SlowGuess/NeurIPS_2023_OCR huggingface dataset:
To prove global convergence of the iterates upon the Kurdyka-Lojasiewicz (KL) property, [Bolte et al., 2018, Theorem 4.1] is based on the hypotheses (a)  $\mathrm{dom}(h) = \mathbb{R}^n$  and  $h$  is strongly convex on  $\mathbb{R}^n$  and (b)  $\nabla h$  and  $\nabla F$  are Lipschitz continuous on any bounded subset of  $\mathbb{R}^n$ . These assumptions are clearly not verified for  $h$  being the Burg's entropy (22) or  $F$  the Poisson data-fidelity term (2). Indeed, in that case,  $\mathrm{dom}(h) = \mathbb{R}_{++}^n$ , and  $h$  is strongly convex only on bounded sets. Moreover  $F$  and  $h$  are not Lipschitz continuous near 0.
However, thanks to the proven decrease of the iterates and as  $\Psi$  is assumed coercive, the iterates remain bounded. We can adopt the following weaker assumptions to ensure that the iterates do not tend to  $+\infty$  or 0.
(i)  $h$  is strongly convex on any bounded convex subset of its domain.
(ii) For all  $\alpha > 0$ ,  $\nabla h$  and  $\nabla F$  are Lipschitz continuous on  $\{\Psi(x) \leq \alpha\}$ .
Under these assumptions, we prove the equivalent of [Bolte et al., 2018, Theorem 4.1].
Theorem 3. Under Assumption 3, 4 and 5, the sequence  $(x^{k})_{k\in \mathbb{N}}$  generated by (18) with  $0 < \tau L < 1$  converges to a critical point of  $\Psi$ .
We follow the proof given in [Bolte et al., 2018] with some updates to adapt to our weaker Assumption 5. In particular, we show in Lemma 3 that the sequence  $(x_{k})_{k\geq 1}$  is still "gradient-like" i.e. it verifies the assumptions H1, H2 and H3 from [Attouch et al., 2013]. Once these conditions are verified, the result directly follows from [Bolte et al., 2018, Theorem 6.2].
Let us first note that by coercivity of  $\Psi$  and decrease of the iterates  $\Psi(x_k)$  (see Proposition 3), the iterates remain in the set
$$
\forall k \geq 1, x _ {k} \in C \left(x _ {0}\right) = \{x \in \operatorname {d o m} (h), \Psi (x) <   \Psi \left(x _ {0}\right) \}. \tag {88}
$$
Lemma 3. The sequence  $(x_{k})_{k\geq 1}$  satisfies the following three conditions:
H1 (Sufficient decrease condition)
$$
\Psi \left(x _ {k}\right) - \Psi \left(x _ {k + 1}\right) \geq a \left\| x _ {k + 1} - x _ {k} \right\| ^ {2}, \tag {89}
$$
H2 (Relative error condition)  $\forall k\geq 1$  , there exists  $\omega_{k + 1}\in \partial \Psi (x_{k + 1})$  such that
$$
\left| \left| \omega_ {k + 1} \right| \right| \leq b \| x _ {k + 1} - x _ {k} \|, \tag {90}
$$
H3 (Continuity condition) Any subsequence  $(x^{k_i})$  converging towards  $x^*$  verifies
$$
\Psi \left(x ^ {k _ {i}}\right)\rightarrow \Psi \left(x ^ {*}\right). \tag {91}
$$
Proof. H1: Sufficient decrease condition. From (85), we get
$$
\Psi \left(x _ {k}\right) - \Psi \left(x _ {k + 1}\right) \geq \left(\frac {1}{\tau} - L\right) D _ {h} \left(x ^ {k + 1}, x ^ {k}\right). \tag {92}
$$
Besides,  $h$  is assumed to be strongly convex on any bounded convex subset of its domain. Furthermore, notice that  $\operatorname{Conv}(C(x_0)) \cap \operatorname{dom}(h)$  is a convex subset of  $\operatorname{dom}(h)$  as intersection of convex sets. Therefore, there is  $\sigma_h > 0$  such that
$$
\forall x, y \in \operatorname {C o n v} \left(C \left(x _ {0}\right)\right) \cap \operatorname {d o m} (h), D _ {h} (x, y) \geq \sigma_ {h} \| x - y \| ^ {2}. \tag {93}
$$
With the convention  $D_h(x,y) = +\infty$  if  $x \notin \operatorname{dom}(h)$  or  $y \notin \operatorname{int} \operatorname{dom}(h)$ ,
$$
\forall x, y \in \operatorname {C o n v} \left(C \left(x _ {0}\right)\right), D _ {h} (x, y) \geq \sigma_ {h} \| x - y \| ^ {2}. \tag {94}
$$
As  $\forall k\geq 1,x_{k}\in C(x_{0})$  ,we get
$$
\Psi \left(x _ {k}\right) - \Psi \left(x _ {k + 1}\right) \geq \sigma_ {h} \left(\frac {1}{\tau} - L\right) \left\| x _ {k + 1} - x _ {k} \right\| ^ {2}, \tag {95}
$$
which proves (H1).
