SlowGuess/NeurIPS_2023_OCR huggingface dataset:
- A subset  $S$  of  $\mathbb{R}^n$  is a semianalytic set if each point of  $\mathbb{R}^n$  admits a neighborhood  $V$  for which there exists a finite number of real analytic functions  $f_{i,j}, g_{i,j}: \mathbb{R}^n \to \mathbb{R}$  such that
$$
S \cap V = \cup_ {j = 1} ^ {p} \cap_ {i = 1} ^ {q} \left\{x \in \mathbb {R} ^ {n}, f _ {i, j} = 0, g _ {i, j} <   0 \right\} \tag {36}
$$
- A subset  $S$  of  ${\mathbb{R}}^{n}$  is a subanalytic set if each point of  ${\mathbb{R}}^{n}$  admits a neighborhood  $V$  for which
$$
S \cap V = \{x \in \mathbb {R} ^ {n}, (x, y) \in U \} \tag {37}
$$
where  $U$  is a bounded semianalytic subset of  $\mathbb{R}^n\times \mathbb{R}^p$  with  $p\geq 1$
- A function  $f: \mathbb{R}^n \to \mathbb{R} \cup \{+\infty\}$  (resp.  $f: \mathbb{R}^n \to \mathbb{R}^m$ ) is called subanalytic if its graph  $\{(x, y) \in \mathbb{R}^n \times \mathbb{R}, y = f(x)\}$  (resp.  $\{(x, y) \in \mathbb{R}^n \times \mathbb{R}^m, y = f(x)\}$ ) is a subanalytic subset of  $\mathbb{R}^n \times \mathbb{R}$  (resp.  $\mathbb{R}^n \times \mathbb{R}^m$ ).
Subanalytic functions encompass both analytic functions and semialgebraic [Coste, 2000] functions [Shiota, 2012]. Stability results of subanalytic functions by sum, product, and composition are given in [Shiota, 2012, Section 1.2].
B More details on the Bregman Denoising Prior
We detail here the calculations realized in Section 3.
We first verify that the Bregman noise conditional probability belongs to the regular exponential family of distributions. Indeed for  $T(y) = \gamma \nabla h(y)$ ,  $\psi(x) = \gamma h(x) - \rho(x)$  and  $p_0(y) = \exp(\gamma h(y) - \gamma \langle \nabla h(y), y \rangle)$
$$
\begin{array}{l} p (y | x) = p _ {0} (y) \exp (\langle x, T (y) \rangle - \psi (x)) \\ = \exp (\gamma h (y) - \gamma \langle \nabla h (y), y \rangle) \exp (\gamma \langle x, \nabla h (y) \rangle - \gamma h (x) + \rho (x)) \tag {38} \\ = \exp (- \gamma (h (x) - h (y) - \langle \nabla h (y), x - y \rangle) + \rho (x)) \\ = \exp (- \gamma D _ {h} (x, y) + \rho (x)) \\ \end{array}
$$
which corresponds to the Bregman noise conditional probability introduced in Equation (6).
The Tweedie formula of the posterior mean is then [Kim and Ye, 2021]
$$
\nabla T (y). \hat {x} _ {M M S E} (y) = - \nabla \log p _ {0} (y) + \nabla \log p _ {Y} (y). \tag {39}
$$
$$
\nabla \log p _ {0} (y) = \gamma \nabla h (y) - \gamma \nabla h (y) - \gamma \nabla^ {2} h (y). y = - \gamma \nabla^ {2} h (y). y, \tag {40}
$$
$$
\gamma \nabla^ {2} h (y). \dot {x} _ {M M S E} (y) = \gamma \nabla^ {2} h (y). y + \nabla \log p _ {Y} (y). \tag {41}
$$
As  $h$  is strictly convex,  $\nabla^2 h(y)$  is invertible and
$$
\hat {x} _ {M M S E} (y) = y + \frac {1}{\gamma} \left(\nabla^ {2} h (y)\right) ^ {- 1}. \nabla \log p _ {Y} (y). \tag {42}
$$
C Proof of Proposition 1
A new characterization of Bregman proximal operators We first derive in Proposition 2 an extension of [Gribonval and Nikolova, 2020, Corollary 5 a)] when  $h$  is strictly convex and thus  $\nabla h$  is invertible. We will then see that Proposition 1 is a direct application of this result.
