SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Notation:  $\mathcal{D}_{\mathrm{GAN}}$
- Generator Function:  $f(u) = u\log (u) - (u + 1)\log (u + 1)$ ,
- Convex Conjugate domain:  $\mathrm{dom}(f^{*}) = \mathbb{R}^{-}$
- Convex Conjugate:  $f^{*}(t) = -\log (1 - \exp (t))$ ,
- Activation:  $v \mapsto -\exp - \log (1 - \exp(v))$ ,
- Optimal discriminator:  $T^{*}(\pmb{x}) = \log \frac{p(\pmb{x})}{p(\pmb{x}) + \widehat{p}(\pmb{x})}$ .
Total Variation:
Notation:  $\mathcal{D}_{\mathrm{TV}}$
- Generator Function:  $f(u) = |u - 1|$ ,
- Convex Conjugate domain:  $\mathrm{dom}(f^{*}) = \left[-\frac{1}{2},\frac{1}{2}\right]$
- Convex Conjugate:  $f^{*}(t) = t$ ,
- Activation:  $v \mapsto \tanh(v)$ ,
- Optimal discriminator:  $T^{*}(\pmb{x}) = \frac{1}{2}\mathrm{sign}\left(\frac{p(\pmb{x})}{\widehat{p}(\pmb{x})} - 1\right)$ .
PR-Divergence:
Notation:  $\mathcal{D}_{\lambda_{-}\mathrm{PR}}$  for  $\lambda >0$
- Generator Function:  $f(u) = \max (\lambda u,1) - \max (\lambda ,1),$
- Convex Conjugate domain:  $\mathrm{dom}(f^{*}) = [0,\lambda ]$
- Convex Conjugate:  $f_{\lambda}^{*}(t) = t / \lambda$  for  $\lambda \leq 1$  and  $f_{\lambda}^{*}(t) = t / \lambda + \lambda - 1$  otherwise,
- Activation:  $v \mapsto \lambda \sigma(v)$ , where  $\sigma$  is the sigmoid function,
- Optimal discriminator:  $T^{*}(\pmb{x}) = \lambda \mathrm{sign}\left(\frac{p(\pmb{x})}{\widehat{p}(\pmb{x})} - 1\right)$ .
A.2 Bregman Divernce
The Bregman Divergence under a strictly convex function  $f: \mathbb{R}^n \to \mathbb{R}$  with a continuously differentiable interior, between two points  $x$  and  $y$  in the interior of the domain of  $f$ , is denoted by  $\mathrm{Breg}_f$  and is defined as:
$$
\operatorname {B r e g} _ {f} (\boldsymbol {x}, \boldsymbol {y}) = f (\boldsymbol {x}) - f (\boldsymbol {y}) - \langle \nabla f (\boldsymbol {y}), \boldsymbol {x} - \boldsymbol {y} \rangle
$$
where  $\nabla f(\pmb{y})$  is the gradient of  $f$  at  $\pmb{y}$ , and  $\langle .,.\rangle$  is the inner product in  $\mathbb{R}^n$ . It follows some properties as a distance metrics:
- Non-Negativity: For any  $\mathbf{x}, \mathbf{y}$  in the interior of the domain of  $f$ , we have  $\mathrm{Breg}_f(\mathbf{x}, \mathbf{y}) \geq 0$ . The equality holds if and only if  $\mathbf{x} = \mathbf{y}$ .
- Convexity: The Bregman Divergence  $\mathrm{Breg}_f(\pmb{x},\pmb{y})$  is convex in its first argument. That is, for any  $\pmb{x}_1,\pmb{x}_2,\pmb{y}$  in the interior of the domain of  $f$  and any  $t\in [0,1]$ , we have:
$$
\operatorname {B r e g} _ {f} (t \boldsymbol {x} _ {1} + (1 - t) \boldsymbol {x} _ {2}, \boldsymbol {y}) \leq t \operatorname {B r e g} _ {f} (\boldsymbol {x} _ {1}, \boldsymbol {y}) + (1 - t) \operatorname {B r e g} _ {f} (\boldsymbol {x} _ {2}, \boldsymbol {y})
$$
- Non-Symmetry: Unlike some other distances or divergences, the Bregman Divergence is not symmetric, meaning that in general,  $\operatorname{Breg}_f(\boldsymbol{x},\boldsymbol{y}) \neq \operatorname{Breg}_f(\boldsymbol{y},\boldsymbol{x})$ .
