Nonsmooth Equilibrium — Examples

The domain is $[0,\mathrm{\infty })$, so it is nonempty and $f$ is proper. Its epigraph is $E=\{(y,s):y\ge 0,s\ge 0\}$, a nonempty closed convex set; [F3] therefore makes $f$ closed and convex. The point $(-1,0)$ lies outside $E$. With $u=(-1,0)$ and $c=0$, every $(y,s)\in E$ satisfies $⟨u,(y,s)⟩=-y\le 0<1=⟨u,(-1,0)⟩$, so this is the strict vertical separator; its height coefficient is zero.

For $p\le 0$ and $y\ge 0$, $py\le 0$, with equality at $y=0$, so $f^{\ast }(p)=0$. For $p>0$, given any real $M$ choose $y>\max \{0,M\mathrm{/}p\}$; then $py>M$, so the defining supremum is $+\mathrm{\infty }$. Therefore $\mathrm{dom}{f}^{\ast }=(-\mathrm{\infty },0]$.

At $x=-1$, the biconjugate is ${\sup }_{p\le 0}(-p-f^{\ast }(p))={\sup }_{p\le 0}(-p)=+\mathrm{\infty }$: for every real bound $M$, the admissible choice $p=-\max \{M,0\}-1$ gives $-p>M$. Thus $f^{\ast \ast }(-1)=f(-1)$.

For each $k\ge 0$ and $y\ge 0$, ${\mathrm{\ell }}_k(y)=-ky\le 0=f(y)$, so ${\mathrm{\ell }}_k$ is an affine minorant. Its value at $-1$ is $k$, and the conjugate choice $p=-k$ has $f^{\ast }(-k)=0$, so the same family realizes arbitrarily large dual values at the point outside the domain. The conclusion is consistent with the full equality in [L2].

The epigraph is $[-1,1]\times [0,\mathrm{\infty })$, a nonempty closed convex rectangle, and the domain $[-1,1]$ is nonempty; hence the indicator is proper, closed, and convex by [F1, F3]. At slope $p=0$, every $y\in [-1,1]$ contributes $0$ to the conjugate supremum, and no other $y$ is in the effective domain, so $f^{\ast }(0)=0$ and every point of the interval attains this supremum.

Fix any $x\in [-1,1]$. For every $y\in \mathbb{R}$, $f(y)\ge 0=f(x)+0(y-x)$: equality holds on the interval and the left side is $+\mathrm{\infty }$ outside it. Hence $0\in \mathrm{\partial }f(x)$; [L1] also gives $f^{\ast }(0)=0=0\cdot x-f(x)$ at each such $x$.

The zero tilt is $g=f$. It is coercive as an extended-real function: for any $M\in \mathbb{R}$, choosing $R=2$ ensures $\mathrm{\mid }x\mathrm{\mid }\ge R$ implies $g(x)=+\mathrm{\infty }>M$. Its minimum is $0$, attained at every $x\in [-1,1]$, so this example satisfies [L2] while exhibiting nonuniqueness and a whole flat supporting face.

For $C=P=2$, there are $n=2+2(2-1)=4$ coordinates and $q=2(2-1)=2$ equations. Since each ${\mu }_i^{\alpha }$ is identically zero, both differences ${\mu }_i^2-{\mu }_i^1$ equal zero at every point of $U$. Thus the zero set is exactly the open set $U$, which is locally parameterized by its four original coordinates.

The derivative of each identically zero equation is the zero covector, so the equation Jacobian has rank zero and no $2\times 2$ minor is invertible. Therefore [L1] does not apply; the example does not contradict its regular dimension count, and it shows why equation count alone does not determine local dimension.

The species-amount vector is $n(\xi )=n^0+\nu \xi =(2-2\xi ,1-\xi ,2\xi {)}^{\mathsf{T}}$. Nonnegativity of its third component requires $\xi \ge 0$, and nonnegativity of its first two components requires $\xi \le 1$; hence the feasible set is exactly $[0,1]$.

The elemental count matrix is $D=\left(\begin{array}{ccc}2& 0& 2\\ 0& 2& 1\end{array}\right)$, with first row counting hydrogen atoms and second row oxygen atoms per molecule. Direct multiplication gives $D\nu =(-4+4,-2+2{)}^{\mathsf{T}}=(0,0{)}^{\mathsf{T}}$, so every change $n(\xi )-n^0=\nu \xi$ preserves both atom totals.

At ${\xi }_{\ast }=0$, only the water component is zero. Its directional change is $(\nu h{)}_3=2h$, so [L1] gives $T_{\mathcal{F}}(0)=\{h:2h\ge 0\}=[0,\mathrm{\infty })$. A negative direction is infeasible because it would make the water amount $2\xi$ negative.

For $\mathrm{\Phi }(n)=n_3$, the induced function is $f(\xi )=\mathrm{\Phi }(n(\xi ))=2\xi$. Its derivative at $0$ is $Df(0)h=2h\ge 0$ for every feasible tangent $h\ge 0$, and $f(\xi )\ge f(0)$ for every $\xi \in [0,1]$. This verifies the boundary first-order condition and minimum directly for this chosen mathematical objective.