A Global Linear and Local Superlinear (Quadratic) Inexact Non-Interior Continuation Method for Variational Inequalities Over General Closed Convex Sets

Abstract

We use the concept of barrier-based smoothing approximations to extend the non-interior continuation method, which was proposed by B. Chen and N. Xiu for nonlinear complementarity problems based on Chen-Mangasarian smoothing functions, to an inexact non-interior continuation method for variational inequalities over general closed convex sets. Newton equations involved in the method are solved inexactly to deal with high dimension problems. The method is proved to have global linear and local superlinear/quadratic convergence under suitable assumptions. We apply the method to non-negative orthants, positive semidefinite cones, polyhedral sets, epigraphs of matrix operator norm cone and epigraphs of matrix nuclear norm cone.

Appendices

Appendix A: Technical proofs

1.1 A.1 Proof of Proposition 3

(i) From Eq. 9 we get

$$\left\lVert {\triangle \tilde{w}^{(k)}} \right\rVert \leq C (\left\lVert {H_{\mu_{k}}(w^{(k)})} \right\rVert + \left\lVert {r_{1}^{(k)}} \right\rVert) \leq C(1+\theta_{1}) {\Psi}_{\mu_{k}}(w^{(k)}) \leq C(1+\theta_{1})\upbeta \mu_{k}. $$

(ii) We have

$$ \begin{array}{@{}rcl@{}} \begin{array}{ll} \left\lVert { H_{0}(w^{(k)})} \right\rVert - {\Psi}_{\mu_{k}}(w^{(k)})& \leq \left\lVert {\phi_{0}(w^{(k)})} \right\rVert - \left\lVert {\phi_{\mu_{k}}(w^{(k)})} \right\rVert\\ & \leq \left\lVert {\phi_{0}(w^{(k)}) - \phi_{\mu_{k}}(w^{(k)})} \right\rVert\\ & = \left\lVert {p_{0}(x^{(k)}-y^{(k)}) - p_{\mu_{k}}(x^{(k)}-y^{(k)})} \right\rVert \leq \sqrt{\vartheta} \mu_{k}, \end{array} \end{array} $$

(48)

where we have used the property $\left \lVert {(a,b)} \right \rVert \leq \left \lVert {a} \right \rVert + \left \lVert {b} \right \rVert$ for the first inequality and Theorem 1 for the last inequality. Inequality (48) with (11) give us

$$ \begin{array}{@{}rcl@{}} \|\triangle \hat{w}\|& \leq C (\left\lVert { H_{0}(w^{(k)})} \right\rVert + \left\lVert {r_{2}^{(k)}} \right\rVert) \leq C(1+\theta_{2})({\Psi}_{\mu_{k}}(w^{(k)}) + \sqrt{\vartheta}\mu_{k} )\\ &\leq C(1+\theta_{2})(\upbeta + \sqrt{\vartheta}) \mu_{k}. \end{array} $$

1.2 A.2 Proof of Theorem 5

1.2.1 A.2.1 Non-negative orthant $\mathbb {R}^{n}_{+}$

Gradient and Hessian of the barrier function are

$$ \nabla f(x)= -\sum\limits_{i=1}^{n}\frac{1}{x_{i}} e_{i}, \quad \nabla^{2} f(x) = \sum\limits_{i=1}^{n} \frac{1}{{x_{i}^{2}}} e_{i} {e_{i}^{T}}, $$

where e_i denote the i −th standard unit vector of $\mathbb {R}^{n}$. The corresponding barrier-based smoothing approximation is $p_{\mu }(z)=\frac {1}{2}\sum \limits _{i=1}^{n}\left ({z_{i}+ \sqrt {{z_{i}^{2}} + 4\mu ^{2}}}\right ) e_{i}$. Its Jacobian is

$$\text{\textbf{J}} p_{\mu}(z)= \frac{1}{2}\textup{Diag}\left( 1 + \frac{z_{i}}{\sqrt{{z_{i}^{2}} + 4\mu^{2}}}|_{i=1,\ldots,n}\right).$$