Proposition 2. Let  $h$  of Legendre type on  $\mathbb{R}^n$ . Let  $\zeta : \operatorname{int}\operatorname{dom}(h) \to \mathbb{R}^n$ . Suppose  $\operatorname{Im}(\zeta) \subset \operatorname{dom}(h)$ . The following properties are equivalent:
(i) There is  $\phi :\mathbb{R}^n\to \mathbb{R}\cup \{+\infty \}$  such that  $\operatorname {Im}(\zeta)\subset \operatorname {dom}(\phi)$  and for each  $y\in int\operatorname {dom}(h)$
$$
\zeta (y) \in \underset {x \in \mathbb {R} ^ {n}} {\arg \min } \left\{D _ {h} (x, y) + \phi (x) \right\}. \tag {43}
$$
(ii) There is a l.s.c  $\psi : \mathbb{R}^n \to \mathbb{R} \cup \{+\infty\}$  proper convex on int  $\operatorname{dom}(h^*)$  such that  $\zeta(\nabla h^*(z)) \in \partial \psi(z)$  for each  $z \in \operatorname{int} \operatorname{dom}(h^*)$ .
When (i) holds,  $\psi$  can be chosen given  $\phi$  with
$$
\psi (z) := \left\{ \begin{array}{l l} \langle \zeta (\nabla h ^ {*} (z)), z \rangle - h (\zeta (\nabla h ^ {*} (z))) - \phi (\zeta (\nabla h ^ {*} (z))) & \text {i f} z \in \operatorname {i n t} \operatorname {d o m} (h ^ {*}) \\ + \infty & \text {o t h e r w i s e .} \end{array} \right. \tag {44}
$$
When (ii) holds,  $\phi$  can be chosen given  $\psi$  with
$$
\phi (x) := \left\{ \begin{array}{l l} \langle \zeta (y), \nabla h (y) \rangle - h (\zeta (y)) - \psi (\nabla h (y)) f o r y \in \zeta^ {- 1} (x) & i f x \in I m (\zeta) \\ + \infty & o t h e r w i s e. \end{array} \right. \tag {45}
$$
Remark 4. [Gribonval and Nikolova, 2020, Corollary 5 a)] with  $\mathcal{Y} = \operatorname{int}\operatorname{dom}(h)$  states that (i) is equivalent to
(iii) There is a l.s.c  $g: \mathbb{R}^n \to \mathbb{R} \cup \{+\infty\}$  convex such that  $\nabla h(\zeta^{-1}(x)) \in \partial g(x)$  for each  $x \in \operatorname{Im}(\zeta)$ .
However (ii) and (iii) are not equivalent. We show here that (ii) implies (iii) but the converse is not true. Let  $\psi$  convex defined from (ii). Thanks to Legendre-Fenchel identity, we have that
$$
\forall z \in \operatorname {i n t} \operatorname {d o m} \left(h ^ {*}\right), \zeta \left(\nabla h ^ {*} (z)\right) \in \partial \psi (z) \tag {46}
$$
$$
\Leftrightarrow \forall z \in i n t \operatorname {d o m} \left(h ^ {*}\right), z \in \partial \psi^ {*} \left(\zeta \left(\nabla h ^ {*} (z)\right)\right) \tag {47}
$$
$$
\Leftrightarrow \forall y \in i n t \operatorname {d o m} (h), \nabla h (y) \in \partial \psi^ {*} (\zeta (y)) \tag {48}
$$
$$
\Rightarrow \forall x \in I m (\zeta), \nabla h \left(\zeta^ {- 1} (x)\right) \in \partial \psi^ {*} (x). \tag {49}
$$
Therefore (ii) implies (iii) with  $g = \psi^{*}$ . However, the last lign is just an implication.
Proof. We follow the same order of arguments from [Gribonval and Nikolova, 2020]. We first prove a general result reminiscent to [Gribonval and Nikolova, 2020, Theorem 3] for a general form of divergence function and then apply this result to Bregman divergences.
