Amenable cones: error bounds without constraint qualifications

Abstract

We provide a framework for obtaining error bounds for linear conic problems without assuming constraint qualifications or regularity conditions. The key aspects of our approach are the notions of amenable cones and facial residual functions. For amenable cones, it is shown that error bounds can be expressed as a composition of facial residual functions. The number of compositions is related to the facial reduction technique and the singularity degree of the problem. In particular, we show that symmetric cones are amenable and compute facial residual functions. From that, we are able to furnish a new Hölderian error bound, thus extending and shedding new light on an earlier result by Sturm on semidefinite matrices. We also provide error bounds for the intersection of amenable cones, this will be used to prove error bounds for the doubly nonnegative cone. At the end, we list some open problems.

A Miscellaneous proofs

1.1 Proof of Proposition 11

(i):

Let ${\mathcal {F}}$ be a face of ${{\mathcal {K}}}^1\times {{\mathcal {K}}}^2$. We have ${\mathcal {F}}= {\mathcal {F}}^1 \times {\mathcal {F}}^2$, where ${\mathcal {F}}^1$ and ${\mathcal {F}}^2$ are faces of ${{\mathcal {K}}}^1$ and ${{\mathcal {K}}}^2$ respectively. From our assumptions in Sect. 2, Eq. (2) and the amenability of ${{\mathcal {K}}}^1$ and ${{\mathcal {K}}}^2$, it follows that there are positive constants $\kappa _1, \kappa _2$ such that

$$\begin{aligned} {\mathrm {dist}\,}((x_1,x_2), {\mathcal {F}}) \le \sqrt{\kappa _1^2 {\mathrm {dist}\,}(x_1, {{\mathcal {K}}}^1)^2 + \kappa _2^2 {\mathrm {dist}\,}(x_2, {{\mathcal {K}}}^2)^2}, \end{aligned}$$

whenever $x_1 \in \mathrm {span}\, {\mathcal {F}}^1$ and $x_2 \in \mathrm {span}\, {\mathcal {F}}^2$. Therefore,

$$\begin{aligned} {\mathrm {dist}\,}((x_1,x_2), {\mathcal {F}})&\le \max \{\kappa _1,\kappa _2\}\sqrt{ {\mathrm {dist}\,}(x_1, {{\mathcal {K}}}^1)^2 + {\mathrm {dist}\,}(x_2, {{\mathcal {K}}}^2)^2}\\&= \max \{\kappa _1,\kappa _2\} {\mathrm {dist}\,}((x_1,x_2), {{\mathcal {K}}}^1\times {{\mathcal {K}}}^2), \end{aligned}$$

whenever $(x_1,x_2) \in \mathrm {span}\,( {\mathcal {F}}^1\times {\mathcal {F}}^2) = (\mathrm {span}\, {\mathcal {F}}^1) \times (\mathrm {span}\, {\mathcal {F}}^2)$.

(ii):

If ${{\mathcal {A}}}$ is the zero map, we are done, since $\{0\}$ is amenable. So, suppose that ${{\mathcal {A}}}$ is a nonzero injective linear map. Then, the faces of ${{\mathcal {A}}}( {{\mathcal {K}}})$ are images of faces of ${{\mathcal {K}}}$ by ${{\mathcal {A}}}$. Accordingly, let ${\mathcal {F}}\mathrel {\unlhd } {{\mathcal {K}}}$. Because ${{\mathcal {K}}}$ is amenable, there is $\kappa$ such that

$$\begin{aligned} {\mathrm {dist}\,}(x, {\mathcal {F}}) \le \kappa {\mathrm {dist}\,}(x, {{\mathcal {K}}}), \quad \forall x \in \mathrm {span}\, {\mathcal {F}}. \end{aligned}$$

(41)

As ${{\mathcal {A}}}$ is a linear map, we have $\mathrm {span}\, {{\mathcal {A}}}( {\mathcal {F}}) = {{\mathcal {A}}}(\mathrm {span}\, {\mathcal {F}})$. Let $\sigma _{\min }, \sigma _{\max }$ denote, respectively, the minimum and maximum singular values of ${{\mathcal {A}}}$. We have

$$\begin{aligned} \sigma _{\min } = \min \{\Vert Ax\Vert \mid \Vert x\Vert = 1 \}, \quad \sigma _{\max } = \max \{\Vert Ax\Vert \mid \Vert x\Vert = 1 \}. \end{aligned}$$

