# On the differentiability of interval functions

Walter F. Mascarenhas  Universidade de São Paulo, Brazil  [
<EMAIL_ADDRESS>Rua do Matão 1010, São Paulo, SP, Brazil, CEP 05508-090

###### Abstract

Two articles published by Information Sciences discuss the derivatives of
interval functions in the sense of Svetoslav Markov. The authors of these
articles tried to characterize for which functions and points such derivatives
exist. Unfortunately, their characterization is inaccurate. This article
describes this inaccuracy and explains how it can be corrected.

###### keywords:

Interval functions , Derivatives

††journal: Information Sciences

url]www.ime.usp.br/ walterfm

##  1 Introduction

In his 1979 article [1], Markov presented a Calculus for interval functions of
a real variable, i.e., functions $F$ of the form
${F}\\!\left(t\right)=[{f}\\!\left(t\right),{g}\\!\left(t\right)]$ for
$f,g:\Omega\subset\mathds{R}{}\rightarrow\mathds{R}{}$ with
${f}\\!\left(t\right)\leq{g}\\!\left(t\right)$ for $t\in\Omega$.
($\mathds{R}{}$ is the set of real numbers.) Assuming that $x$ is in the
interior of $\Omega$, as we will ways do, he uses the operator $\ominus$,
defined by

$[\underline{a},\overline{a}]\ominus[\underline{b},\overline{b}]=\left[\,{\min}\\!\left(\underline{a}-\underline{b},\overline{a}-\overline{b}\right),{\max}\\!\left(\underline{a}-\underline{b},\overline{a}-\overline{b}\right)\right],$

the usual division operator $/$ for intervals and numbers, and the Hausdorff
metric

${\mathrm{dist}}\\!\left([\underline{a},\overline{a}],[\underline{b},\overline{b}]\right)={\max}\\!\left(\,{\left|{\,\underline{a}-\underline{b}\,}\right|},\,{\left|{\,\overline{a}-\overline{b}\,}\right|}\,\right),$

to define the derivative of $F$ at $x$ as

${F}^{\prime}\\!\left(x\right):=\lim_{t\rightarrow
x}\frac{{F}\\!\left(t\right)\ominus{F}\\!\left(x\right)}{t-x}.$ (1)

The relevance of Markov’s ${F}^{\prime}\\!\left(x\right)$, in theory and in
practice, was latter discussed in [2, 3, 4], and readers unaware of it should
read these articles and some of their references.

Markov showed that if $f$ and $g$ are differentiable at $x$ then

${F}^{\prime}\\!\left(x\right)=\left[\,{\min}\\!\left({f}^{\prime}\\!\left(x\right),{g}^{\prime}\\!\left(x\right)\right),\,{\max}\\!\left({f}^{\prime}\\!\left(x\right),{g}^{\prime}\\!\left(x\right)\right)\,\right],$

and he also showed that ${F}^{\prime}\\!\left(x\right)$ may exist in cases in
which ${f}^{\prime}\\!\left(x\right)$ or ${g}^{\prime}\\!\left(x\right)$ does
not exist. To illustrate this fact, in page 331 of [1], he considered the
directional derivatives

${f}^{\prime}_{-}\\!\left(x\right)=\lim_{t\uparrow
0}\frac{{f}\\!\left(t\right)-{f}\\!\left(x\right)}{t-x}\hskip
28.45274pt\mathrm{and}\hskip
28.45274pt{f}^{\prime}_{+}\\!\left(x\right)=\lim_{t\downarrow
0}\frac{{f}\\!\left(t\right)-{f}\\!\left(x\right)}{t-x}$

and claimed that if ${f}^{\prime}_{\pm}\\!\left(x\right)$ and
${g}^{\prime}_{\pm}\\!\left(x\right)$ exist then
${F}^{\prime}\\!\left(x\right)$ exists when

