Semidiscrete Ritz error splitting and energy estimate

The fixed-map product rule [F4] gives $(P_{h}u{)}^{\mathrm{\prime }}=P_h{u}^{\mathrm{\prime }}$, so ${\rho }^{\mathrm{\prime }}=P_h{u}^{\mathrm{\prime }}-u^{\mathrm{\prime }}\in L^2(0,T;V)$. For a $C^1$ family it gives instead ${\rho }^{\mathrm{\prime }}=P_h^{\mathrm{\prime }}u+P_h{u}^{\mathrm{\prime }}-u^{\mathrm{\prime }}$. Also ${\theta }^{\mathrm{\prime }}=u_h^{\mathrm{\prime }}-(P_{h}u{)}^{\mathrm{\prime }}$. The residual is an element of $L^2(0,T;V_h^{\ast })$: for each fixed $v_h$, approximate the strongly measurable $\rho$ by simple $V$-valued functions. Scalar measurability of $a(t;w,v_h)$ for each fixed $w$ makes the corresponding simple-function pairings measurable, and the bound [F6] lets their pointwise limits pass to $a(t;\rho (t),v_h)$. The first term $({\rho }^{\mathrm{\prime }},v_h{)}_H$ is measurable as well. Because $V_h$ is finite-dimensional, measurability on a finite basis gives strong measurability of the $V_h^{\ast }$-valued residual. Moreover, $\mathrm{\parallel }{\mathcal{R}}_h(t){\mathrm{\parallel }}_{V_h^{\ast }}\le c_i^2\mathrm{\parallel }{\rho }^{\mathrm{\prime }}(t){\mathrm{\parallel }}_V+M\mathrm{\parallel }\rho (t){\mathrm{\parallel }}_V,$ so it is square-integrable.

Subtract [F1] from [F2]. Since $u_h-u=\theta +\rho$ and $u_h^{\mathrm{\prime }}-u^{\mathrm{\prime }}={\theta }^{\mathrm{\prime }}+{\rho }^{\mathrm{\prime }}$, the result for each $v_h$ is $({\theta }^{\mathrm{\prime }},v_h{)}_H+a(t;\theta ,v_h)=-({\rho }^{\mathrm{\prime }},v_h{)}_H-a(t;\rho ,v_h)=-{\mathcal{R}}_h(t)(v_h)\mathrm{.}$ This uses only the variational equations; no commutation of the full evolution operator with $P_h$ is assumed.

Put $v_h=\theta (t)$ in step 2.1 and use [F5]. Coercivity from [F6] and Cauchy–Schwarz/Young give $\frac{1}{2}\mathrm{\parallel }\theta (t){\mathrm{\parallel }}_H^2-\frac{1}{2}\mathrm{\parallel }\theta (0){\mathrm{\parallel }}_H^2+\frac{\alpha }{2}{\int }_0^t\mathrm{\parallel }\theta (s){\mathrm{\parallel }}_V^{2}ds\le \frac{1}{2\alpha }{\int }_0^t\mathrm{\parallel }{\mathcal{R}}_h(s){\mathrm{\parallel }}_{V_h^{\ast }}^{2}ds\mathrm{.}$ Multiplying by $2$ proves the asserted energy bound.

In the time-independent Ritz case, [F3] gives $a(\rho (t),v_h)=a(R_{h}u(t)-u(t),v_h)=0$ for every $v_h$. By [F4], ${\rho }^{\mathrm{\prime }}=R_h{u}^{\mathrm{\prime }}-u^{\mathrm{\prime }}$, which yields the special residual stated above. In the time-dependent case this cancellation has not been assumed, so step 2.1 retains the complete residual, including $P_h^{\mathrm{\prime }}u$ when the comparison varies.