- Additivity: If  $f(\pmb{x}) = f_1(x_1) + \dots + f_d(x_d)$  where each  $f_i$  is a convex function, then the Bregman Divergence decomposes into a sum of Bregman Divergences, i.e.,  $\mathrm{Breg}_f(\pmb{x},\pmb{y}) = \mathrm{Breg}_{f_1}(x_1,y_1) + \dots + \mathrm{Breg}_{f_d}(x_d,y_d)$ .
- Taylor Approximation: Bregman Divergence is essentially the error of the first-order Taylor approximation of  $f$  around the point  $y$  at the point  $x$ .
- Connection with Dual Functions: If  $f$  is convex, then it has a convex conjugate (or dual function)  $f^{*}$ . The Bregman Divergence  $\mathrm{Breg}_f(\pmb{x},\pmb{y})$  is related to the Bregman Divergence  $\mathrm{Breg}_{f^*}(\nabla f(\pmb{y}),\nabla f(\pmb{x}))$  between the gradients of  $f$  at  $\pmb{x}$  and  $\pmb{y}$ , where the gradient map  $\nabla f$  serves as a Legendre transformation between the primal and dual spaces.
B Proof and supplementary for Section 4
B.1 Proof for Theorem 4.3
We have to prove that  $\alpha (\lambda)$  can be written as a function of an  $f$  -divergence for any  $\lambda \in \mathbb{R}^{+}$ . First we can develop the expression of  $\alpha (\lambda)$ :
$$
\begin{array}{l} \alpha (\lambda) = \int_ {\mathcal {X}} \min  \left(\lambda p (\boldsymbol {x}), \widehat {p} (\boldsymbol {x})\right) \mathrm {d} \boldsymbol {x} (16) \\ = \int_ {\mathcal {X}} \widehat {p} (\boldsymbol {x}) \min  \left(\lambda \frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})}, 1\right) \mathrm {d} \boldsymbol {x} (17) \\ \end{array}
$$
For this integral to be considered as an  $f$ -divergence, we need  $f$  to be first convex lower semicontinuous and then to satisfy  $f(1) = 0$ . However, for every  $a, b \in \mathbb{R}$ , the min satisfies  $\min(a, b) = a + b - \max(a, b)$ . Therefore,
$$
\begin{array}{l} \alpha (\lambda) = \int_ {\mathcal {X}} \widehat {p} (\boldsymbol {x}) \left[ \lambda \frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})} + 1 - \max  \left(\lambda \frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})}, 1\right) \right] d \boldsymbol {x} (18) \\ = \lambda \int_ {\mathcal {X}} p (\boldsymbol {x}) \mathrm {d} \boldsymbol {x} + 1 - \int_ {\mathcal {X}} \max  \left(\lambda \frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})}, 1\right) \mathrm {d} \boldsymbol {x} (19) \\ = \lambda + 1 - \int_ {\mathcal {X}} \widehat {p} (\boldsymbol {x}) \max  \left(\lambda \frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})}, 1\right) \mathrm {d} \boldsymbol {x} (20) \\ \end{array}
$$
Thus, we can take  $f(u) = \max(\lambda u, 1) - \max(\lambda, 1)$  such that  $f(1) = 0$ . The precision becomes:
$$
\begin{array}{l} \alpha (\lambda) = \lambda + 1 - \int_ {\mathcal {X}} \widehat {p} (\boldsymbol {x}) f \left(\left(\frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})}\right) - \max  (\lambda , 1) \int_ {\mathcal {X}} \widehat {p} (\boldsymbol {x}) \mathrm {d} \boldsymbol {x} \right. (21) \\ = \min  (\lambda , 1) - \int_ {\mathcal {X}} \widehat {p} (\boldsymbol {x}) f \left(\frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})}\right) d \boldsymbol {x} = \min  (\lambda , 1) - \mathcal {D} _ {\lambda - \mathrm {P R}} (P, \widehat {P}). (22) \\ \end{array}
$$
Consequently,  $\alpha (\lambda)$  can be written as a function of an  $f$  -divergence  $\mathcal{D}_{\lambda \text{-PR}}$  with  $f(u) = \max (\lambda u,1) - \max (\lambda ,1)$ .