${f}^{\prime}_{-}\\!\left(x\right)={g}^{\prime}_{+}\\!\left(x\right)\hskip
28.45274pt\mathrm{and}\hskip
28.45274pt{g}^{\prime}_{-}\\!\left(x\right)={f}^{\prime}_{+}\\!\left(x\right).$
(2)

The authors of [2] and [3] misunderstood page 331 of [1] and this
misunderstanding lead them to “prove” a “theorem” which in our notation reads
like

###### Inaccurate Theorem 1

(inaccurate) Theorem 6 in [2], (inaccurate) Theorem 1 in [3]. The function
${F}\\!\left(t\right)=[{f}\\!\left(t\right),{g}\\!\left(t\right)]$ has a
derivative ${F}^{\prime}\\!\left(x\right)$ at $x$ if AND ONLY IF one of the
following cases hold:

  * (a)

The functions $f$ and $g$ are differentiable at $x$.

  * (b)

The derivatives ${f}^{\prime}_{\pm}\\!\left(x\right)$ and
${g}^{\prime}_{\pm}\\!\left(x\right)$ exist and Equation
$\left({\ref{dpm}}\right)$ holds. $\blacktriangle$

Markov did not state the “ONLY IF” part of this “theorem.” In fact, the next
lemma shows that this part of “Theorem 1” is false:

###### Lemma 1

Let ${\mathds{Q}}{}$ be the set of rational numbers. The function
${F}\\!\left(t\right)=[{f}\\!\left(t\right),{g}\\!\left(t\right)]$ for

${f}\\!\left(t\right):=\left\\{\begin{array}[]{r}t\ \mathrm{if}\
t\in{\mathds{Q}}{}\cap\left({-1,1}\right),\\\ 0\ \mathrm{if}\
t\in\left({-1,1}\right)\setminus{}{\mathds{Q}}{}.\end{array}\right.\hskip
28.45274pt\mathrm{and}\hskip
28.45274pt{g}\\!\left(t\right):=\left\\{\begin{array}[]{r}1\ \mathrm{if}\
t\in{\mathds{Q}}{}\cap\left({-1,1}\right),\\\ t+1\ \mathrm{if}\
t\in(-1,1)\setminus{}{\mathds{Q}}{}.\end{array}\right.$

is such that ${F}^{\prime}\\!\left(0\right)=[0,1]$, but
${f}^{\prime}_{\pm}\\!\left(0\right)$ and ${g}^{\prime}_{\pm}\left(0\right)$
do not exist. $\blacktriangle$

The “IF” part of Theorem 1, which is correct, follows from the next theorem:

###### Theorem 2

If the right derivatives ${f}^{\prime}_{+}\\!\left(x\right)$ and
${g}^{\prime}_{+}\\!\left(x\right)$ exist then the right derivative

${F}^{\prime}_{+}\\!\left(x\right):=\lim_{t\downarrow
0}\frac{{F}\\!\left(t\right)\ominus{F}\\!\left(x\right)}{t-x}$

exists and

${F}^{\prime}_{+}\\!\left(x\right)=[{\min}\\!\left({f}^{\prime}_{+}\\!\left(x\right),{g}^{\prime}_{+}\\!\left(x\right)\right),{\max}\\!\left({f}^{\prime}_{+}\\!\left(x\right),{g}^{\prime}_{+}\\!\left(x\right)\right)].$
(3)

Similarly, if the left derivatives ${f}^{\prime}_{-}\\!\left(x\right)$ and
${g}^{\prime}_{-}\\!\left(x\right)$ exist then the left derivative

${F}^{\prime}_{-}\\!\left(x\right):=\lim_{t\uparrow
0}\frac{{F}\\!\left(t\right)\ominus{F}\\!\left(x\right)}{t-x}$

exists and

${F}^{\prime}_{-}\\!\left(x\right)=[{\min}\\!\left({f}^{\prime}_{-}\\!\left(x\right),{g}^{\prime}_{-}\\!\left(x\right)\right),{\max}\\!\left({f}^{\prime}_{-}\\!\left(x\right),{g}^{\prime}_{-}\\!\left(x\right)\right)].\hfill\blacktriangle{}$
(4)

Theorem 2 can be proved with the same technique used by Markov to prove his
Theorem 3. We omit this proof for the sake of brevity, and because our goal is
to provide conditions under which we can fix the “ONLY IF” part of “Theorem
1.”

