Improvements of Plachky-Steinebach theorem

TL;DR

This paper extends the Plachky-Steinebach theorem to broader conditions, allowing for non-differentiability and discontinuities in the generalized log-moment generating function, thereby broadening its applicability in probability theory.

Contribution

It provides new conditions under which the Plachky-Steinebach theorem holds, including cases with non-differentiability, discontinuities, and without the need for strict convexity.

Findings

The theorem holds for intervals based on the right derivative of the generalized log-moment generating function.

The result extends to the case when , with conditions on the limit points and continuity.

The theorem applies to general nets of probability measures, even with non-strict convexity.

Abstract

We show that the conclusion of Plachky-Steinebach theorem holds true for intervals of the form $\left]\overline{L}_r'(\lambda),y\right[$, where $\overline{L}_r'(\lambda)$ is the right derivative (but not necessarily a derivative) of the generalized log-moment generating function $\overline{L}$ at some $\lambda> 0$ and $y\in\ \left]\overline{L}_r'(\lambda),+\infty\right]$, under the only two following conditions: $(a)$ $\overline{L}'_r(\lambda)$ is a limit point of the set $\left\{\overline{L}'_r(t):t>\lambda\right\}$, $(b)$ $\overline{L}(t_i)$ is a limit for a suitable sequence $(t_i)$. By replacing $\overline{L}_r'(\lambda)$ by $\overline{L}_r'(\lambda^+)$, the above result extends verbatim to the case $\lambda=0$ (replacing $(a)$ by the right continuity of $\overline{L}$ at zero when $\overline{L}_r'(0^+)=-\infty$). No hypothesis is made on $\overline{L}_{]-\infty,\lambda[}$ (e.g. $\overline{L}_{]-\infty,\lambda[}$ may be the constant $+\infty$ when $\lambda=0$); $\lambda\ge 0$ may be a non-differentiability point of $\overline{L}$ and moreover a limit point of non-differentiability points of $\overline{L}$; $\lambda=0$ may be a left and right discontinuity point of $\overline{L}$. The map $\overline{L}_{\mid ]\lambda,\lambda+\varepsilon[}$ may fail to be strictly convex for all $\varepsilon>0$. If we drop the assumption $(b)$, then the same conclusion holds with upper limits in place of limits. The foregoing is valid for general nets $(\mu_\alpha,c_\alpha)$ of Borel probability measures and powers and replacing the intervals $\left]\overline{L}_r'(\lambda^+),y\right[$ by $]x_\alpha,y_\alpha[$ or $[x_\alpha,y_\alpha]$, where $(x_\alpha,y_\alpha)$ is any net such that $(x_\alpha)$ converges to $\overline{L}_r'(\lambda^+)$ and $\liminf_\alpha y_\alpha>\overline{L}_r'(\lambda^+)$.

Equations

$$\overline{L}(t)=\underset{\alpha }{\mathrm{limsup}}{c}_{\alpha }\log {\int }_{\mathbb{R}}{e}^{c_{\alpha }^{-1}tx}{\mu }_{\alpha }(dx);$$

$$\underset{n}{\lim }{n}^{-1}\log {\mu }_n(]x_n,+\infty [)=L(\lambda )-\lambda L^{\prime }(\lambda ).$$

$${\overline{L}}_r^{\prime }({\lambda }^{+})<{\overline{L}}_r^{\prime }({\lambda }_i)<+\infty ,$$

$$\limsup_{\alpha}c_{\alpha}\log\mu_{\alpha}([x_{\alpha},y_{\alpha}])=\limsup_{\alpha}c_{\alpha}\log\mu_{\alpha}(]x_{\alpha},y_{\alpha}[)=\overline{L}(\lambda^{+})-\lambda\overline{L}_{r}^{\prime}(\lambda^{+})\\ =\left\{\begin{array}[]{ll}-\overline{L}^{*}(\overline{L}_{r}^{\prime}(\lambda^{+}))=\inf_{t>0}\{\overline{L}(t)-t\overline{L}_{r}^{\prime}(\lambda^{+})\}&if\ \overline{L}_{r}^{\prime}(\lambda^{+})\neq-\infty,\\ -\lim_{\alpha}\overline{L}^{*}(x_{\alpha})=0&if\ \overline{L}_{r}^{\prime}(\lambda^{+})=-\infty.\end{array}\right.$$

$$\limsup_{\alpha}c_{\alpha}\log\mu_{\alpha}([x_{\alpha,z},y_{\alpha,z}])=\limsup_{\alpha}c_{\alpha}\log\mu_{\alpha}(]x_{\alpha,z},y_{\alpha,z}[)=\overline{L}(t_{z}^{+})-t_{z}z\\ =\left\{\begin{array}[]{ll}-\overline{L}^{*}(z)=\inf_{t>0}\{\overline{L}(t)-tz\}&if\ z\neq-\infty\\ -\lim_{\alpha}\overline{L}^{*}(x_{\alpha,z})=0&if\ z=-\infty\ \ ({\it i.e.\/}\ \lambda=0\ and\ z=\overline{L}_{r}^{\prime}(0^{+})=-\infty).\end{array}\right.$$

$$\underset{\alpha }{\lim }{c}_{\alpha }\log {\mu }_{\alpha }([x_{\alpha },+\infty [)=0\ne \overline{L}(0^{+}),$$

$$\tilde{\lambda }=\sup \{t>\lambda :{\overline{L}}_{\mid ]\lambda ,t]}\text{is affine}\}=\underset{i}{\lim }{\lambda }_i$$

$${\overline{L}}_r^{\prime }({\lambda }^{+})<{\overline{L}}_r^{\prime }({\lambda }_i)<{\overline{L}}_r^{\prime }({\tilde{\lambda }}^{+}),$$

$${\overline{L}}_r^{\prime }({\lambda }^{+})={\overline{L}}_r^{\prime }({\tilde{\lambda }}^{+})<{\overline{L}}_r^{\prime }({\lambda }_i)<{\overline{L}}_r^{\prime }(\tilde{\lambda }+\epsilon )$$

$$\forall x\in \mathbb{R},l_0(x)=-\underset{\epsilon \to 0,\epsilon >0}{\lim }\underset{\alpha }{\mathrm{limsup}}{c}_{\alpha }\log {\mu }_{\alpha }(]x-\epsilon ,x+\epsilon [);$$

$$\forall x\in \mathbb{R},l_0(x)\ge {\overline{L}}^{\ast }(x).$$

$$\forall \lambda \in \mathbb{R}\setminus \{0\},\overline{L}(\lambda )\ge \underset{y\in \mathbb{R}}{\sup }\{\lambda y-l_0(y)\}.$$

$$\forall (\lambda ,x)\in {\mathbb{R}}^2,\overline{L}(\lambda )-\lambda x\ge \underset{y\in \mathbb{R}}{\sup }\{\lambda y-l_0(y)\}-\lambda x\ge -l_0(x),$$

$$\forall x\in \mathbb{R},l_0(x)\ge \underset{\lambda \in \mathbb{R}}{\sup }\{\lambda x-\overline{L}(\lambda )={\overline{L}}^{\ast }(x).$$

$$\overline{L}(t)=\underset{x\in \mathbb{R}}{\sup }\{tx-l_0(x)\}=t{x}_t-l_0(x_t)$$

$$\forall t\in ]\lambda ,\lambda +\epsilon [,\underset{M\to +\infty }{\lim }\underset{\alpha }{\mathrm{limsup}}{c}_{\alpha }\log {\int }_{[M/t,+\infty [}{e}^{c_{\alpha }^{-1}tz}{\mu }_{\alpha }(dz)=-\infty$$

$$\overline{L}(t)=\underset{x\le M/t}{\sup }\{tx-l_0(x)\}.$$

$$\overline{L}(t)\ge t{x}_n-l_0(x_n)>\overline{L}(t)-1/n\ge \overline{L}(t)-1$$

$$\overline{L}(t)=\underset{m}{\lim }(t{x}_{n_m}-l_0(x_{n_m}))=tx-\underset{m}{\lim }{l}_0(x_{n_m})\le tx-l_0(x),$$

$$\forall t\in ]\lambda ,\lambda +\epsilon [,l_0({\overline{L}}_r^{\prime }(t))={\overline{L}}^{\ast }({\overline{L}}_r^{\prime }(t)).$$

$$\sup \partial \overline{L}(t)={\overline{L}}_r^{\prime }(t)\in \mathbb{R}$$

$$\forall \epsilon \in ]0,\lambda +{\epsilon }_0-t[,\partial \overline{L}(t+\epsilon )=\left[{\overline{L}}_l^{\prime }(t+\epsilon ),{\overline{L}}_r^{\prime }(t+\epsilon )\right].$$

$$\overline{L}(t+\epsilon )=(t+\epsilon )x_{t+\epsilon }-l_0(x_{t+\epsilon })$$

$$\overline{L}(t+\epsilon )\le (t+\epsilon )x_{t+\epsilon }-{\overline{L}}^{\ast }(x_{t+\epsilon })\le {\overline{L}}^{\ast \ast }(t+\epsilon );$$

$$\forall \epsilon \in ]0,\lambda +{\epsilon }_0-t[,x=\sup \partial \overline{L}(t)\le x_{t+\epsilon }\le {\overline{L}}_r^{\prime }(t+\epsilon )$$

$$\begin{gathered} x=\underset{\epsilon \to 0^{+}}{\lim }\frac{\overline{L}(t+\epsilon )-\overline{L}(t)}{\epsilon }=\underset{\epsilon \to 0^{+}}{\lim }\frac{(t+\epsilon )x_{t+\epsilon }-l_0(x_{t+\epsilon })-\overline{L}(t)}{\epsilon } \\ =x+\underset{\epsilon \to 0^{+}}{\lim }\frac{t{x}_{t+\epsilon }-l_0(x_{t+\epsilon })-\overline{L}(t)}{\epsilon }, \end{gathered}$$

$$0=\underset{\epsilon \to 0^{+}}{\lim }(t{x}_{t+\epsilon }-l_0(x_{t+\epsilon })-\overline{L}(t))\le tx-l_0(x)-\overline{L}(t)\le 0,$$

$$\underset{i}{\lim }{\overline{L}}^{\ast }({\overline{L}}_r^{\prime }(t_i))=-\overline{L}(0^{+})\in [0,+\infty [.$$

$$\underset{i}{\lim }{\overline{L}}^{\ast }({\overline{L}}_r^{\prime }(t_i))={\overline{L}}^{\ast }({\overline{L}}_r^{\prime }(0^{+}))=\underset{t>0}{\sup }\{t{\overline{L}}_r^{\prime }(0^{+})-\overline{L}(t)\}.$$

$${\overline{L}}^{\ast }({\overline{L}}_r^{\prime }(t_i))=t_i{\overline{L}}_r^{\prime }(t_i)-\overline{L}(t_i)=\underset{t>0}{\sup }\{t{\overline{L}}_r^{\prime }(t_i)-\overline{L}(t)\},$$

Full text

${\dagger}$ Partially supported by FONDECYT grant 1120493.

