is immediate. If $b>a$, step 2.1 gives $(b^2-a^2)\mathrm{/}2\le \mathrm{\Delta }trb$; dividing by $(a+b)\mathrm{/}2>0$ yields $b-a\le 2\mathrm{\Delta }trb\mathrm{/}(a+b)\le 2\mathrm{\Delta }tr$. This proves the forced estimate in all cases. When $q^n=0$, it gives $b\le a$; iteration over any finite number of steps preserves the same mesh-independent constant $1$.