The most general condition on $f$ and $g$ which we could find in order to fix
“Theorem 1” is the existence of functions
$\alpha,\beta,c,d:\Omega\rightarrow\mathds{R}{}$ such that $\alpha,\beta$ and
$c$ are continuous in $\Omega$,
${\alpha}\\!\left(t\right)\neq{\beta}\\!\left(t\right)$ for $t\in\Omega$,
$\alpha$ and $\beta$ are bounded, $d$ is differentiable at $x$, with
${d}\\!\left(x\right)={d}^{\prime}\\!\left(x\right)=0$ and

${\alpha}\\!\left(t\right){f}\\!\left(t\right)+{\beta}\\!\left(t\right){g}\\!\left(t\right)={c}\\!\left(t\right)+{d}\\!\left(t\right).$
(5)

This condition is reasonable and applies to most practical situations. For
instance, it holds if $f$ is continuous in $\Omega$, in which case we can take
$\alpha=1$, $\beta=0$, $c=f$ and $d=0$. It also holds when the length of
${F}\\!\left(t\right)$, ${g}\\!\left(t\right)-{f}\\!\left(t\right)$, is
continuous, in which case we can take $\alpha=-1$, $\beta=1$, $c=g-f$ and
$d=0$. We formalize these results in the next theorem, which is proved with
other results in Section Proofs.

###### Theorem 3

Consider functions $\alpha,\beta,c,d:\Omega\rightarrow\mathds{R}{}$ and
$\mu>0$ such that ${d}\\!\left(x\right)=0$ and
${\left|{{\alpha}\\!\left(t\right)}\right|}+{\left|{{\beta}\\!\left(t\right)}\right|}\leq\mu$
and
$\mu{\left|{\,{\alpha}\\!\left(t\right)-{\beta}\\!\left(t\right)\,}\right|}\geq
1$ for all $t\in\Omega$. If ${F}^{\prime}_{+}\\!\left(x\right)$ exists,
Equation $\left({\ref{cond}}\right)$ holds for $t\in\Omega\cap(x,+\infty)$,
$\alpha,\beta$ and $c$ are continuous in $\Omega\cap(x,+\infty)$ and
${d}^{\prime}_{+}\\!\left(0\right)=0$, then
${f}^{\prime}_{+}\\!\left(x\right)$ and ${g}^{\prime}_{+}\\!\left(x\right)$
exist. If ${F}^{\prime}_{-}\\!\left(x\right)$ exists, Equation
$\left({\ref{cond}}\right)$ holds for $t\in\Omega\cap(-\infty,x)$,
$\alpha,\beta$ and $c$ are continuous in $\Omega\cap(-\infty,x)$ and
${d}^{\prime}_{-}\\!\left(0\right)=0$, then
${f}^{\prime}_{-}\\!\left(x\right)$ and ${g}^{\prime}_{-}\\!\left(x\right)$
exist. $\blacktriangle$

The last two theorems lead to a proper “ONLY IF” part for “Theorem 1”:

###### Corollary 4

If ${F}^{\prime}\\!\left(x\right)$ exists and there exist functions
$\alpha,\beta,c,d:\Omega\rightarrow\mathds{R}{}$ such that $\alpha$, $\beta$
and $c$ are continuous,
${\alpha}\\!\left(t\right)\neq{\beta}\\!\left(t\right)$ for $t\in\Omega$,
${d}\\!\left(x\right)={d}^{\prime}\\!\left(x\right)=0$ and $\alpha f+\beta
g=c+d$ then the directional derivatives ${f}^{\prime}_{\pm}\\!\left(x\right)$
and ${g}^{\prime}_{\pm}\\!\left(x\right)$ exist and

