An Exp Model with Spatially Adaptive Regularization Parameters for Multiplicative Noise Removal

Abstract

This article proposes a total variation (TV) based model with local constraints for heavy multiplicative noise removal. The local constraint involves multiple local windows rather than one local window as in Chen and Cheng (IEEE Trans Image Process 21(4):1650–1662, 2012), and the proposed model is an extension model of Lu et al. (Appl Comput Harmon Anal 41(2):518–539, 2016) that incorporates a spatially adaptive regularization parameter, which enables us to handle heavy multiplicative noise as well as to sufficiently denoise in homogeneous regions while preserving small details and edges. In addition, convergence analysis such as the existence and uniqueness of a solution for our model is also provided. We also derive an optimization algorithm from the first-order optimality characterization of our model. Furthermore, we utilize a proximal linearized alternating direction algorithm for efficiently solving our subproblem. Numerical results are shown to validate the effectiveness of our model, with comparisons with several existing TV based models.

Appendices

Appendix

The Proof of Lemma 1

Proof

Let us consider a minimizing sequence $\{u_n\}$ of the problem (45).

Define $\tilde{\gamma } = \inf (\log (t^2_{\alpha ,\beta } f))$, $\tilde{\xi } = \sup (\log (t^2_{\alpha ,\beta } f))$, and

$$\begin{aligned} V(u)=V_0(u) +J(u) \quad \text {and} \quad V_0(u) = \int _{\varOmega } \lambda (x)q(u)\,dx. \end{aligned}$$

(54)

For fixed $x\in \varOmega ,$ let $v_0(s) = \lambda (x) \left\{ s+f(x)e^{-s}+\alpha \left( \sqrt{\frac{e^s}{f(x)}}-\beta \right) ^2 \right\}$. Then, the function $v_0(s)$ has the unique minimum value at $s = \log (t^2_{\alpha ,\beta }f(x))$. In addition, v(s) decreases on $(-\infty , \log (t^2_{\alpha ,\beta }f(x)]$ and increases on $[\log (t^2_{\alpha ,\beta }f(x)), \infty )$. This implies that if $M_1 \geqslant \log (t^2_{\alpha ,\beta }f(x))$ and $M_2 \leqslant \log (t^2_{\alpha ,\beta }f(x))$, then $v_0(\min (s, M_1)) \leqslant v_0(s)$ and $v_0(\max (s,M_2)) \leqslant v_0(s).$ This yields that

$$\begin{aligned} V_0(\inf (u,\tilde{\xi })) \leqslant V_0(u) \quad \text {and}\quad V_0(\sup (u,\tilde{\gamma })) \leqslant V_0(u). \end{aligned}$$

Moreover, by Lemma 1 in [39], we have that $J(\inf (u,\tilde{\xi })) \leqslant J(u)$ and $J(\sup (u,\tilde{\gamma })) \leqslant J(u)$, which yields that

$$\begin{aligned} V(\inf (u,\tilde{\xi })) \leqslant V(u) \quad \text {and}\quad V(\sup (u,\tilde{\gamma })) \leqslant V(u). \end{aligned}$$

(55)

In a word, we can assume that $\tilde{\gamma } \leqslant u_n \leqslant \tilde{\xi }$, we have $\{u_n\}\in L^1(\varOmega )$. Since $\{u_n\}$ is a minimizing sequence of V, $V(u_n)$ is bounded. Further, from the (32), $V_0(u_n) \geqslant K$ for some constant K. Then, $J(u_n)$ is bounded, and thus $u_n$ is bounded in BV($\varOmega$). By the compactness of BV$(\varOmega )$, there exists a subsequence $\{u_{n_k}\}$ of $\{u_n\}$ such that $u_{n_k} \rightarrow u$ in BV($\varOmega$)-weak$^{*}$ and $u_{n_k} \rightarrow u$ in $L^1(\varOmega )$-strong. Then, we can obtain that $\tilde{\gamma } \leqslant u \leqslant \tilde{\xi }$. Finally, owing to the lower semi-continuity of $J(\cdot )$ and Fatou’s lemma, u is a solution of (45). $\square$

The Proof of Lemma 2

Proof

Let $h(u):=\nabla F(u) =\lambda \left( 1-\frac{f}{e^u}+\alpha \left( \frac{e^u}{f}-\beta \sqrt{\frac{e^u}{f}} \right) \right)$. By the mean value theorem, we have that for all $u,v \in \mathcal {U}$,

$$\begin{aligned} \Vert \nabla F(u) - \nabla F(v)\Vert _2 = \Vert h(u)-h(v) \Vert _2 = |h'(\tilde{u})| \Vert u-v \Vert _2 \leqslant \Vert \lambda \Vert _{\infty } L_q \Vert u-v \Vert _2, \end{aligned}$$

(56)

where $\tilde{u} = t u+(1-t)v$ for some $t \in [0,1]$, and $L_q = \sup _{\tilde{w} \in \mathcal {U}} |q'(\tilde{w})|$. By basic calculation, we can find an upper bound for $L_q$ as follows:

$$\begin{aligned} L_q= & {} \sup \left| fe^{-u}+\alpha \left( \frac{e^u}{f}-\frac{\beta }{2}\sqrt{\frac{e^u}{f}} \right) \right| \leqslant \sup |fe^{-u}| +\alpha \sup \left| \frac{e^u}{f}-\frac{\beta }{2}\sqrt{\frac{e^u}{f}} \right| \nonumber \\\leqslant & {} \frac{f_{\max }}{t_{\alpha ,\beta }^2 f_{\min }} + \alpha \max \left( \frac{ \beta ^2}{16}, \frac{t_{\alpha ,\beta }^2 f_{\max }}{f_{\min }}- \frac{\beta }{2} \sqrt{\frac{t_{\alpha ,\beta }^2 f_{\max }}{f_{\min }}},\frac{t_{\alpha ,\beta }^2 f_{\min }}{f_{\max }}- \frac{\beta }{2} \sqrt{\frac{t_{\alpha ,\beta }^2 f_{\min }}{f_{\max }}} \right) .\nonumber \\ \end{aligned}$$

(57)

Define $\tilde{L} := \frac{f_{\max }}{t_{\alpha ,\beta }^2 f_{\min }} + \alpha \max \left( \frac{\beta ^2}{16}, \frac{t_{\alpha ,\beta }^2 f_{\max }}{f_{\min }}- \frac{\beta }{2} \sqrt{\frac{t_{\alpha ,\beta }^2 f_{\max }}{f_{\min }}},\frac{t_{\alpha ,\beta }^2 f_{\min }}{f_{\max }}- \frac{\beta }{2} \sqrt{\frac{t_{\alpha ,\beta }^2 f_{\min }}{f_{\max }}} \right)$. Then, we have

$$\begin{aligned} \Vert \nabla _u F(u) - \nabla _u F(v)\Vert _2 \leqslant \Vert \lambda \Vert _{\infty } \tilde{L} \Vert u-v \Vert _2, \end{aligned}$$

(58)

which proves the Lipschitz continuity of $\nabla F$.$\square$