They are both positive since ${{\mathcal {A}}}$ is injective and nonzero. Now, let $x \in \mathrm {span}\, {\mathcal {F}}$, then

$$\begin{aligned} {\mathrm {dist}\,}( {{\mathcal {A}}}(x), {{\mathcal {A}}}( {\mathcal {F}}))&\le \sigma _{\max } {\mathrm {dist}\,}(x, {\mathcal {F}})&\\&\le {\kappa }{\sigma _{\max }} {\mathrm {dist}\,}(x, {{\mathcal {K}}})&\text {(From } (41))\\&\le \frac{{\kappa }{\sigma _{\max }}}{\sigma _{\min }} {\mathrm {dist}\,}( {{\mathcal {A}}}(x), {{\mathcal {A}}}( {{\mathcal {K}}})).&\end{aligned}$$

$\square$

1.1.1 Proof of Proposition 12

Proof

$(i) \Rightarrow (ii)$

Let $x,u \in {\mathcal {E}}$ be such that $x+u\in \mathrm {span}\, {\mathcal {F}}$ and $\Vert u\Vert = {\mathrm {dist}\,}(x,\mathrm {span}\, {\mathcal {F}})$. Since ${\mathrm {dist}\,}(\cdot , {{\mathcal {K}}})$ and ${\mathrm {dist}\,}(\cdot , {\mathcal {F}})$ are sublinear functions, we have that (4) implies that

$$\begin{aligned} {\mathrm {dist}\,}(x, {\mathcal {F}})&\le {\mathrm {dist}\,}(-u, {\mathcal {F}}) + {\mathrm {dist}\,}(x+u, {\mathcal {F}}) \nonumber \\&\le {\mathrm {dist}\,}(x,\mathrm {span}\, {\mathcal {F}}) + \kappa ( {\mathrm {dist}\,}(x+u, {{\mathcal {K}}})) \nonumber \\&\le {\mathrm {dist}\,}(x,\mathrm {span}\, {\mathcal {F}}) + \kappa ( {\mathrm {dist}\,}(x, {{\mathcal {K}}})+ {\mathrm {dist}\,}(x,\mathrm {span}\, {\mathcal {F}}))\nonumber \\&\le (1+\kappa )( {\mathrm {dist}\,}(x, {{\mathcal {K}}})+ {\mathrm {dist}\,}(x,\mathrm {span}\, {\mathcal {F}})), \quad \forall x \in {\mathcal {E}}. \end{aligned}$$

(42)

Here we used the fact that ${\mathrm {dist}\,}(-u, {\mathcal {F}}) \le \Vert -u\Vert$, since $0 \in {\mathcal {F}}$. This shows that $(i) \Rightarrow (ii)$.

$(ii) \Rightarrow (i)$ Since ${{\mathcal {K}}}$ and $\mathrm {span}\, {\mathcal {F}}$ intersect at 0 subtransversally, there is $\delta > 0$ such that

$$\begin{aligned} {\mathrm {dist}\,}(z, {\mathcal {F}}) \le \kappa ( {\mathrm {dist}\,}(z, {{\mathcal {K}}})+ {\mathrm {dist}\,}(z,\mathrm {span}\, {\mathcal {F}})), \quad \forall z \text { with } \Vert z\Vert \le \delta . \end{aligned}$$

Therefore, if $x \in {\mathcal {E}}$ is nonzero, we have

$$\begin{aligned} {\mathrm {dist}\,}\left( \delta \frac{x}{\Vert x\Vert }, {\mathcal {F}}\right) \le \kappa \left( {\mathrm {dist}\,}(\delta \frac{x}{\Vert x\Vert }, {{\mathcal {K}}}\right) + {\mathrm {dist}\,}\left( \delta \frac{x}{\Vert x\Vert },\mathrm {span}\, {\mathcal {F}})\right) . \end{aligned}$$

Now, we recall that if C is a convex cone, then ${\mathrm {dist}\,}(\alpha x, C) = \alpha {\mathrm {dist}\,}(x,C)$ for every positive $\alpha$. We conclude that

$$\begin{aligned} {\mathrm {dist}\,}(x, {\mathcal {F}}) \le \kappa ( {\mathrm {dist}\,}(x, {{\mathcal {K}}})+ {\mathrm {dist}\,}(x,\mathrm {span}\, {\mathcal {F}})), \quad \forall x \in {\mathcal {E}}. \end{aligned}$$

Therefore, if $x \in \mathrm {span}\, {\mathcal {F}}$, then ${\mathrm {dist}\,}(x, {\mathcal {F}}) \le \kappa {\mathrm {dist}\,}(x, {{\mathcal {K}}})$.

$(i) \Rightarrow (iii)$ The inequality in (42) shows that

$$\begin{aligned} {\mathrm {dist}\,}(x, {\mathcal {F}}) \le (2+2\kappa )\max ( {\mathrm {dist}\,}(x, {{\mathcal {K}}}), {\mathrm {dist}\,}(x,\mathrm {span}\, {\mathcal {F}})), \quad \forall x \in {\mathcal {E}}. \end{aligned}$$

Therefore, ${{\mathcal {K}}}$ and $\mathrm {span}\, {\mathcal {F}}$ are boundedly linearly regular.

$(iii) \Rightarrow (ii)$ Let $U = \{x \in {\mathcal {E}}\mid \Vert x\Vert \le 1 \}$. Then, there exists $\kappa _U$ such that

$$\begin{aligned} {\mathrm {dist}\,}(x, {\mathcal {F}})&\le \kappa _U\max ( {\mathrm {dist}\,}(x, {{\mathcal {K}}}), {\mathrm {dist}\,}(x,\mathrm {span}\, {\mathcal {F}})) \\&\le \kappa _U ( {\mathrm {dist}\,}(x, {{\mathcal {K}}})+ {\mathrm {dist}\,}(x,\mathrm {span}\, {\mathcal {F}})), \quad \forall x \in U. \end{aligned}$$

Therefore, ${{\mathcal {K}}}$ and $\mathrm {span}\, {\mathcal {F}}$ intersect subtransversally at 0. $\square$

1.1.2 Proof of Proposition 17

• 1. Suppose that $x \in \mathrm {span}\, {{\mathcal {K}}}$ satisfies the inequalities

  $$\begin{aligned} {\mathrm {dist}\,}(x, {{\mathcal {K}}}) \le \epsilon , \quad \langle x , z \rangle \le \epsilon , \quad {\mathrm {dist}\,}(x, \mathrm {span}\, {\mathcal {F}}) \le \epsilon . \end{aligned}$$

  (43)

  Note that

  $$\begin{aligned} {\mathcal {F}}\cap \{z\}^{\perp } = ( {\mathcal {F}}^{1} \cap \{z_1\}^\perp ) \times ( {\mathcal {F}}^{2} \cap \{z_2\}^\perp ). \end{aligned}$$

  Also, due to our assumptions (Sect. 2.1), we have

  $$\begin{aligned} \Vert x-y\Vert ^2 = \Vert x_1-y_1\Vert ^2 + \Vert x_2-y_2\Vert ^2 \end{aligned}$$

  for every $x,y \in {\mathcal {E}}^1\times {\mathcal {E}}^2$. Thus we have the following implications:

  $$\begin{aligned} {\mathrm {dist}\,}(x, {{\mathcal {K}}}) \le \epsilon \quad&\Rightarrow \quad {\mathrm {dist}\,}(x_1, {{\mathcal {K}}}^1) \le \epsilon , \quad {\mathrm {dist}\,}(x_2, {{\mathcal {K}}}^2) \le \epsilon \end{aligned}$$

  (44)
  $$\begin{aligned} {\mathrm {dist}\,}(x,\mathrm {span}\, {\mathcal {F}}) \le \epsilon \quad&\Rightarrow \quad {\mathrm {dist}\,}(x_1,\mathrm {span}\, {\mathcal {F}}^1) \le \epsilon , \quad {\mathrm {dist}\,}(x_2,\mathrm {span}\, {\mathcal {F}}^2) \le \epsilon \end{aligned}$$

  (45)

  The first step is showing that there are positive constants $\kappa _1$ and $\kappa _2$ such that for all $x \in {\mathcal {E}}^1\times {\mathcal {E}}^2$, we also have

  $$\begin{aligned} x \text { satisfies } (43)&\Rightarrow \quad \langle x_1 , z_1 \rangle \le \kappa _1\epsilon \quad \text {and} \quad \langle x_2 , z_2 \rangle \le \kappa _2\epsilon . \end{aligned}$$

  (46)

  Suppose x satisfies (43). By (45), we have ${\mathrm {dist}\,}(x_1, \mathrm {span}\,{ {\mathcal {F}}^1}) \le \epsilon$. Therefore, there exists $y_1 \in {\mathcal {E}}^1$ such that $x_1 + y_1 \in \mathrm {span}\,{ {\mathcal {F}}^1}$ and $\Vert y_1\Vert \le \epsilon$. Due to (44) and the amenability of ${{\mathcal {K}}}^1$, there exists ${\hat{\kappa }}_1$ (not depending on $x_1$) such that

  $$\begin{aligned} {\mathrm {dist}\,}(x_1 + y_1, {\mathcal {F}}^1) \le {\hat{\kappa }}_1 {\mathrm {dist}\,}(x_1 + y_1, {{\mathcal {K}}}^1) \le 2\epsilon {\hat{\kappa }}_1. \end{aligned}$$

  Therefore, there exists $v_1 \in {\mathcal {E}}^1$ such that $\Vert v_1\Vert \le 2\epsilon {\hat{\kappa }}_1$ and

  $$\begin{aligned} x_1 + y_1 + v_1 \in {\mathcal {F}}^1. \end{aligned}$$

  In a completely analogous manner, there is a constant ${\hat{\kappa }}_2> 0$ and there are $y_2,v_2 \in {\mathcal {E}}^2$ such

  $$\begin{aligned} x_2 + y_2 + v_2 \in {\mathcal {F}}^2, \end{aligned}$$

  with $\Vert y_2\Vert \le \epsilon$ and $\Vert v_2\Vert \le 2\epsilon {\hat{\kappa }}_2$. It follows that

  $$\begin{aligned} \langle (x_1+y_1+v_1,x_2+y_2+v_2) , (z_1,z_2) \rangle \le M\epsilon , \end{aligned}$$

  for $M = 1 + \Vert z_1\Vert + 2{\hat{\kappa }}_1 + \Vert z_2\Vert + 2{\hat{\kappa }}_2$. Since $\langle x_1+y_1+v_1 , z_1 \rangle \ge 0$ and $\langle x_2+y_2+v_2 , z_2 \rangle \ge 0$, we get

  $$\begin{aligned} \langle x_i+y_i+v_i , z_i \rangle \le M\epsilon , \end{aligned}$$

  for $i = 1,2$. We then conclude that

  $$\begin{aligned} \langle x , z \rangle \le \epsilon \quad \Rightarrow \quad \langle x_i , z_i \rangle \le \kappa _i\epsilon , \end{aligned}$$

  (47)

  whenever x satisfies (44) and (45), where $\kappa _i = M+ \Vert z_i\Vert + \Vert z_i\Vert 2{\hat{\kappa }}_i$. Now, let $\psi _{ {\mathcal {F}}_1,z_1}$ and $\psi _{ {\mathcal {F}}_2,z_2}$ be arbitrary facial residual functions for ${\mathcal {F}}_1,z_1$ and ${\mathcal {F}}_2,z_2$, respectively. We positive rescale $\psi _{ {\mathcal {F}}_1,z_1}$ and $\psi _{ {\mathcal {F}}_2,z_2}$ so that

  $$\begin{aligned} {\mathrm {dist}\,}(x_i, {{\mathcal {K}}}) \le \epsilon ,\quad \langle x_i , z_i \rangle \le \kappa _i\epsilon , \quad {\mathrm {dist}\,}(x, \mathrm {span}\, {\mathcal {F}}_i ) \le \epsilon \end{aligned}$$

  implies ${\mathrm {dist}\,}(x_i,{\hat{ {\mathcal {F}}}}_i) \le \psi _{ {\mathcal {F}}_i,z_i}(\epsilon ,\Vert x_i\Vert )$, for $i = 1,2$.

  Finally, from (44), (45), (47) and using the fact that $\psi _{ {\mathcal {F}}_1,z_1}$ and $\psi _{ {\mathcal {F}}_2,z_2}$ are monotone nondecreasing on the second argument we conclude that whenever x satisfies (43) we have

  $$\begin{aligned} {\mathrm {dist}\,}(x,{\hat{ {\mathcal {F}}}})&= \sqrt{ {\mathrm {dist}\,}(x_1,{\hat{ {\mathcal {F}}}}^1)^2 + {\mathrm {dist}\,}(x_2,{\hat{ {\mathcal {F}}}}^2)^2 }\\&\le { {\mathrm {dist}\,}(x_1,{\hat{ {\mathcal {F}}}}^1)} + { {\mathrm {dist}\,}(x_2,{\hat{ {\mathcal {F}}}}^2)} \\&\le \psi _{ {\mathcal {F}}_1,z_1}(\epsilon ,\Vert x\Vert ) + \psi _{ {\mathcal {F}}_2,z_2}(\epsilon ,\Vert x\Vert ). \end{aligned}$$

  Therefore, $\psi _{ {\mathcal {F}}_1,z_1}+\psi _{ {\mathcal {F}}_2,z_2}$ is a facial residual function for ${\mathcal {F}},z$.

• 2. The proposition is true if ${{\mathcal {A}}}$ is the zero map, so suppose that ${{\mathcal {A}}}$ is a nonzero injective linear map. First, we observe that

  $$\begin{aligned} ( {{\mathcal {A}}}( {\mathcal {F}}))\cap \{z\}^\perp = {{\mathcal {A}}}( {\mathcal {F}}\cap \{ {{\mathcal {A}}}^\top z \}^\perp ). \end{aligned}$$

  Let ${\hat{ {\mathcal {F}}}} = {\mathcal {F}}\cap \{ {{\mathcal {A}}}^\top z \}^\perp$. Let $\psi _{ {\mathcal {F}}, {{\mathcal {A}}}^\top z}$ be a facial residual function for ${\mathcal {F}}$ and ${{\mathcal {A}}}^\top z$. Let $\sigma _{\min }$ denote the minimum singular value of ${{\mathcal {A}}}$. We note that $\sigma _{\min }$ is positive because ${{\mathcal {A}}}$ is injective. We positive rescale $\psi _{ {\mathcal {F}}, {{\mathcal {A}}}^\top z}$ so that whenever x satisfies

  $$\begin{aligned} {\mathrm {dist}\,}(x, {{\mathcal {K}}}) \le \frac{1}{\sigma _{\min }}\epsilon , \quad \langle x , {{\mathcal {A}}}^\top z \rangle \le \epsilon , \quad {\mathrm {dist}\,}(x, \mathrm {span}\, {\mathcal {F}}) \le \frac{1}{\sigma _{\min }}\epsilon \end{aligned}$$

  we have:

  $$\begin{aligned} {\mathrm {dist}\,}(x, {\hat{ {\mathcal {F}}}}) \le \psi _{ {\mathcal {F}}, {{\mathcal {A}}}^\top z} (\epsilon , \Vert x\Vert ). \end{aligned}$$

  Then, we have the following implications:

  $$\begin{aligned} {\mathrm {dist}\,}( {{\mathcal {A}}}(x), {{\mathcal {A}}}( {{\mathcal {K}}})) \le \epsilon \quad&\Rightarrow \quad {\mathrm {dist}\,}(x, {{\mathcal {K}}}) \le \frac{1}{\sigma _{\min }}\epsilon \\ \langle {{\mathcal {A}}}(x) , z \rangle \le \epsilon \quad&\Leftrightarrow \quad \langle x , {{\mathcal {A}}}^\top z \rangle \le \epsilon \\ {\mathrm {dist}\,}( {{\mathcal {A}}}(x),\mathrm {span}\, {{\mathcal {A}}}( {\mathcal {F}})) \le \epsilon \quad&\Rightarrow \quad { {\mathrm {dist}\,}(x,\mathrm {span}\, {\mathcal {F}}) \le \frac{1}{\sigma _{\min }}\epsilon }\\ {\mathrm {dist}\,}( {{\mathcal {A}}}(x), {{\mathcal {A}}}({\hat{ {\mathcal {F}}}})) \le \sigma _{\max }\psi _{ {\mathcal {F}}, {{\mathcal {A}}}^\top z} (\epsilon , \Vert {{\mathcal {A}}}x\Vert /\sigma _{\min }) \quad&\Leftarrow \quad {\mathrm {dist}\,}(x, {\hat{ {\mathcal {F}}}}) \le \psi _{ {\mathcal {F}}, {{\mathcal {A}}}^\top z} (\epsilon , \Vert x\Vert ), \end{aligned}$$

  where $\sigma _{\max }$ is the maximum singular value of ${{\mathcal {A}}}$. This shows that we can use

  $$\begin{aligned} {\tilde{\psi }} _{ {{\mathcal {A}}}( {\mathcal {F}}),z}(\epsilon , \Vert {{\mathcal {A}}}x\Vert ) = \sigma _{\max }\psi (\epsilon , \Vert {{\mathcal {A}}}x\Vert /\sigma _{\min }) \end{aligned}$$

  as a facial residual function for ${{\mathcal {A}}}( {\mathcal {F}})$ and z. $\square$