Lemma 1. Let  $a: \mathcal{Y} \subseteq \mathbb{R}^n \to \mathbb{R} \cup \{+\infty\}$ ,  $b: \mathbb{R}^n \to \mathbb{R} \cup \{+\infty\}$ ,  $A: \mathcal{Y} \to \mathcal{Z}$  bijection from  $\mathcal{Y}$  to  $\mathcal{Z}$  (with  $\mathcal{Z} \subset \mathbb{R}^n$ ). Consider  $\zeta: \mathcal{Y} \to \mathbb{R}^n$ . Let  $D(x, y) := a(y) - \langle x, A(y) \rangle + b(x)$ . Suppose  $\operatorname{Im}(\zeta) \subset \operatorname{dom}(b)$ . The following properties are equivalent:
(i) There is  $\phi :\mathbb{R}^n\to \mathbb{R}\cup \{+\infty \}$  such that  $\operatorname {Im}(\zeta)\subset \operatorname {dom}(\phi)$  and for each  $y\in \mathcal{V}$
$$
\zeta (y) \in \underset {x \in \mathbb {R} ^ {n}} {\arg \min } \{D (x, y) + \phi (x) \}. \tag {50}
$$
(ii) There is a l.s.c  $\psi : \mathbb{R}^n \to \mathbb{R} \cup \{+\infty\}$  proper convex such that  $\zeta(A^{-1}(z)) \in \partial \psi(z)$  for each  $z \in \mathcal{Z}$ .
$$
\psi (z) := \left\{ \begin{array}{l l} \langle \zeta (A ^ {- 1} (z)), z \rangle - b (\zeta (A ^ {- 1} (z))) - \phi (\zeta (A ^ {- 1} (z))) & i f z \in \mathcal {Z} \\ + \infty & o t h e r w i s e. \end{array} \right. \tag {51}
$$
$$
\phi (x) := \left\{ \begin{array}{l l} \langle \zeta (y), A (y) \rangle - b (\zeta (y)) - \psi (A (y)) f o r y \in \zeta^ {- 1} (x) & i f x \in I m (\zeta) \\ + \infty & o t h e r w i s e. \end{array} \right. \tag {52}
$$
Before proving the Lemma, we can directly see that Proposition 2 is the specialization of Lemma 1 with Bregman divergences. Given  $h$  of Legendre-type, the divergence  $D(x,y) = a(y) - \langle x,A(y)\rangle + b(x)$  becomes the Bregman divergence  $D_{h}(x,y)$  defined in (5) for  $\mathcal{V} = \operatorname{int}\operatorname{dom}(h)$ ,  $\mathcal{Z} = \operatorname{int}\operatorname{dom}(h^{*})$ ,  $A(y) := \nabla h(y)$ ,  $a(y) := \langle \nabla h(y),y\rangle - h(y)$  and  $b(x) := h(x)$  if  $x \in \operatorname{dom}(h)$  and  $+\infty$  otherwise. For  $h$  of Legendre-type,  $\nabla h$  is a bijection from  $\operatorname{int}\operatorname{dom}(h)$  to  $\operatorname{int}\operatorname{dom}(h^{*})$  and  $(\nabla h)^{-1} = \nabla h^{*}$ .
Proof. We now prove Lemma 1. We follow the same arguments as the proof of [Gribonval and Nikolova, 2020, Theorem 3(c)].  $(\mathbf{i})\Rightarrow (\mathbf{ii})$  ..