The projection of z onto $\mathbb {R}^{n}_{+}$ is ${\Pi }_{\mathbb {R}^{n}_{+}}(z)=[z]_{+}$. We observe that the projector is differentiable at z^∗ if and only if $z_{i}^{*}\ne 0, \forall i=1,\ldots ,n$. On the other hand, a pair (x^∗,y^∗) is strictly complimentary if and only if $x_{i}^{*}+y_{i}^{*} >0$ for all i = 1,…,n, see [9]. Furthermore, $x^{*}+y^{*}={\Pi }_{\mathbb {R}^{n}_{+}}(z^{*})+{\Pi }_{\mathbb {R}^{n}_{+}}(-z^{*})$. Hence, it is easy to see that differentiability of the projector at z^∗ is equivalent to strict complimentarity of (x^∗,y^∗).

Now let $z_{i}^{*}\ne 0$ for i = 1,…,n, then we observe that the Jacobian Jp_μ(z) converges to

$$\text{\textbf{J}} {\Pi}_{\mathbb{R}^{n}_{+}}(z^{*})=\frac{1}{2}\textup{Diag}\left( 1 + \frac{z_{i}^{*}}{|z_{i}^{*}|}|_{i=1,\ldots,n}\right)$$

when $(z,\mu )\rightarrow (z^{*},0)$. Since the map (z,μ)↦Jp_μ(z) is continuously differentiable at (z^∗,0), thus is locally Lipschitz at this point. Consequently, $\left \lVert {\text {\textbf {J}} p_{\mu }(z)-T^{*}}\right \rVert = O(\left \lVert {(z-z^{*},\mu )}\right \rVert )$. On the other hand, $\left \lVert {\textup {D} p_{\mu }(z)-\textup {D} {\Pi }_{\mathbb {R}^{n}_{+}}(z^{*})}\right \rVert =\left \lVert {\text {\textbf {J}} p_{\mu }(z)-\text {\textbf {J}} {\Pi }_{\mathbb {R}^{n}_{+}}(z^{*})}\right \rVert$. Thus, we get the result.

1.2.2 A.2.2 Positive semidefinite cone $\mathbb {S}_{+}^{n}$

We have ∇f(x) = −x^− 1.

From the equation x + μ²∇f(x) = z, we deduce that the corresponding barrier-based smoothing approximation is

$$ p_{\mu}(z)=\frac{1}{2}\left( z+ (z^{2} + 4\mu^{2} I)^{1/2}\right). $$

Denote $\mathfrak {g}: u\in \mathbb {R} \mapsto \mathfrak {g}(u)= u+ \sqrt {u^{2} + 4\mu ^{2}}$, $\mathfrak {g}^{\prime }(u)=1+ \frac {u}{\sqrt {u^{2} + 4\mu ^{2}}}$ and $\mathfrak {g}^{(1)}$ is a matrix whose (i,j)-th entry with respect to a vector d is

$$ \begin{array}{@{}rcl@{}} (\mathfrak{g}^{(1)}(d))_{ij}&=&\left\{ \begin{array}{ll} \frac{\mathfrak{g}(d_{i})-\mathfrak{g}(d_{j})}{d_{i} - d_{j}} &\text{if} d_{i}\ne d_{j}\\ \mathfrak{g}^{\prime}(d_{i}) & \text{if} d_{i} = d_{j} \end{array} \right.\\ & =& \left\{ \begin{array}{ll}1+ \frac{\sqrt{{d_{i}^{2}} + 4\mu^{2}} - \sqrt{{d_{j}^{2}} + 4\mu^{2}}}{d_{i} - d_{j}} &\text{if} d_{i}\ne d_{j}\\ 1+ \frac{d_{i}}{\sqrt{{d_{i}^{2}} + 4\mu^{2}}} & \text{if} d_{i} = d_{j} \end{array}\right.\\ & = &1+\frac{d_{i} + d_{j}}{\sqrt{{d_{i}^{2}} + 4\mu^{2}} + \sqrt{{d_{j}^{2}} + 4\mu^{2}}}. \end{array} $$