Now we prove the converse. Suppose there exists a strictly decreasing linear $^4$  function  $h:[0,1]\to \mathbb{R}^+$  and an  $f$ -divergence  $\mathcal{D}_f$  such that  $h(\alpha_{\lambda}(P\parallel \widehat{P})) = \mathcal{D}_f(P\| \widehat{P})$  for all  $P,\widehat{P}\in \mathcal{P}(\mathcal{X})$ .
For  $P = \widehat{P}$ , we get from the definition of  $\alpha_{\lambda}$  that  $\alpha_{\lambda}(P\| P) = \min (\lambda ,1)$ . Hence,
$$
0 = \mathcal {D} _ {f} (P \| P) = h (\alpha_ {\lambda} (P \| P)) = h (\min  (\lambda , 1)).
$$
<sup>4</sup>We omitted the critical constraint of  $h$  being linear in the original statement of the theorem in our submission by mistake. We apologize for this oversight. We stress again that the first part of the Theorem, which is the most important part, remains completely unaffected.
Combining the above with the fact that  $h$  is a strictly decreasing linear function, we see that for any fixed  $\lambda$ ,  $h$  must be of the form,  $h(u) = c_{\lambda}(\min(\lambda, 1) - u)$ , where  $c_{\lambda} > 0$  is a constant. Now,
$$
\mathcal {D} _ {f} (P \| \widehat {P}) = h (\alpha_ {\lambda} (P \| \widehat {P})) = c _ {\lambda} (\min (\lambda , 1) - \alpha_ {\lambda} (P \| \widehat {P})) = c _ {\lambda} \mathcal {D} _ {\lambda - \mathrm {P R}} (P \| \widehat {P}),
$$
where the last equality follows from the first part of the theorem, which shows that  $\alpha_{\lambda}(P\| \widehat{P}) = \min (\lambda ,1) - \mathcal{D}_{\lambda -\mathrm{PR}}(P\| \widehat{P})$ . Rewriting the above inequality, we get the following.
$$
\mathcal {D} _ {\lambda - \mathrm {P R}} (P \| \widehat {P}) = \frac {1}{c _ {\lambda}} \mathcal {D} _ {f} (P \| \widehat {P}) = \mathcal {D} _ {\frac {1}{c _ {\lambda}} f} (P \| \widehat {P}).
$$
By the uniqueness theorem of  $f$ -divergence  $f(u) = \frac{c_1}{c_\lambda} f_\lambda(u) + c_2(u - 1)$  for some constants  $c_1, c_2 \in \mathbb{R}$ .
B.2 Proof of Proposition 4.2
If the generator function  $f$  of the Precision-Recall Divergence is  $f(u) = \max(\lambda u, 1) - \max(\lambda, 1)$  then its Fenchel conjugate function is:
$$
f ^ {*} (t) = \sup  _ {u \in \operatorname {d o m} (f)} \left\{t u - f (u) \right\} = \max  (\lambda , 1) + \sup  _ {u \in \mathbb {R} ^ {+}} \left\{t u - \max  (\lambda u, 1) \right\} \tag {23}
$$
If  $t > \lambda$  or  $\lambda < 0$ , then the  $\sup_{u \in \mathbb{R}^+} \{tu - \max(\lambda u, 1)\} = \infty$  for respectively  $u \to \infty$  and  $u \to -\infty$ . The domain of  $f^*$  is thus restricted to  $[0, \lambda]$ . Thus for  $0 \leq t \leq \lambda$ , the supremum is obtained for  $u = 1 / \lambda$  since 0 is in the sub-differential of the function in  $1 / \lambda$  as Figure 6(b).
(a) Function  $f(u)$  for different values of  $\lambda$ .