$\displaystyle{F}^{\prime}\\!\left(x\right)$ $\displaystyle=$
$\displaystyle\left[\,{\min}\\!\left({f}^{\prime}_{-}\\!\left(x\right),{g}^{\prime}_{-}\\!\left(x\right)\right),\,{\max}\\!\left({f}^{\prime}_{-}\\!\left(x\right),{g}^{\prime}_{-}\\!\left(x\right)\right)\,\right]$
(6) $\displaystyle=$
$\displaystyle\left[\,{\min}\\!\left({f}^{\prime}_{+}\\!\left(x\right),{g}^{\prime}_{+}\\!\left(x\right)\right),\,{\max}\\!\left({f}^{\prime}_{+}\\!\left(x\right),{g}^{\prime}_{+}\\!\left(x\right)\right)\,\right].$

In particular, if ${F}^{\prime}\\!\left(x\right)$ exists and $f$ or $g$ or
$g-f$ is continuous, then the directional derivatives
${f}^{\prime}_{\pm}\\!\left(x\right)$ and
${g}^{\prime}_{\pm}\\!\left(x\right)$ exist and Equation
$\left({\ref{ufa}}\right)$ holds. $\blacktriangle$

Combining this corollary with the next lemma, we end this introduction with
the simplest version of the ONLY IF part of “Theorem 1” that we could find:

###### Lemma 5

If ${F}^{\prime}\\!\left(x\right)$ exists then ${f}\\!\left(t\right)$ and
${g}\\!\left(t\right)$ are continuous at $x$. $\blacktriangle$

###### Corollary 6

If ${F}^{\prime}\\!\left(x\right)$ exists for all $x$ in an open interval
$(a,b)$ then the directional derivatives ${f}^{\prime}_{\pm}\\!\left(x\right)$
and ${g}^{\prime}_{\pm}\\!\left(x\right)$ exist for all $x\in(a,b)$ and
satisfy Equation $\left({\ref{ufa}}\right)$. $\blacktriangle$

The Denjoy-Young-Saks family of theorems show that, in Corollary 6, the set
${\cal{S}}$ of $x\in(a,b)$ at which $f$ or $g$ is not differentiable is
“tiny.” Formally, the Lebesgue measure of ${\cal{S}}$ is zero and it is of
first category (in Baire’s sense.) Actually, it is $\sigma$-porous (see [5].)

Thanks to the results above, it is quite likely that if the inaccurate “ONLY
IF” part of “Theorem 1” was ever used then it caused no harm, because their
hypothesis would hold and ensure the validity of the conclusions derived from
“Theorem 1.” However, we still believe that the minor technical point
discussed in this article deserves the attention of the potential users of
“Theorem 1.”

##  2 Proofs

Proof of Lemma 1 We show that

${q}\\!\left(t\right)=\frac{{F}\\!\left(t\right)\ominus{}{F}\\!\left(0\right)}{t}=[0,1]$
(7)

for $t\in(-1,1)\setminus\left\\{{0}\right\\}$, by considering the four cases
below:

  * 1.

if $t>0$ and $t\in{\mathds{Q}}{}$ then

${q}\\!\left(t\right)=\frac{[{f}\\!\left(t\right),{g}\\!\left(t\right)]\ominus[{f}\\!\left(0\right),{g}\\!\left(0\right)]}{t}=\frac{[t,1]\ominus[0,1]}{t}$
$=\frac{[{\min}\\!\left(t-0,1-1\right),{\max}\\!\left(t-0,1-1\right)]}{t}=\frac{[0,t]}{t}=[0,1].$

  * 2.

