Upper and Lower Derivatives of Real-Valued Functions

# Upper and Lower Derivatives of Real-Valued Functions

 Definition: Let $f$ be a real-valued function defined on an open interval $I$. The Upper Derivative of $f$ at $x \in I$ is defined as $\displaystyle{\overline{D} f(x) = \limsup_{h \to 0} \frac{f(x + h) - f(x)}{h}}$. The Lower Derivative of $f$ at $x \in I$ is defined as $\displaystyle{\underline{D} f(x) = \liminf_{h \to 0} \frac{f(x + h) - f(x)}{h}}$.

Note that $\limsup_{x \to a} f(x) = \lim_{\delta \to 0^+} \sup \left \{ f(x) : 0 < |x - a| < \delta \right \}$ and $\liminf_{x \to 0} f(x) = \lim_{\delta \to 0^+} \inf \left \{ f(x) : 0 < |x - a| < \delta \right \}$. Therefore the upper and lower derivatives of $f$ at $x$ can also be written as:

(1)
\begin{align} \quad \overline{D} f(x) = \lim_{\delta \to 0^+} \sup \left \{ \frac{f(x + h) - f(x)}{h} : 0 < |h| < \delta \right \} \quad \mathrm{and} \quad \underline{D} f(x) = \lim_{\delta \to 0^+} \inf \left \{ \frac{f(x + h) - f(x)}{h} : 0 < |h| < \delta \right \} \end{align}

From the definition above, it should be noted that for every $x \in I$:

(2)
\begin{align} \quad \underline{D} f(x) \leq \overline{D} f(x) \end{align}

The points $x \in I$ for which the upper and lower derivatives of $f$ at $x$ are equal will be of our interest.

 Definition: Let $f$ be a real-valued function defined on an open interval $I$. Then $f$ is said to be Differentiable at $x \in I$ if the upper and lower derivatives of $f$ at $x$ are finite and $\overline{D} f(x) = \underline{D} f(x)$.

We now prove a very important result.

