A Nonconvex Nonsmooth Image Prior Based on the Hyperbolic Tangent Function

Abstract

In this paper, we propose a nonconvex and nonsmooth image prior based on the hyperbolic tangent function and apply it as a regularization term for image restoration and reconstruction problems. Theoretically, we analyze the properties of the function and the minimizers of its associated proximal problem. Since the proximal problem has no closed-form solution, we propose a derivative-free Nelder-Mead simplex based selection algorithm to find the global minimizer. To reduce the computational cost, we only solve a small 1D problem and then use the 1D solution template as a look-up table to interpolate high-dimension data. Moreover, we consider a variational model based on the proposed image prior. Then we use the alternating direction method of multipliers algorithm on the nonconvex model to derive efficient numerical algorithms. Various experiments on image denoising, image deblurring, and image reconstruction demonstrate that the proposed nonconvex prior is competitive with the existing priors. In particular, it outperforms others in recovering piecewise constant images.

Appendices

Appendices

A. proof of Theorem 1

Proof

 

• (a) For $z=0$, the conclusion holds since $x_{opt}^{(z=0,\lambda ,a)} =0.$ For $z\ne 0$, without loss of generality, we only consider the case of $z>0$. If $x>z>0$, since $\phi '(x;a)>0$, we have

  $$\begin{aligned} H'(x;z,\lambda ,a) =\lambda \phi '(x;a)+x-z>0. \end{aligned}$$

  (27)

  Then $H(x;z,\lambda ,a)$ is monotonically increasing for $x\in [z,+\infty ]$. If $x<0$, we have $\phi '(x;a)<0$ and $H'(x;z,\lambda ,a)<0$. Then $H(x;z,\lambda ,a)$ is monotonically decreasing for $x\in [-\infty ,0]$. Note that $H(x;z,\lambda ,a)$ is a continuous function when $x\in [0,z]$. Therefore, the global minimizer of $H(x;z,\lambda ,a)$ must be in [0, z].

• (b) By definition of H, we get $H(x;z,\lambda ,a)=H(-x;-z,\lambda ,a)$. Since $x_{opt}^{(z,\lambda ,a)}$ is the minimizer of $H(x;z,\lambda ,a)$, we have

  $$\begin{aligned} H(x_{opt}^{(z,\lambda ,a)};z,\lambda ,a)\le H(x;z,\lambda ,a), \forall x\in R. \end{aligned}$$

  Equivalently,

  $$\begin{aligned} H(-x_{opt}^{(z,\lambda ,a)};-z,\lambda ,a)\le H(-x;-z,\lambda ,a), \forall -x\in R. \end{aligned}$$

  Hence

  $$\begin{aligned} H(-x_{opt}^{(z,\lambda ,a)};-z,\lambda ,a)\le H(x;-z,\lambda ,a), \forall x\in R, \end{aligned}$$

  i.e., $x_{opt}^{(-z,\lambda ,a)} = -x_{opt}^{(z,\lambda ,a)}$.

• (c) Denote

  $$\begin{aligned} x_i\triangleq x_{opt}^{(z_i,\lambda ,a)} \in \arg \min _x H(x;z_i,\lambda ,a), i = 1,2. \end{aligned}$$

  Then we get

  $$\begin{aligned} \frac{1}{2}(x_1-z_1)^2+\lambda \phi (x_1;a)\le & {} \frac{1}{2}(x_2-z_1)^2+\lambda \phi (x_2;a),\\ \frac{1}{2}(x_2-z_2)^2+\lambda \phi (x_2;a)\le & {} \frac{1}{2}(x_1-z_2)^2+\lambda \phi (x_1;a). \end{aligned}$$

Adding the two inequalities together yields

$$\begin{aligned}{} & {} \frac{1}{2}(x_1-z_1)^2+\lambda \phi (x_1;a)+\frac{1}{2}(x_2-z_2)^2+\lambda \phi (x_2;a)\\{} & {} \quad \le \frac{1}{2}(x_2-z_1)^2+\lambda \phi (x_2;a)+ \frac{1}{2}(x_1-z_2)^2+\lambda \phi (x_1;a). \end{aligned}$$