if $t<0$ and $t\in{\mathds{Q}}{}$ then

${q}\\!\left(t\right)=\frac{[{f}\\!\left(t\right),{g}\\!\left(t\right)]\ominus[{f}\\!\left(0\right),{g}\\!\left(0\right)]}{t}=\frac{[t,1]\ominus[0,1]}{t}$
$=\frac{[{\min}\\!\left(t-0,1-1\right),{\max}\\!\left(t-0,1-1\right)]}{t}=\frac{[t,0]}{t}=[0,1].$

  * 3.

if $t>0$ and $t\not{\in}{\mathds{Q}}{}$ then

$\frac{[{f}\\!\left(t\right),{g}\\!\left(t\right)]\ominus[{f}\\!\left(0\right),{g}\\!\left(0\right)]}{t}=\frac{[0,t+1]\ominus[0,1]}{t}$
$=\frac{[{\min}\\!\left(0-0,t+1-1\right),{\max}\\!\left(0-0,t+1-1\right)]}{t}=\frac{[0,t]}{t}=[0,1].$

  * 4.

if $t<0$ and $t\not{\in}{\mathds{Q}}{}$ then

${q}\\!\left(t\right)=\frac{[{f}\\!\left(t\right),{g}\\!\left(t\right)]\ominus[{f}\\!\left(0\right),{g}\\!\left(0\right)]}{t}=\frac{[0,t+1]\ominus[0,1]}{t}$
$=\frac{[{\min}\\!\left(0-0,t+1-1\right),{\max}\\!\left(0-0,t+1-1\right)]}{t}=\frac{[t,0]}{t}=[0,1].\hfill\blacksquare{}$

Proof of Lemma 5 We show that if $f$ is discontinuous at $x$ then
${F}^{\prime}\\!\left(x\right)$ does not exist (and the same holds for $g$.)
In fact, in this case there exists a sequence $t_{k}$ which converges to $x$
such that ${f}\\!\left(t_{k}\right)$ does not converge to
${f}\\!\left(x\right)$. This implies that

$\lim_{k\rightarrow\infty}{\left|{\frac{{f}\\!\left(t_{k}\right)-{f}\\!\left(x\right)}{t_{k}-x}}\right|}=+\infty$

and the limit in the definition $\left({\ref{md}}\right)$ of
${F}^{\prime}\\!\left(x\right)$ does not exist. $\blacksquare$

Proof of Theorem 3. We only present the proof for the right derivatives. Since
$x$ is in the interior of $\Omega$, there exists $\delta>0$ such that
$[x,x+\delta)\subset\Omega$. Writing ${F}^{\prime}_{+}\\!\left(x\right)$ as
$[\underline{a},\overline{a}]$ we have that

$\underline{a}=\lim_{t\downarrow
x}\frac{{\min}\\!\left({f}\\!\left(t\right)-{f}\\!\left(x\right),\,{g}\\!\left(t\right)-{g}\\!\left(x\right)\right)}{t-x}$
(8)

and

$\overline{a}=\lim_{t\downarrow
x}\frac{{\max}\\!\left({f}\\!\left(t\right)-{f}\\!\left(x\right),\,{g}\\!\left(t\right)-{g}\\!\left(x\right)\right)}{t-x}.$
(9)

Let us then define the set

${\cal{P}}:=\left\\{{t\in(x,x+\delta)\ \mathrm{with}\
{f}\\!\left(t\right)-{f}\\!\left(x\right)\leq{g}\\!\left(t\right)-{g}\\!\left(x\right)}\right\\}.$

If ${\cal{P}}$ is empty, let us define $i_{p}:=x+\delta$. Otherwise, let
$i_{p}$ be its infimum. If $i_{p}>x$ then

$\frac{{\min}\\!\left({f}\\!\left(t\right)-{f}\\!\left(x\right),\,{g}\\!\left(t\right)-{g}\\!\left(x\right)\right)}{t-x}=\frac{{g}\\!\left(t\right)-{g}\\!\left(x\right)}{t-x}$

for $t\in(x,i_{p})$ and Equation $\left({\ref{equd}}\right)$ implies that
${g}^{\prime}_{+}\\!\left(x\right)$ exists and is equal to $\underline{a}$.
Similarly, for $t\in(x,i_{p})$,