H2: Relative error condition. Given (76), the optimality condition for the update of  $x_{k + 1}$  is
$$
0 \in \partial \mathcal {R} \left(x ^ {k + 1}\right) + \nabla F \left(x ^ {k}\right) + \frac {1}{\tau} \left(\nabla h \left(x ^ {k + 1}\right) - \nabla h \left(x ^ {k}\right)\right). \tag {96}
$$
$$
\omega_ {k + 1} = \nabla F \left(x ^ {k + 1}\right) - \nabla F \left(x ^ {k}\right) + \frac {1}{\tau} \left(\nabla h \left(x ^ {k}\right) - \nabla h \left(x ^ {k + 1}\right)\right) \tag {97}
$$
$$
\omega_ {k + 1} \in \partial \Psi (x _ {k + 1}) = \partial \mathcal {R} \left(x ^ {k + 1}\right) + \nabla F \left(x ^ {k + 1}\right) \tag {98}
$$
$$
\left| \left| \omega_ {k + 1} \right| \right| \leq \left| \left| \nabla F \left(x ^ {k + 1}\right) - \nabla F \left(x ^ {k}\right) \right| \right| + \frac {1}{\tau} \left| \left| \nabla h \left(x ^ {k}\right) - \nabla h \left(x ^ {k + 1}\right) \right| \right|. \tag {99}
$$
By assumption,  $\nabla F$  and  $\nabla h$  are Lipschitz continuous on  $C(x_0) = \{\Psi (x) < \Psi (x_0)\}$ . As seen before,  $\forall k\geq 1,x_{k}\in C(x_{0})$ . Thus, there is  $b > 0$  such that
$$
\left| \left| \omega_ {k + 1} \right| \right| \leq \left| \left| \nabla F \left(x ^ {k + 1}\right) - \nabla F \left(x ^ {k}\right) \right| \right| + \frac {1}{\tau} \left| \left| \nabla h \left(x ^ {k}\right) - \nabla h \left(x ^ {k + 1}\right) \right| \right| \leq b \left| \left| x ^ {k + 1} - x ^ {k} \right| \right|. \tag {100}
$$
H3: Continuity condition. Let a subsequence  $(x^{k_i})$  converging towards  $x^*$ . Using the optimality in the update of  $x_{k}$  we have
$$
\mathcal {R} \left(x ^ {k}\right) + \langle x ^ {k} - x ^ {k - 1}, \nabla F \left(x ^ {k - 1}\right) \rangle + \frac {1}{\tau} D _ {h} \left(x ^ {k}, x ^ {k - 1}\right) \tag {101}
$$
$$
\leq \mathcal {R} \left(x ^ {*}\right) + \left\langle x ^ {*} - x ^ {k - 1}, \nabla F \left(x ^ {k - 1}\right) \right\rangle + \frac {1}{\tau} D _ {h} \left(x ^ {*}, x ^ {k - 1}\right) \tag {102}
$$
$$
\Leftrightarrow \mathcal {R} \left(x ^ {k}\right) \leq \mathcal {R} \left(x ^ {*}\right) + \left\langle x ^ {*} - x ^ {k - 1}, \nabla F \left(x ^ {k - 1}\right) \right\rangle + \frac {1}{\tau} D _ {h} \left(x ^ {*}, x ^ {k - 1}\right) - \frac {1}{\tau} D _ {h} \left(x ^ {k}, x ^ {k - 1}\right). \tag {103}
$$
From (95) and the fact that  $(\Psi(x_k))_k$  converges, we have that  $||x^k - x^{k-1}|| \to 0$ . Thus  $(x^{k_i - 1})_i$  also converges towards  $x^*$ . In addition, since  $h$  is continuously differentiable,  $D_h(x^*, x^{k-1}) = h(x^*) - h(x^{k-1}) - \langle \nabla h(x^{k-1}), x^* - x^{k-1} \rangle \to 0$ . Passing to the limit in (103), we get
$$
\lim  _ {i \rightarrow + \infty} \sup  _ {x ^ {k _ {i}}} \mathcal {R} \left(x ^ {k _ {i}}\right) \leq \mathcal {R} \left(x ^ {*}\right). \tag {104}
$$
By lower semicontinuity of  $\mathcal{R}$  and continuity of  $F$ , we get the desired result:
$$
\mathcal {R} \left(x ^ {k _ {i}}\right) + F \left(x ^ {k _ {i}}\right)\rightarrow \mathcal {R} \left(x ^ {*}\right) + F \left(x ^ {*}\right). \tag {105}
$$
Backtracking The convergence actually requires to control the NoLip constant. In order to avoid small step sizes, we adapt the backtracking strategy of [Beck, 2017, Chapter 10] to the BPG algorithm.