Let z = q Diag(λ_f(z))q^T, then $\textup {D} p_{\mu }(z)[h]= \frac {1}{2} q \left [ \mathfrak {g}^{(1)} (\lambda _{f}(z)) \circ (q^{T} h q)\right ] q^{T}.$ Projection of z onto $\mathbb {S}^{n}_{+}$ is ${\Pi }_{\mathbb {S}^{n}_{+}}(z)=q\textup {Diag}\left ({[\lambda _{f}(z)]_{+}} \right )q^{T}$. We see that ${\Pi }_{\mathbb {S}^{n}_{+}}(\cdot )$ is differentiable at z^∗ if and only if all eigenvalues $\lambda _{i}^{*}$, for i = 1,…,n, of z^∗ are non-zeroes. Furthermore, strict complementarity of (x^∗,y^∗) is equivalent to the condition that all eigenvalues of x^∗ + y^∗ is positive. We now let $z^{*}=\hat {q}\textup {Diag}(\lambda _{f}(z^{*})) \hat {q}^{T}$ be the eigenvalue decomposition of z^∗. Then,

$$ \begin{array}{@{}rcl@{}} x^{*}+y^{*}&=&{\Pi}_{\mathbb{S}^{n}_{+}}(z^{*})+ {\Pi}_{\mathbb{S}^{n}_{+}}(-z*)\\ &=&\hat{q}\textup{Diag}\left( {[\lambda_{f}(z^{*})]_{+}} \right) \hat{q}^{T}+ \hat{q}\textup{Diag}\left( {[-\lambda_{f}(z^{*})]_{+}} \right) \hat{q}^{T}\\ &=&\hat{q} \textup{Diag}\left( {[\lambda_{f}(z^{*})]_{+} + [-\lambda_{f}(z^{*})]_{+}} \right)\hat{q}^{T}. \end{array} $$

Hence differentiability of ${\Pi }_{\mathbb {S}^{n}_{+}}(\cdot )$ at z^∗ is equivalent to strict complementarity of (x^∗,y^∗).

Now we consider z^∗ whose eigenvalues are non-zeros. Let (z,μ) go to (z^∗,0), then λ_f(z) converges to λ^∗. Let $\bar {q}$ be a limit point of q. We then have $z^{*}=\bar {q} \textup {Diag}(\lambda ^{*}) \bar {q}^{T}$ with $\bar {q} \in \mathcal {O}^{n}(z^{*}).$ We deduce from $\lambda _{i}^{*} \ne 0$, i = 1,…,n that

$$(\mathfrak{g}^{(1)} (\lambda_{f}(z)))_{ij}= 1 + \frac{\lambda_{i} + \lambda_{j}}{\sqrt{{\lambda_{i}^{2}} + 4\mu^{2}} + \sqrt{{\lambda_{j}^{2}} + 4\mu^{2}}}\to 1+ \frac{\lambda_{i}^{*} + \lambda_{j}^{*}}{|\lambda_{i}^{*}| + | \lambda_{j}^{*}|}.$$

Therefore, by Theorem 2, when (z,μ) → (z^∗,0), where z^∗ are differential points of ${\Pi }_{\mathbb {S}^{n}_{+}}(\cdot )$, Dp_μ(z) converges to $\textup {D} {\Pi }_{\mathbb {S}^{n}_{+}}(z^{*})$ with

$$ \begin{array}{@{}rcl@{}} \textup{D} {\Pi}_{\mathbb{S}^{n}_{+}}(z^{*})[h]=\frac{1}{2} \bar{q} \left[ \bar{\mathfrak{g}}^{(1)} (\lambda^{*}) \circ (\bar{q}^{T} h \bar{q})\right] \bar{q}^{T}, \end{array} $$