(b) Function  $u\mapsto ut - \max (\lambda u,1)$  for values of  $t$  between 0 and  $\lambda$
Figure B.6: Graphical illustration of  $f_{\lambda}$  and  $f_{\lambda}^{*}$ . Both are piece-wise linear
Consequently the Fenchel conjugate of  $f$  is:
$$
\forall t \in [ 0, \lambda ], \quad f ^ {*} (t) = \max  (\lambda , 1) + t \lambda - 1 = \left\{ \begin{array}{l l} t / \lambda & \text {i f} \lambda \leq 1, \\ t / \lambda - 1 + \lambda & \text {o t h e r w i s e .} \end{array} \right. \tag {24}
$$
Finally, the optimal discriminator  $T^{\mathrm{opt}}$  by taking the derivative of  $f$  in  $\frac{p(\boldsymbol{x})}{\widehat{p}(\boldsymbol{x})}$ , we get:
$$
T ^ {\text {o p t}} (\boldsymbol {x}) = \nabla f \left(\frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})}\right) = \left\{ \begin{array}{l l} \lambda & \text {i f} \frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})} \leq 1 / \lambda , \\ 0 & \text {o t h e r w i s e .} \end{array} \right. \tag {25}
$$
Then we can compute the compute the reverse  $\mathcal{D}_{\lambda_{\mathrm{PR}}}$
$$
\begin{array}{l} \mathcal {D} _ {\lambda - \mathrm {P R}} (\widehat {P} \| P) = \int_ {\chi} p (\boldsymbol {x}) f _ {\lambda} \left(\frac {\widehat {p} (\boldsymbol {x})}{p (\boldsymbol {x})}\right) \mathrm {d} \boldsymbol {x} (26) \\ = \int_ {\mathcal {X}} \max  (\lambda \widehat {p} (\boldsymbol {x}), p (\boldsymbol {x})) - p (\boldsymbol {x}) \max  (\lambda , 1) \mathrm {d} \boldsymbol {x} (27) \\ = \lambda \left(\int_ {\mathcal {X}} \max  (\widehat {p} (\boldsymbol {x}), p (\boldsymbol {x}) / \lambda) \mathrm {d} \boldsymbol {x} - \max  (1, 1 / \lambda)\right) (28) \\ = \lambda \int_ {\mathcal {X}} \widehat {p} (\boldsymbol {x}) \max  \left(1, \frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})} / \lambda\right) - \widehat {p} (\boldsymbol {x}) \max  \left(1, 1 / \lambda\right) d \boldsymbol {x} (29) \\ = \lambda \int_ {\mathcal {X}} \widehat {p} (\boldsymbol {x}) f _ {1 / \lambda} \left(\frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})}\right) d \boldsymbol {x} (30) \\ = \lambda \mathcal {D} _ {\frac {1}{\lambda} - \mathrm {P R}} (P \| \widehat {P}). (31) \\ \end{array}
$$
With this results, we can show that:
$$
\begin{array}{l} \mathcal {D} _ {\mathrm {T V}} (P \| \widehat {P}) = \int_ {\mathcal {X}} | p (\boldsymbol {x}) - \widehat {p} (\boldsymbol {x}) | \mathrm {d} \boldsymbol {x} (32) \\ = \int_ {\mathcal {X}} \max  (p (\boldsymbol {x}) - \widehat {p} (\boldsymbol {x}), 0) + \max  (\widehat {p} (\boldsymbol {x}) - p (\boldsymbol {x}), 0) \mathrm {d} \boldsymbol {x} (33) \\ \end{array}
$$