$$
\rho (z) = \left\{ \begin{array}{l l} \langle \zeta (A ^ {- 1} (z)), z \rangle - b (\zeta (A ^ {- 1} (z))) - \phi (\zeta (A ^ {- 1} (z))) & \text {i f} z \in \mathcal {Z} \\ + \infty & \text {e l s e .} \end{array} \right. \tag {53}
$$
As we assume  $\operatorname{Im}(\zeta) \subset \operatorname{dom}(b)$  and  $\operatorname{Im}(\zeta) \subset \operatorname{dom}(\psi)$ , we have  $\rho: \mathbb{R}^n \to \mathbb{R} \cup \{+\infty\}$  and  $\operatorname{dom}(\rho) = \mathcal{Z}$ . Let  $z \in \mathcal{Z}$  and  $y = A^{-1}(z)$ . From (i),  $\zeta(y)$  is the global minimizer of  $x \to D(x, y) + \phi(x)$  or of  $x \to -\langle x, A(y) \rangle + b(x) + \phi(x)$  and  $\forall z' \in \mathcal{Z}, y' = A^{-1}(z')$ ,
$$
\begin{array}{l} \rho \left(z ^ {\prime}\right) - \rho (z) = \left\langle \zeta \left(A ^ {- 1} \left(z ^ {\prime}\right)\right), z ^ {\prime} \right\rangle - b \left(\zeta \left(A ^ {- 1} \left(z ^ {\prime}\right)\right)\right) - \phi \left(\zeta \left(A ^ {- 1} \left(z ^ {\prime}\right)\right)\right) \\ - \left\langle \zeta (A ^ {- 1} (z)), z \right\rangle + b (\zeta (A ^ {- 1} (z))) + \phi (\zeta (A ^ {- 1} (z))) \\ = \left\langle \zeta \left(y ^ {\prime}\right), A \left(y ^ {\prime}\right) \right\rangle - b \left(\zeta \left(y ^ {\prime}\right)\right) - \phi \left(\zeta \left(y ^ {\prime}\right)\right) - \left\langle \zeta (y), A (y) \right\rangle + b (\zeta (y)) + \phi (\zeta (y)) \tag {54} \\ = \langle \zeta (y), A \left(y ^ {\prime}\right) - A (y) \rangle + \langle \zeta \left(y ^ {\prime}\right), A \left(y ^ {\prime}\right) \rangle - b \left(\zeta \left(y ^ {\prime}\right)\right) - \phi \left(\zeta \left(y ^ {\prime}\right)\right) \\ - \langle \zeta (y), A \left(y ^ {\prime}\right) \rangle + b (\zeta (y)) + \phi (\zeta (y)) \\ \geq \langle \zeta (y), A (y ^ {\prime}) - A (y) \rangle = \langle \zeta (A ^ {- 1} (z)), z ^ {\prime} - z \rangle . \\ \end{array}
$$
By definition of the subdifferential, this shows that
$$
\zeta (A ^ {- 1} (z)) \in \partial \rho (z). \tag {55}
$$
Let  $\tilde{\rho}$  the lower convex envelope of  $\rho$  (pointwise supremum of all the convex l.s.c functions below  $\rho$ ).  $\tilde{\rho}$  is proper convex l.s.c. and  $\forall z \in \mathcal{Z}, \partial \rho(z) \neq \emptyset$ . By [Gribonval and Nikolova, 2020, Proposition 3],  $\forall z \in \mathcal{Z}, \rho(z) = \tilde{\rho}(z)$  and  $\partial \rho(z) = \partial \tilde{\rho}(z)$ . Thus for  $\psi = \tilde{\rho}$ , we get (ii).
(ii)  $\Rightarrow$  (i):
Define  $\eta :\mathcal{V}\to \mathbb{R}$  by
$$
\eta (y) := \langle \zeta (y), A (y) \rangle - \psi (A (y)). \tag {56}
$$
The previous definition is valid because by (ii),  $\operatorname{Im}(A) = \mathcal{Z} \subset \operatorname{dom}(\partial \psi) \subset \operatorname{dom}(\psi)$  and therefore  $\psi(A(y)) < +\infty$ . By (ii),  $\forall z, z' \in \mathcal{Z}$ ,
$$
\psi (z) - \psi \left(z ^ {\prime}\right) \geq \langle \zeta \left(A ^ {- 1} \left(z ^ {\prime}\right)\right), z - z ^ {\prime} \rangle \tag {57}
$$
which gives  $\forall y, y' \in \mathcal{Y}$ ,
$$
\left. \psi (A (y)) - \psi (A \left(y ^ {\prime}\right)) \geq \langle \zeta \left(y ^ {\prime}\right), A (y) - A \left(y ^ {\prime}\right) \rangle . \right. \tag {58}
$$
This yields
$$
\begin{array}{l} \eta \left(y ^ {\prime}\right) - \eta (y) = \left\langle \zeta \left(y ^ {\prime}\right), A \left(y ^ {\prime}\right) \right\rangle - \psi \left(A \left(y ^ {\prime}\right)\right) - \left\langle \zeta (y), A (y) \right\rangle + \psi (A (y)) \tag {59} \\ \geq \langle \zeta (y ^ {\prime}) - \zeta (y), A (y) \rangle . \\ \end{array}
$$
We define  $\theta :\mathbb{R}^n\to \mathbb{R}\cup \{+\infty \}$  obeying  $\mathrm{dom}(\theta) = \mathrm{Im}(\zeta)$  with
$$
\theta (x) := \left\{ \begin{array}{l l} \eta (y) \text {f o r} y \in \zeta^ {- 1} (x) & \text {i f} x \in I m (\zeta) \\ + \infty & \text {o t h e r w i s e .} \end{array} \right. \tag {60}
$$
For  $y, y' \in \zeta^{-1}(x)$ , as  $\zeta(y') = \zeta(y)$ , we have by (59)  $\eta(y') - \eta(y) \geq 0$  and  $\eta(y) - \eta(y') \geq 0$  and thus  $\eta(y') = \eta(y)$ . The definition of  $\theta$  is thus independent of the choice of  $y \in \zeta^{-1}(x)$ .