 Lemma 1: Let $f$ be an increasing function on $[a, b]$. Then for all $\alpha \in \mathbb{R}$ with $\alpha > 0$ we have that: a) $m^* \left ( \{ x \in (a, b) : \overline{D} f(x) \geq \alpha \} \right ) \leq \frac{1}{\alpha} [f(b) - f(a)]$. b) $m^* \left \{ \{ x \in (a, b) : \overline{D} f(x) = \infty \} \right ) = 0$.
• Proof of a) Let $\alpha \in \mathbb{R}$, $\alpha > 0$ and define the following set:
(3)
\begin{align} \quad E_{\alpha} = \{ x \in (a, b) : \overline{D} f(x) \geq \alpha \} \end{align}
• Let $\alpha'$ be such that $0 < \alpha' < \alpha$ and define the collection $\mathcal I$ of intervals as:
(4)
\begin{align} \quad \mathcal I = \{ [c, d] \subseteq (a, b) : f(d) - f(c) \geq \alpha' (d - c) \} \end{align}
• We claim that $\mathcal I$ is a Vitali cover of $E_{\alpha}$. To prove this, let $\epsilon > 0$ be given and let $x \in E_{\alpha}$. Then by definition, $x \in (a, b)$ and $\overline{D} f(x) \geq \alpha > \alpha'$. So:
(5)
\begin{align} \quad \overline{D} f(x) = \inf_{\delta > 0} \sup \left \{ \frac{f(x + h) - f(x)}{h} : 0 < |h| < \delta \right \} > \alpha' \end{align}
• So for all $\delta > 0$ we have that:
(6)
\begin{align} \quad \sup_{h} \left \{ \frac{f(x + h) - f(x)}{h} : 0 < |h| < \delta \right \} > \alpha' \end{align}
• So for all $\delta > 0$ there exists an $h$ with $0 < |h| < \delta$ such that $\displaystyle{\frac{f(x + h) - f(x)}{h} > \alpha'}$.
• Let $\delta^* = \min \{ x - a, b - x, \epsilon \}$. Then for $h$ such that $0 < |h| < \delta^*$ we have that:
(7)
\begin{align} \quad \frac{f(x + h) - f(x)}{h} > \alpha' \end{align}
• If $h > 0$ we let $c = x$ and $d = x + h$. Then from above we get:
(8)
\begin{align} \quad \frac{f(d) - f(c)}{d - c} & > \alpha' \\ \quad f(d) - f(x) & > \alpha'(d - c) \end{align}
• If $h < 0$ we let $c = x + h$ and $d = x$. Then from above we get:
(9)
\begin{align} \quad \frac{f(x + h) - f(x)}{h} & > \alpha' \\ \quad \frac{f(c) - f(d)}{c - d} & > \alpha' \\ \quad f(c) - f(d) & < \alpha'(c - d) \\ \quad f(d) - f(c) & > \alpha'(d - c) \end{align}
• In either case, $[c, d] \in \mathcal I$, $x \in [c, d]$ and furthermore, $l([c, d]) = |h| < \delta < \epsilon$. So indeed, $\mathcal I$ is a Vitali cover of $E_{\alpha}$.
• Now let $\epsilon > 0$ be given. By the Vitali cover lemma, since $m^*(E_{\alpha}) \leq m^*((a, b)) = b - a < \infty$, and $\mathcal I$ is a Vitali cover of $E_{\alpha}$ there exists a finite mutually disjoint collection of intervals $\{ [c_1, d_1], [c_2, d_2], ..., [c_n, d_n] \}$ from $\mathcal I$ such that:
(10)
\begin{align} \quad m^* \left ( E_{\alpha} \setminus \bigcup_{k=1}^{n} [c_k, d_k] \right ) < \epsilon \end{align}
• Now we have that:
(11)
\begin{align} \quad E_{\alpha} \left ( \subseteq E_{\alpha} \setminus \bigcup_{k=1}^{n} [c_k, d_k] \right ) \cup \left ( \bigcup_{k=1}^{n} [c_k, d_k] \right ) \end{align}
• By the monotonicity property of the Lebesgue outer measure we have that:
(12)
\begin{align} \quad m^*(E_{\alpha}) & \leq m^* \left ( E_{\alpha} \setminus \bigcup_{k=1}^{n} [c_k, d_k] \right ) + m^* \left ( \bigcup_{k=1}^{n} [c_k, d_k] \right ) \\ & < \epsilon + \sum_{k=1}^{n} m^*([c_k, d_k]) \\ & < \epsilon + \sum_{k=1}^{n} l([c_k, d_k]) \\ & < \epsilon + \sum_{k=1}^{n} (d_k - c_k) \end{align}
• Now since $[c_k, d_k] \in \mathcal I$ for all $k \in \{ 1, 2, ..., n \}$ we have by the definition of $\mathcal I$ that for each $k$:
(13)
\begin{align} \quad f(d_k) - f(c_k) \geq \alpha' (d_k - c_k) \quad \Leftrightarrow \quad (d_k - c_k) \leq \frac{1}{\alpha'} [f(d_k) - f(c_k)] \end{align}
• Therefore:
(14)
\begin{align} \quad m^*(E_{\alpha}) < \epsilon + \frac{1}{\alpha'} \sum_{k=1}^{n} [f(d_k) - f(c_k)] \end{align}
• Since $f$ is an increasing function we have that:
(15)
\begin{align} \quad m^* ( \{ x \in (a, b) : \overline{D} f(x) \geq \alpha \}) < \epsilon + \frac{1}{\alpha'} [f(b) - f(a)] \end{align}
• By letting $\epsilon \to 0$ and $\alpha' \to \alpha$ we get:
(16)
\begin{align} \quad m^* ( \{ x \in (a, b) : \overline{D} f(x) \geq \alpha \}) \leq \frac{1}{\alpha} [f(b) - f(a)] \quad \blacksquare \end{align}
• Proof of b) We note that
(17)
\begin{align} \quad \{ x \in (a, b) : \overline{D} f(x) = \infty \} &= \bigcap_{\alpha} \{ x \in (a, b) : \overline{D} f(x) > \alpha \} \\ \end{align}
• For any fixed $\alpha^* \in \mathbb{R}$ we have that $\displaystyle{\{ x \in (a, b) : \overline{D} f(x) \geq \alpha^* \} \supseteq \bigcap_{\alpha} \{ x \in (a, b) : \overline{D} f(x) > \alpha \}}$. So for each $\alpha^* \in \mathbb{R}$ with $\alpha^* > 0$ we have that:
(18)
\begin{align} \quad m^* ( \{ x \in (a, b) : \overline{D} f(x) = \infty \} ) \leq m^* ( \{ x \in (a, b) : \overline{D} f(x) \geq \alpha^* \}) \leq \frac{1}{\alpha^*} [f(b) - f(a)] \end{align}
• By letting $\alpha^* \to \infty$ we get that:
(19)
\begin{align} \quad m^* ( \{ x \in (a, b) : \overline{D} f(x) = \infty \} ) = 0 \quad \blacksquare \end{align}