It is equivalent to

$$\begin{aligned} \frac{1}{2}(x_1-z_1)^2+\frac{1}{2}(x_2-z_2)^2 \le \frac{1}{2}(x_2-z_1)^2+ \frac{1}{2}(x_1-z_2)^2. \end{aligned}$$

(28)

Equation (28) can be simplified as

$$\begin{aligned} (x_2-x_1)(z_2-z_1)\ge 0. \end{aligned}$$

Since $z_1<z_2$, we get $x_1\le x_2$, that means, $x_{opt}^{(z_1,\lambda ,a)}\le x_{opt}^{(z_2,\lambda ,a)}$. $\square$

B. proof of Theorem 2

Proof

By Eqs. (7) and (10), we can rewrite $H''(x;z,\lambda ,a)$ as a function of $\theta =\tanh \left( \frac{|x|}{a}\right)$:

$$\begin{aligned} F(\theta ) = \frac{2\lambda }{a^2}\theta (\theta ^2-1)+1. \end{aligned}$$

(29)

Note that when $x\in (0,+\infty )$, $\theta (x)\in (0,1)$ is monotonically increasing. Since

$$\begin{aligned} F'(\theta ) = \frac{2\lambda }{a^2}(3\theta ^2-1), \end{aligned}$$

(30)

we have $F'<0$ for $\theta \in (0,\frac{1}{\sqrt{3}})$, while $F'>0$ for $\theta \in (\frac{1}{\sqrt{3}},+\infty )$. It implies that F is monotonically decreasing for $\theta \in (0,\frac{1}{\sqrt{3}})$ and monotonically increasing for $\theta \in (\frac{1}{\sqrt{3}},+\infty )$. Hence F attains its minima at $\theta = \frac{1}{\sqrt{3}}$ with function value

$$\begin{aligned} F(\theta )|_{\theta =\frac{1}{\sqrt{3}}}=1-\frac{4\lambda }{3\sqrt{3}a^2}. \end{aligned}$$

(31)

• (a) If $\lambda <\frac{3\sqrt{3}a^2}{4}$, by Eq. (31), we have $H''(x;z,\lambda ,a)>0$ when $x>0$. Since $H(x;z,\lambda ,a)$ is continuous at $x=0$ and $H(x;z,\lambda ,a)$ is strictly convex in $(0,+\infty )$, together with Theorem 1(a), we deduce that $H(x;z,\lambda ,a)$ has a unique (global) minimizer in [0, z]. Since $H'(x=z;z,\lambda ,a)=\lambda \phi '(z;\lambda ,a)>0$, H is monotonically increasing in a local neighbourhood of $x=z$. Therefore the global minimizer of H is in [0, z).

• (b) If $\lambda =\frac{3\sqrt{3}a^2}{4}$, by Eq. (31), $H''(x;z,\lambda ,a)\ge 0$ and

  $$\begin{aligned} H''(x;z,\lambda ,a)=0\ \text{ iff } \ x=a\pm \text{ atanh }\left( \frac{1}{\sqrt{3}}\right) , \end{aligned}$$

  where $\text{ atanh }$ is the inverse hyperbolic tangent function. It implies that $H'(x;z,\lambda ,a)$ is strictly increasing in $(0,+\infty )$. Note that $H'(x\rightarrow 0^+;z,\lambda ,a)=\frac{\lambda }{a}-z$ and $H'(x=z;z,\lambda ,a)>0$. If $\frac{\lambda }{a}-z\ge 0$, then H is monotonically increasing and such that the global minimizer is $x=0$. Otherwise, $H'(x\rightarrow 0^+;z,\lambda ,a)<0$. By the zero point theorem of continuous function H, we get that $H'(x;z,\lambda ,a)=0$ has a unique solution in (0, z), which is the unique (global) minimizer.

