= Solution
Using the <dual pairing>, a <monotone operator> satisfies $\langle A(u)-A(v),u-v\rangle\geq0$ for all $u,v\in M$. It is a <strongly monotone operator> if some $m>0$ gives
$$
\boxed{\langle A(u)-A(v),u-v\rangle\geq m\|u-v\|^2\quad(u,v\in M).}
$$
A <nonlinear coercive operator> with respect to $u_0\in M$ satisfies $\langle A(u),u-u_0\rangle/\|u-u_0\|\to+\infty$ as $\|u-u_0\|\to\infty$ in $M$.
Strong monotonicity, with $v=u_0$, gives
$$
\langle A(u),u-u_0\rangle\geq m\|u-u_0\|^2+\langle A(u_0),u-u_0\rangle
\geq m\|u-u_0\|^2-\|A(u_0)\|_{X^*}\|u-u_0\|.
$$
After division, the lower bound is $m\|u-u_0\|-\|A(u_0)\|_{X^*}\to+\infty$, proving coercivity. Reflexivity and separability are not needed for this particular estimate.
Back to article page