For  $x' \in \operatorname{Im}(\zeta)$ ,  $x' = \zeta(y')$ . Using the previous inequality with  $\eta$ , we get
$$
\begin{array}{l} \theta \left(x ^ {\prime}\right) - \theta (\zeta (y)) = \theta (\zeta \left(y ^ {\prime}\right)) - \theta (\zeta (y)) \\ = \eta (y ^ {\prime}) - \eta (y) (61) \\ \geq \langle \zeta (y ^ {\prime}) - \zeta (y), A (y) \rangle (61) \\ = \left\langle x ^ {\prime} - \zeta (y), A (y) \right\rangle , \\ \end{array}
$$
that is to say,  $\forall x^{\prime}\in Im(\zeta)$
$$
\theta \left(x ^ {\prime}\right) - \langle x ^ {\prime}, A (y) \rangle \geq \theta (\zeta (y)) - \langle \zeta (y), A (y) \rangle . \tag {62}
$$
Given the definition of  $\theta$ , this is also true for  $x' \notin \operatorname{Im}(\zeta)$ .
We set  $\phi = \theta - b$ . As  $\operatorname{Im}(\zeta) \subset \operatorname{dom}(b)$ ,  $b(\zeta(y)) < +\infty$  and  $\phi: \mathbb{R}^n \to \mathbb{R} \cup \{+\infty\}$ . Adding  $a(y)$  on both sides, we get
$$
\begin{array}{l} \forall x ^ {\prime} \in \mathbb {R} ^ {n}, b (x ^ {\prime}) + \phi (x ^ {\prime}) - \langle x ^ {\prime}, A (y) \rangle + a (y) \geq b (\zeta (y)) + \phi (\zeta (y)) - \langle \zeta (y), A (y) \rangle + a (y). \\ \Leftrightarrow \forall x ^ {\prime} \in \mathbb {R} ^ {n}, \phi (x ^ {\prime}) + D (x ^ {\prime}, y) \geq \phi (\zeta (y)) + D (\zeta (y), y) \tag {63} \\ \Leftrightarrow \zeta (y)\in \operatorname *{arg  min}_{x}\phi (x) + D(x,y) \\ \end{array}
$$
As this is true for all  $y \in \mathcal{V}$ , we get the desired result.
Eventually, Proposition 1 is a direct application of Proposition 2 with  $\zeta = \mathcal{B}_{\gamma}:\mathbb{R}^n\to \mathbb{R}^n$  defined on  $int\operatorname {dom}(h)$  by
$$
\mathcal {B} _ {\gamma} (y) = \nabla \left(\psi_ {\gamma} \circ \nabla h ^ {*}\right) \circ \nabla h (y). \tag {64}
$$
The function  $\mathcal{B}_{\gamma}$  verifies  $\forall z\in int\mathrm{dom}(h^{*})$
$$
\mathcal {B} _ {\gamma} \left(\nabla h ^ {*} (z)\right) = \nabla \left(\psi_ {\gamma} \circ \nabla h ^ {*}\right) (z) \tag {65}
$$
and  $\psi = \psi_{\gamma} \circ \nabla h^{*}$  is assumed convex on  $int \, \mathrm{dom}(h^{*})$ . From Proposition 2, we get that there is  $\phi_{\gamma}: \mathbb{R}^n \to \mathbb{R} \cup \{+\infty\}$  such that for each  $y \in int \, \mathrm{dom}(h)$
$$
\mathcal {B} _ {\gamma} (y) \in \arg \min  \left\{D _ {h} (x, y) + \phi_ {\gamma} (x) \right\}. \tag {66}
$$
Moreover, for  $y \in \operatorname{int} \operatorname{dom}(h)$ ,
$$
\begin{array}{l} \mathcal {B} _ {\gamma} (y) = \nabla \left(\psi_ {\gamma} \circ \nabla h ^ {*}\right) \circ \nabla h (y) (67) \\ = \nabla^ {2} h ^ {*} (\nabla h (y)). \nabla \psi_ {\gamma} \circ \nabla h ^ {*} \circ \nabla h (y) (68) \\ = \nabla^ {2} h ^ {*} (\nabla h (y)). \nabla \psi_ {\gamma} (y). (69) \\ \end{array}
$$
As  $h$  is assumed strictly convex on  $int\operatorname{dom}(h)$ , for  $y \in int\operatorname{dom}(h)$ , the Hessian of  $h$ , denoted as  $\nabla^2 h(y)$  is invertible. By differentiating
$$
\nabla h ^ {*} (\nabla h (y)) = y \tag {70}
$$
$$
\nabla^ {2} h ^ {*} (\nabla h (y)) = \left(\nabla^ {2} h (y)\right) ^ {- 1}, \tag {71}
$$
$$
\mathcal {B} _ {\gamma} (y) = \left(\nabla^ {2} h (y)\right) ^ {- 1}. \nabla \psi_ {\gamma} (y). \tag {72}
$$
With the definition (11), we directly get
$$
\mathcal {B} _ {\gamma} (y) = \left(\nabla^ {2} h (y)\right) ^ {- 1}. \nabla \psi_ {\gamma} (y) = y - \left(\nabla^ {2} h (y)\right) ^ {- 1}. \nabla g _ {\gamma} (y). \tag {73}
$$
Finally, Proposition 2 also indicates that  $\phi_{\gamma}$  can be chosen given  $\psi_{\gamma}$  with
$$
\phi_ {\gamma} (x) := \left\{ \begin{array}{l l} \langle \mathcal {B} _ {\gamma} (y), \nabla h (y) \rangle - h (\mathcal {B} _ {\gamma} (y)) - \psi_ {\gamma} \circ \nabla h ^ {*} (\nabla h (y)) & \text {f o r} y \in \mathcal {B} _ {\gamma} ^ {- 1} (x) \\ + \infty & \text {o t h e r w i s e .} \end{array} \right. \tag {74}
$$
This gives  $\forall y\in int\operatorname {dom}(h)$
$$
\begin{array}{l} \phi_ {\gamma} \left(\mathcal {B} _ {\gamma} (y)\right) = \left\langle \mathcal {B} _ {\gamma} (y), \nabla h (y) \right\rangle - h \left(\mathcal {B} _ {\gamma} (y)\right) - \psi_ {\gamma} \circ \nabla h ^ {*} (\nabla h (y)) \\ = \left\langle \mathcal {B} _ {\gamma} (y) - y, \nabla h (y) \right\rangle - h \left(\mathcal {B} _ {\gamma} (y)\right) + h (y) + \left\langle y, \nabla h (y) \right\rangle - h (y) - \psi_ {\gamma} (y) \tag {75} \\ = - D _ {h} \left(\mathcal {B} _ {\gamma} (y), y\right) + \langle y, \nabla h (y) \rangle - h (y) - \psi_ {\gamma} (y) \\ = - D _ {h} \left(\mathcal {B} _ {\gamma} (y), y\right) + g _ {\gamma} (y). \\ \end{array}
$$
D The Bregman Proximal Gradient (BPG) algorithm
D.1 Convergence analysis of the nonconvex BPG algorithm
We study in this section the convergence of the BPG algorithm
$$
x ^ {k + 1} \in T _ {\tau} \left(x _ {k}\right) = \underset {x \in \mathbb {R} ^ {n}} {\arg \min } \left\{\mathcal {R} (x) + \langle x - x ^ {k}, \nabla F \left(x ^ {k}\right) \rangle + \frac {1}{\tau} D _ {h} \left(x, x ^ {k}\right) \right\}. \tag {76}
$$
for minimizing  $\Psi = F + \mathcal{R}$  with nonconvex functions  $F$  and/or  $\mathcal{R}$ .
For the rest of the section, we take the following general assumptions.
Assumption 3.
(i)  $h:\mathbb{R}^n\to \mathbb{R}$  is of Legendre-type.
(ii)  $F:\mathbb{R}^n\to \mathbb{R}$  is proper,  $\mathcal{C}^1$  on int  $\mathrm{dom}(h)$ , with  $\mathrm{dom}(h)\subset \mathrm{dom}(F)$ .
(iii)  $\mathcal{R}:\mathbb{R}^n\to \mathbb{R}$  is proper, lower semi-continuous with  $\mathrm{dom}\mathcal{R}\cap int\mathrm{dom}(h)\neq \emptyset$
(iv)  $\Psi = F + \mathcal{R}$  is lower-bounded, coercive and verifies the Kurdyka-Lojasiewicz (KL) property (defined in Appendix A.3).
(v) For  $x \in \operatorname{int} \operatorname{dom}(h)$ ,  $T_{\tau}(x)$  is nonempty and included in  $\operatorname{int} \operatorname{dom}(h)$ .
Note that, since  $\mathcal{R}$  is nonconvex, the mapping  $T_{\tau}$  is not in general single-valued.
Assumption (v) is required for the algorithm to be well-posed. As shown in [Bauschke et al., 2017, Bolte et al., 2018], one sufficient condition for  $T_{\tau}(x) \neq \emptyset$  is the supercoercivity of function  $h + \lambda \mathcal{R}$  for all  $\lambda > 0$ , that is  $\lim_{||x|| \to +\infty} \frac{h(x) + \lambda \mathcal{R}(x)}{||x||} = +\infty$ . As  $T_{\tau}(x) \subset \operatorname{dom}(h)$ ,  $T_{\tau}(x) \subset \operatorname{int} \operatorname{dom}(h)$  is true when  $\operatorname{dom}(h)$  is open (which is the case for Burg's entropy for example).
The convergence of the BPG algorithm in the nonconvex setting is studied by the authors of [Bolte et al., 2018]. Under the main assumption that  $Lh - F$  is convex on  $int\operatorname{dom}(h)$ , they show first the sufficient decrease property (and thus convergence) of the function values, and second, global convergence of the iterates. However, as we will develop in Appendix D.3,  $Lh - F$  is not convex on the full domain of  $int\operatorname{dom}(h)$  but only on the compact subset  $\operatorname{dom}(\mathcal{R})$ .
One can verify that all the iterates (76) belong to the convex set
$$
T _ {\tau} (x) \subset \operatorname {C o n v} (\operatorname {d o m} \mathcal {R}) \cap \operatorname {i n t} \operatorname {d o m} (h), \tag {77}
$$
where  $\operatorname{Conv}(E)$  stands for the convex envelope of  $E$ . We argue that it is enough to assume  $Lh - F$  convex on this convex subset with the following assumption.
Assumption 4. There is  $L > 0$  such that,  $Lh - F$  is convex on  $\operatorname{Conv}(\operatorname{dom}\mathcal{R}) \cap \operatorname{int}\operatorname{dom}(h)$ .
We can now prove a result similar than [Bolte et al., 2018, Proposition 4.1].
Proposition 3. Under Assumptions 3 and 4, let  $(x^{k})_{k\in \mathbb{N}}$  be a sequence generated by (18) with  $0 < \tau L < 1$ . Then the following properties hold
(i)  $(\Psi (x^{k}))_{k\in \mathbb{N}}$  is non-increasing and converges.
(ii)  $\sum_{k}D_{h}(x^{k + 1},x^{k}) <   \infty$  and  $\min_{0\leq k\leq K}D_h(x^{k + 1},x^k) = O(1 / K).$
Proof. We adapt here the proof from [Bolte et al., 2018] to the case where  $Lh - F$  is not globally convex but only convex on the convex subset  $\mathrm{Conv}(\mathrm{dom}\mathcal{R}) \cap int\mathrm{dom}(h)$ .
Sufficient decrease property We first show that the sufficient decrease property of  $\Psi(x_k)$  holds. This is true because the following characterisation of  $\mathcal{C}^1$  convex functions holds on a convex subset. We recall this classical proof for the sake of completeness.
Lemma 2. Let  $f: \mathcal{X} \to \mathbb{R}^n$  be of class  $\mathcal{C}^1$ , then  $f$  is locally convex on  $C$  a convex subset of  $\mathcal{X} = \text{dom}(f)$  if and only if  $\forall x, y \in C$ ,  $D_f(x, y) = f(x) - f(y) - \langle \nabla f(y), x - y \rangle \geq 0$ .