• (c) Assume $\lambda >\frac{3\sqrt{3}a^2}{4}$. By Eq. (29), the equation

  $$\begin{aligned} F(\theta ) = 0 \end{aligned}$$

  (32)

is equivalent to the standard cubic equation

$$\begin{aligned} \theta ^3+p\theta +q=0 \end{aligned}$$

(33)

where $p=-1, q=\frac{a^2}{2\lambda }$. This equation can be solved by using Cardano’s formula. Denote the discriminant as

$$\begin{aligned} \Delta =\left( \frac{q}{2}\right) ^{2}+\left( \frac{p}{3}\right) ^{3}=\frac{a^4}{16\lambda ^2}-\frac{1}{27}. \end{aligned}$$

Since $\lambda >\frac{3\sqrt{3}a^2}{4}$, we have $\Delta <0$ such that the equation Eq. (33) has three real roots:

$$\begin{aligned} \theta _{1}= & {} 2 \root 3 \of {r} \cos \left( \alpha +240^{\circ }\right) \\ \theta _{2}= & {} 2 \root 3 \of {r} \cos \alpha \\ \theta _{3}= & {} 2 \root 3 \of {r} \cos \left( \alpha +120^{\circ }\right) \end{aligned}$$

where

$$\begin{aligned} r= & {} \sqrt{-\left( \frac{p}{3}\right) ^{3}}=\sqrt{\frac{1}{27}},\\ \alpha= & {} \frac{1}{3} \arccos \left( -\frac{q}{2 r}\right) =\frac{1}{3}\arccos \left( -\frac{3\sqrt{3}a^2}{4\lambda }\right) . \end{aligned}$$

Since $\alpha _0 = \arccos \left( -\frac{3\sqrt{3}a^2}{4\lambda }\right) \in [\frac{\pi }{2},\pi ]$, then $\alpha =\frac{1}{3}\alpha _0 \in [\frac{\pi }{6},\frac{\pi }{3}]$. It is straightforward to verify that $\theta _1>0, \theta _2>0, \theta _3<0$ and $\theta _1<\theta _2$. By definition, $\theta = \tanh \left( \frac{|x|}{a}\right) >0$. So we need only consider the roots $\theta _i,i=1,2$. Moreover, since

$$\begin{aligned}{} & {} F\left( \theta \rightarrow 0^+\right) =1>0,\\{} & {} F\left( \theta =\frac{1}{\sqrt{3}}\right) =1-\frac{4\lambda }{3\sqrt{3}a^2}<0,\\{} & {} F\left( \theta \rightarrow 1\right) = 1>0, \end{aligned}$$

we deduce that $\theta _1\in \left( 0,\frac{1}{\sqrt{3}}\right)$ and $\theta _2\in \left( \frac{1}{\sqrt{3}},1\right)$. Using the monotonicity of $\theta (x)$, we get that $H''(x;z,\lambda ,a)=0$ has two different solutions

$$\begin{aligned} {\tilde{x}}_i=\pm a \text{ atanh }(\theta _i), i=1,2. \end{aligned}$$

(34)

which are inflection points of $H(x;z,\lambda ,a)$.

Note that finding the zero points of $H'$ can be regarded as finding the intersection points of two curves:

$$\begin{aligned} y_1(x) = \lambda \phi '(x;a)+x,\ y_2(x)=z. \end{aligned}$$

It is obvious that when $x>0$,

$$\begin{aligned} y_1(x)\ge x, \lim \limits _{x\rightarrow 0^+} y_1(x)=\frac{\lambda }{a}, \lim \limits _{x\rightarrow +\infty } \frac{ y_1(x)}{x}=1. \end{aligned}$$

Hence the curve $y_1(x)$ is above $y=x$, and its asymptote is $y=x$.