$\frac{{\max}\\!\left({f}\\!\left(t\right)-{f}\\!\left(x\right),\,{g}\\!\left(t\right)-{g}\\!\left(x\right)\right)}{t-x}=\frac{{f}\\!\left(t\right)-{f}\\!\left(x\right)}{t-x}$

and Equation $\left({\ref{eqod}}\right)$ implies that
${f}^{\prime}_{+}\\!\left(x\right)=\overline{a}$. Therefore, if $i_{p}>x$ then
we are done.

Analogously, if the set

${\cal{Q}}:=\left\\{{t\in(x,x+\delta)\ \mathrm{with}\
{f}\\!\left(t\right)-{f}\\!\left(x\right)\geq{g}\\!\left(t\right)-{g}\\!\left(x\right)}\right\\}.$

is empty or if its infimum $i_{q}$ is greater than $x$ then the left
derivatives ${f}^{\prime}_{+}\\!\left(x\right)$ and
${g}^{\prime}_{+}\\!\left(x\right)$ exist, and we are done. Therefore, we can
assume that ${\cal{P}}$ and ${\cal{Q}}$ are not empty and $i_{p}=i_{q}=x$. We
claim that in this case $\underline{a}=\overline{a}$, and prove this claim by
showing that

$\overline{a}-\underline{a}={\left|{\,\overline{a}-\underline{a}\,}\right|}\leq
2\mu\left({1+\mu}\right)\epsilon$ (10)

for all $\epsilon>0$. Let us then prove Equation
$\left({\ref{tiny_da}}\right)$. Equations $\left({\ref{equd}}\right)$ and
$\left({\ref{eqod}}\right)$ and the hypothesis that
${d}\\!\left(x\right)={d}^{\prime}_{+}\\!\left(x\right)=0$ imply that there
exists $t_{\epsilon}\in(x,x+\delta)$ such that if $t\in(x,t_{\epsilon})$ then

$\displaystyle{\left|{{d}\\!\left(t\right)}\right|}\leq\epsilon\left({t-x}\right),$
(11)
$\displaystyle{\left|{\,\frac{{\min}\\!\left({f}\\!\left(t\right)-{f}\\!\left(x\right),\,{g}\\!\left(t\right)-{g}\\!\left(x\right)\right)}{t-x}-\underline{a}\,}\right|}\leq\epsilon,$
(12)
$\displaystyle{\left|{\,\frac{{\max}\\!\left({f}\\!\left(t\right)-{f}\\!\left(x\right),\,{g}\\!\left(t\right)-{g}\\!\left(x\right)\right)}{t-x}-\overline{a}\,}\right|}\leq\epsilon.$
(13)

Consider ${\cal{P}}_{\epsilon}:={\cal{P}}\cap(x,t_{\epsilon})$ and
${\cal{Q}}_{\epsilon}:={\cal{Q}}\cap(x,t_{\epsilon})$. These sets are not
empty because $\inf{\cal{P}}=\inf{\cal{Q}}=x$. Calling by
$\overline{{\cal{P}}}_{\epsilon}$ and $\overline{{\cal{Q}}}_{\epsilon}$ their
closures in $(x,t_{\epsilon})$, we have that
$(x,t_{\epsilon})={\cal{P}}_{\epsilon}\cup{\cal{Q}}_{\epsilon}=\overline{{\cal{P}}}_{\epsilon}\cup\overline{{\cal{Q}}}_{\epsilon}$.
Since $(x,t_{\epsilon})$ is connected,
$\overline{{\cal{P}}}_{\epsilon}\cap\overline{{\cal{Q}}}_{\epsilon}$ is not
empty. Taking
$t\in\overline{{\cal{P}}}_{\epsilon}\cap\overline{{\cal{Q}}}_{\epsilon}$,
there exist sequences $p_{k}\in{\cal{P}}_{\epsilon}$ and
$q_{k}\in{\cal{Q}}_{\epsilon}$ such that
$\lim_{k\rightarrow\infty}p_{k}=\lim_{k\rightarrow\infty}q_{k}=t$. The
definitions of ${\cal{P}}$ and ${\cal{Q}}$ and Equations
$\left({\ref{bnd}}\right)$ and $\left({\ref{bndq}}\right)$ yield