Given  $\gamma \in (0,1)$ ,  $\eta \in [0,1)$  and an initial stepsize  $\tau_0 > 0$ , the following backtracking update rule on  $\tau$  is applied at each iteration  $k$ :
$$
\text {w h i l e} \quad \Psi \left(x _ {k}\right) - \Psi \left(T _ {\tau} \left(x _ {k}\right)\right) <   \frac {\gamma}{\tau} D _ {h} \left(T _ {\tau} \left(x _ {k}\right), x _ {k}\right), \quad \tau \longleftarrow \eta \tau . \tag {106}
$$
Proposition 4. At each iteration of the algorithm, the backtracking procedure (106) is finite and with backtracking, the convergence results of Proposition 3 and Theorem 3 still hold.
Proof. For a given stepsize  $\tau$ , we showed in equation (85) that
$$
\Phi (x _ {k}) - \Phi (T _ {\tau} (x _ {k})) \geq \left(\frac {1}{\tau} - L\right) D _ {h} (T _ {\tau} (x _ {k}), x _ {k}). \tag {107}
$$
Taking  $\tau < \frac{1 - \gamma}{L}$ , we get  $\frac{1}{\tau} - L > \frac{\gamma}{\tau}$  so that
$$
\Phi \left(x _ {k}\right) - \Phi \left(T _ {\tau} \left(x _ {k}\right)\right) > \frac {\gamma}{\tau} D _ {h} \left(T _ {\tau} \left(x _ {k}\right), x _ {k}\right). \tag {108}
$$
Hence, when  $\tau < \frac{1 - \gamma}{L}$ , the sufficient decrease condition is satisfied and the backtracking procedure  $(\tau \longleftarrow \eta \tau)$  must end.
Replacing the former sufficient decrease condition (107) with (108), the rest of the proofs from Proposition 3 and Theorem 3 are identical.
D.2 Proof of Theorem 1
Proof. We recall the B-RED algorithm
$$
\left(\mathbf {B} - \mathbf {R E D}\right) \quad x ^ {k + 1} \in T _ {\tau} \left(x _ {k}\right) = \underset {x \in \mathbb {R} ^ {n}} {\arg \min } \left\{i _ {C} (x) + \langle x - x ^ {k}, \nabla F _ {\lambda , \gamma} \left(x ^ {k}\right) \rangle + \frac {1}{\tau} D _ {h} \left(x, x ^ {k}\right) \right\}. \tag {109}
$$
It corresponds to the BPG algorithm (14) with  $F = F_{\lambda, \gamma} = \lambda f + g_{\gamma}$  and  $\mathcal{R} = i_C$ .
Theorem 1 is a direct application of Proposition 4 that is to say of the convergence results of Proposition 3 and Theorem 3 with backtracking. Given Assumptions 1 and 2, we verify that Assumptions 3, 4 and 5 are verified:
Assumption 3.  $\mathcal{R} = i_C$  verifies  $\mathrm{dom}(\mathcal{R})\cap \mathrm{int}\mathrm{dom}(h) = C\cap \mathrm{int}\mathrm{dom}(h)\neq \emptyset$ . Moreover,  $\mathcal{R}$  is semi-algebraic and thus subanalytic as the indicator function of a closed semi-algebraic set.  $g_{\gamma}$  and  $f$
being both assumed subanalytic and lower-bounded, by [Shiota, 2012, I.2.1.9], their sum  $\lambda f + g_{\sigma}$  is then subanalytic (up to adding a constant to make  $f$  and  $g_{\sigma}$  non-negative).  $\Psi = F_{\lambda,\gamma} + i_C$  is then also subanalytic, and thus KL.  $\Psi$  is also lower-bounded and coercive as  $f$  is lower-bounded and coercive and  $g_{\gamma}$  is lower-bounded. Finally, for  $x \in \operatorname{int} \operatorname{dom}(h)$ ,  $T_{\tau}(x)$  is non-empty as  $h + \lambda i_C$  is supercoercive.