Assume $y_1(x)$ and $y_2(x)$ are tangent, then the tangent point satisfies the following two equations:

$$\begin{aligned} y_1(x) = y_2(x), \ y_1'(x) = y_2'(x), \end{aligned}$$

(35)

which are equivalent to

$$\begin{aligned} H'(x;z,\lambda ,a)=0, \ H''(x;z,\lambda ,a)=0. \end{aligned}$$

(36)

According to the second equation in Eq. (36), the tangent points are just the inflection points ${\tilde{x}}_i$ of H. Define two thresholding values:

$$\begin{aligned} z_i =y_1({\tilde{x}}_i) =\lambda \phi '({\tilde{x}}_i;a)+{\tilde{x}}_i. \end{aligned}$$

Since $y_1'(x)=H''(x;z,\lambda ,a)$ is positive in $(0,{\tilde{x}}_1)\cup ({\tilde{x}}_2,+\infty )$, and negative in $({\tilde{x}}_1,{\tilde{x}}_2)$, it implies that $y_1(x)$ is strictly increasing in $(0,{\tilde{x}}_1)$ and $({\tilde{x}}_2,+\infty )$, and strictly decreasing in $({\tilde{x}}_1,{\tilde{x}}_2)$. Hence we deduce that $y_1(x)$ attains its local maximum at ${\tilde{x}}_1$ and its local minimum at ${\tilde{x}}_2$. Meanwhile, we have $z_1> z_2$. Since $y_1(x)$ is increasing in $(0,{\tilde{x}}_1)$, $y_1(x\rightarrow 0^+)=\frac{\lambda }{a}<y_1({\tilde{x}}_1)=z_1$. Therefore, there are only two cases that can be occurred, i.e., $\lambda /a\le z_2$ or $\lambda /a> z_2$, as displayed in Fig. 23a and b, respectively. The horizontal lines in Fig. 23 denote $y_2(x)=z$ with different z values, and the marked points with magenta boundaries are local minimizers. In the following, we analyze the critical and extreme points of $H(x;z,\lambda ,a)$ in detail.

Fig. 23

The plots of $y_1(x)=\lambda \phi '(x;a)+x$ and $y_2(x)=z$. The points with margeta circles are local minimizers

Case 1: Assume $\lambda /a\le z_2$, then

• $z\le \lambda /a$: $y_1(x)$ and $y_2(x)$ have no intersection point, i.e., $H'$ is always positive. Hence H attains its global minimum at $x=0$. See the line $y_2=\lambda /a$ and the gray dotted line below it in Fig. 23a, for instance.

• $\lambda /a<z<z_2$: $y_1(x)$ and $y_2(x)$ have one positive intersection point such that $H'=0$, which is the unique local minimizer. See the gray dotted line between $y_2=\lambda /a$ and $y_2=z_2$ in Fig. 23a, for instance.

• $z=z_2$: $y_1(x)$ and $y_2(x)$ have two positive intersection points, and the smaller one is a local minimizer. See the line $y_2=z_2$ in Fig. 23a.

• $z_2<z<z_1$: $y_1(x)$ and $y_2(x)$ have three positive intersection points. Using the monotonicity of $y_1(x)$, we deduce that the middle one is a local maximizer and the other two are local minimizers. See the gray dotted line between $y_2=z_2$ and $y_2=z_1$ in Fig. 23a, for example.

• $z=z_1$: $y_1(x)$ and $y_2(x)$ have two positive intersection points, and the bigger one is a local minimizer. See the line $y_2=z_1$ in Fig. 23a.

• $z>z_1$: $y_1(x)$ and $y_2(x)$ have one positive intersection point, which is a local minimizer. Since $y_1(x)\rightarrow x$ as $x\rightarrow \infty$, the local minimizer of $H(x;z,\lambda ,a)$approximates z as x goes to infinity. See the gray dotted line above $y_2=z_1$ in Fig. 23a, for example.

Case 2: Assume $\lambda /a> z_2$, then

• $z\le z_2$: $y_1(x)$ and $y_2(x)$ have no intersection point, i.e., $H'$ is always positive. Hence H attains its global minimum at $x=0$. See the line $y_2=z_2$ and the gray dotted line below it in Fig. 23b, for example.

• $z=z_2$: $y_1(x)$ and $y_2(x)$ have one positive intersection point, which is not an extreme point. Hence H attains its global minimum at $x=0$. See the line $y_2=z_2$ in Fig. 23b.