Then since  $\mathcal{D}_{1\text{-PR}}(P\| \widehat{P}) = \int_{\mathcal{X}}\max (\widehat{p} (\boldsymbol {x}),p(\boldsymbol {x})) - p(\boldsymbol {x})\mathrm{d}\boldsymbol {x} = \int_{\mathcal{X}}\max (\widehat{p} (\boldsymbol {x}) - p(\boldsymbol {x}),0)\mathrm{d}\boldsymbol{x}$  and  $\mathcal{D}_{1\text{-PR}}(P\| \widehat{P}) = \mathcal{D}_{1\text{-PR}}(\widehat{P}\| P)$ , we have:
$$
\begin{array}{l} \mathcal {D} _ {\mathrm {T V}} (P \| \widehat {P}) = \mathcal {D} _ {1 - \mathrm {P R}} (P \| \widehat {P}) + \mathcal {D} _ {1 - \mathrm {P R}} (\widehat {P} \| P) (34) \\ = 2 \mathcal {D} _ {1 - \mathrm {P R}} (P \| \widehat {P}). (35) \\ \end{array}
$$
From now on, assume the support of  $P$  and  $\widehat{P}$  coincide. For any  $T:\mathcal{X}\to \mathbb{R}$
$$
\begin{array}{l} \mathcal {D} _ {f, T} ^ {\mathrm {d u a l}} (P \| \widehat {P}) = \mathbb {E} _ {x \sim P} [ d (x) ] - \mathbb {E} _ {x \sim Q} [ f ^ {*} (d (x)) ] \\ = \mathbb {E} _ {x \sim Q} \left[ \frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})} d (x) - f ^ {*} (d (x)) \right] \\ \end{array}
$$
Let  $T^{\mathrm{opt}}\in \arg \sup \mathcal{D}_{f,T}^{\mathrm{dual}}(P\| \widehat{P})$  . For any  $T:\mathcal{X}\to \mathbb{R}$
$$
\begin{array}{l} \mathcal {D} _ {f} (P \| \widehat {P}) - D _ {f, T} ^ {\text {d u a l}} (P \| \widehat {P}) = \mathcal {D} _ {f, T ^ {\text {o p t}}} ^ {\text {d u a l}} (P \| \widehat {P}) - \mathcal {D} _ {f, T} ^ {\text {d u a l}} (P \| \widehat {P}) \\ = \mathbb {E} _ {\widehat {P}} \left[ \frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})} \left(T ^ {\mathrm {o p t}} (\boldsymbol {x}) - T (\boldsymbol {x})\right) - f ^ {*} \left(T ^ {\mathrm {o p t}} (\boldsymbol {x})\right) + f ^ {*} \left(T (\boldsymbol {x})\right) \right] \\ \end{array}
$$
It is known that for all  $x\in \mathcal{X}$  we have  $\nabla f^{*}(T^{\mathrm{opt}}(\pmb {x})) = \frac{p(\pmb{x})}{\widehat{p}(\pmb{x})}$
$$
\mathcal {D} _ {f} (P \| \widehat {P}) - \mathcal {D} _ {f, T} ^ {\mathrm {d u a l}} (P \| \widehat {P}) = \mathbb {E} _ {\widehat {P}} \left[ \nabla f ^ {*} \left(T ^ {\mathrm {o p t}} (\boldsymbol {x})\right) \left(T ^ {\mathrm {o p t}} (\boldsymbol {x}) - T (\boldsymbol {x})\right) - f ^ {*} \left(T ^ {\mathrm {o p t}} (\boldsymbol {x})\right) + f ^ {*} \left(T (\boldsymbol {x})\right) \right]
$$
Recall that for any continuously differentiable strictly convex function  $f$ , the Bregman divergence of  $f$  is  $\mathrm{Breg}_f(a, b) = f(a) - f(b) - \langle \nabla f(b), a - b \rangle$ . So we have
$$
\mathcal {D} _ {f} (P \| \widehat {P}) - \mathcal {D} _ {f, T} ^ {\mathrm {d u a l}} (P \| \widehat {P}) = \mathbb {E} _ {\widehat {P}} \left[ \operatorname {B r e g} _ {f ^ {*}} \left(T (\boldsymbol {x}), T ^ {\mathrm {o p t}} (\boldsymbol {x})\right) \right]
$$
Let us now use the following property:  $\mathrm{Breg}_f(a,b) = \mathrm{Breg}_{f^*}(a^*,b^*)$  where  $a^* = \nabla f(a)$  and  $b^{*} = \nabla f(b)$ .
