Let $(\Omega, \mathcal{F}, \mu)$ be a probability space. For the sake of simplicity, let $\Omega$ be $\mathbb{R}$. In what follows, the measurability always refers to Borel measurability.

Let $f \colon \mathbb{R}_+ \times \Omega \to \mathbb{R}_+$ be a function such that:

(i) For each $z \in \Omega$, the function $k \mapsto f(k,z)$ is concave, increasing, and continuously differentiable, while $z \mapsto f(k,z)$ is Borel measurable for each $k \in \mathbb{R}_+$; (ii) $\lim_{k \downarrow 0} f'(k,z) >0$ for each $z \in \Omega$. Here and below, $f'(k,z)$ denotes the partial derivative of $f$ with respect to $k$; and (iii) $f(0,z)=0$ for all $z \in \Omega$.

Let $v \colon \mathbb{R}_+ \to \mathbb{R}_+$ be a bounded, strictly concave and strictly increasing function, and be continuously differentiable on $(0, \infty)$.

Define a function $g $ by

\begin{align*} g(k) := \left( \int_{\Omega} \left[ v\left( f( k, z ) \right) \right]^\alpha \mu (\mathrm{d}z) \right)^{1/\alpha}, \qquad (0<\alpha <1). \end{align*}

Question: In fact, since $g$ is concave (it has been proved), we know that the right-hand and the left-hand derivatives of $g$ exist.

I aim to show that $g$ is differentiable on $(0, \eta)$ for any fixed constant $\eta >0$, and to show the derivative of $g$ which I conjecture is \begin{align*} g'(k) = \left( \int_{\Omega} \left[ v\left( f( k, z ) \right) \right]^\alpha \mu (\mathrm{d}z) \right)^{\frac{1}{\alpha} -1 } \int_{\Omega} \left[ v\left( f( k, z ) \right) \right]^{\alpha -1} v'\left( f( k, z ) \right) f'( k, z ) \mu( \mathrm{d} z) \end{align*} for all $0 < k < \eta$.

Here, $g'(k) = \dfrac{\mathrm{d}}{\mathrm{d} k} g(k)$, $v'\left( f( k, z ) \right) := \dfrac{\mathrm{d}}{\mathrm{d} f }v( f( k, z ))$, and $f'(k, z) := \dfrac{\partial}{\partial k} f(k,z)$.

My attempt:

The above stated derivative of $g$ is just my conjecture and I am not sure if the right-hand derivative of $g$ is equal to its left-hand derivative, thus I wish to verify it. My attempt is making use of the limit definition to show the left-hand and right-hand derivatives of $g$ are the same and equal to the above stated formula.

In fact, I even got stuck in finding the left-hand side and the right-hand side derivatives of $g$. But I thought it might suffice to show that the left-hand side derivative $g’_-(k) := \lim_{h \to 0^-} \dfrac{g(k+h)-g(k)}{h}$ is less than the conjecture formulation that stated above, and to show that the right-hand side derivative $g’_+(k) := \lim_{h \to 0^+}\dfrac{g(k+h)-g(k)}{h}$ is greater than the conjecture formulation. Then, by concavity of $g$, we have $g’_-(k) \geq g’_+(k)$ and hence, $g’_-(k)= g’_+(k)=g’(k)$ as desired. In this connection, I think the problem becomes how to establish the relation between the right-hand derivative and conjecture formula, and relation between the left-hand derivative and conjecture formula.

Could anyone give me some guidance and help me out please?

Thank you very much in advance!

  • $\begingroup$ Thank you @ArnaudMortier . $\endgroup$ – Paradiesvogel May 27 '18 at 2:52
  • $\begingroup$ "while $z \mapsto f(k,z)$ is Borel measurable for each $k \in \mathbb{R}_+$;" that's weird: you didn't say anything about $\Omega$ being a topological space $\endgroup$ – zhw. May 30 '18 at 2:33
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    $\begingroup$ Can't you just use the measure theory formulation of Leibniz rule? Or are you looking for its the explicit proof/derivations? $\endgroup$ – Vlad May 30 '18 at 2:39
  • $\begingroup$ @zhw. Thanks for your comment. I’ve updated my post now. $\endgroup$ – Paradiesvogel May 30 '18 at 2:41
  • $\begingroup$ Thanks for your reply @Vlad . In fact, I’m looking for a method to solve out the derivative of the differentiable function $g$. And such $g$ is quite complicated so that I’m not sure if the conventional chain rule applies. $\endgroup$ – Paradiesvogel May 30 '18 at 2:45

To use the Liebnitz rule, nicely explained by Vlad, we need

$$\tag 1 \int_\Omega \partial_k (v^\alpha \circ f(k,z))\,d \mu (z) <\infty.$$

Think about this for a bit: If we had $\Omega, f, v,$ such that the integrand is not bounded for some $k,$ we could fashion a probability measure $\mu$ on $\Omega$ for which $(1)$ fails. So boundedness is something we need to look for.