Assumption 4. By summing convex functions, using Assumption 1(iii) and Assumption 2(ii),  $L = \lambda L_{f} + L_{\gamma}$  verifies  $Lh - (\lambda f + g_{\gamma})$  convex on  $\operatorname{Conv}(\operatorname{dom} \mathcal{R}) \cap \operatorname{int} \operatorname{dom}(h) = C \cap \operatorname{int} \operatorname{dom}(h)$ .
Assumption 5. As  $g_{\gamma}$  is assumed to have globally Lipschitz continuous gradient, this follows directly from Assumption 1(iv).
D.3 Proof of Theorem 2
$$
(\mathbf {B} - \mathbf {P n P}) \quad x ^ {k + 1} = \mathcal {B} _ {\gamma} \circ \nabla h ^ {*} (\nabla h - \lambda \nabla f) (x _ {k}). \tag {110}
$$
It corresponds to the BPG algorithm (14) with  $F = \lambda f$  and  $\mathcal{R} = \phi_{\gamma}$ .
Theorem 2 is a direct application of the convergence results of Proposition 3 and Theorem 3. We now denote  $\Psi = \lambda f + \psi_{\gamma}$ . Given Assumptions 1, we verify that Assumptions 3, 4 and 5 are verified.
Assumption 3. For  $\mathcal{R} = \phi_{\gamma}$ , we have  $Im(\mathcal{B}_{\gamma}) \subset \operatorname{dom}(\phi_{\gamma})$  and as for  $y \in \operatorname{int} \operatorname{dom}(h)$ ,  $Im(\mathcal{B}_{\gamma}) \subset \operatorname{int} \operatorname{dom}(h)$  (by equation 13), we get  $\operatorname{dom}(\mathcal{R}) \cap \operatorname{int} \operatorname{dom}(h) \neq \emptyset$ .  $\phi_{\gamma}$  and  $f$  being both assumed subanalytic and lower-bounded, by [Shiota, 2012, I.2.1.9], their sum  $\lambda f + \phi_{\sigma}$  is then subanalytic (up to adding a constant to make  $f$  and  $\phi_{\sigma}$  non-negative) and thus KL.  $\Psi$  is coercive because  $f$  is coercive and  $\phi_{\gamma}$  is lower-bounded. Finally,  $T_{\tau}(x)$  well-posed is ensured via Proposition 1.
Assumption 4. This is the L-smad property of  $f$  given by Assumption 1.
Assumption 5. This is directly given by Assumption 1(iv).
E Application to Poisson Inverse Problems
E.1 Burg's entropy Bregman noise model
For  $h$  being the Burg's entropy (22), the Bregman noise model (6) writes for  $x, y \in \mathbb{R}_{++}^{n}$  as
$$
\begin{array}{l} p (y | x) = \exp (- \gamma D _ {h} (x, y) + \rho (x)) \\ = \exp (\rho (x)) \exp \Big (- \gamma (h (x) - h (y) - \langle \nabla h (y), x - y \rangle) \Big) \\ = \exp (\rho (x)) \exp \left(- \gamma \left(\sum_ {i = 1} ^ {n} - \log \left(x _ {i}\right) + \log \left(y _ {i}\right) - 1 + \frac {x _ {i}}{y _ {i}}\right)\right) \tag {111} \\ = \exp (\rho (x) + n \gamma) \prod_ {i = 1} ^ {n} \left(\frac {x _ {i}}{y _ {i}}\right) ^ {\gamma} \exp \left(- \gamma \frac {x _ {i}}{y _ {i}}\right). \\ \end{array}
$$
E.2 Inverse Gamma Bregman denoiser
Derivation of (29) We first derive here the condition for convexity of  $\eta_{\gamma} \coloneqq \psi_{\gamma} \circ \nabla h^{*}$  introduced in Section 5.1.
$$
\nabla \eta_ {\gamma} (x) = \nabla^ {2} h ^ {*} (x). \nabla \psi_ {\gamma} (\nabla h ^ {*} (x)) \tag {112}
$$
$$
\nabla^ {2} \eta_ {\gamma} (x) = \left(\nabla^ {2} h ^ {*} (x)\right) ^ {2}. \nabla^ {2} \psi_ {\gamma} (\nabla h ^ {*} (x)) + \nabla^ {3} h ^ {*} (x). \nabla \psi_ {\gamma} (\nabla h ^ {*} (x)) \tag {113}
$$
$$
\nabla^ {2} \eta_ {\gamma} (\nabla h (y)) = \left(\nabla^ {2} h ^ {*} (\nabla h (y))\right) ^ {2}. \nabla^ {2} \psi_ {\gamma} (y) + \nabla^ {3} h ^ {*} (\nabla h (y)). \nabla \psi_ {\gamma} (y). \tag {114}
$$
For Burg's entropy and  $y = \nabla h^{*}(x) = -\frac{1}{x}$  with  $x < 0$ , this writes
$$
\nabla^ {2} \eta_ {\gamma} (x) = y ^ {4} \nabla^ {2} \psi_ {\gamma} (y) + 2 y ^ {3} \operatorname {D i a g} (\nabla \psi_ {\gamma} (y)). \tag {115}
$$
$$
\psi_ {\gamma} (y) = - h (y) - g _ {\gamma} (y) - 1, \tag {116}
$$
$$
\nabla \psi_ {\gamma} (y) = - \nabla h (y) - \nabla g _ {\gamma} (y) = \frac {1}{y} - \nabla g _ {\gamma} (y) \tag {117}
$$
$$
\nabla^ {2} \psi_ {\gamma} (y) = - \nabla^ {2} h (y) - \nabla^ {2} g _ {\gamma} (y) = - \frac {1}{y ^ {2}} - \nabla^ {2} g _ {\gamma} (y), \tag {118}
$$
$$
\begin{array}{l} \nabla^ {2} \eta_ {\gamma} (x) = y ^ {2} \left(y ^ {2} \nabla^ {2} \psi_ {\gamma} (y) + 2 y D i a g (\nabla \psi_ {\gamma} (y))\right) \\ = y ^ {2} \left(- 1 - y ^ {2} \nabla^ {2} g _ {\gamma} (y) + 2 - 2 y \nabla g _ {\gamma} (y)\right) \tag {119} \\ = y ^ {2} \left(1 - y ^ {2} \nabla^ {2} g _ {\gamma} (y) - 2 y \nabla g _ {\gamma} (y)\right) \\ \end{array}
$$
For  $\eta_{\gamma}$  to be convex, a necessary condition is that  $\forall y\in \mathbb{R}_{+ + }^{n}$  and  $d\in \mathbb{R}^n$
$$
\langle \nabla^ {2} \eta_ {\gamma} (y) d, d \rangle \geq 0 \tag {120}
$$
$$
\text {i . e .} \quad \langle y ^ {4} \nabla^ {2} g _ {\gamma} (y) d, d \rangle \leq \sum_ {i = 1} ^ {n} \left(y ^ {2} \left(1 - 2 y \nabla g _ {\gamma} (y)\right)\right) _ {i} d _ {i} ^ {2} \tag {121}
$$
Training details For  $N_{\gamma}$ , we use the DRUNet architecture from [Zhang et al., 2021] with 2 residual blocks at each scale and softplus activation functions. We condition the network  $N_{\gamma}$  on  $\gamma$  similarly to what is done for DRUNet. We stack to 3 color channels of the input image an additional channel containing an image with constant pixel value equal to  $1 / \gamma$ . We use the same training dataset as [Zhang et al., 2021]. Training is performed with ADAM during 1200 epochs. The learning rate is initialized with learning rate  $10^{-4}$  and is divided by 2 at epochs 300, 600 and 900.
Validation of the convexity hypothesis Given the trained Bregman Score Denoiser  $\mathcal{B}_{\gamma} = y - y^{2}\nabla g_{\gamma}(y)$ , we propose to verify validity of the hypothesis of convexity from Proposition 1 i.e. the convexity of  $\eta_{\gamma}:x\to \psi_{\gamma}\circ \nabla h^{*}(x) = \psi_{\gamma}\left(\frac{-1}{x}\right)$  on int  $\mathrm{dom}(h^*) = \mathbb{R}_{- - }^n$ .  $\psi_{\gamma}$  is defined in (11). We showed above that for  $z\in \mathbb{R}_{- - }^{n}$  and  $y = -1 / z$ ,  $\forall d\in \mathbb{R}^n$
$$
\left\langle \nabla^ {2} \eta_ {\gamma} (z) d, d \right\rangle = \sum_ {i = 1} ^ {n} \left(y ^ {2} \left(1 - 2 y \nabla g _ {\gamma} (y)\right)\right) _ {i} d _ {i} ^ {2} - \left\langle y ^ {4} \nabla^ {2} g _ {\gamma} (y) d, d \right\rangle \tag {122}
$$
In order to verify this assumption around the image manifold, we represent in Figure 4
$$
\begin{array}{l} c (\gamma , \xi) = \min  _ {x \in \left\{X _ {i} \right\}, d \in \left\{D _ {i} \right\}} \left\langle \nabla^ {2} \eta_ {\gamma} \left(\frac {- 1}{y _ {\xi}}\right) d, d \right\rangle (123) \\ = \min  _ {x \in \left\{X _ {i} \right\}, d \in \left\{D _ {i} \right\}} \sum_ {i = 1} ^ {n} \left(y _ {\xi} ^ {2} \left(1 - 2 y _ {\xi} \nabla g _ {\gamma} \left(y _ {\xi}\right)\right)\right) _ {i} d _ {i} ^ {2} - \left\langle y _ {\xi} ^ {4} \nabla^ {2} g _ {\gamma} \left(y _ {\xi}\right) d, d \right\rangle (124) \\ \end{array}
$$
where the minimum is taken over
-  $\{X_{i}\}_{i}$ : the images from an evaluation dataset of clean images, here the 68 images from CSBD68.
-  $\{D_i\}$ : 100 random images realizations of a uniform random variable  $D_i \sim \mathcal{U}[0,1]^n$ .
and  $y_{\xi}$  is obtained from  $x$  by interpolating between  $\hat{y}_{\xi} \sim p_{\xi}(y|x)$  a noisy version of  $x$  and  $\mathcal{B}_{\gamma}(\hat{y}_{\xi})$  with  $\alpha \sim \mathcal{U}[0,1]$  as
$$
y _ {\xi} = \alpha \hat {y} _ {\xi} + (1 - \alpha) \mathcal {B} _ {\gamma} (\hat {y} _ {\xi}). \tag {125}
$$
In the figures, we represent  $c(\gamma ,\xi)$  with respect to:
- in the  $x$ -axis: the denoiser strength, i.e. the  $\gamma$  parameter of the denoiser.
- in the  $y$ -axis: the noise level  $\xi$  in the input image. Figure 4(a), the noise model is Inverse Gamma noise  $p_{\xi}(y|x) = \mathcal{I}\mathcal{G}(\xi - 1, \xi x)$  i.e. the noise model used for training the denoiser. In figure 4(b), the noise model is Poisson  $p_{\xi}(y|x) = \mathcal{P}(\xi x)$  i.e. the noise model of the degradation on which our plug-and-play algorithms are evaluated, and in Figure 4(c), the noise model is Gaussian  $p_{\xi}(y|x) = \mathcal{N}(x, \xi^2\mathrm{Id})$ .
We observe that for a large variety of noise levels  $\gamma$  and even far away from the image manifold i.e. for large noise, we verify the convexity assumption.
(a) Inverse Gamma noise
(b) Poisson noise
$p_{\xi}(\hat{y}_{\xi}|x) = \mathcal{I}\mathcal{G}(\xi -1,\xi x).$
$p_{\xi}(\hat{y}_{\xi}|x) = \mathcal{I}\mathcal{G}(\xi x).$
(c) Gaussian noise
$p_{\xi}(\hat{y}_{\xi}|x) = \mathcal{I}\mathcal{G}(x,\xi^{2}\mathrm{Id})$
Figure 4: Plot of  $c(\gamma, \xi) = \min_{y_{\xi}, d} \left\langle \nabla^2 \eta_{\gamma} \left( \frac{-1}{y_{\xi}} \right) d, d \right\rangle$  w.r.t the denoiser parameter  $\gamma$  and the noise level in the input image  $\xi$ . From  $\{x_i\}$  the natural images from the CBSD68 testset and 100 random  $d \sim \mathcal{U}[0,1]^n$ , the image  $y_{\xi}$  is obtained from  $x \in \{x_i\}$  via the interpolation  $y = \alpha \hat{y}_{\xi} + (1 - \alpha) \mathcal{B}_{\gamma}(\hat{y}_{\xi})$  where  $\hat{y}_{\xi}$  is a noisy version of  $x$  sampled with (a) Inverse Gamma (b) Poisson (c) Gaussian noise distributions.  $\mathcal{B}_{\gamma}$  is the trained Bregman Score Denoiser.
E.3 The MMSE denoiser verifies the convexity hypothesis from Proposition 1
Lemma 4. For  $h$  Burg's entropy, and thus for the noise model (23), with the MMSE denoiser  $\hat{x}_{MMSE}(y) = E[x|y]$ , we have  $\eta_{\gamma}:x\to \psi_{\gamma}\circ \nabla h^{*}$  is strictly convex on int  $\mathrm{dom}(h^*)$ , where  $\psi_{\gamma}$  is defined in (11) from  $g_{\gamma} = -\frac{1}{\gamma}\log p_Y$ .
Proof. Recall that the MMSE denoiser corresponds to the Bregman Score Denoiser
$$
\mathcal {B} _ {\gamma} (y) = \nabla \left(\psi_ {\gamma} \circ \nabla h ^ {*}\right) \circ \nabla h (y) = \nabla \eta_ {\gamma} \circ \nabla h (y) \tag {126}
$$
in the particular case where we have exactly
$$
g _ {\gamma} (y) = - \frac {1}{\gamma} \log p _ {Y} (y). \tag {127}
$$
Note that in the Euclidean case (i.e. for the  $L^2$  Bregman divergence  $h = \frac{1}{2} ||.|^2$  and  $p_Y = p_\sigma = p*\mathcal{N}(0,\sigma^2\mathrm{Id})$ ), it is shown in Gribonval [2011] to be verified by the MMSE denoiser. We now show the same result for Burg's entropy Bregman potential. For  $x\in \mathbb{R}_{++}^{n}$
$$
h (x) = - \sum_ {i = 1} ^ {n} \log \left(x _ {i}\right), \tag {128}
$$
Recall that, in this case, the Bregman noise model writes
$$
p (y | x) = \exp (\rho (x) + n \gamma) \prod_ {i = 1} ^ {n} \left(\frac {x _ {i}}{y _ {i}}\right) ^ {\gamma} \exp \left(- \gamma \frac {x _ {i}}{y _ {i}}\right) = \alpha (x) \prod_ {i = 1} ^ {n} \left(y _ {i}\right) ^ {- \gamma} \exp \left(- \gamma \frac {x _ {i}}{y _ {i}}\right). \tag {129}
$$
where  $\alpha (x) = \exp (\rho (x) + n\gamma)\prod_{i = 1}^{n}x_{i}^{\gamma} > 0$  . We verify that  $\forall x\in int\mathrm{dom}(h^{*}) = \mathbb{R}_{- - }^{n}$
$$
\langle \nabla^ {2} \eta_ {\gamma} (x) d, d \rangle > 0. \tag {130}
$$
We showed in (119) that,  $\forall y\in \mathbb{R}_{++}^{n}$
$$
\nabla^ {2} \eta_ {\gamma} (\nabla h (y)) = \operatorname {D i a g} \left(y ^ {2} \left(1 - 2 y \nabla g _ {\gamma} (y)\right)\right) - y ^ {4} \nabla^ {2} g _ {\gamma} (y). \tag {131}
$$
Similar to Gribonval [2011], for simplicity, we first write the proof in the single variable case  $(n = 1)$ .
We have  $\nabla g_{\gamma}(y) = \frac{-1}{\gamma}\frac{p_{Y}^{\prime}(y)}{p_{Y}(y)}$  and  $\nabla^2 g_{\gamma}(y) = \frac{1}{\gamma}\frac{[p_{Y}^{\prime}(y)]^2 - p_{Y}^{\prime\prime}(y)p_{y}(y)}{p_{Y}^2(y)}$ , and thus
$$
\begin{array}{l} \eta_ {\gamma} ^ {\prime \prime} (x) = y ^ {2} \left(1 + \frac {2 y}{\gamma} \frac {p _ {Y} ^ {\prime} (y)}{p _ {Y} (y)} - \frac {y ^ {2}}{\gamma} \frac {[ p _ {Y} ^ {\prime} (y) ] ^ {2} - p _ {Y} ^ {\prime \prime} (y) p _ {y} (y)}{p _ {Y} ^ {2} (y)}\right) (132) \\ = \frac {y ^ {2}}{\gamma p _ {Y} ^ {2} (y)} \left(\gamma p _ {Y} ^ {2} (y) + 2 y p _ {Y} ^ {\prime} (y) p _ {Y} (y) - y ^ {2} \left[ p _ {Y} ^ {\prime} (y) \right] ^ {2} + y ^ {2} p _ {Y} ^ {\prime \prime} (y) p _ {y} (y)\right) (133) \\ \end{array}
$$
$$
p _ {Y} (y) = \int_ {z} \alpha (z) p _ {X} (z) y ^ {- \gamma} \exp \left(- \gamma \frac {z}{y}\right) d z = \int_ {x} u (z, y) d z. \tag {134}
$$
$$
p _ {Y} ^ {\prime} (y) = \frac {- \gamma}{y} p _ {Y} (y) + \frac {\gamma}{y ^ {2}} \int_ {z} z u (z, y) d z = \frac {- \gamma}{y} p _ {Y} (y) + \frac {\gamma}{y ^ {2}} I ^ {(1)} (y) \tag {135}
$$
where we denote  $I^{(1)}(y) = \int_{z} z u(z, y) dz$ .
$$
p _ {Y} ^ {\prime \prime} (y) = \left(\gamma^ {2} + \gamma\right) \frac {1}{y ^ {2}} p _ {Y} (y) - 2 \left(\gamma^ {2} + \gamma\right) \frac {1}{y ^ {3}} I ^ {(1)} (y) + \gamma^ {2} \frac {1}{y ^ {4}} I ^ {(2)} (y), \tag {136}
$$
where we denote  $I^{(2)}(y) = \int_{z}z^{2}u(z,y)dz$
Eventually, we get, after simplification,
$$
\begin{array}{l} \eta_ {\gamma} ^ {\prime \prime} (x) = \frac {y ^ {2}}{\gamma p _ {Y} ^ {2} (y)} \frac {\gamma^ {2}}{y ^ {2}} \left(I ^ {(2)} (y) p _ {Y} (y) - \left[ I ^ {(1)} (y) \right] ^ {2}\right) (137) \\ = \frac {\gamma}{p _ {Y} ^ {2} (y)} \int_ {z} \int_ {z ^ {\prime}} \left(\frac {z ^ {2}}{2} + \frac {z ^ {\prime 2}}{2} - z z ^ {\prime}\right) u (z, y) u \left(z ^ {\prime}, y\right) d z d z ^ {\prime} (138) \\ = \frac {\gamma}{p _ {Y} ^ {2} (y)} \int_ {z} \int_ {z ^ {\prime}} \frac {\left(z - z ^ {\prime}\right) ^ {2}}{2} u (z, y) u \left(z ^ {\prime}, y\right) d z d z ^ {\prime} (139) \\ = \frac {\gamma y ^ {- 2 \gamma}}{p _ {Y} ^ {2} (y)} \int_ {z} \int_ {z ^ {\prime}} \frac {(z - z ^ {\prime}) ^ {2}}{2} \exp \left(- \gamma \frac {z + z ^ {\prime}}{y}\right) \alpha (z) \alpha \left(z ^ {\prime}\right) p _ {X} (z) p _ {X} \left(z ^ {\prime}\right) d z d z ^ {\prime} \geq 0 (140) \\ \end{array}
$$
As  $\alpha(x) > 0$ , the previous term is 0 if and only if  $p_X(z)p_X(z') = 0$  when  $z \neq z'$ , which would imply  $p_X(z) = 0 \forall z$ . This is impossible since  $p_X$  is a proper pdf.
The extension for  $n > 1$  is straightforward and  $(z - z')^2$  becomes  $\langle z - z', d \rangle^2$ .
E.4 Convergence of B-RED and B-PnP algorithm for Poisson inverse problems
In this section we verify that the Burg's entropy  $h$  in (22) and the Poisson data likelihood  $f$  defined in (2) verify the assumptions required for convergence of B-RED and B-PnP. We remind the expression of  $f$  and  $h$ .
$$
h (x) = - \sum_ {i = 1} ^ {n} \log \left(x _ {i}\right), \tag {141}
$$