$\displaystyle{\left|{\,{d}\\!\left(p_{k}\right)\,}\right|}\leq\epsilon\left({p_{k}-x}\right)\hskip
28.45274pt\mathrm{and}\hskip
28.45274pt{\left|{\,{d}\\!\left(q_{k}\right)\,}\right|}\leq\epsilon\left({q_{k}-x}\right),$
$\displaystyle{\left|{\,\frac{{f}\\!\left(p_{k}\right)-{f}\\!\left(x\right)}{p_{k}-x}-\underline{a}\,}\right|}\leq\epsilon\hskip
14.22636pt\mathrm{and}\hskip
14.22636pt{\left|{\,\frac{{f}\\!\left(q_{k}\right)-{f}\\!\left(x\right)}{q_{k}-x}-\overline{a}\,}\right|}\leq\epsilon,$
$\displaystyle{\left|{\,\frac{{g}\\!\left(q_{k}\right)-{g}\\!\left(x\right)}{q_{k}-x}-\underline{a}\,}\right|}\leq\epsilon\hskip
14.22636pt\mathrm{and}\hskip
14.22636pt{\left|{\,\frac{{g}\\!\left(p_{k}\right)-{g}\\!\left(x\right)}{p_{k}-x}-\overline{a}\,}\right|}\leq\epsilon.$

These equations imply that the sequences ${f}\\!\left(p_{k}\right)$,
${f}\\!\left(q_{k}\right)$, ${g}\\!\left(p_{k}\right)$,
${g}\\!\left(q_{k}\right)$, ${d}\\!\left(p_{k}\right)$ and
${d}\\!\left(q_{k}\right)$ are bounded and, by taking subsequences if
necessary, we can assume that they have limits $f_{p},f_{q},g_{p}$, $g_{q}$,
$d_{p}$ and $d_{q}$ respectively. We also have that

$\displaystyle{\alpha}\\!\left(p_{k}\right){f}\\!\left(p_{k}\right)+{\beta}\\!\left(p_{k}\right){g}\\!\left(p_{k}\right)={c}\\!\left(p_{k}\right)+{d}\\!\left(p_{k}\right),$
$\displaystyle{\alpha}\\!\left(p_{k}\right){f}\\!\left(q_{k}\right)+{\beta}\\!\left(p_{k}\right){g}\\!\left(q_{k}\right)={c}\\!\left(q_{k}\right)+{d}\\!\left(q_{k}\right).$

Taking the limit $k\rightarrow\infty$ in the equations above, and recalling
that $\alpha,\beta$ and $c$ are continuous, we get

$\displaystyle{\left|{\,d_{p}\,}\right|}\leq\epsilon\left({t-x}\right)\hskip
29.87547pt\mathrm{and}\hskip
17.07182pt{\left|{\,d_{q}\,}\right|}\leq\epsilon\left({t-x}\right),\hskip
17.07182pt$ (14)
$\displaystyle{\left|{\,\frac{f_{p}-{f}\\!\left(x\right)}{t-x}-\underline{a}\,}\right|}\leq\epsilon\hskip
14.22636pt\mathrm{and}\hskip
14.22636pt{\left|{\,\frac{f_{q}-{f}\\!\left(x\right)}{t-x}-\overline{a}\,}\right|}\leq\epsilon,$
(15)
$\displaystyle{\left|{\,\frac{g_{q}-{g}\\!\left(x\right)}{t-x}-\underline{a}\,}\right|}\leq\epsilon\hskip
14.22636pt\mathrm{and}\hskip
14.22636pt{\left|{\,\frac{g_{p}-{g}\\!\left(x\right)}{t-x}-\overline{a}\,}\right|}\leq\epsilon.$
(16)
$\displaystyle{\alpha}\\!\left(t\right)f_{p}+{\beta}\\!\left(t\right)g_{p}={c}\\!\left(t\right)+d_{p}\hskip
14.22636pt\mathrm{and}\hskip
14.22636pt{\alpha}\\!\left(t\right)f_{q}+{\beta}\\!\left(t\right)g_{q}={c}\\!\left(t\right)+d_{q}.$
(17)

Combining equations $\left({\ref{dd}}\right)$ and $\left({\ref{dc}}\right)$ we
obtain that

${\left|{\,{\alpha}\\!\left(t\right)\frac{f_{p}-f_{q}}{t-x}+{\beta}\\!\left(t\right)\frac{g_{p}-g_{q}}{t-x}\,}\right|}\leq
2\epsilon.$ (18)

Equations $\left({\ref{d3}}\right)$ and $\left({\ref{d4}}\right)$ yield

${\left|{\,\frac{f_{p}-f_{q}}{t-x}+\left({\overline{a}-\underline{a}}\right)\,}\right|}\leq
2\epsilon\hskip 14.22636pt\mathrm{and}\hskip
14.22636pt{\left|{\,\frac{g_{p}-g_{q}}{t-x}-\left({\overline{a}-\underline{a}}\right)\,}\right|}\leq
2\epsilon,$

and these inequalities yield

${\left|{\,{\alpha}\\!\left(t\right)\frac{f_{p}-f_{q}}{t-x}+{\beta}\\!\left(t\right)\frac{g_{p}-g_{q}}{t-x}+\left({{\alpha}\\!\left(t\right)-{\beta}\\!\left(t\right)}\right)\left({\overline{a}-\underline{a}}\right)\,}\right|}\leq
2\left({{\left|{{\alpha}\\!\left(t\right)}\right|}+{\left|{{\beta}\\!\left(t\right)}\right|}}\right)\epsilon.$
(19)

Combining inequalities $\left({\ref{foo}}\right)$ and
$\left({\ref{bar}}\right)$ we complete the proof of Equation
$\left({\ref{tiny_da}}\right)$.

We have now shown that $\underline{a}=\overline{a}$, and define
$a:=\underline{a}=\overline{a}$. Let us now prove that
${f}^{\prime}_{+}\\!\left(x\right)=a$. In fact, given $\epsilon>0$, Equations
$\left({\ref{equd}}\right)$ and $\left({\ref{eqod}}\right)$ imply that there
exists $t_{\epsilon}\in(x,x+\delta)$ such that if $t\in(x,t_{\epsilon})$ then

$-\epsilon\leq\frac{{\min}\\!\left({f}\\!\left(t\right)-{f}\\!\left(x\right),{g}\\!\left(t\right)-{g}\\!\left(x\right)\right)}{t-x}-a\leq\frac{{f}\\!\left(t\right)-{f}\\!\left(x\right)}{t-x}-a$
$\leq\frac{{\max}\\!\left({f}\\!\left(t\right)-{f}\\!\left(x\right),{g}\\!\left(t\right)-{g}\\!\left(x\right)\right)}{t-x}-a\leq\epsilon.$

Therefore, if $t\in(x,t_{\epsilon})$ then

${\left|{\,\frac{{f}\\!\left(t\right)-{f}\\!\left(x\right)}{t-x}-a\,}\right|}\leq\epsilon,$

and this shows that ${f}^{\prime}_{+}\\!\left(x\right)=a$. Analogously,
${g}^{\prime}_{+}\\!\left(x\right)=a$ and we are done. $\blacksquare$

## References

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