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.

For the log-cosh energy on a Sobolev space, the elementary inequalities $0\le \log \cosh t\le t^2/2$ follow by integrating $|\tanh t|\le |t|$ from zero. Thusso the energy is well-defined on $H^1$. For a fixed $v$, its proposed derivative is linear in the direction $h$ and bounded, because $|\tanh v|\le |v|$ givesTo justify differentiation, use $(\log \cosh t{)}^{\prime \prime }={\mathrm{sech}}^{2}t\in (0,1]$. Taylor's integral remainder pointwise, combined with the exact quadratic gradient expansion, givesEvery term is integrable by the displayed bounds. The remainder is quadratic in the norm, so the functional is even Fréchet differentiable, and in particular Gâteaux differentiable with the required derivative. This also proves that $A(v)$ belongs to the stated dual space.

For $v,w\in H^1$,Both terms are nonnegative, and the integrand in the second is strictly positive wherever $v\ne w$, since $\tanh$ is strictly increasing. Distinct Sobolev elements differ on a set of positive measure, so the expression is positive. Hence $A$ is strictly monotone.

This gives strict monotonicity without coercivity on H1: it is not coercive on $H^1$. For constant functions $v_t=t$ with $t\to +\infty$, the norm is $t|\Omega {|}^{1/2}$ andwhich is finite. In contrast it is coercive on $H_0^1$: if $\Vert v{\Vert }_2\le C_P\Vert \nabla v{\Vert }_2$, thenDividing by the norm proves growth to infinity. On this smaller space the operator is understood by restricting its dual action to zero-trace test functions. Removing the constant directions by the boundary condition is what restores coercivity.

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.

One form of Schauder's theorem says that if $M$ is a nonempty closed bounded convex subset of a Banach space and $T:M\to M$ is continuous with relatively compact image, then $T$ has a fixed point.

For a Schauder solution of a small Hammerstein integral equation, take $X=C([0,1])$ with the supremum norm and $M=\{u:\Vert u{\Vert }_{\infty }\le R\}$. This set is nonempty, closed, bounded and convex. PutBoth maxima are finite by continuity on compact sets. Define $(T_{\mu }u)(x)=\mu {\int }_0^{1}G(x,y)f(y,u(y))dy$. Continuity and uniform continuity of the kernel give $T_{\mu }u\in X$, and for $0<\mu \le {\mu }_0$,so it is a self-map. This choice of ${\mu }_0$ also handles the case when either maximum is zero without division by zero.

If $u_n\to u$ uniformly in $M$, uniform continuity of $f$ on $[0,1]\times [-R,R]$ makes $f(\cdot ,u_n(\cdot ))\to f(\cdot ,u(\cdot ))$ uniformly. The integral estimate then proves continuity of $T_{\mu }$ in the supremum norm. Moreoverwhich tends to zero with $|x-x^{\prime }|$ uniformly in $u\in M$. Thus the image is uniformly bounded and equicontinuous. Arzelà-Ascoli states that such a family on a compact interval has compact closure in the supremum norm. Hence $T_{\mu }(M)$ is relatively compact, and Schauder gives a fixed point for each $\mu$ in the specified interval. Therefore a continuous solution with supremum norm at most $R$ exists. The argument imposes no Lipschitz hypothesis on $f$ and makes no uniqueness assertion.

Brouwer's fixed-point theorem says that every continuous self-map of a nonempty closed Euclidean ball has a fixed point. It does not promise convergence of repeated application. For example $f(x)=1-x$ with starting point $x_1=0$ gives the nonconvergent alternating sequence $0,1,0,1,\dots$, although $1/2$ is a fixed point.

For an invariant ball from sublinear growth, chooseThen $A{R}^{\alpha }\le R/2$ and $B\le R/2$. For every $|x|\le R$ the given estimate yields $|f(x)|\le A{R}^{\alpha }+B\le R$. Thus $f$ restricts to a continuous self-map of the closed ball of this radius, and Brouwer gives a fixed point. The exponent $\alpha <1$ ensures that the linear radius eventually dominates its sublinear growth; no differentiability or contraction estimate is needed.