Coercivity makes a small form step a strict contraction

On a nonzero Hilbert space the named constants satisfy $\alpha \le M$: choosing $u\mathrm{\ne }0$ and dividing $u$ by its norm, $\alpha \le \mathrm{Re}a(u,u)\le \mathrm{\mid }a(u,u)\mathrm{\mid }\le M\mathrm{\parallel }u{\mathrm{\parallel }}^2=M$, The identity $1-2\rho \alpha +{\rho }^2{M}^2=M^2(\rho -\alpha \mathrm{/}{M}^2{)}^2+1-{\alpha }^2\mathrm{/}{M}^2$ makes the radicand nonnegative; it may vanish when $\alpha =M$ and $\rho =\alpha \mathrm{/}{M}^2$.

Contraction estimate: for $z\in H$ and $0<\rho <2\alpha \mathrm{/}{M}^2$ the expansion of [F2] together with [F1] gives $\mathrm{\parallel }z-\rho Az{\mathrm{\parallel }}^2=\mathrm{\parallel }z{\mathrm{\parallel }}^2-2\rho \mathrm{Re}(Az,z)+{\rho }^2\mathrm{\parallel }Az{\mathrm{\parallel }}^2\le (1-2\rho \alpha +{\rho }^2{M}^2)\mathrm{\parallel }z{\mathrm{\parallel }}^2=q_{\rho }^2\mathrm{\parallel }z{\mathrm{\parallel }}^2,$ so $\mathrm{\parallel }(I-\rho A)z\mathrm{\parallel }\le q_{\rho }\mathrm{\parallel }z\mathrm{\parallel }$; taking $z=u-v$ and using $T_{\rho ,w}(u)-T_{\rho ,w}(v)=(I-\rho A)(u-v)$ gives the contraction estimate for every $w$.

The constant lies in $[0,1)$: the radicand is a quadratic in $\rho$ with minimum $1-{\alpha }^2\mathrm{/}{M}^2$ at $\rho =\alpha \mathrm{/}{M}^2$, which is nonnegative because $\alpha \le M$ by step 1.1; at the endpoints $\rho =0$ and $\rho =2\alpha \mathrm{/}{M}^2$ it equals $1$, and for $0<\rho <2\alpha \mathrm{/}{M}^2$ either directly $1-2\rho \alpha +{\rho }^2{M}^2<1$ or by the strict minimum unless $\rho =\alpha \mathrm{/}{M}^2$ and $\alpha =M$, in which case the radicand vanishes and $q_{\rho }=0$; in every case $0\le q_{\rho }<1$. Hence $I-\rho A$, and with it $T_{\rho ,w}$, is a strict contraction with constant $q_{\rho }$.