(49)

where $\bar {\mathfrak {g}}^{(1)}(\lambda ^{*})_{ij}= 1+ \frac {\lambda _{i}^{*} + \lambda _{j}^{*}}{|\lambda _{i}^{*}| + | \lambda _{j}^{*}|}$. Note that Formula (49) is independent of the choice $\bar {q}$. Finally, similarly to the case $\mathbb {R}_{+}^{n}$, we have $\left \lVert {\textup {D} p_{\mu }(z) - \textup {D} {\Pi }_{\mathbb {S}^{n}_{+}}(z^{*})}\right \rVert =O(\left \lVert {z-z^{*},\mu }\right \rVert )$ since (z,μ)↦Dp_μ(z) is locally Lipschitz around (z^∗,0).

1.3 A.3 Proof of Proposition 8

We note that $\mathcal {F}_{I^{*}}$ is the unique neighbour face of z^∗ and $z^{*} \in \textup {int} (\mathcal {F}_{I} + \mathcal {N}_{I})$ as the projector is differentiable at z^∗ (see Proposition 7). When $(z,\mu )\rightarrow (z^{*},0)$, we have $x\rightarrow {\Pi }_{K}(z^{*})=\bar {z}^{*},$ satisfying $A_{i}\bar {z}^{*} =b_{i}, \forall i\in \mathcal {I}^{*}$ and $A_{i}\bar {z}^{*} > b_{i}, \forall i\not \in \mathcal {I}^{*}.$ From Eq. 22 we have

$$ \begin{array}{@{}rcl@{}} z^{*} -\bar{z}^{*} = \lim\limits_{(z,\mu)\rightarrow (z^{*},0)} (x-z) = \lim\limits_{(z,\mu)\rightarrow (z^{*},0)} \sum\limits_{i\in \mathcal{I}^{*}} \frac{\mu^{2}}{A_{i} x-b_{i}} (-{A_{i}^{T}}) \end{array} $$

(50)

If there exists $j\in \mathcal {I}^{*}$ and a subsequence $(z,\mu )_{k}\rightarrow (z^{*},0)$ such that $\frac {{\mu _{k}^{2}}}{A_{j} x_{k} - b_{j}} \rightarrow 0$ then we take the limit of this subsequence in (50) to get

$$z^{*} -\bar{z}^{*} = \lim\limits_{k\rightarrow \infty} \sum\limits_{i\in \mathcal{I}^{*}\setminus\{j\}} \frac{{\mu_{k}^{2}}}{A_{i} x_{k}-b_{i}} (-{A_{i}^{T}}) \in \textup{cone} \{ -{A_{i}^{T}} : i\in \mathcal{I}^{*}\setminus\{j\}\} = \mathcal{N}_{\mathcal{I}^{*}\setminus\{j\}} $$

On the other hand, $\bar {z}^{*}\in \mathcal {F}_{\mathcal {I}^{*}\setminus \{j\}}$ as $A_{i}\bar {z}^{*}=b_{i} \forall i\in \mathcal {I}^{*}\setminus \{j\}$. Hence $z^{*} = \bar {z}^{*} + z^{*} -\bar {z}^{*} \in \mathcal {F}_{I^{*}\setminus \{j\}} + \mathcal {N}_{\mathcal {I}^{*}\setminus \{j\}}$, which implies $\mathcal {I}^{*}\setminus \{j\}$ is a neighbour face of z^∗. This contradicts to the fact $\mathcal {I}^{*}$ is the unique neighbour face of z^∗. Therefore, for all $i\in \mathcal {I}^{*}, \frac {\mu ^{2}}{A_{i} x -b_{i}}$ only have nonzero limit points. The result follows then.

1.4 A.4 Proof of Proposition 9

By re-indexing z^∗ if necessary, we can assume that π is the identity permutation. For each i ∈{1,…,n}, if $|x_{i}^{*}| < t^{*}$ then

$$x_{i}^{*} = \lim x_{i} = \lim (z_{i}-\frac{2\mu^{2}}{t^{2}-{x_{i}^{2}}}x_{i}) = z_{i}^{*}. $$

Together with sgn(x_i) = sgn(z_i) and t > |x_i|, we deduce that $x_{i}^{*}=\textup {sgn}(z_{i}^{*}) t^{*} \text {or} z_{i}^{*} \text {for} i=1,\ldots ,n.$ Moreover, $|x_{i}| < {\min \limits } \{t,|z_{i}| \}$ further implies that

$$x_{i}^{*}=\left\{\begin{array}{ll} \textup{sgn}(z_{i}^{*}) t^{*} &\text{if} t^{*} < |z_{i}^{*}|\\ z_{i}^{*} &\text{if} t^{*} > |z_{i}^{*}|. \end{array}\right. $$

Thus, there exists a unique positive integer k^∗ such that

$$x_{i}^{*} =\left\{\begin{array}{ll}\textup{sgn}(z_{i}^{*}) t^{*} &\text{for} i=1,\ldots, \mathbf k^{*},\\ z_{i}^{*} &\text{for} i=\mathbf k^{*}+1,\ldots,n, \end{array}\right. $$

and $|z_{\mathbf k^{*}}^{*} |> t^{*} \geq |z_{\mathbf k^{*}+1}^{*}|$. Summing up (n + 1) equations in (26) gives

$$z_{o} + \sum\limits_{i=1}^{n}|z_{i}| = t+ \sum\limits_{i=1}^{n}|x_{i}| - \mu^{2} \sum\limits_{i=1}^{n} \frac{2(t-|x_{i}|)}{t^{2} - {x_{i}^{2}}} = t+ \sum\limits_{i=1}^{n} |x_{i}| - \mu^{2} \sum\limits_{i=1}^{n} \frac{2}{t+ |x_{i}|}. $$

If t^∗ > 0 then taking limnit gives, $z_{o}^{*}+\sum \limits _{i=1}^{n} |z_{i}^{*}| = t^{*} + \sum \limits _{i=1}^{n}|x_{i}^{*}|$, and hence

$$(\mathbf k^{*}+1)t^{*} = t^{*} + \sum\limits_{i=1}^{\mathbf k^{*}}|x_{i}^{*}| = z_{o}^{*} + \sum\limits_{i=1}^{n}|z_{i}^{*}| - \sum\limits_{i=1}^{n}|x_{i}^{*}|=z_{o}^{*} + \sum\limits_{i=1}^{\mathbf k^{*}}|z_{i}^{*}|. $$

If t^∗ = 0 then |x_i| < t implies that $x_{i}^{*}=0$ for i = 1,…,n. Thus

$$ \begin{array}{@{}rcl@{}} z_{o}^{*} + \sum\limits_{i=1}^{n}|z_{i}^{*}| &= \lim (z_{o}+\sum\limits_{i=1}^{n}|z_{i}|) = \lim (t+ \sum\limits_{i=1}^{n}|x_{i}| - \mu^{2}\sum\limits_{i=1}^{n}\frac{2}{t+|x_{i}|}\\ & \leq \lim (t+\sum\limits_{i=1}^{n}|x_{i}|) = t^{*} + \sum\limits_{i=1}^{n}|x_{i}^{*}|=0. \end{array} $$

Subsequently, $z_{o}^{*} + \sum \limits _{i=1}^{\mathbf k^{*}}|z_{i}^{*}| \leq z_{o}^{*} + \sum \limits _{i=1}^{n}|z_{i}^{*}|\leq 0$. Hence,

$$t^{*} = \max \left\{\frac{1}{\mathbf k^{*}+1}(z_{o}^{*} + \sum\limits_{i=1}^{\mathbf k^{*}} | z^{*}_{\pi(i)}|),0 \right\}.$$

Appendix B: Example

We consider the second order cone in $\mathbb {R}^{3}$

$$K_{2}=\{ (t,z): z \in \mathbb{R}^{2}, t\in \mathbb{R}_{+}, \left\lVert {z} \right\rVert \leq t\}. $$

Firstly, we use the barrier $f^{(1)}(t,z)=-\log (t^{2}-\|z\|^{2}).$ Denote $\mathbf M=t^{2} - {z_{1}^{2}} - {z_{2}^{2}}$. The gradient of f⁽¹⁾ is

$$ \nabla f^{(1)}(t,z)=\left( {-2t \mathbf M^{-1}, 2z_{1} \mathbf M^{-1}, 2z_{2} \mathbf M^{-1}} \right).$$

The smoothing approximation $p^{(1)}_{\mu }(t^{o},z^{o})=(t,z)$ regarding to f⁽¹⁾(t,z) is computed by

$$ \left\{\begin{array}{ll} t -\mu^{2} (2 t \mathbf M^{-1})&=t^{o}\\ z_{1} + \mu^{2} (2 z_{1} \mathbf M^{-1}) &= {z_{1}^{o}}\\ z_{2} + \mu^{2} (2 z_{2} \mathbf M^{-1})&={z_{2}^{o}}. \end{array}\right. $$

(51)

The unique solution of (51) is

$$\left\{\begin{array}{ll} t=\frac{1}{4}\left( {2t^{o}+\sqrt{(t^{o}-\left\lVert {z^{o}} \right\rVert)^{2} + 8\mu^{2}} + \sqrt{(t^{o}+\left\lVert {z^{o}} \right\rVert)^{2} + 8\mu^{2}}} \right)\\ z_{1}=\frac{1}{4} \frac{{z_{1}^{o}}}{\left\lVert {z^{o}} \right\rVert} \left( {2\left\lVert {z^{o}} \right\rVert + \sqrt{(t^{o}+\left\lVert {z^{o}} \right\rVert)^{2} + 8\mu^{2}} - \sqrt{(t^{o}-\left\lVert {z^{o}} \right\rVert)^{2} + 8\mu^{2}} } \right)\\ z_{2}=\frac{1}{4} \frac{{z_{2}^{o}}}{\left\lVert {z^{o}} \right\rVert} \left( {2\left\lVert {z^{o}} \right\rVert + \sqrt{(t^{o}+\left\lVert {z^{o}} \right\rVert)^{2} + 8\mu^{2}} - \sqrt{(t^{o}-\left\lVert {z^{o}} \right\rVert)^{2} + 8\mu^{2}} } \right). \end{array}\right. $$

Denote $s_{1}=\sqrt {(t^{o}-\left \lVert {z^{o}} \right \rVert )^{2} + 8\mu ^{2}}, s_{2}=\sqrt {(t^{o}+\left \lVert {z^{o}} \right \rVert )^{2} + 8\mu ^{2}}$. We have

$$\textbf{J} p_{\mu}^{(1)}(t^{o},{z_{1}^{o}},{z_{2}^{o}})=\left( \begin{array}{llll} d_{11} & d_{12} & d_{13}\\ d_{21} & d_{22} & d_{23} \\ d_{31} & d_{32} & d_{33}, \end{array}\right) $$

where

$$ \begin{array}{@{}rcl@{}} d_{11}&=&\frac{1}{2}+\frac{t^{o}(t^{o}-\left\lVert {z^{o}} \right\rVert)}{4s_{1}}+\frac{t^{o}(t^{o}+\left\lVert {z^{o}} \right\rVert)}{4s_{2}}, \\ d_{12}=d_{21}&=&-\frac{(t^{o}-\left\lVert{z^{o}}\right\rVert){z_{1}^{o}}}{4\left\lVert{z^{o}}\right\rVert s_{1}}+\frac{(t^{o}+\left\lVert{z^{o}}\right\rVert){z_{1}^{o}}}{4\left\lVert{z^{o}} \right\rVert s_{2}},\\ d_{13}=d_{31}&=&-\frac{(t^{o}-\left\lVert{z^{o}}\right\rVert){z_{2}^{o}}}{4\left\lVert{z^{o}}\right\rVert s_{1}}+\frac{(t^{o}+\left\lVert{z^{o}}\right\rVert){z_{2}^{o}}}{4\left\lVert{z^{o}}\right\rVert s_{2}},\\ d_{22}&=&\frac{1}{2}+\frac{({z_{1}^{o}})^{2}}{\left\lVert{z^{o}}\right\rVert^{2}} \left( {\frac{t^{o}+\left\lVert{z^{o}}\right\rVert}{s_{2}} + \frac{t^{o}-\left\lVert{z^{o}}\right\rVert}{s_{1}}}\right) +\frac{({z_{2}^{o}})^{2}(s_{2}-s_{1})}{\left\lVert{z^{o}}\right\rVert^{2}},\\ d_{23}=d_{32}&=&\frac{{z_{1}^{o}}}{\left\lVert{z^{o}}\right\rVert}\left( {\frac{(t^{o}+\left\lVert{z^{o}}\right\rVert){z_{2}^{o}}}{s_{2}} + \frac{(t^{o}-\left\lVert{z^{o}}\right\rVert){z_{2}^{o}}}{s_{1}}}\right),\\ d_{33}&=&\frac{1}{2}+\frac{({z_{2}^{o}})^{2}}{\left\lVert{z^{o}}\right\rVert^{2}} \left( {\frac{t^{o}+\left\lVert{z^{o}}\right\rVert}{s_{2}} + \frac{t^{o}-\left\lVert{z^{o}}\right\rVert}{s_{1}}}\right) +\frac{({z_{1}^{o}})^{2}(s_{2}-s_{1})}{\left\lVert{z^{o}}\right\rVert^{2}}. \end{array} $$

We choose $(t^{o},{z_{1}^{o}},{z_{2}^{o}})$ such that $(t^{o},{z_{1}^{o}},{z_{2}^{o}}) \to (t^{*},z_{1}^{*},z_{2}^{*})=(0,1,0)$; then $s_{1}\to 1, s_{2} \to 1, \left \lVert {z^{o}} \right \rVert \to 1$. We imply

$$\lim\limits_{(t^{o},{z_{1}^{o}},{z_{2}^{o}},\mu)\rightarrow (0,1,0,0)} \text{\textbf{J}} p_{\mu}^{(1)}(t^{o},{z_{1}^{o}},{z_{2}^{o}}) = \left( \begin{array}{ccc} 1/2 & 1/2 & 0\\ 1/2 & 1/2 & 0\\ 0 & 0 & 1/2 \end{array}\right), $$

which equals to $\text {\textbf {J}}{\Pi }_{K_{2}}(0,1,0)$ by Theorem 2. Now we use another barrier

$$ f^{(2)} (u,v_{1},v_{2}) = -\log(u^{2}-\|v\|^{2})-\log (u-v_{1}) - \log (u+v_{1}).$$

Denote M_μ = u² − v² − w². Gradient ∇f⁽²⁾(u,v₁,v₂) of f⁽²⁾ is

$$ \left( {-2u \mathbf M_{\mu}^{-1} -\frac{1}{u-v_{1}} - \frac{1}{u+v_{1}}, 2v_{1} \mathbf M_{\mu}^{-1} +\frac{1}{u-v_{1}} - \frac{1}{u+v_{1}} , 2v_{2} \mathbf M_{\mu}^{-1}} \right). $$