• $z_2<z\le \lambda /a$: $y_1(x)$ and $y_2(x)$ have two positive intersection point such that $H'=0$. The smaller one is a local maximizer, and the smaller one is a local minimizer. $x=0$ is also a candidate of extreme point. See the gray dotted line between $y_2=z_2$ and $y_2 = \lambda /a$ in Fig. 23b, for instance.

• $\lambda /a<z<z_1$: $y_1(x)$ and $y_2(x)$ have three positive intersection points. Using the monotonicity of $y_1(x)$, we deduce that the middle one is a local maximizer and the other two are local minimizers. As an example, see the gray dotted line between $y_2=z_1$ and $y_2=\lambda /a$ in Fig. 23b.

• $z=z_1$: $y_1(x)$ and $y_2(x)$ have two positive intersection points, and the bigger one is a local minimizer. See the line $y_2=z_1$ in Fig. 23b.

• $z>z_1$: $y_1(x)$ and $y_2(x)$ have one positive intersection point, which is a local minimizer. Since the asymptote of $y_1(x)$ is $y=x$, the local minimizer of $H(x;z,\lambda ,a)$ approximates z as x goes to infinity. For instance, see the gray dotted line above $y_2=z_1$ in Fig. 23b.

Based on the above analysis, we get that there are one or two local minimizers of H, which are in $[0,{\tilde{x}}_1)$ or $[{\tilde{x}}_2,z]$. Moreover, if there exist two local minimizers, then one is in $[0,{\tilde{x}}_1)$, and the other is in $[{\tilde{x}}_2,z]$. This completes the proof. $\square$

C. proof of Theorem 3

Proof

Assume ${\varvec{x}}^{*}= t{\varvec{z}}^0$ for some $t>0$. For any ${\varvec{x}}$ satisfies $\Vert {\varvec{x}}\Vert _{2}=\left\| {\varvec{x}}^{*}\right\| _{2}=t$, we have

$$\begin{aligned} \begin{aligned}&H({\varvec{x}};{\varvec{z}},\lambda ,a)-H\left( {\varvec{x}}^{*};{\varvec{z}},\lambda ,a\right) \\&\quad =\lambda \tanh \left( \frac{\Vert {\varvec{x}}\Vert _2}{a}\right) -\lambda \tanh \left( \frac{\Vert {\varvec{x}}^*\Vert _2}{a}\right) +\frac{1}{2}\left( \Vert {\varvec{x}}-{\varvec{z}}\Vert _{2}^{2} -\left\| {\varvec{x}}^*-{\varvec{z}}\right\| _{2}^{2}\right) \\&\quad =\frac{1}{2}\left( \Vert {\varvec{x}}\Vert _{2}^{2}+\Vert {\varvec{z}}\Vert _{2}^{2}-2\langle {\varvec{x}},{\varvec{z}}\rangle - \left\| {\varvec{x}}^{*}\right\| _{2}^{2}-\Vert {\varvec{z}}\Vert _{2}^{2}+2\left\langle {\varvec{x}}^{*}, {\varvec{z}}\right\rangle \right) \\&\quad =\left\langle {\varvec{x}}^{*}-{\varvec{x}}, {\varvec{z}}\right\rangle \\&\quad =\left\langle t{\varvec{z}}^0-t{\varvec{x}}^0, t{\varvec{z}}^0\right\rangle \ge 0 \end{aligned} \end{aligned}$$

Hence, the solution ${\varvec{x}}^{*}$ of Eq. (15) must be parallel with ${\varvec{z}}$, i.e., ${\varvec{x}}^{*} = t^*{\varvec{z}}^0, t^*\ge 0$.

Let ${\varvec{x}}=t{\varvec{z}}^0$. Then the data fitting term can be rewritten as

$$\begin{aligned} \frac{1}{2}\Vert {\varvec{x}}-{\varvec{z}}\Vert _2^2 =\frac{1}{2}\Vert t{\varvec{z}}^0-\Vert {\varvec{z}}\Vert _2{\varvec{z}}^0\Vert _2^2= \frac{1}{2}(t-\Vert {\varvec{z}}\Vert _2)^2. \end{aligned}$$

Hence problem Eq. (15) can be transformed into problem Eq. (16). This completes the proof. $\square$