$$
\begin{array}{l} \mathcal {D} _ {f} (P \| \widehat {P}) - \mathcal {D} _ {f, T} ^ {\text {d u a l}} (P \| \widehat {P}) = \mathbb {E} _ {\widehat {P}} \left[ \operatorname {B r e g} _ {f} \left(\nabla f ^ {*} (T (\boldsymbol {x})), \nabla f ^ {*} (T ^ {\mathrm {o p t}} (\boldsymbol {x}))\right) \right] \\ = \mathbb {E} _ {\widehat {P}} \left[ \operatorname {B r e g} _ {f} \left(\nabla f ^ {*} (T (\boldsymbol {x})), \frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})}\right) \right] \\ \end{array}
$$
Let us define  $r(\pmb{x}) = \nabla f^{*}T(\pmb{x})$  as our estimator of  $p(\pmb{x}) / \widehat{p}(\pmb{x})$ . So finally, we have
$$
\mathcal {D} _ {f} (P \| \widehat {P}) - \mathcal {D} _ {f, T} ^ {\mathrm {d u a l}} (P \| \widehat {P}) = \mathbb {E} _ {\widehat {P}} \left[ \operatorname {B r e g} _ {f} \left(r (\boldsymbol {x}), \frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})}\right) \right]
$$
B.4 Proof of Theorem 5.3
Now assume that  $f$  is  $\mu$ -strongly convex, then  $\mathrm{Breg}_f(a,b) \geq \frac{\mu}{2} \|a - b\|^2$  If  $\mathbb{E}_{\hat{P}}\left[\mathrm{Breg}_f\left(r(\boldsymbol{x}), \frac{p(\boldsymbol{x})}{\widehat{p}(\boldsymbol{x})}\right)\right] \leq \epsilon$  and if  $f$  is  $\mu$ -strongly convex, then
$$
\mathbb {E} _ {\hat {P}} \left[ \left(r (\boldsymbol {x}) - \frac {p (\boldsymbol {x})}{\hat {p} (\boldsymbol {x})}\right) ^ {2} \right] \leq \frac {2 \epsilon}{\mu}. \tag {36}
$$
Consider an arbitrary f-divergence  $\mathcal{D}_g(P\| \widehat{P}) = \int g\left(\frac{dP}{\mathrm{d}\widehat{P}}\right)\mathrm{d}\widehat{P}$ . Define  $\mathcal{D}_{g,T}^{\mathrm{primal}}(P\| \widehat{P}) = \int g(r(\pmb {x}))\mathrm{d}\widehat{P}$ . Then,
$$
\begin{array}{l} \left| \mathcal {D} _ {g} (P \| \widehat {P}) - \mathcal {D} _ {g, T} ^ {\text {p r i m a l}} (P \| \widehat {P}) \right| = \left| \mathbb {E} _ {\widehat {P}} \left[ g \left(\frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})}\right) - g \left(r (\boldsymbol {x})\right) \right] \right| \\ \leq \mathbb {E} _ {\widehat {P}} \left[ \left| g \left(\frac {p (\boldsymbol {x})}{\widehat {p} (\boldsymbol {x})}\right) - g (r (\boldsymbol {x})) \right| \right] \\ \stackrel {(a)} {\leq} \mathbb {E} _ {\widehat {P}} [ \sigma | e (x) | ] \\ = \sigma \mathbb {E} _ {\hat {P}} \left[ \left| e (x) \right| \right] \\ \stackrel {(b)} {\leq} \sigma \sqrt {\mathbb {E} _ {\hat {P}} [ e (x) ^ {2} ]} \\ \stackrel {(c)} {\leq} \sigma \sqrt {\frac {2 \epsilon}{\mu}}, \\ \end{array}
$$
where  $(a)$  follows from the  $\sigma$ -Lipschitz assumption on  $g$ ,  $(b)$  follows from Jensen's inequality and finally,  $(c)$  follows from equation (36). s
B.5 Proof of Theorem 8