Let $p_{\mu }^{(2)}(t^{*},z_{1}^{*},z_{2}^{*})=p_{\mu }^{(2)}(0,1,0)=(u,v_{1},v_{2})$, which is defined by

$$\left\{\begin{array}{lll} u - \mu^{2} \left( {2 u \mathbf M_{\mu}^{-1} +\frac{1}{u-v_{1}} + \frac{1}{u+v_{1}}} \right) =0 \\ v_{1} + \mu^{2} \left( { 2 v_{1} \mathbf M_{\mu}^{-1} + \frac{1}{u-v_{1}} -\frac{1}{u+v_{1}}} \right) = 1\\ v_{2} + \mu^{2} (2 v_{2} \mathbf M_{\mu}^{-1}) = 0. \end{array}\right. $$

The third equation implies v₂ = 0, hence $\mathbf M_{\mu }=u^{2} - {v_{1}^{2}} =(u-v_{1})(u+v_{1}).$ Thus, the first and the second equation imply

$$ \left\{\begin{array}{lll} u+v_{1} - \frac{4\mu^{2}}{u+v_{1}}=1\\ u-v_{1} -\frac{4\mu^{2}}{u-v_{1}}=-1, \end{array}\right. $$

which give $u=\frac {1}{2}\sqrt {1+ 16\mu ^{2}}, v_{1}=\frac {1}{2}, \mathbf M_{\mu } = 4\mu ^{2}$. Denote $a=\sqrt {1+ 16\mu ^{2}}$. Hessian matrix ∇²f⁽²⁾(u,v₁,v₂) of the barrier f⁽²⁾ at (u,v₁,v₂) is

$$ \begin{array}{@{}rcl@{}} \left( \begin{array}{cccccccc} \frac{1+8\mu^{2}}{16\mu^{4}}+\frac{4}{(a-1)^{2}}+\frac{4}{(a+1)^{2}} & -\frac{a}{16\mu^{4}} - \frac{4}{(a-1)^{2}} +\frac{4}{(a+1)^{2}} & 0\\ -\frac{a}{16\mu^{4}} - \frac{4}{(a-1)^{2}} +\frac{4}{(a+1)^{2}} &\frac{1+8\mu^{2}}{16\mu^{4}} + \frac{4}{(a-1)^{2}} +\frac{4}{(a+1)^{2}} & 0\\ 0&0 & \frac{1}{2\mu^{2}} \end{array}\right)=\left( \begin{array}{cccc} \frac{1+8\mu^{2}}{8\mu^{4}}& \frac{-a}{8 \mu^{4}}&0\\ \frac{-a}{8 \mu^{4}} &\frac{1+8\mu^{2}}{8\mu^{4}} & 0\\ 0 & 0 & \frac{1}{2\mu^{2}}. \end{array}\right) \end{array} $$

We remind that $\text {\textbf {J}} p^{(2)}_{\mu }(0,1,0)=[I + \mu ^{2} \nabla ^{2} f^{(2)}(u,v_{1},v_{2})]^{-1}=\left (\begin {array}{ccc} 1/2 & \frac {1}{2a} &0 \\ \frac {1}{2a} & 1/2&0\\ 0 & 0 & 2/3 \end {array}\right ).$ Thus, $\lim \limits _{\mu \rightarrow 0}\text {\textbf {J}} p^{(2)}_{\mu }(0,1,0)$ equals $\left (\begin {array}{ccc} 1/2&1/2&0\\ 1/2&1/2&0\\ 0&0&2/3 \end {array}\right ).$ It does not coincide with the Jacobian of ${\Pi }_{K_{2}}$ at (0,1,0). This shows that the limit $\lim \limits _{(t^{o},{z_{1}^{o}},{z_{2}^{o}},\mu )\rightarrow (0,1,0,0)}\textup {D} p^{(2)}_{\mu }(t^{o},z^{o})$ does not exist.