1. Introduction and statement of the results

Let $(\mu_{\alpha},c_{\alpha})$ be a net where $\mu_{\alpha}$ is a Borel probability measure on ${\mathbb{R}}$, $c_{\alpha}>0$ and $(c_{\alpha})$ converges to zero. Let $\overline{L}$ be the generalized log-moment generating function associated with $(\mu_{\alpha},c_{\alpha})$, defined for each $t\in{\mathbb{R}}$ by

[TABLE]

when the above upper limit is a limit we write $L(t)$ in place of $\overline{L}(t)$. Let $\lambda>0$ and assume that the derivative map $L^{\prime}$ exists and is strictly monotone on a neighbourhood of $\lambda$ (in particular, $L(t)$ exists in ${\mathbb{R}}$ for all $t$ in this neighbourhood). When $(\mu_{\alpha},c_{\alpha})$ is a sequence of the form $(\mu_{n},n^{-1})$, Plachky-Steinebach theorem ([10]) asserts that for each sequence $(x_{n})$ of real numbers converging to $L^{\prime}(\lambda)$ we have

[TABLE]

(Note that $L(\lambda)-\lambda L^{\prime}(\lambda)=-\overline{L}^{*}(L^{\prime}(\lambda))=\inf_{t>0}\{\overline{L}(t)-tL^{\prime}(\lambda)\}$, where $\overline{L}^{*}$ denotes the Legendre-Fenchel transform of $\overline{L}$.)

In this paper, we improve the above result in various ways: First, we weaken all the hypotheses by (a) replacing the differentiability of $\overline{L}$ on a neighbourhood of $\lambda$ by the condition that $\overline{L}^{\prime}_{r}(\lambda)$ is a limit point of the set $\left\{\overline{L}^{\prime}_{r}(t):t>\lambda\right\}$, where $\overline{L}^{\prime}_{r}$ denotes the right derivative map of $\overline{L}$, (b) removing entirely the strict convexity hypothesis, (c) allowing $\lambda=0$ (with the only hypothesis of right continuity of $\overline{L}$ at zero when $\overline{L}^{\prime}_{r}(0^{+})=-\infty$), (d) requiring the existence of $L$ only on a suitable sequence, (e) allowing general nets $(\mu_{\alpha},c_{\alpha},x_{\alpha})$ in place of the sequence $(\mu_{n},n^{-1},x_{n})$; second, we strengthen the conclusion by allowing a more general class of intervals in the left hand side of (2); third, we generalize all the above by giving a version with upper limits.

For each $\lambda\geq 0$, our result covers all situations except when the condition mentioned in $(a)$ or in $(c)$ fails, i.e. excluding the following ones (cf. Lemma 1):

• $(i)$

The map ${\overline{L}^{\prime}_{r}}_{\mid]\lambda,+\infty[}$ takes only infinite values (equivalently, either $\overline{L}_{\mid[0,+\infty[}$ is improper, or $\overline{L}(t)=+\infty$ for all $t>\lambda$).

• $(ii)$

There exists $T\in\ ]\lambda,+\infty]$ such that $\overline{L}_{\mid]\lambda,T[}$ is affine, with $\overline{L}^{\prime}_{r}(\lambda^{+})<\overline{L}^{\prime}_{r}(T)<+\infty$ when $T<+\infty$.

• $(iii)$

$\lambda=0$, $\overline{L}_{r}^{\prime}(0^{+})=-\infty$ and $\overline{L}$ is right discontinuous at zero.

(In particular, in the most common case where $\overline{L}_{[\lambda,\lambda+\varepsilon[}$ is real-valued and continuous for some $\varepsilon>0$, only $(ii)$ is excluded.) Equivalently, Theorem 1 applies in (and only in) all the cases where $\overline{L}^{\prime}_{r}(\lambda^{+})=\lim_{i}\overline{L}^{\prime}_{r}(\lambda_{i})$ for some sequence $(\lambda_{i})$ fulfilling eventually

[TABLE]

with $\overline{L}(0^{+})=0$ when $\lambda=0$ and $\overline{L}_{r}^{\prime}(0^{+})=-\infty$ (note that the above condition is satisfied under the hypotheses of Plachky-Steinebach theorem); then, there is a unique real number ${\tilde{\lambda}}\geq\lambda$ fulfillling $\overline{L}^{\prime}_{r}(\lambda^{+})=\overline{L}^{\prime}_{r}({\tilde{\lambda}}^{+})$ and $\overline{L}^{\prime}_{r}(t)>\overline{L}^{\prime}_{r}(\lambda^{+})$ for all $t>{\tilde{\lambda}}$; in fact, ${\tilde{\lambda}}=\sup\{t>\lambda:\overline{L}_{\mid]\lambda,t]}\ \textnormal{is affine}\}=\lim_{i}\lambda_{i}$ for every sequence as above (cf. Lemma 1).

Let us emphasize that no hypothesis is made on the map $\overline{L}_{\mid]-\infty,\lambda[}$. The proof deals only with the properties of the right derivative map ${\overline{L}^{\prime}_{r}}_{\mid[\lambda,\lambda+\varepsilon[}$; in particular, the following special situations may arise:

• •

$\lambda$ may be a non-differentiability point of $\overline{L}$, and moreover a limit point of non-differentiability points of $\overline{L}$ (e.g. when $\lambda={\tilde{\lambda}}$ and eventually $\lambda_{i}$ is a non-differentiability point of $\overline{L}$).

• •

$\overline{L}$ may be left discontinuous at $\lambda$ when $\lambda=0$ (which is the case when $\overline{L}_{r}^{\prime}(0)=-\infty$) and moreover right discontinuous at [math] (when $\overline{L}_{r}^{\prime}(0)=-\infty<\overline{L}^{\prime}_{r}(0^{+})$); cf. Remark 4.

• •

$\overline{L}_{\mid]\lambda,\lambda+\varepsilon[}$ may not be strictly convex for all $\varepsilon>0$ (e.g. when $\lambda={\tilde{\lambda}}$ and eventually $\overline{L}_{\mid[\lambda_{i+1},\lambda_{i}]}$ is affine); when $\lambda<{\tilde{\lambda}}$, $\overline{L}$ is not strictly convex in any neighbourhood of $\lambda$ or ${\tilde{\lambda}}$.

• •

As regards the strong version with limits, for each $t<\lambda$, $\overline{L}(t)$ may not be a limit.

The foregoing contrasts sharply with the hypotheses of Plachky-Steinebach theorem, which require $\lambda>0$ and some $\varepsilon>0$ such that $L(t)$ exists in ${\mathbb{R}}$ for all $t\in\ ]\lambda-\varepsilon,\lambda+\varepsilon[$ and $L_{\mid]\lambda-\varepsilon,\lambda+\varepsilon[}$ is strictly convex and differentiable.

Theorem 1 111Some words of caution about the notations: First, the net $(\mu_{\alpha})$ is considered as a net of Borel measures on $[-\infty,+\infty]$ in order to give a sense to $\mu_{\alpha}([x_{\alpha},y_{\alpha}])$ when $x_{\alpha}=-\infty$ or $y_{\alpha}=+\infty$; second, when $\overline{L}_{r}^{\prime}(\lambda^{+})=-\infty$, we adopt the convention $0\cdot\overline{L}^{\prime}_{r}(\lambda^{+})=0$; third, in order to simplify the wording of the theorem, we do not distinguish with the notations between the cases $\lambda>0$ and $\lambda=0$, so that we use $\overline{L}_{r}^{\prime}(\lambda^{+})$ and $\overline{L}(\lambda^{+})$, although $\overline{L}_{r}^{\prime}(\lambda^{+})=\overline{L}_{r}^{\prime}(\lambda)$ and $\overline{L}(\lambda^{+})=\overline{L}(\lambda)$ when $\lambda>0$, and $\overline{L}_{r}^{\prime}(\lambda^{+})=-\infty$ implies $\lambda=0$ (cf. Lemma 2). below constitutes the first general result giving the upper limit (and a fortiori, the limit) of $(c_{\alpha}\log\mu_{\alpha}([x_{\alpha},y_{\alpha}]))$ or $(c_{\alpha}\log\mu_{\alpha}(]x_{\alpha},y_{\alpha}[)$ in terms of $\overline{L}^{*}$, when $\lim_{\alpha}x_{\alpha}$ is not a value of the derivative map of the generalized log-moment generating function (cf. Remark 5).

Theorem 1**.**

Let $\lambda\geq 0$. When $\overline{L}_{r}^{\prime}(\lambda^{+})>-\infty$ we assume that $\overline{L}^{\prime}_{r}(\lambda^{+})$ is a limit point of the set $\left\{\overline{L}^{\prime}_{r}(t):t>\lambda\right\}$; when $\overline{L}_{r}^{\prime}(\lambda^{+})=-\infty$ we assume that $\overline{L}(\lambda)=\overline{L}(\lambda^{+})$. For each net $(x_{\alpha},y_{\alpha})$ in $[-\infty,+\infty]^{2}$ such that $(x_{\alpha})$ converges to $\overline{L}_{r}^{\prime}(\lambda^{+})$ and $\liminf_{\alpha}y_{\alpha}>\overline{L}_{r}^{\prime}(\lambda^{+})$ we have

[TABLE]

Furthermore, if $L(t_{i})$ exists for a sequence $(t_{i})$ in $]{\tilde{\lambda}},+\infty[$ converging to ${\tilde{\lambda}}$, then the above upper limits are limits (with ${\tilde{\lambda}}=\sup\{t>\lambda:\overline{L}_{\mid]\lambda,t]}\ \textit{is affine}\}$).

It is possible that for some $\varepsilon>0$, $\lambda$ is the only point in $[0,\lambda+\varepsilon]$ where Theorem 1 applies (e.g. when $\lambda={\tilde{\lambda}}$, $\overline{L}$ is not differentiable at $\lambda$, and eventually $\lambda_{i}$ appearing in (3) is a non-differentiability point of $\overline{L}$ and $\overline{L}_{[\lambda_{i+1},\lambda_{i}]}$ is affine; although this may seem to be an extreme case, we show in Appendix A that there are plenty of such examples); in the light of the above, Theorem 1 may be thought of as a pointwise version of Plachky-Steinebach theorem.

Hereafter, we focus on the opposite situation, motivated by the following observation: If $L$ exists as a differentiable and strictly convex map on some open interval containing $\lambda>0$ and zero (says $]-\varepsilon,\lambda+\varepsilon[$), then the hypotheses of Plachky-Steinebach theorem hold for all $t\in\ ]0,\lambda+\varepsilon[$ (in place of $\lambda$), and thus it applies to every sequence converging to $L^{\prime}(t)$ for all $t\in\ ]0,\lambda+\varepsilon[$; since ${\overline{L}^{\prime}}_{\mid]0,\lambda+\varepsilon[}$ is in this case an increasing homeomorphism, the limit obtained, as a function of $L^{\prime}(t)$, is exactly $-{L^{*}}_{\mid]L_{r}^{\prime}(0),L_{l}^{\prime}(\lambda+\varepsilon)[}$ (where $\overline{L}^{\prime}_{l}$ denotes the left derivative map of $\overline{L}$); a weak version of this (with the constant sequence $(x_{n})\equiv L^{\prime}(t)$) has been widely used in the context of dynamical systems (e.g. [8], Lemma 13.2; [2], Lemma 6.2; [3], Corollary 4.3; [4], Theorem 2.2; [1], Proposition 7).

A much stronger result can be derived from Theorem 1, as shows the following Corollary 1; aside the extension with upper limits and the possibility to consider nets in place of sequences, the improvements are obtained first, by allowing a more general class of intervals in the conclusion, and second, by weakening the hypotheses as follows:

• •

The differentiability of $\overline{L}_{\mid]-\varepsilon,\lambda+\varepsilon[}$ is replaced by the differentiability of $\overline{L}_{\mid]\lambda,\lambda+\varepsilon[}$.

• •

The strict convexity of $\overline{L}_{\mid]-\varepsilon,\lambda+\varepsilon[}$ is replaced by the condition that ${\overline{L}^{\prime}}_{\mid]\lambda,\lambda+\varepsilon[}$ does not attain its supremum (equivalently, for each $t\in\ ]\lambda,\lambda+\varepsilon[$ the map $\overline{L}_{\mid[t,\lambda+\varepsilon[}$ is not affine).

(Note first, that this condition is obviously fulfilled when $\overline{L}_{\mid]t,\lambda+\varepsilon[}$ is strictly convex for some $t\in[\lambda,\lambda+\varepsilon[$, and second, it is far weaker than strict convexity: There may be an infinite countable set $\{S_{i}:i\in{\mathbb{N}}\}$ of non-empty mutually disjoint intervals included in $]\lambda,\lambda+\varepsilon[$ on which $\overline{L}$ is affine; when $\sup_{i}\sup S_{i}=\lambda+\varepsilon$, the above condition implies $\sup S_{i}<\lambda+\varepsilon$ for all $i\in{\mathbb{N}}$.)

• •

The possibility to take $\lambda=0$, including the cases $\overline{L}^{\prime}(0^{+})=-\infty$ and $\overline{L}^{\prime}(0)=-\infty<\overline{L}^{\prime}(0^{+})$.

• •

The version with limits only requires the existence of $L$ on a sequence $(t_{i})$ in $]\tilde{t},+\infty[$ converging to $\tilde{t}$, for all $t\in[\lambda,\lambda+\varepsilon[$; note that the set $[\lambda,\lambda+\varepsilon[\setminus\{\tilde{t}:t\in[\lambda,\lambda+\varepsilon[\}$ contains all the open intervals on which $\overline{L}$ is affine.

Corollary 1**.**

*Let $\lambda\geq 0$ and let $\varepsilon>0$. We assume that $\overline{L}_{\mid]\lambda,\lambda+\varepsilon[}$ is differentiable and $\sup\left\{\overline{L}^{\prime}(t):t\in\ ]\lambda,\lambda+\varepsilon[\right\}$ is not a maximum; when $\lambda=0$ and $\overline{L}_{r}^{\prime}(0^{+})=-\infty$, we further assume that $\overline{L}$ is right continuous at zero.

• a)

For each $z\in\left[\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)\right[$, there exists $t_{z}\in[\lambda,\lambda+\varepsilon[$ fulfilling $\overline{L}_{r}^{\prime}(t_{z}^{+})=z$, and for each such $t_{z}$, for each net $(x_{\alpha,z})$ in $[-\infty,+\infty]$ converging to $z$, and for each net $(y_{\alpha,z})$ in $[-\infty,+\infty]$ fulfilling $\liminf_{\alpha}y_{\alpha,z}>z$, we have

[TABLE]

Furthermore, if $L(t_{z,i})$ exists for a sequence $(t_{z,i})$ in $]\tilde{t_{z}},+\infty[$ converging to $\tilde{t_{z}}$, then the above upper limits are limits (with $\tilde{t_{z}}=\sup\{t>t_{z}:\overline{L}_{\mid]t_{z},t]}\ \textit{is affine}\}$).

• b)

The map $\left[\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)\right[\ni z\rightarrow\overline{L}(t_{z}^{+})-t_{z}z$ is $]-\infty,0]$-valued, strictly decreasing, continuous, unbounded below if and only if $\overline{L}_{l}^{\prime}(\lambda+\varepsilon)=+\infty$, and vanishes at $\overline{L}_{r}^{\prime}(\lambda^{+})$ if and only if either $\lambda>0$ and $\overline{L}$ is differentiable at $\lambda$ and linear on $[0,\lambda]$, or $\lambda=0$ and $\overline{L}$ is right continuous at zero; its restriction to $\left]\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)\right[$ is strictly concave, and it is furthermore differentiable when $\overline{L}_{\mid]{\tilde{\lambda}},\lambda+\varepsilon[}$ is strictly convex (with ${\tilde{\lambda}}=\sup\{t>\lambda:\overline{L}_{\mid]\lambda,t]}\ \textit{is affine}\}$).

Remark 1*.*

We have $\overline{L}_{r}^{\prime}(\lambda^{+})=-\infty$ only if $\lambda=0$, in which case the hypothesis of right continuity of $\overline{L}$ at zero in Theorem 1 (and Corollary 1) cannot be removed; indeed, suppose that $\overline{L}(0^{+})\neq 0$, and let $c\in\ ]0,1[$. Since for each index $\alpha$ the measure $\mu_{\alpha}$ is tight, there exists a net $(x_{\alpha})$ of real numbers converging to $-\infty$ such that $\mu_{\alpha}([x_{\alpha},-x_{\alpha}])>c$, hence

[TABLE]

and the conclusion does not hold with $(x_{\alpha})$. The hypothesis for the case $\overline{L}_{r}^{\prime}(\lambda^{+})>-\infty$ also holds when $\overline{L}_{r}^{\prime}(\lambda^{+})=-\infty$ (cf. Lemma 1$b)$).

Remark 2*.*

The upper limits in Theorem 1 vanish if and only if one of the following cases occurs:

• •

$\lambda>0$, $\overline{L}$ is differentiable at $\lambda$ and linear on $[0,\lambda]$;

• •

$\lambda=0$ and $\overline{L}$ is right continuous at zero.

Remark 3*.*

Under the hypotheses of Theorem 1, the equality $\overline{L}^{*}(\overline{L}_{r}^{\prime}(\lambda^{+}))=\lim_{\alpha}\overline{L}^{*}(x_{\alpha})$ fails when eventually $x_{\alpha}$ does not belong to the effective domain of $\overline{L}^{*}$, which happens if and only if $L_{\mid]-\infty,\lambda[}$ is linear with slope $\overline{L}_{r}^{\prime}(\lambda^{+})$ (equivalently, $\overline{L}_{r}^{\prime}(\lambda^{+})$ is the left end-point of the effective domain of $\overline{L}^{*}$) and eventually $x_{\alpha}<\overline{L}_{r}^{\prime}(\lambda^{+})$.

Remark 4*.*

Regarding the proof as well as the wording of Theorem 1, the main difference between the cases $\lambda>0$ and $\lambda=0$ stems from the fact that when the map $\overline{L}_{\mid[0,+\infty[}$ is proper (which is implied by the hypotheses, cf. Lemma 1), [math] is the only nonnegative real number $\lambda$ in the effective domain of $\overline{L}_{\mid[0,+\infty[}$ for which:

• •

$\overline{L}^{\prime}_{r}(\lambda)$ or $\overline{L}^{\prime}_{r}(\lambda^{+})$ may take the value $-\infty$;

• •

$\overline{L}^{\prime}_{r}(\lambda)$ may differ from $\overline{L}^{\prime}_{r}(\lambda^{+})$;

• •

$\overline{L}^{\prime}_{r}(\lambda^{+})<+\infty$ and $\overline{L}$ may be right discontinuous at $\lambda$.

(cf. Lemma 2 and Lemma 3; for instance, the conclusion in the last assertion of Lemma 2 is not true for $\lambda=0$). Theorem 1 and Corollary 1 include the case $\lambda=0$ and $\overline{L}^{\prime}_{r}(0)=-\infty<\overline{L}^{\prime}_{r}(0^{+})$, which implies the left and right discontinuities of $\overline{L}$ at zero (cf. Lemma 3); the case $\lambda=0$ and $\overline{L}^{\prime}_{r}(0^{+})=-\infty$ implies the left discontinuity of $\overline{L}$ at zero (in both cases, we have $\overline{L}_{]-\infty,0[}\equiv+\infty$).

Remark 5*.*

The standard version of Gärtner-Ellis theorem is unworkable when $\overline{L}$ is not differentiable at $\lambda$: Indeed, the main hypothesis (i.e. essentially smoothness) implies the differentiability of $\overline{L}$ on the interior of its effective domain (beside requiring the existence of $L$ on an open interval containing [math]); the same applies to the variant of Gärtner-Ellis theorem given by Theorem 5.1 of [9] (which allows $L$ to exists only on some open interval not necessarily containing [math]) since it also requires the essential smoothness. Corollary 1 of [5] strengthens both above versions, but, although weaker than essential smoothness, the general hypothesis is still a global condition requiring the existence of $L$ on some open interval and relating the range of the one-sided derivatives of $\overline{L}$ with the effective domain of $\overline{L}^{*}$, and thus it is of no use here.

2. Proofs

Recall that $\overline{L}$ is a $[-\infty,+\infty]$-valued convex function ([7], [6]), and such a function is said to be proper when it is $]-\infty,+\infty]$-valued and takes at least one finite value; note that $\overline{L}(0)=0$. We denote by $\overline{L}^{*}$ and $\overline{L}_{r}^{\prime}$ (resp. $\overline{L}_{l}^{\prime}$) respectively the Legendre-Fenchel transform and right (resp. left) derivative map of $\overline{L}$, where for each $t\geq 0$ we put $\overline{L}_{r}^{\prime}(t)=+\infty$ (resp. $\overline{L}_{r}^{\prime}(t)=-\infty$) when $\overline{L}(s)=+\infty$ for all $s>t$ (resp. $\overline{L}(s)=-\infty$ for some $s>t$); note that $\overline{L}^{*}$ is $[0,+\infty]$-valued since $\overline{L}(0)=0$. The basic properties of the map ${\overline{L}^{\prime}_{r}}_{\mid]0,+\infty[}$ are summarized in Lemmas 1$a)$, Lemma 2 and Lemma 3; in particular, ${\overline{L}^{\prime}_{r}}_{\mid]0,+\infty[}$ is non-decreasing so that the quantity $\overline{L}^{\prime}_{r}(\lambda^{+})=\lim_{t\rightarrow\lambda,t>\lambda}\overline{L}^{\prime}_{r}(t)$ is well-defined for all $\lambda\geq 0$.

Let $\lambda\geq 0$. All the lemmas below hold for any net $(\mu_{\alpha},c_{\alpha})$ as in (1); the hypotheses of Theorem 1 are made only in the last part of the proof.

Lemma 1**.**

• $a)$

The map $\overline{L}_{\mid[0,+\infty[}$ is improper if and only if there exists $T\in\ ]0,+\infty]$ such that $\overline{L}_{\mid]0,T[}={\overline{L}^{\prime}_{r}}_{\mid[0,T[}\equiv-\infty$ and $\overline{L}_{\mid]T,+\infty[}={\overline{L}^{\prime}_{r}}_{\mid[T,+\infty[}\equiv+\infty$.

• $b)$

If $\overline{L}_{\mid[0,+\infty[}$ is proper, then one and only one of the following cases holds:

• $(i)$

There exists a sequence $\left(\overline{L}^{\prime}_{r}(\lambda_{i})\right)$ in $\left]\overline{L}^{\prime}_{r}(\lambda^{+}),+\infty\right[$ converging to $\overline{L}^{\prime}_{r}(\lambda^{+})$.

• $(ii)$

There exists $\varepsilon>0$ such that $\overline{L}_{\mid]\lambda,\lambda+\varepsilon]}$ is affine and $\overline{L}^{\prime}_{r}(\lambda^{+})<\overline{L}^{\prime}_{r}(\lambda+\varepsilon)<+\infty$.

• $(iii)$

$\overline{L}_{\mid]\lambda,+\infty[}$* is affine.*

• $(iv)$

$\overline{L}(\lambda)\in{\mathbb{R}}$* and $\overline{L}(t)=+\infty$ for all $t>\lambda$;*

• $(v)$

$\overline{L}(t)=+\infty$* for all $t\geq\lambda$.*

In particular, $(i)$ or $(ii)$ or $(iii)$ holds if and only if there exists $\varepsilon>0$ such that $\overline{L}_{\mid[\lambda,\lambda+\varepsilon]}$ is real-valued, bounded, and $\overline{L}_{\mid]\lambda,\lambda+\varepsilon]}$ (resp. $\overline{L}_{\mid[\lambda,\lambda+\varepsilon]}$ when $\lambda>0$) continuous. Furthermore, $(i)$ holds if and only if there exists ${\tilde{\lambda}}\in[\lambda,+\infty[$ fulfilling $\overline{L}^{\prime}_{r}(\lambda^{+})=\overline{L}^{\prime}_{r}({\tilde{\lambda}}^{+})$ and $\overline{L}^{\prime}_{r}(t)>\overline{L}^{\prime}_{r}(\lambda^{+})$ for all $t>{\tilde{\lambda}}$; such a ${\tilde{\lambda}}$ is unique and given by

[TABLE]

for every sequence $(\lambda_{i})$ as in $(i)$.

Proof.

$a)$ It is a direct consequence of the definitions together with the convexity of $\overline{L}$ and the fact that $\overline{L}(0)=0$.

$b)$ Assume that $\overline{L}_{\mid[0,+\infty[}$ is proper. If $\overline{L}(\lambda)=+\infty$, then $\overline{L}(t)=+\infty$ for all $t\geq\lambda$ (because $\overline{L}$ is convex and $\overline{L}(0)=0$) and $(v)$ holds. Assume now that $\overline{L}(\lambda)<+\infty$. Then $\overline{L}(\lambda)\in{\mathbb{R}}$ because $\overline{L}_{\mid[0,+\infty[}$ is proper. Since $\overline{L}$ is convex and $\overline{L}(0)=0$, we have either $\overline{L}(t)=+\infty$ for all $t>\lambda$ and $(iv)$ holds, or $\overline{L}_{\mid[\lambda,\lambda+\delta[}$ is real-valued for some $\delta>0$, in which case $\overline{L}^{\prime}_{r}(t)\in{\mathbb{R}}$ for all $t\in\ ]\lambda,\lambda+\delta[$. Assume that $(i)$ does not hold, i.e. $\overline{L}^{\prime}_{r}(\lambda^{+})<\inf\left\{\overline{L}^{\prime}_{r}(t)\in\ \left]\overline{L}^{\prime}_{r}(\lambda^{+}),+\infty\right[:t>\lambda\right\}$. If $\left\{\overline{L}^{\prime}_{r}(t)\in\ \left]\overline{L}^{\prime}_{r}(\lambda^{+}),+\infty\right[:t>\lambda\right\}=\emptyset$, then $\overline{L}^{\prime}_{r}(t)=\overline{L}^{\prime}_{r}(\lambda^{+})$ for all $t>\lambda$, and $(iii)$ holds. If $\left\{\overline{L}^{\prime}_{r}(t)\in\ \left]\overline{L}^{\prime}_{r}(\lambda^{+}),+\infty\right[:t>\lambda\right\}\neq\emptyset$, then there exists $\varepsilon\in\ ]0,\delta[$ such that $\overline{L}^{\prime}_{r}(\lambda^{+})<\overline{L}^{\prime}_{r}(\lambda+\varepsilon)$ and $\overline{L}^{\prime}_{r}(\lambda^{+})=\overline{L}^{\prime}_{r}(t)$ for all $t\in\ ]\lambda,\lambda+\varepsilon[$, so that $(ii)$ holds. The first assertion is proved.

Assume that either $(i)$, $(ii)$, or $(iii)$ holds. There exists $\varepsilon>0$ such that ${\overline{L}^{\prime}_{r}}_{\mid]\lambda,\lambda+\varepsilon[}$ is real-valued, hence $\overline{L}_{\mid[\lambda,\lambda+\varepsilon[}$ is real-valued. The map $\overline{L}_{\mid]\lambda,\lambda+\varepsilon/2]}$ (resp. $\overline{L}_{\mid[\lambda,\lambda+\varepsilon/2]}$ when $\lambda>0$) is continuous since $]\lambda,\lambda+\varepsilon/2]$ (resp. $[\lambda,\lambda+\varepsilon/2]$ when $\lambda>0$) belongs to the interior of the effective domain of $\overline{L}_{\mid[0,+\infty[}$. The boundness of $\overline{L}_{\mid[\lambda,\lambda+\varepsilon/2]}$ follows from the continuity when $\lambda>0$, and the boundness of $\overline{L}_{\mid]\lambda,\lambda+\varepsilon/2]}$ when $\lambda=0$ follows from the convexity and the fact that $\overline{L}(0)=0$. Conversely, if $\overline{L}_{\mid[\lambda,\lambda+\varepsilon[}$ is real-valued for some $\varepsilon>0$, then the first assertion implies that either $(i)$, $(ii)$, or $(iii)$ holds. The second assertion is proved.

Assume that there exists $s\in[\lambda,+\infty[$ fufilling $\overline{L}^{\prime}_{r}(\lambda^{+})=\overline{L}^{\prime}_{r}(s^{+})$ and $\overline{L}^{\prime}_{r}(t)>\overline{L}^{\prime}_{r}(\lambda^{+})$ for all $t>s$. We have $\overline{L}^{\prime}_{r}(\lambda^{+})=\lim_{t\rightarrow s,t>s}\overline{L}^{\prime}_{r}(t)$ hence $(i)$ holds.

Assume that $(i)$ holds. Put ${\tilde{\lambda}}=\sup\{t>\lambda:\overline{L}_{\mid]\lambda,t]}\ \textnormal{is affine}\}$. If ${\tilde{\lambda}}=+\infty$, then $\overline{L}^{\prime}_{r}(t)=\overline{L}^{\prime}_{r}(\lambda^{+})$ for all $t>\lambda$, which contradicts $(i)$, hence ${\tilde{\lambda}}\in[\lambda,+\infty[$. Let $t>{\tilde{\lambda}}$. Since $\overline{L}_{\mid]\lambda,t]}$ is not affine, we have $\overline{L}^{\prime}_{r}(t)>\overline{L}^{\prime}_{r}(\lambda^{+})$. Suppose that $\overline{L}^{\prime}_{r}(\lambda^{+})<\overline{L}^{\prime}_{r}({\tilde{\lambda}}^{+})$; by $(i)$ there exists $\overline{L}^{\prime}_{r}(\lambda_{i})$ such that

[TABLE]

which implies $\lambda<\lambda_{i}<{\tilde{\lambda}}$ and contradicts the fact that $\overline{L}_{]\lambda,{\tilde{\lambda}}[}$ is affine; therefore, $\overline{L}^{\prime}_{r}(\lambda^{+})=\overline{L}^{\prime}_{r}({\tilde{\lambda}}^{+})$.

Let $(\lambda_{i})$ be a sequence as in $(i)$. For each $\varepsilon>0$ we have eventually

[TABLE]

hence eventually ${\tilde{\lambda}}<\lambda_{i}<{\tilde{\lambda}}+\varepsilon$, which implies ${\tilde{\lambda}}=\lim_{i}\lambda_{i}$.

The equalities $\overline{L}^{\prime}_{r}(\lambda^{+})=\overline{L}^{\prime}_{r}({\tilde{\lambda}}^{+})=\overline{L}^{\prime}_{r}(s^{+})$ implies $\overline{L}_{\mid]\lambda,s]}$ affine, hence $s\leq{\tilde{\lambda}}$ by definition of ${\tilde{\lambda}}$. Since $\overline{L}^{\prime}_{r}(t)>\overline{L}^{\prime}_{r}(\lambda^{+})=\overline{L}^{\prime}_{r}({\tilde{\lambda}}^{+})$ for all $t>s$, we have $\overline{L}^{\prime}_{r}(s^{+})\geq\overline{L}^{\prime}_{r}({\tilde{\lambda}}^{+})$ hence $s\geq{\tilde{\lambda}}$; therefore, $s={\tilde{\lambda}}$. The proof of the third assertion is complete. ∎

Lemma 2**.**

The following statements are equivalent:

• $(i)$

$\overline{L}_{\mid[0,+\infty[}$* is proper;*

• $(ii)$

${\overline{L}^{\prime}_{r}}_{\mid]0,+\infty[}$* is $]-\infty,+\infty]$-valued, non-decreasing and right continuous.*

When the above holds and $\lambda>0$, we have ${\overline{L}^{\prime}_{r}}(\lambda)\in{\mathbb{R}}$ if and only if $\overline{L}(t)<+\infty$ for some $t>\lambda$.

Proof.

$(ii)\Rightarrow(i)$ follows from the definitions since $\overline{L}^{\prime}_{r}$ takes the value $-\infty$ when $\overline{L}_{\mid[0,+\infty[}$ is improper (cf. Lemma 1). Assume that $(i)$ holds. Since $\overline{L}$ is convex and $\overline{L}(0)=0$, the map ${\overline{L}^{\prime}_{r}}_{\mid]0,+\infty[}$ is $]-\infty,+\infty]$-valued. Let $\lambda>0$. If $\overline{L}^{\prime}_{r}(\lambda)=+\infty$, then $\lambda$ fulfils one of the last two cases of Lemma 1, so that $\overline{L}^{\prime}_{r}$ takes only the value $+\infty$ on $[\lambda,+\infty[$ hence is non-decreasing and right continuous on $[\lambda,+\infty[$. Assume that $\overline{L}^{\prime}_{r}(\lambda)<+\infty$. Then $\lambda$ fulfils one of the first three cases of Lemma 1; in particular, there exists $\varepsilon>0$ such that $\overline{L}_{\mid[\lambda,\lambda+\varepsilon]}$ is real-valued and continuous. Extend $\overline{L}_{\mid[\lambda,\lambda+\varepsilon]}$ to a lower semi-continuous function $f$ on ${\mathbb{R}}$ by putting $f(x)=+\infty$ for all $x\in{\mathbb{R}}\setminus[\lambda,\lambda+\varepsilon]$. The right derivative map of $f$ is non-decreasing and right continuous on ${\mathbb{R}}$ by Theorem 24.1 of [11] so that ${\overline{L}^{\prime}_{r}}$ is non-decreasing on $[\lambda,\lambda+\varepsilon]$ and right continuous on $[\lambda,\lambda+\varepsilon[$; therefore, $(ii)$ holds and the first assertion is proved. If $\overline{L}(t)<+\infty$ for some $t>\lambda$, then ${\overline{L}^{\prime}_{r}}(\lambda)<+\infty$ (by definition) hence ${\overline{L}^{\prime}_{r}}(\lambda)\in{\mathbb{R}}$ since $\lambda>0$ and $\overline{L}_{\mid[0,+\infty[}$ is proper; the converse is obvious. ∎

Lemma 3**.**

The following statements are equivalent:

• $(i)$

$\overline{L}^{\prime}_{r}$* is not right continuous at [math];*

• $(ii)$

$\overline{L}_{\mid[0,+\infty[}$* is not lower semi-continuous at [math], $\overline{L}^{\prime}_{r}(0)=-\infty$ and $\overline{L}_{r}^{\prime}(0^{+})>-\infty$.*

• $(iii)$

$\overline{L}_{\mid[0,+\infty[}$* is not lower semi-continuous at [math] and $\overline{L}_{r}^{\prime}(0^{+})\in{\mathbb{R}}$.*

• $(iv)$

$\overline{L}^{\prime}_{r}(0)=-\infty$* and $\overline{L}_{r}^{\prime}(0^{+})>-\infty$.*

The above equivalence hold verbatim replacing $\overline{L}_{r}^{\prime}(0^{+})>-\infty$ by $\overline{L}_{r}^{\prime}(0^{+})\in{\mathbb{R}}$. In particular, $\overline{L}^{\prime}_{r}(0)\in{\mathbb{R}}$ if and only if $\overline{L}^{\prime}_{r}(0)=\overline{L}^{\prime}_{r}(0^{+})\in{\mathbb{R}}$.

Proof.

Assume that $(i)$ holds. By Lemma 1$a)$ the map $\overline{L}_{\mid[0,+\infty[}$ is proper, and by Lemma 1$b)$ there exists $\varepsilon>0$ such that $\overline{L}_{\mid[0,\varepsilon]}$ is real-valued and $\overline{L}_{\mid]0,\varepsilon]}$ is continuous. If $\overline{L}_{\mid[0,+\infty[}$ is lower semi-continuous at [math], then the extension of $\overline{L}_{\mid[0,\varepsilon]}$ to a lower semi-continuous function on ${\mathbb{R}}$ yields the right continuity of $\overline{L}^{\prime}_{r}$ at [math] (by Theorem 24.1 of [11]), which gives a contradiction; therefore, $\overline{L}_{\mid[0,+\infty[}$ is not lower semi-continuous at [math]; since $\overline{L}$ is convex and $\overline{L}(0)=0$ it follows that $\overline{L}(0^{+})$ exists as a negative number, and consequently, $\overline{L}_{r}^{\prime}(0)=-\infty$. Since $\overline{L}^{\prime}_{r}(t)\in{\mathbb{R}}$ for all $t\in\ ]0,\varepsilon[$, and ${\overline{L}^{\prime}_{r}}_{\mid]0,+\infty[}$ is non-decreasing by Lemma 2, $\overline{L}_{r}^{\prime}(0^{+})$ exists in $[-\infty,+\infty[$, hence $\overline{L}_{r}^{\prime}(0^{+})>-\infty$ (since $\overline{L}^{\prime}_{r}$ is not right continuous at [math] by hypothesis) and $(ii)$ holds.

Assume that $(ii)$ holds. Then, $\overline{L}_{r}^{\prime}(0^{+})\in{\mathbb{R}}$ (otherwise $\overline{L}_{r}^{\prime}(0^{+})=+\infty$ would imply $\overline{L}_{r}^{\prime}(0)=+\infty$ and a contradiction) hence $(iii)$ holds.

Assume that $(iii)$ holds. By Lemma 1 there exists $\varepsilon>0$ such that $\overline{L}_{\mid[0,\varepsilon]}$ is real-valued and $\overline{L}_{\mid]0,\varepsilon]}$ is continuous, hence (since $\overline{L}$ is convex and $\overline{L}(0)=0$), $\overline{L}(0^{+})$ exists as a negative number, which implies $\overline{L}_{r}^{\prime}(0)=-\infty$, and gives $(iv)$. Since the implication $(iv)\Rightarrow(i)$ is obvious, the proof of the first two assertions is complete; the last assertion is a direct consequence of them. ∎

Let $l_{0}$ be the function defined on ${\mathbb{R}}$ by

[TABLE]

note that $l_{0}$ is $[0,+\infty]$-valued and lower semi-continuous.

Lemma 4**.**

We have

[TABLE]

Proof.

Since for each real number $\lambda\neq 0$ the set $\{x\in{\mathbb{R}}:a\leq e^{\lambda x}\leq b\}$ is compact for all $(a,b)\in{\mathbb{R}}^{2}$ with $a\leq b$, Theorem 1 of [6] yields

[TABLE]

Since $l_{0}$ is a $[0,+\infty]$-valued function and $\overline{L}(0)=0$, the above inequality is true with $\lambda=0$ so that

[TABLE]

hence

[TABLE]

∎

Lemma 5**.**

Assume that $\overline{L}_{\mid[0,+\infty[}$ is proper and $\overline{L}(t)<+\infty$ for some $t>\lambda$. There exists $\varepsilon>0$ such that for each $t\in\ ]\lambda,\lambda+\varepsilon[$ we have

[TABLE]

for some $x_{t}\in{\mathbb{R}}$.

Proof.

By Lemma 1 there exists $\varepsilon>0$ such that $\overline{L}_{\mid[\lambda,\lambda+\varepsilon]}$ is real-valued, hence

[TABLE]

(Lemma 4.3.8 of [7]). Part (b) of Theorem 1 of [6] yields for each $t\in\ ]\lambda,\lambda+\varepsilon]$ and for each $M$ large enough

[TABLE]

Let $t\in\ ]\lambda,\lambda+\varepsilon[$. For each integer $n\geq 1$ there exists $x_{n}\leq M/t$ such that

[TABLE]

hence $x_{n}\in\left[\overline{L}(t)/t-1/t,M/t\right]$. Therefore, the sequence $(x_{n})$ has a subsequence $(x_{n_{m}})$ converging to some $x\in\left[\overline{L}(t)/t-1/t,M/t\right]$, so that letting $n\rightarrow+\infty$ in (5) yields

[TABLE]

where the last inequality follows from the lower semi-continuity of $l_{0}$. From the above expression and (4), we get $\overline{L}(t)=tx-l_{0}(x)$, which proves the lemma. ∎

Lemma 6**.**

Assume that $\overline{L}_{\mid[0,+\infty[}$ is proper and $\overline{L}(t)<+\infty$ for some $t>\lambda$. There exists $\varepsilon>0$ such that

[TABLE]

When $\overline{L}^{\prime}_{r}(\lambda^{+})\neq-\infty$, the above equality is true with $t=\lambda$, and $\overline{L}_{r}^{\prime}(\lambda^{+})$ in place of $\overline{L}_{r}^{\prime}(t)$.

Proof.

By Lemma 1 there exists $\varepsilon_{0}>0$ such that $\overline{L}_{\mid[\lambda,\lambda+\varepsilon_{0}]}$ is real-valued bounded and $\overline{L}_{\mid]\lambda,\lambda+\varepsilon_{0}]}$ continuous. Let $t\in\ ]\lambda,\lambda+\varepsilon_{0}[$. For each $\varepsilon\in[0,\lambda+\varepsilon_{0}-t[$ let $\partial\overline{L}(t+\varepsilon)$ denote the set of subgradients of $\overline{L}$ at $t+\varepsilon$. Note that

[TABLE]

and

[TABLE]

For each $\varepsilon\in\ ]0,\lambda+\varepsilon_{0}-t[$, Lemma 5 yields some $x_{t+\varepsilon}\in{\mathbb{R}}$ such that

[TABLE]

hence by Lemma 4,

[TABLE]

since $\overline{L}\geq\overline{L}^{**}$ we obtain $l_{0}(x_{t+\varepsilon})=\overline{L}^{*}(x_{t+\varepsilon})$; in particular, $x_{t+\varepsilon}\in\partial\overline{L}(t+\varepsilon)$. Putting $x=\overline{L}_{r}^{\prime}(t)$ we have

[TABLE]

hence $\lim_{\varepsilon\rightarrow 0}x_{t+\varepsilon}=x$ (because $\lim_{\varepsilon\rightarrow 0}\overline{L}^{\prime}_{r}(t+\varepsilon)=x$ by Lemma 2) and

[TABLE]

which implies

[TABLE]

where the first inequality follows from the lower semi-continuity of $l_{0}$, and the second inequality follows from Lemma 5; therefore, $\overline{L}(t)=tx-l_{0}(x)$ hence $l_{0}(x)=\overline{L}^{*}(x)$ since $\overline{L}(t)=tx-\overline{L}^{*}(x)$ (because $x\in\partial\overline{L}(t)$); this proves the first assertion.

When $\overline{L}^{\prime}_{r}(\lambda^{+})\neq-\infty$, the hypotheses imply $\overline{L}^{\prime}_{r}(\lambda^{+})\in{\mathbb{R}}$ and the last assertion follows noting that the above proof works verbatim with $t=\lambda$ and $\overline{L}_{r}^{\prime}(\lambda^{+})$ (resp. $\overline{L}(\lambda^{+})$) in place of $\overline{L}_{r}^{\prime}(t)$ (resp. $\overline{L}(t)$). ∎

Lemma 7**.**

Assume that $\overline{L}_{\mid[0,+\infty[}$ is proper and $\overline{L}(t)<+\infty$ for some $t>0$. Let $(t_{i})$ be a sequence of positive numbers converging to [math]. We have

[TABLE]

If furthermore $\overline{L}^{\prime}_{r}(0^{+})>-\infty$, then

[TABLE]

Proof.

By Lemma 1 (applied with $\lambda=0$), there exists $\varepsilon>0$ such that $\overline{L}_{\mid]0,\varepsilon]}$ is real-valued, bounded and continuous, hence eventually $\overline{L}^{\prime}_{r}(t_{i})\in{\mathbb{R}}$,

[TABLE]

and $\overline{L}(0^{+})$ exists in $]-\infty,0]$; furthermore, $\overline{L}^{\prime}_{r}(0^{+})$ exists in $[-\infty,+\infty[$ by Lemma 2. Since $\overline{L}^{*}$ is continuous on its effective domain, and $\overline{L}(0^{+})$ is a non-positive real number, (6) ensures the existence of $\lim_{i}\overline{L}^{*}(\overline{L}^{\prime}_{r}(t_{i}))$ in $[0,+\infty[$.

First assume that $\overline{L}^{\prime}_{r}(0^{+})>-\infty$. Then $\overline{L}^{\prime}_{r}(0^{+})\in{\mathbb{R}}$ and (6) yields

[TABLE]

which proves the first assertion and the first equality of the second assertion; we have

[TABLE]

and the second equality of the second assertion follows.

Assume that $\overline{L}^{\prime}_{r}(0^{+})=-\infty$. For each $i$ large enough there exists $\lambda_{i}>0$ such that $\lambda_{i}\overline{L}^{\prime}_{r}(t_{i})=\overline{L}(\lambda_{i})$, hence

[TABLE]

and

[TABLE]

so that the first assertion follows from (6) since $-\overline{L}(0^{+})\geq 0$. ∎

Lemma 7 shows that when $\overline{L}^{\prime}_{r}(0^{+})=-\infty$, the map $\overline{L}^{*}$ extends by continuity to a $[-\overline{L}(0^{+}),+\infty]$-valued map on $[-\infty,+\infty[$ by putting $\overline{L}^{*}(-\infty)=-\overline{L}(0^{+})$; in what follows, we implicitely use this extension.

Lemma 8**.**

We assume that $\overline{L}_{\mid]\lambda,\lambda+\varepsilon[}$ is differentiable for some $\varepsilon>0$.

• $a)$

${\overline{L}^{\prime}}_{\mid]\lambda,\lambda+\varepsilon[}$* extends to a non-decreasing continuous surjection between $[\lambda,\lambda+\varepsilon]$ and $\left[\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)\right]$.*

• $b)$

${\overline{L}^{*}}_{\mid[\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)[}$* is $[0,+\infty[$-valued and extends to a $[0,+\infty]$-valued, strictly increasing and continuous map on $\left[\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)\right]$, which takes the value $+\infty$ at $\overline{L}_{l}^{\prime}(\lambda+\varepsilon)$ if and only if $\overline{L}_{l}^{\prime}(\lambda+\varepsilon)=+\infty$; it vanishes at $\overline{L}_{r}^{\prime}(\lambda^{+})$ if and only if either $\lambda>0$, $\overline{L}$ is differentiable at $\lambda$ and linear on $[0,\lambda]$, or $\lambda=0$ and $\overline{L}$ is right continuous at zero. The map ${\overline{L}^{*}}_{\mid]\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)[}$ is positive and strictly convex; if furthermore, ${\tilde{\lambda}}<\lambda+\varepsilon$ and $\overline{L}_{\mid]{\tilde{\lambda}},\lambda+\varepsilon[}$ is strictly convex, then ${\overline{L}^{*}}_{\mid]\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)[}$ is differentiable (with ${\tilde{\lambda}}=\sup\{t>\lambda:\overline{L}_{\mid]\lambda,t]}\ \textit{is affine}\}$).*

Proof.

$a)$ Since $\overline{L}_{\mid]\lambda,\lambda+\varepsilon[}$ is differentiable, the map ${\overline{L}^{\prime}}_{\mid]\lambda,\lambda+\varepsilon[}$ is continuous by Corollary 25.5.1 of [11]. Since $\overline{L}^{\prime}_{l}(\lambda+\varepsilon)=\lim_{t\rightarrow\varepsilon,t<\varepsilon}\overline{L}^{\prime}(\lambda+t)$, the map ${\overline{L}^{\prime}}_{\mid]\lambda,\lambda+\varepsilon[}$ is a non-decreasing continuous surjection onto $\left]\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)\right[$, which can be extended by continuity to a non-decreasing continuous surjection between $[\lambda,\lambda+\varepsilon]$ and $\left[\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)\right]$.

$b)$ Each $t\in\ ]\lambda,\lambda+\varepsilon[$ is a subgradient of $\overline{L}^{*}$ at $\overline{L}^{\prime}(t)$ so that ${\overline{L}^{*}}_{\mid]\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)[}$ is $[0,+\infty[$-valued and strictly increasing, hence ${\overline{L}^{*}}_{\mid[\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)[}$ is $[0,+\infty[$-valued and strictly increasing; in particular, ${\overline{L}^{*}}_{\mid]\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)[}$ is positive. Since $\overline{L}^{*}$ is lower semi-continuous, it is continuous on its effective domain so that ${\overline{L}^{*}}_{\mid]\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)[}$ is continuous, and thus extends to a $[0,+\infty]$-valued, strictly increasing and continuous map on $\left[\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)\right]$; this extended map can takes the value $+\infty$ only at $\overline{L}_{l}^{\prime}(\lambda+\varepsilon)$; since $\overline{L}^{*}$ is lower semi-continuous, $\overline{L}^{*}(\overline{L}_{l}^{\prime}(\lambda+\varepsilon))$ is finite when $\overline{L}_{l}^{\prime}(\lambda+\varepsilon)$ is finite; conversely, if $\overline{L}_{l}^{\prime}(\lambda+\varepsilon)=+\infty$, then $\left]\overline{L}_{r}^{\prime}(\lambda^{+}),+\infty\right[$ is included in the effective domain of $\overline{L}^{*}$, hence $\lim_{t\rightarrow\lambda+\varepsilon,t<\lambda+\varepsilon}\overline{L}^{*}(\overline{L}_{r}^{\prime}(t))=+\infty$. Let $g$ be the extension of $\overline{L}_{\mid]\lambda,\lambda+\varepsilon[}$ to ${\mathbb{R}}$ defined by putting

[TABLE]

so that $g$ is convex, lower semi-continuous, differentiable on the interior of its effective domain, but not sub-differentiable at each point in the complement of the interior of its effective domain; therefore, the Legendre-Fenchel transform $g^{*}$ of $g$ is strictly convex on the interior of its effective domain ([12], Theorem 11.13), hence on $]g_{r}^{\prime}(\lambda^{+}),g_{l}^{\prime}(\lambda+\varepsilon)[$. Since $g^{\prime}(t)=\overline{L}^{\prime}(t)$ for all $t\in\ ]\lambda,\lambda+\varepsilon[$, we obtain ${g^{*}}_{\mid]g_{r}^{\prime}(\lambda^{+}),g_{l}^{\prime}(\lambda+\varepsilon)[}=\overline{L}^{*}_{\mid]\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)[}$, which proves the strict convexity property. The first part of the first assertion, and the first part of the second assertion are proved. The continuity and strict increasingness imply that $\inf_{[\overline{L}^{\prime}_{r}(\lambda^{+}),\overline{L}_{r}^{\prime}(\lambda+\varepsilon)[}\overline{L}^{*}$ is a minimum, which is attained at the unique point $\overline{L}^{\prime}_{r}(\lambda^{+})$; this proves the second part of the first assertion since $\overline{L}^{*}(\overline{L}^{\prime}_{r}(\lambda^{+}))=\lambda\overline{L}_{r}^{\prime}(\lambda^{+})-\overline{L}(\lambda^{+})$ (using Lemma 7 when $\lambda=0$). Assume furthermore that $\overline{L}_{\mid]\lambda,\lambda+\varepsilon[}$ is strictly convex. The map $g$ is not sub-differentiable at some $t\in[\lambda,\lambda+\varepsilon]$ if and only if either $t=\lambda=0$ and $g^{\prime}_{r}(0^{+})=-\infty$, or $t=\lambda+\varepsilon$ and $g^{\prime}_{l}(\lambda+\varepsilon)=+\infty$, hence ]\lambda,\lambda+\varepsilon[\ \subset\{t\in[\lambda,\lambda+\varepsilon]:g\ \textnormal{is sub-differentiable at t}\}. Therefore, the map $g1_{[\lambda,\lambda+\varepsilon]}+\infty 1_{{\mathbb{R}}\setminus[\lambda,\lambda+\varepsilon]}$ is strictly convex on every convex subset of the set \{t\in[\lambda,\lambda+\varepsilon]:g1_{[\lambda,\lambda+\varepsilon]}+\infty 1_{{\mathbb{R}}\setminus[\lambda,\lambda+\varepsilon]}\ \textnormal{is sub-differentiable at t}\}, hence its Legendre-Fenchel transform is differentiable on the interior of its effective domain ([12], Theorem 11.13); consequently, $\overline{L}^{*}_{\mid]\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)[}$ is strictly convex. Assume that ${\tilde{\lambda}}<\lambda+\varepsilon$ and $\overline{L}_{\mid]{\tilde{\lambda}},\lambda+\varepsilon[}$ is strictly convex. The above proof and the proof of part $a)$ work verbatim replacing $]\lambda,\lambda+\varepsilon[$ by $]{\tilde{\lambda}},\lambda+\varepsilon[$. Since $\overline{L}$ is differentiable at ${\tilde{\lambda}}$, we have $\overline{L}^{\prime}_{r}(\lambda^{+})=\overline{L}^{\prime}_{r}({\tilde{\lambda}}^{+})$, which proves the second part of the second assertion. ∎

Proof of Theorem 1. Put $x=\overline{L}_{r}^{\prime}(\lambda^{+})$. The hypotheses imply that $\overline{L}_{\mid[0,+\infty[}$ is proper (otherwise, by Lemma 1$a)$, $\overline{L}^{\prime}_{r}(t)\in\{-\infty,+\infty\}$ for all $t>0$, which contradicts the hypothesis when $x>-\infty$, and $\overline{L}_{\mid]0,t[}=-\infty$ for all $t>0$ small enough, which contradicts the hypothesis when $x=-\infty$), and the case $(i)$ of Lemma 1$b)$ holds; in particular, $x<+\infty$ and there exists $\varepsilon>0$ such that $\overline{L}_{\mid[\lambda,\lambda+\varepsilon[}$ is real-valued and $\overline{L}_{\mid]\lambda,\lambda+\varepsilon]}$ is continuous. Let $(x_{\alpha},y_{\alpha})$ be a net in $[-\infty,+\infty]^{2}$ such that $\lim_{\alpha}x_{\alpha}=x$ and $\liminf_{\alpha}y_{\alpha}>x$. Put $y=\liminf_{\alpha}y_{\alpha}$.

$\bullet$ First assertion, the case $x\in{\mathbb{R}}$: For each $t\in\ ]\lambda,\lambda+\varepsilon[$, Chebyshev’s inequality yields eventually

[TABLE]

hence

[TABLE]

and letting $t\rightarrow\lambda$,

[TABLE]

where the last equality follows from Lemma 7 when $\lambda=0$. Suppose that

[TABLE]

The hypothesis together with the continuity of $\overline{L}^{*}$ on its effective domain implies the existence of $t>\lambda$ and $\delta>0$ such that

[TABLE]

and

[TABLE]

Since eventually

[TABLE]

we have eventually

[TABLE]

hence

[TABLE]

which contradicts (8) since $l_{0}(\overline{L}^{\prime}_{r}(t))=\overline{L}^{*}(\overline{L}^{\prime}_{r}(t))$ by Lemma 6; therefore, we have

[TABLE]

which together with (7) proves the first three equalities of the first assertion; the last equality is obvious when $\lambda>0$ (definition of $\overline{L}^{*}$), and follows from Lemma 7 when $\lambda=0$.

$\bullet$ First assertion, the case $x=-\infty$: Lemma 2 implies $\lambda=0$. Let $(t_{i})$ be a sequence of positive numbers converging to [math], so that eventually

[TABLE]

by Lemma 6. Since $\lim_{i}\overline{L}^{\prime}_{r}(t_{i})=-\infty$, there exists $\delta>0$ such that $x_{\alpha}+\delta<\overline{L}^{\prime}_{r}(t_{i})<y_{\alpha}-\delta$ eventually with respect to $i$ and eventually with respect to $\alpha$, which together with the above equality yields

[TABLE]

where the third equality is given by Lemma 7, and the last equality follows from the hypothesis of right continuity of $\overline{L}$ at zero. The first two equalities of the first assertion follow from the above expression together with Lemma 7 (recall that by convention, $0\cdot(-\infty)=0$). For each $i\in{\mathbb{N}}$, $t_{i}$ is a subgradient of $\overline{L}^{*}$ at $\overline{L}^{\prime}_{r}(t_{i})$ so that $\overline{L}^{*}$ is non-decreasing on $\left]-\infty,\overline{L}^{\prime}_{r}(t_{i})\right]$; since $\lim_{\alpha}x_{\alpha}=-\infty$, eventually $x_{\alpha}$ belongs to the effective domain of $\overline{L}^{*}$ and fullfils

[TABLE]

hence

[TABLE]

letting $i\rightarrow+\infty$ gives $\lim_{\alpha}\overline{L}^{*}(x_{\alpha})=0$, which proves the last two equalities of the first assertion. The proof of the first assertion is complete.

$\bullet$ Second assertion: Let $(t_{i})$ be a sequence in $]{\tilde{\lambda}},+\infty[$ converging to ${\tilde{\lambda}}$ such that $L(t_{i})$ exists for all $i\in{\mathbb{N}}$. Let $(\mu_{\beta},c_{\beta},x_{\beta},y_{\beta})$ be a subnet of $(\mu_{\alpha},c_{\alpha},x_{\alpha},y_{\alpha})$. For each $t\in{\mathbb{R}}$ we put

[TABLE]

Since $\overline{L}^{(\mu_{\beta},c_{\beta})}(t_{i})=L(t_{i})$ for all $i\in{\mathbb{N}}$, we have

[TABLE]

hence

[TABLE]

where the last equality follow from Lemma 1$b)$. The inequality ${\overline{L}^{(\mu_{\beta},c_{\beta})}}^{\prime}_{r}(\lambda^{+})\leq{\overline{L}^{(\mu_{\beta},c_{\beta})}}^{\prime}_{r}({\tilde{\lambda}}^{+})$ together with (10) implies

[TABLE]

We have $\overline{L}^{\prime}_{r}(t_{i})>\overline{L}^{\prime}_{r}({\tilde{\lambda}}^{+})$ for all $i\in{\mathbb{N}}$ (by definition of ${\tilde{\lambda}}$) and

[TABLE]

so that $(\overline{L}^{\prime}_{r}(t_{i}))$ has a strictly decreasing subsequence $(\overline{L}^{\prime}_{r}(t_{i_{j}}))$ converging to $\overline{L}^{\prime}_{r}({\tilde{\lambda}}^{+})$, which implies eventually $t_{i_{j+1}}<t_{i_{j}}<t_{i_{j-1}}$, hence

[TABLE]

and

[TABLE]

Put ${\tilde{\lambda}}^{(\mu_{\beta},c_{\beta})}=\sup\{t>\lambda:\overline{L}^{(\mu_{\beta},c_{\beta})}_{\mid]\lambda,t]}\ \textnormal{is affine}\}$. The inequality $\overline{L}^{(\mu_{\beta},c_{\beta})}\leq\overline{L}$ together with (9) implies

[TABLE]

Claim 1**.**

When ${\overline{L}^{(\mu_{\beta},c_{\beta})}}^{\prime}_{r}(\lambda^{+})>-\infty$, the hypothesis of Theorem 1 holds with the net $(\mu_{\beta},c_{\beta})$ in place of $(\mu_{\alpha},c_{\alpha})$. Furthermore, we have $\overline{L}^{(\mu_{\beta},c_{\beta})}(\lambda^{+})=\overline{L}(\lambda^{+})$ and ${\overline{L}^{(\mu_{\beta},c_{\beta})}}^{\prime}_{r}(\lambda^{+})=x$.

Proof of Claim 1. Assume ${\overline{L}^{(\mu_{\beta},c_{\beta})}}^{\prime}_{r}(\lambda^{+})>-\infty$. Since $x>-\infty$ by (11) we have

[TABLE]

where the second equality follows from (9) and the last equality follows from (10); therefore, all the above inequalities are equalities, which gives

[TABLE]

which together with (9) implies

[TABLE]

Since $\overline{L}^{(\mu_{\beta},c_{\beta})}\leq\overline{L}$ with $\overline{L}$ convex, (17) implies

[TABLE]

We have

[TABLE]

hence by (12) and (13), all the above inequalities are equalities, which together with (10) yields

[TABLE]

where the strict inequality follows from (14). The above expression together with (10) and (18) gives

[TABLE]

which proves the first assertion of the claim.

The first equality of the second assertion is given by (16); the second equality of the second assertion is given by (10) when $\lambda={\tilde{\lambda}}$. Assume $\lambda<{\tilde{\lambda}}$. Then, (17) and (18) yield

[TABLE]

Since $\overline{L}$ is differentiable at ${\tilde{\lambda}}$ when ${\tilde{\lambda}}>\lambda$, we have

[TABLE]

(where (16) is used to obtain the first inequality), hence by (20) all the above inequalities are equalities, and the second equality of the second assertion follows.

Claim 2**.**

When ${\overline{L}^{(\mu_{\beta},c_{\beta})}}^{\prime}_{r}(\lambda^{+})=-\infty$, the hypothesis of Theorem 1 holds with the net $(\mu_{\beta},c_{\beta})$ in place of $(\mu_{\alpha},c_{\alpha})$; in particular, $\overline{L}^{(\mu_{\beta},c_{\beta})}(\lambda^{+})=\overline{L}(\lambda^{+})$ and ${\overline{L}^{(\mu_{\beta},c_{\beta})}}^{\prime}_{r}(\lambda^{+})=x$.

Proof of Claim 2. The hypothesis implies $\lambda=0$ and $\tilde{\lambda}^{(\mu_{\beta},c_{\beta})}=0$, hence $\tilde{\lambda}=0$ by (15); therefore, $x=-\infty$ by (10), which implies $\overline{L}(\lambda^{+})=0$ by hypothesis of Theorem 1; this last equality together with (9) yields $\overline{L}^{(\mu_{\beta},c_{\beta})}(\lambda^{+})=0$, which proves the claim.

By Claim 1 and Claim 2 we can apply the first assertion of Theorem 1 with the net $(\mu_{\beta},c_{\beta},x_{\beta},y_{\beta})$ in place of $(\mu_{\alpha},c_{\alpha},x_{\alpha},y_{\alpha})$, which gives

[TABLE]

Since the right hand side of the above last equality does not depend on $(\mu_{\beta},c_{\beta},x_{\beta},y_{\beta})$, and $(\mu_{\beta},c_{\beta},x_{\beta},y_{\beta})$ is an arbitrary subnet of $(\mu_{\alpha},c_{\alpha},x_{\alpha},y_{\alpha})$, the second assertion follows.

Proof of Corollary 1. Let $t\in[\lambda,\lambda+\varepsilon[$ and put ${\tilde{t}}=\sup\{s>t:\overline{L}_{\mid]t,s]}\ \textnormal{is affine}\}$ (note that ${\tilde{t}}$ is a maximum since $\overline{L}_{\mid]\lambda,\lambda+\varepsilon[}$ is continuous). Since (by hypothesis) $\sup\left\{\overline{L}^{\prime}(t):t\in\ ]\lambda,\lambda+\varepsilon[\right\}$ is not a maximum, we have ${\tilde{t}}<\lambda+\varepsilon$. For each $\delta>0$, the map $\overline{L}_{\mid[{\tilde{t}},{\tilde{t}}+\delta[}$ is not affine (otherwise, since $\overline{L}$ is differentiable at ${\tilde{t}}$, the slope of $\overline{L}_{\mid[{\tilde{t}},{\tilde{t}}+\delta[}$ would be the same as the one of $\overline{L}_{\mid]t,{\tilde{t}}]}$, which would contradict the definition of ${\tilde{t}}$). Therefore, we have $\overline{L}^{\prime}(s)>\overline{L}^{\prime}({\tilde{t}}^{+})$ for all $s>{\tilde{t}}$; since $\overline{L}^{\prime}(t^{+})=\overline{L}^{\prime}({\tilde{t}}^{+})$ (because $\overline{L}_{\mid]t,{\tilde{t}}]}$ is affine and $\overline{L}$ is differentiable at ${\tilde{t}}$) we get $\overline{L}^{\prime}(s)>\overline{L}^{\prime}(t^{+})$ for all $s>{\tilde{t}}$. Consequently, by the last assertion of Lemma 1 $b)$, the case $(i)$ of Lemma 1$b)$ holds, hence the hypotheses of Theorem 1 holds for $t$ (i.e. with $\lambda=t$ in Theorem 1), and in particular when $\overline{L}^{\prime}_{r}(t^{+})>-\infty$. If $\overline{L}^{\prime}_{r}(t^{+})=-\infty$, then $t=\lambda=0$ by Lemma 2; since in this case $\overline{L}$ is assumed to be right continuous at zero, the hypotheses of Theorem 1 holds for $t$. Lemma 8$a)$ ensures that for each $z\in\left[\overline{L}_{r}^{\prime}(\lambda^{+}),\overline{L}_{l}^{\prime}(\lambda+\varepsilon)\right[$ there exists $t_{z}\in[\lambda,\lambda+\varepsilon[$ such that $z=\overline{L}_{r}^{\prime}(t_{z}^{+})$, hence part $a)$ follows applying Theorem 1 for all $t\in[\lambda,\lambda+\varepsilon[$. Part $b)$ follows from Lemma 8 $b)$.

Appendix

Let $\lambda\geq 0$, let $\varepsilon>0$ and let $f$ be a real-valued strictly convex and continuous function on $[0,\lambda+\varepsilon]$ such that $f(0)=0$. In what follows, we show how from such a function $f$ it is possible to build a generalized log-moment generating function $L_{f}$ with effective domain $[0,\lambda+\varepsilon]$, such that $L_{f}(t)$ is a limit for all $t\in{\mathbb{R}}$, $L_{f}$ is not differentiable at $\lambda$, ${L_{f}}_{\mid[0,\lambda]}$ is affine, $\lambda$ is a limit of a decreasing sequence $(\lambda_{i})_{i\geq 1}$ of non-differentiability points of ${L_{f}}_{\mid]\lambda,\lambda+\varepsilon[}$ and ${L_{f}}_{\mid[\lambda_{i},\lambda_{i-1}]}$ is affine for all $i\geq 1$ (with $\lambda_{0}=\lambda+\varepsilon$). Therefore, the hypotheses of Theorem 1 holds for $\lambda$, but the usual version of Plachky-Steinebach theorem does not apply (either because $\overline{L}$ is not differentiable in a neighbourhood of $\lambda$, or because $\overline{L}$ is not strictly convex in a neighbourhood of $\lambda$); however, its conclusion remains true (i.e. both conclusions of Theorem 1 hold); the case $\lambda=0$ and ${L_{f}}^{\prime}_{r}(0^{+})=-\infty$ is included.

Note that $\lambda$ is the only point in $[0,\lambda+\varepsilon]$ to which Theorem 1 applies, since (a) every $t\in[0,\lambda[\ \cup\ ]\lambda,\lambda+\varepsilon[$ fulfils the case $(ii)$ of Lemma 1$b)$, hence the excluded case $(ii)$ mentioned in Section 1, and (b) $L^{\prime}_{r}(\lambda+\varepsilon)=+\infty$, i.e. $\lambda+\varepsilon$ fulfils the case $(iv)$ of Lemma 1, hence the excluded case $(i)$ mentioned in Section 1.

First, we extend $f$ to a convex lower semi-continuous function on ${\mathbb{R}}$ by putting $f(t)=+\infty$ for all $t\not\in[0,\lambda+\varepsilon]$. Put $\lambda_{0}=\lambda+\varepsilon$. Let $(\lambda_{i})_{i\geq 1}$ be a decreasing sequence in $]\lambda,\lambda+\varepsilon[$ converging to $\lambda$. Draw a line segment $D_{i}$ between $(\lambda_{i},f(\lambda_{i})$ and $(\lambda_{i+1},f(\lambda_{i+1}))$ for all $i\in{\mathbb{N}}$, and draw a line segment $D$ between and $(0,0)$ and $(\lambda,f(\lambda)$. Let $L_{f}$ be the function whose graph coincides with the graph of $f$ (resp. $D_{i}$, $D$) on $]-\infty,0[\ \cup\ ]\lambda_{0},+\infty[$ (resp. $[\lambda_{i+1},\lambda_{i}]$ for all $i\in{\mathbb{N}}$, $[0,\lambda]$). Clearly, $L_{f}$ is a convex function on ${\mathbb{R}}$, continuous on its effective domain $[0,\lambda+\varepsilon]$, affine on $[0,\lambda]$ and $[\lambda_{i+1},\lambda_{i}]$ for all $i\in{\mathbb{N}}$, and fulfils $L_{f}(0)=0$. The strict convexity of $f$ ensures for each $i\geq 1$ the existence of some $t_{i}\in\ ]\lambda_{i+1},\lambda_{i}[$ such that

[TABLE]

hence

[TABLE]

and ${L_{f}}^{\prime}_{l}(\lambda)<{L_{f}}^{\prime}_{r}(\lambda)$ when ${L_{f}}^{\prime}_{r}(\lambda)>-\infty$. Therefore, $L_{f}$ is not differentiable at $\lambda$, and not differentiable at $\lambda_{i}$ for all $i\in{\mathbb{N}}$, but (21) and (22) show that $L_{f}$ fulfils the hypotheses of Theorem 1.

It remains to show that $L_{f}$ is a generalized log-moment generating function such that $L_{f}(t)$ is a limit for all $t\in{\mathbb{R}}$. Let $\{z_{k}:k\geq 1\}$ be a countable set dense in the effective domain $\textnormal{dom }L_{f}^{*}$ of $L_{f}^{*}$, and put

[TABLE]

Since $L_{f}$ is lower semi-continuous, we have $L_{f}=L_{f}^{**}$ ([11], Theorem 23.5) i.e.

[TABLE]

hence

[TABLE]

and letting $n\rightarrow+\infty$,

[TABLE]

Since the set $\{z_{k}:k\geq 1\}$ is dense in $\textnormal{dom\ }L_{f}^{*}$ and $L_{f}^{*}$ is continuous on $\textnormal{dom\ }L_{f}^{*}$, we get for each $t\in{\mathbb{R}}$ and for each $z\in\textnormal{dom }L_{f}^{*}$,

[TABLE]

hence

[TABLE]

which together with (23) gives

[TABLE]

Since

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for all $(t,n)\in{\mathbb{R}}\times{\mathbb{N}}\setminus\{0\}$, it follows from (24) that the generalized log-moment generating function associated with $(\mu_{n,f},n^{-1})$ is a limit for all $t\in{\mathbb{R}}$, and coincides with $L_{f}$. Consequently, both conclusions of Theorem 1 hold, i.e. for every sequence $(x_{n})$ converging to ${L_{f}}_{r}^{\prime}(\lambda)$, and for every sequence $(y_{n})$ fulfilling $\liminf_{n}y_{n}>{L_{f}}_{r}^{\prime}(\lambda)$, we have

[TABLE]