Proof.  $\Rightarrow$  Let  $x, y \in C$ , for all  $t \in (0,1)$ ,  $x + t(y - x) \in C$  and by convexity of  $f$
$$
f (y + t (x - y)) \leq f (y) + t (f (x) - f (y)), \tag {78}
$$
$$
\frac {f (y + t (x - y)) - f (y)}{t} \leq f (x) - f (y) \tag {79}
$$
$$
\langle \nabla f (y), x - y \rangle = \lim  _ {t \rightarrow 0 ^ {+}} \frac {f (y + t (x - y)) - f (y)}{t} \leq f (x) - f (y). \tag {80}
$$
$\Leftarrow$  Let  $x, y \in C, t \in (0,1)$  and  $z = y + t(x - y) \in C$ . We have
$$
f (y) \geq f (z) + \langle \nabla f (z), y - z \rangle \tag {81}
$$
$$
f (x) \geq f (z) + \langle \nabla f (z), x - z \rangle . \tag {82}
$$
Combining both equations gives
$$
\begin{array}{l} t f (x) + (1 - t) f (y) \leq t (f (z) + \langle \nabla f (z), x - z \rangle + (1 - t) (f (z) + \langle \nabla f (z), y - z \rangle) \\ = f (z) + \left\langle \nabla f (z), t x + (1 - t) y - z \right\rangle \tag {83} \\ = f (t x + (1 - t) y). \\ \end{array}
$$
Therefore, we have  $Lh - F$  convex on  $C$ , if and only if,  $\forall x,y\in C$ ,  $D_{Lh - F}(x,y)\geq 0$ , i.e.  $D_F(x,y)\leq LD_h(x,y)$ .
The rest of the proof is identical to the one of [Bolte et al., 2018] and we recall it here. Given the optimality conditions in (76), all the iterates  $x_{k} \in C$  satisfy
$$
\mathcal {R} \left(x ^ {k + 1}\right) + \left\langle x ^ {k + 1} - x ^ {k}, \nabla F \left(x _ {k}\right) \right\rangle + \frac {1}{\tau} D _ {h} \left(x ^ {k + 1}, x ^ {k}\right) \leq \mathcal {R} \left(x ^ {k}\right) \tag {84}
$$
using  $D_F(x,y)\leq LD_h(x,y)$  ,we get
$$
\begin{array}{l} \left(\mathcal {R} \left(x ^ {k + 1}\right) + F \left(x ^ {k + 1}\right)\right) - \left(\mathcal {R} \left(x ^ {k}\right) + F \left(x ^ {k}\right)\right) \leq - \frac {1}{\tau} D _ {h} \left(x ^ {k + 1}, x ^ {k}\right) + L D _ {h} \left(x ^ {k + 1}, x ^ {k}\right) \\ = \left(L - \frac {1}{\tau}\right) D _ {h} \left(x ^ {k + 1}, x ^ {k}\right) \leq 0 \tag {85} \\ \end{array}
$$
which, together with the fact that  $\Psi$  is lower bounded, proves (i). Summing the previous inequality from  $k = 0$  to  $K - 1$  gives
$$
0 \leq \sum_ {k = 0} ^ {K - 1} D _ {h} \left(x ^ {k + 1}, x ^ {k}\right) \leq \frac {\tau}{1 - \tau L} \left(\Psi \left(x _ {0}\right) - \Psi \left(x _ {K}\right)\right) \leq \frac {\tau}{1 - \tau L} \left(\Psi \left(x _ {0}\right) - \inf  _ {x \in C} \Psi (x)\right) <   + \infty . \tag {86}
$$
Thus  $(D_h(x^{k+1}, x^k))_k$  is summable and converges to 0 when  $k \to +\infty$ . Finally
$$
\min  _ {0 \leq k \leq K} D _ {h} \left(x ^ {k + 1}, x ^ {k}\right) \leq \frac {1}{K + 1} \sum_ {k = 0} ^ {K} D _ {h} \left(x ^ {k + 1}, x ^ {k}\right) \leq \frac {1}{K + 1} \frac {\tau}{1 - \tau L} \left(\Psi \left(x _ {0}\right) - \inf  _ {x \in C} \Psi (x)\right). \tag {87}
$$