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For $u,w\in X$, the forcing terms cancel, givingThe scalar function $s\mapsto s^7$ is strictly increasing on all of $\mathbb{R}$, so the second integrand is nonnegative. For the first one, use the quartic gradient monotonicity inequality. Put $d=\xi -\eta$ and $m=(\xi +\eta )/2$ for real vectors $\xi ,\eta$. Expanding givesConsequentlyThe last strict inequality uses the Poincaré inequality: if the weak gradient of $u-w\in W_0^{1,4}(\Omega )$ vanishes, then $u=w$. This proves that $A$ is a strictly monotone operator.

Using the dual pairing, monotonicity meansStrict monotonicity requires strict positivity whenever $u\ne v$. Coercivity with respect to zero, in the growth sense used for monotone-operator existence theory, meansThis is a condition on all sequences escaping to infinity, not just each fixed direction with a bound depending on that direction. Strict monotonicity concerns pairs of distinct arguments; it does not by itself imply this uniform large-norm growth.

For the log-cosh obstacle variational inequality, the admissible set is convex: for any $v\in M$, $u+t(v-u)$ remains above the obstacle and in $H_0^1$ for $0\le t\le 1$. Minimality and the right derivative at zero therefore giveThe one-sided sign comes from admissibility, which permits motion toward every $v\in M$ but does not permit an arbitrary negative variation at a contact point.

Now fix $\varphi \in C_c^{\infty }(D)$. Smoothness on $D$ and continuity of the obstacle give a strictly positive minimum of $u-g$ on the compact support of $\varphi$. Hence $u+t\varphi$ is admissible for both signs of sufficiently small $t$. Applying the inequality to both variations yieldsIntegration by parts gives $-\Delta u+\tanh u=0$ distributionally on $D$, and smoothness makes it pointwise. ThusOn the contact set the variational inequality, rather than unrestricted stationarity, remains the governing statement. Convexity of the energy also makes the inequality sufficient for minimization; strict monotonicity implies at most one minimizer, although existence was already stipulated in this part.

A strictly monotone operator $A:X\to X^{\ast }$ on a real Banach space satisfies strict operator monotonicity for every pair of distinct inputs. It is injective: $Au=Av$ would make the pairing zero and hence force $u=v$. In particular $Au=f$ has at most one solution for each $f\in X^{\ast }$.

Let $\Omega$ be a nonempty bounded Lipschitz domain, and define $A:H^1(\Omega )\to H^1(\Omega {)}^{\ast }$ byThis operator is strictly monotone: the difference pairing is the sum of $\Vert \nabla (v-w){\Vert }_2^2$ and the nonnegative integral $(\tanh v-\tanh w)(v-w)$, which is strictly positive wherever $v\ne w$. The bounded-domain hypothesis also makes nonzero constant functions admissible in $H^1$.

For constant $v=t$,Thus $A$ is not a nonlinear coercive operator in the large-norm sense. On $H_0^1(\Omega )$, the Poincaré inequality instead giveswhich proves coercivity on that zero-boundary space. The example separates being a strictly monotone operator from coercivity and identifies the uncontrolled constant direction.