Let  $c: \mathbb{R}^+ \mapsto \mathbb{R}$  be a  $\mathcal{C}^2$  function and take  $u_{\min}$  and  $u_{\max}$ . The goal is to express  $f(u)$  for all  $u \in [u_{\min}, u_{\max}]$  as a weighted average of  $f_{\lambda}^{\mathrm{PR}}$ :
$$
\forall u \in \mathbb {R} _ {*} ^ {+}, \int_ {1 / u _ {\max }} ^ {1 / u _ {\min }} c ^ {\prime \prime} (\lambda) f _ {\lambda} ^ {\mathrm {P R}} (u) \mathrm {d} \lambda = \int_ {1 / u _ {\max }} ^ {1 / u _ {\min }} c ^ {\prime \prime} (\lambda) \left[ \max  (\lambda u, 1) - \max  (\lambda , 1) \right] \mathrm {d} \lambda \tag {37}
$$
First, let us assume that  $u_{\min} \leq 1$  and  $u_{\max} \geq 1$ , then the terms can be decomposed and the integral split to evaluate the max:
$$
\begin{array}{l} \int_ {1 / u _ {\max }} ^ {1 / u _ {\min }} c ^ {\prime \prime} (\lambda) f _ {\lambda} ^ {\mathrm {P R}} (u) \mathrm {d} \lambda = \int_ {1 / u _ {\max }} ^ {1 / u _ {\min }} c ^ {\prime \prime} (\lambda) \max  (\lambda u, 1) \mathrm {d} \lambda - \int_ {1 / u _ {\max }} ^ {1 / u _ {\min }} c ^ {\prime \prime} (\lambda) \max  (\lambda , 1) \mathrm {d} \lambda (38) \\ = \int_ {1 / u _ {\max }} ^ {1 / u} c ^ {\prime \prime} (\lambda) \max  (\lambda u, 1) d \lambda + \int_ {1 / u} ^ {1 / u _ {\min }} c ^ {\prime \prime} (\lambda) \max  (\lambda u, 1) d \lambda \\ - \int_ {1 / u _ {\max }} ^ {1} c ^ {\prime \prime} (\lambda) \max  (\lambda , 1) d \lambda - \int_ {1} ^ {1 / u _ {\min }} c ^ {\prime \prime} (\lambda) \max  (\lambda , 1) d \lambda (39) \\ = \int_ {1 / u _ {\max }} ^ {1 / u} c ^ {\prime \prime} (\lambda) \mathrm {d} \lambda + \int_ {1 / u} ^ {1 / u _ {\min }} c ^ {\prime \prime} (\lambda) \lambda u \mathrm {d} \lambda - \int_ {1 / u _ {\max }} ^ {1} c ^ {\prime \prime} (\lambda) \mathrm {d} \lambda - \int_ {1} ^ {1 / u _ {\min }} c ^ {\prime \prime} (\lambda) \lambda \mathrm {d} \lambda . (40) \\ \end{array}
$$
By integrating by parts, we have:  $\int_{1 / u_{\mathrm{max}}}^{1 / u_{\mathrm{min}}}c^{\prime \prime}(\lambda)\lambda \mathrm{d}\lambda = [c^{\prime}(\lambda)\lambda ]_{1 / u_{\mathrm{max}}}^{1 / u_{\mathrm{min}}} - \int_{1 / u_{\mathrm{max}}}^{1 / u_{\mathrm{min}}}c^{\prime}(\lambda)\mathrm{d}\lambda$  so it satisfies:
$$
\begin{array}{l} \int_ {1 / u _ {\max }} ^ {1 / u _ {\min }} c ^ {\prime \prime} (\lambda) f _ {\lambda} ^ {\mathrm {P R}} (u) \mathrm {d} \lambda = \int_ {1 / u _ {\max }} ^ {1 / u} c ^ {\prime \prime} (\lambda) \mathrm {d} \lambda + u [ c ^ {\prime} (\lambda) \lambda ] _ {1 / u} ^ {1 / u _ {\min }} - u \int_ {1 / u} ^ {1 / u _ {\min }} c ^ {\prime} (\lambda) \mathrm {d} \lambda (41) \\ - \int_ {1 / u _ {\max }} ^ {1} c ^ {\prime \prime} (\lambda) d \lambda - [ c ^ {\prime} (\lambda) \lambda ] _ {1} ^ {1 / u _ {\min }} + \int_ {1} ^ {1 / u _ {\min }} c ^ {\prime} (\lambda) d \lambda \\ = \left[ c ^ {\prime} (\lambda) \right] _ {1 / u _ {\max }} ^ {1 / u} + u \left[ c ^ {\prime} (\lambda) \lambda \right] _ {1 / u} ^ {1 / u _ {\min }} - u \left[ c (\lambda) \right] _ {1 / u} ^ {1 / u _ {\min }} (42) \\ - \left[ c ^ {\prime} (\lambda) \right] _ {1 / u _ {\max}} ^ {1} - \left[ c ^ {\prime} (\lambda) \lambda \right] _ {1} ^ {1 / u _ {\min}} + \left[ c (\lambda) \right] _ {1} ^ {1 / u _ {\min}} \\ = c ^ {\prime} \left(\frac {1}{u}\right) - c ^ {\prime} (0) + u c ^ {\prime} \left(\frac {1}{u _ {\min }}\right) - u c ^ {\prime} \left(\frac {1}{u}\right) \frac {1}{u} - u c \left(\frac {1}{u _ {\min }}\right) + u c \left(\frac {1}{u}\right) - c ^ {\prime} (1) \\ + c ^ {\prime} (0) - c ^ {\prime} \left(\frac {1}{u _ {\operatorname* {m i n}}}\right) \frac {1}{u _ {\operatorname* {m i n}}} + c ^ {\prime} (1) \times 1 + c \left(\frac {1}{u _ {\operatorname* {m i n}}}\right) - c (1) (43) \\ = \left[ c ^ {\prime} \left(\frac {1}{u _ {\operatorname* {m i n}}}\right) \frac {1}{u _ {\operatorname* {m i n}}} - c \left(\frac {1}{u _ {\operatorname* {m i n}}}\right) \right] (u - 1) + u c \left(\frac {1}{u}\right) - c (1). (44) \\ \end{array}
$$
We would like  $\int_{1 / u_{\mathrm{max}}}^{1 / u_{\mathrm{min}}}c^{\prime \prime}(\lambda)f_{\lambda}^{\mathrm{PR}}(u)\mathrm{d}\lambda$  to be equal to  $f$  between  $u_{\mathrm{min}}$  and  $u_{\mathrm{max}}$ . But since two  $f$ -divergences generated by  $f$  and  $g$  are equals if there is a  $c\in \mathbb{R}$  such that  $f(u) = g(u) + c(u - 1)$ , the Divergence generated by  $\int_{1 / u_{\mathrm{max}}}^{1 / u_{\mathrm{min}}}c^{\prime \prime}(\lambda)f_{\lambda}^{\mathrm{PR}}(u)\mathrm{d}\lambda$  is equal to the divergence generated by  $uc\left(\frac{1}{u}\right) - c(1)$ . Therefore, we require  $c$  to satisfy:
$$
\forall u \in [ u _ {\min }, u _ {\max } ], \quad f (u) = u c \left(\frac {1}{u}\right) - c (1).
$$
By differentiating with respect to  $u$ , we have:
$$
f ^ {\prime} (u) = \lim  _ {\lambda \rightarrow \infty} \left[ c ^ {\prime} (\lambda) \lambda - c (\lambda) \right] + c \left(\frac {1}{u}\right) - \frac {1}{u} c ^ {\prime} \left(\frac {1}{u}\right). \tag {45}
$$
And finally:
$$
\begin{array}{l} f ^ {\prime \prime} (u) = - \frac {1}{u ^ {2}} c \left(\frac {1}{u}\right) + \frac {1}{u ^ {2}} c ^ {\prime} \left(\frac {1}{u}\right) + \frac {1}{u ^ {3}} c ^ {\prime \prime} \left(\frac {1}{u}\right) (46) \\ = \frac {1}{u ^ {3}} c ^ {\prime \prime} \left(\frac {1}{u}\right). (47) \\ \end{array}
$$
Consequently, with  $\lambda = 1 / u$ , we have that:
$$
c ^ {\prime \prime} (\lambda) = \frac {1}{\lambda^ {3}} f ^ {\prime \prime} \left(\frac {1}{\lambda}\right). \tag {48}
$$
With such a results, with  $m = \min_{\mathcal{X}}\left(\frac{\widehat{p}(\mathbf{x})}{p(\mathbf{x})}\right)$  and  $M = \max_{\mathcal{X}}\left(\frac{\widehat{p}(\mathbf{x})}{p(\mathbf{x})}\right)$ , we can write any  $f$ -divergence as: