Calculus Rules of the Generalized Concave Kurdyka–Łojasiewicz Property

Abstract

In this paper, we propose several calculus rules for the generalized concave Kurdyka–Łojasiewicz (KL) property, which generalize Li and Pong’s results for KL exponents. The optimal concave desingularizing function has various forms and may be nondifferentiable. Our calculus rules do not assume desingularizing functions to have any specific form nor differentiable, while the known results do. Several examples are also given to show that our calculus rules are applicable to a broader class of functions than the known ones.

Appendix

1.1 A Supplementary Lemmas

Lemma A.1

Let $m\in {\mathbb {N}}$ obey $m\ge 2$. For each $i\in \{1,\ldots ,m\}$, Let $h_i:(0,\infty )\rightarrow {\mathbb {R}}_+$. Suppose that for each i the function $\varphi _i:[0,\infty )\rightarrow {\mathbb {R}}_+$ given by $\varphi _i(t)=\int _0^th_i(s)\textrm{d}s$ for $t\in (0,\infty )$ and $\varphi _i(0)=0$, is finite and right-continuous at 0. Then, the function $\varphi :[0,\infty )\rightarrow {\mathbb {R}}_+$,

$$\begin{aligned} (\forall 0<t<\eta )~t\mapsto \int _0^t\max _{1\le i\le m}h_i(s)\textrm{d}s, \end{aligned}$$

and $\varphi (0)=0$, is finite and right-continuous at 0.

Proof

Let $m=2$. Then, by using the inequality $\max \{\alpha ,\beta \}\le \alpha +\beta$ for $\alpha ,\beta \ge 0$, one has $\max \{h_1(s),h_2(s)\}\le h_1(s)+h_2(s)$. Note that $\varphi (t)=\lim _{u\rightarrow 0^+} \int _u^t\max \{ h_1(s),h_2(s)\}\textrm{d}s$ for $t>0$. Hence, we have for $t>0$

$$\begin{aligned} \varphi (t){} & {} \le \lim _{u\rightarrow 0^+}\int _u^th_1(s)+h_2(s)\textrm{d}s=\varphi _1(t)+\varphi _2(t)-\lim _{u\rightarrow 0^+}\left[ \varphi _1(u)+\varphi _2(u)\right] \\{} & {} =\varphi _1(t)+\varphi _2(t)<\infty , \end{aligned}$$

where the last equality is implied by the right-continuity of $\varphi _1$ and $\varphi _2$ at 0. Taking $t\rightarrow 0^+$, one gets $\lim _{t\rightarrow 0^+}\varphi (t)=0$. The desired result then follows from a simple induction. $\blacksquare$

Lemma A.2

Let $\eta \in (0,\infty ]$ and let $\varphi \in \Phi _\eta$ be concave. Then, the following hold:

• (i) Let $t>0$. Then $\varphi (t)=\lim _{u\rightarrow 0^+}\int _u^t\varphi _-^\prime (s)\textrm{d}s=\int _0^t\varphi _-^\prime (s)\textrm{d}s$.

• (ii) The function $t\mapsto \varphi _-^\prime (t)$ is decreasing and $\varphi _-^\prime (t)>0$ for $t\in (0,\eta )$. If in addition $\varphi (t)$ is strictly concave, then $t\mapsto \varphi _-^\prime (t)$ is strictly decreasing.

• (iii) For $0\le s<t<\eta$, $\varphi ^\prime _-(t)\le \frac{\varphi (t)-\varphi (s)}{t-s}$.

Proof

(i) Invoking Lemma 2.5(ii) yields

$$\begin{aligned} \varphi (t)=\lim _{u\rightarrow 0^+}\big (\varphi (t)-\varphi (u)\big )=\lim _{u\rightarrow 0^+}\int _{u}^t\varphi _-^\prime (s)\textrm{d}s<\infty , \end{aligned}$$

where the first equality holds because $\varphi$ is right-continuous at 0 with $\varphi (0)=0$. Let $(u_n)_{n\in {\mathbb {N}}}$ be a decreasing sequence with $u_1<t$ such that $u_n\rightarrow 0^+$ as $n\rightarrow \infty$. For each n, define $h_n:(0,t]\rightarrow {\mathbb {R}}_+$ by $h_n(s)=\varphi _-^\prime (s)$ if $s\in (u_n,t]$ and $h_n(s)=0$ otherwise. Then, the sequence $(h_n)_{n\in {\mathbb {N}}}$ satisfies: (a) $h_n\le h_{n+1}$ for every $n\in {\mathbb {N}}$; (b) $h_n(s)\rightarrow \varphi _-^\prime (s)$ pointwise on (0, t); (c) The integral $\int _0^th_n(s)\textrm{d}s=\int _{u_n}^t\varphi _-^\prime (s)\textrm{d}s=\varphi (t)-\varphi (u_n)\le \varphi (t)-\varphi (0)<\infty$ for every $n\in {\mathbb {N}}$. Hence, the monotone convergence theorem implies that

$$\begin{aligned} \lim _{u\rightarrow 0^+}\int _{u}^t\varphi _-^\prime (s)\textrm{d}s=\lim _{n\rightarrow \infty }\int _{u_n}^t\varphi _-^\prime (s)\textrm{d}s=\lim _{n\rightarrow \infty }\int _0^th_n(s)\textrm{d}s=\int _0^t\varphi _-^\prime (s)\textrm{d}s. \end{aligned}$$

(ii) According to Lemma 2.5 (i), the function $t\mapsto \varphi _-^\prime (t)$ is decreasing. Suppose that $\varphi _-^\prime (t_0)=0$ for some $t_0\in (0,\eta )$. Then, by the monotonicity of $\varphi _-^\prime$ and (i), we would have $\varphi (t)-\varphi (t_0)=\int _{t_0}^t\varphi _-^\prime (s)\textrm{d}s\le (t-t_0)\varphi _-^\prime (t_0)=0$ for $t>t_0$, which contradicts to the assumption that $\varphi$ is strictly increasing.

(iii) For $0<s<t<\eta$, applying Lemma 2.5 (iii) to the convex function $-\varphi$ yields that $-\varphi (s)+\varphi (t)\ge -\varphi _-^\prime (t)(s-t)\Leftrightarrow \varphi _-^\prime (t)\le \big (\varphi (t)-\varphi (s)\big )/(t-s)$. The desired inequality then follows from the right-continuity of $\varphi$ at 0. $\blacksquare$

Lemma A.3

Let $\eta \in (0,\infty ]$ and let $h:(0,\eta )\rightarrow {\mathbb {R}}_+$ be a positive-valued decreasing function. Define $\varphi (t)=\int _0^th(s)\textrm{d}s$ for $t\in (0,\eta )$ and set $\varphi (0)=0$. Suppose that $\varphi (t)<\infty$ for $t\in (0,\eta )$. Then, $\varphi$ is a strictly increasing concave function on $[0,\eta )$ with

$$\begin{aligned} \varphi _-^\prime (t)\ge h(t) \end{aligned}$$

for $t\in (0,\eta )$, and right-continuous at 0. If in addition h is a continuous function, then $\varphi$ is $C^1$ on $(0,\eta )$.

Proof Let $0<t_0<t_1<\eta$. Then $\varphi (t_1)-\varphi (t_0)=\int _{t_0}^{t_1}h(s)\textrm{d}s\ge (t_1-t_0)\cdot h(t_1)>0$, which means $\varphi$ is strictly increasing. Applying [18, Theorem 6.79], one concludes that $\varphi (t)\rightarrow \varphi (0)=0$ as $t\rightarrow 0^+$. $\blacksquare$

1.2 B Properties of the Generalized Concave KL Property and Its Associated Exact Modulus

In this subsection, we recall some pleasant properties of the generalized concave KL property and its associated exact modulus for the sake of completeness.

Proposition B.1

[19, Proposition 2] Let $f:{\mathbb {R}}^n\rightarrow {\overline{{\mathbb {R}}}}$ be proper lsc and let ${\bar{x}}\in {\text {dom}}\partial f$. Let U be a nonempty neighborhood of ${\bar{x}}$ and $\eta \in (0,\infty ]$. Let $\varphi \in \Phi _\eta$ and suppose that f has the generalized concave KL property at ${\bar{x}}$ with respect to U, $\eta$ and $\varphi$. Then, the exact modulus of the generalized concave KL property of f at ${\bar{x}}$ with respect to U and $\eta$, denoted by ${\tilde{\varphi }}$, is well-defined and satisfies

$$\begin{aligned} {\tilde{\varphi }}(t)\le \varphi (t),~\forall t\in [0,\eta ). \end{aligned}$$

Moreover, the function f has the generalized concave KL property at ${\bar{x}}$ with respect to U, $\eta$ and ${\tilde{\varphi }}$. Furthermore, the exact modulus ${\tilde{\varphi }}$ satisfies

$$\begin{aligned}{} & {} {\tilde{\varphi }}=\inf \\{} & {} \quad \big \{\varphi \in \Phi _\eta :\varphi \,\,\text {is a concave desingularizing function of} f \text {at} \,\,{\bar{x}} \text { with respect to}\,\, U \text { and } \eta \big \}. \end{aligned}$$

Example B.2

[19, Example 1] Let $\rho >0$. Consider the function given by

$$\begin{aligned} f(x)= {\left\{ \begin{array}{ll} 2\rho |x|-3\rho ^2/2,&{}\text {if }|x|>\rho ;\\ |x|^2/2, &{}\text {if }|x|\le \rho . \end{array}\right. } \end{aligned}$$

Then, the function

$$\begin{aligned} {\tilde{\varphi }}(t)={\left\{ \begin{array}{ll}\sqrt{2t},&{}\text {if}0\le t\le \rho ^2/2;\\ t/(2\rho )+3\rho /4, &{}\text {if}t>\rho ^2/2,\end{array}\right. } \end{aligned}$$

is the exact modulus of the generalized concave KL property of f at ${\bar{x}}=0$ with respect to $U={\mathbb {R}}$ and $\eta =\infty$.

The next proposition restates an example from [5, Section 1] in our extended framework. We include a detailed proof for the sake of completeness.

Proposition B.3

Define $f:{\mathbb {R}}\rightarrow {\mathbb {R}}$ by $f(x)=e^{-1/x^2}$ for $x\ne 0$ and $f(0)=0$. Then, the following hold:

• (i) The function ${\tilde{\varphi }} (t)=\sqrt{-1/\ln (t)}$ for $t>0$ and ${\tilde{\varphi }} (0)=0$ is the exact modulus of generalized KL property of f at ${\bar{x}}=0$ with respect to $U=(-\sqrt{2/3},\sqrt{2/3})$ and $\eta =\exp (-3/2)$.

• (ii) For every $c>0$ and $\theta \in [0,1)$, the function $\varphi (t)=c\cdot t^{1-\theta }$ cannot be a desingularizing function of the generalized concave KL property of f at 0 with respect to any neighborhood $U^\prime \ni 0$ and $\eta ^\prime \in (0,\infty ]$.

Proof

• (i) For $0<s\le \exp (-3/2)$, $s\le \exp (-1/x^2)\Leftrightarrow |x|\ge \sqrt{-1/\ln {s}}$. Thus

  $$\begin{aligned} h(s)&=\sup \{|f^\prime (x)|^{-1}: x\in U\cap [0<f-f({\bar{x}})<\eta ], s\le f(x)-f({\bar{x}})\}\\&=\sup \{|2x^{-3}\exp (-1/x^2)|^{-1}: \sqrt{-1/\ln {s}}\le |x|<2/3\}=(-\ln (s))^{-3/2}/(2s). \end{aligned}$$

  Hence, ${\tilde{\varphi }} (t)=\sqrt{-1/\ln (t)}$ for $t>0$.

• (ii) Suppose to the contrary that there were $c>0$ and $\theta \in [0,1)$ such that f has the generalized concave KL property at 0 with respect to some $U^\prime \ni 0$ and $\eta ^\prime >0$ and $\varphi (t)$. Taking the intersection if necessary, assume without loss of generality that $U^\prime \cap [0<f<\eta ^\prime ]\subseteq (-\sqrt{2/3},\sqrt{2/3})\cap [0<f<e^{-3/2}]$. Then, it is easy to see that f is convex and $C^1$ on $U^\prime \cap [0<f<\eta ^\prime ]$ with ${\tilde{\varphi }}(t)$ being the associated exact modulus. Hence, Proposition B.1 implies that

  $$\begin{aligned} {\tilde{\varphi }}(t)\le \varphi (t)=c\cdot t^{1-\theta },~\forall t\in (0,\min \{\eta ^\prime ,e^{-3/2}\}). \end{aligned}$$

  (20)

  Let $s>0$. Then, $s={\tilde{\varphi }}(t)\Leftrightarrow t=e^{-1/s^2}$, which further implies that

  $$\begin{aligned} \limsup _{t\rightarrow 0^+}\frac{{\tilde{\varphi }}(t)}{t^{1-\theta }}=\limsup _{s\rightarrow 0^+}\frac{s}{e^{-(1-\theta )/s^2}}=\limsup _{s\rightarrow 0^+}\frac{e^{(1-\theta )/s^2}}{s^{-1}}=\infty , \end{aligned}$$

  which contradicts to (20). $\blacksquare$