Let $\mathcal U$ be the collection of functions $u\in C^1(0,\infty)$ that are positive, increasing, and concave. Let $\mathcal U_b$ be the set of bounded functions $u\in \mathcal U.$

Lemma: Suppose $u\in \mathcal U.$ Then

$$u'(x) \le \frac{u(x)}{x}\,\,\text {for all } x>0.$$

If in addition $u\in \mathcal U_b,$ then

$$u'(x)x \le \|u\|_\infty.$$

I'll leave the proof of this to you for now. It's fairly simple, but ask questions if you like.

Let's turn to $(1).$ We are given $v\in \mathcal U_b.$ This implies $v^\alpha \in \mathcal U_b.$ Thus by the lemma, we have

$$ (v^\alpha)'(x) x \le \|v^\alpha\|_\infty.$$

We also have $f(k,z) \in \mathcal U$ as a function of $k$ for each fixed $z.$ Thus the lemma gives

$$f'(k,z) \le \frac{f(k,z)}{k}.$$

It follows that

$$\tag 2(v^\alpha)'(f(k,z)) f'(k,z) \le (v^\alpha)'(f(k,z)) \frac{f(k,z)}{k} \le \frac{\|v^\alpha\|_\infty}{k}.$$

But the left side of $(2)$ is exactly the integrand in $(1)$. Thus for $k \ge k_0$ we have the uniform bound $\|v^\alpha\|_\infty/k_0$ on the integrands. Of course the constant function equal to to this bound belongs to $L^1(\mu),$ since we're in a probablity space. This allows us to use Leibnitz on each interval $(k_0,\infty),$ and gives the desired result.

  • $\begingroup$ What do you mean by "this needs to hold for [...] every probability space $(\Omega,\mu)$"? Aren't $(\Omega,\mu)$ fixed once and for all? The definition of the map $f$ requires it to be fixed, doesn't it? $\endgroup$ – Arnaud Mortier Jun 4 '18 at 20:26
  • $\begingroup$ Thanks @zhw. In fact, by concavity of the function $[ v ( f ( \cdot , z))]^\alpha$ for each $z$, we can directly have that for any $k>0$, $ (v^\alpha)’ (f(k,z)) f’(k,z) \leq (v^\alpha)’ ( f( k_0,z)) f’(k_0,z)=: M(z)$ for each $z$ whenever $k_0 \in (0, k)$. Evidently, $M(z)$ is bounded and thus $\mu$-integrable. I think the difficulty is how to show that $g$ is differentiable, otherwise we cannot directly apply the Chain rule and Leibniz rule. $\endgroup$ – Paradiesvogel Jun 4 '18 at 20:46
  • $\begingroup$ @ArnaudMortier I should have said "probability measure". Notice if there existed $\Omega, f, v$ such that the slice integrand is unbounded, you can cook up a $\mu$ that yields nonintegrability. That would be a counterexample. $\endgroup$ – zhw. Jun 4 '18 at 20:50
  • $\begingroup$ @Paradiesvogel No, my answer shows $g$ is differentiable by Leibniz. $\endgroup$ – zhw. Jun 4 '18 at 20:51
  • $\begingroup$ @zhw. Would you mind to explain more details about how the differentiability of $g$ is proven via Leibniz rule please? I thought the role of Leibniz rule just gives us the interchange between the differential operator and integral operator. Perhaps I am missing something out. Many thanks again:) $\endgroup$ – Paradiesvogel Jun 4 '18 at 21:05

Perhaps I am missing something out, but it seems to me that you can obtain your final result using Leibniz's rule as shown below.

$$ \begin{align*} g(k) := \left( \int_{\Omega} \big[ v\left( f( k, z ) \right) \big]^\alpha \mu (\mathrm{d}z) \right)^{\frac{1}{\alpha}} =I^\frac{1}{\alpha}\left(k\right)\qquad (0<\alpha <1). \end{align*} $$ where $$ I(k)=\int_{\Omega} v^\alpha \left( f( k, z ) \right) \mu (\mathrm{d}z) $$

Thus, if we apply regular chain rule we get

$$ \frac{\partial }{\partial k}\left[g\left(k\right)\right] =\frac{\partial }{\partial k}\left[I^\frac{1}{\alpha}\left(k\right)\right] =\frac{1}{\alpha}I^{\frac{1}{\alpha}-1}\left(k\right) \frac{\partial }{\partial k}\left[I \left(k\right)\right] $$

Then, using Measure theory form of Leibniz rule we write

$$ \frac{\partial }{\partial k}\left[I \left(k\right)\right] = \frac{\partial }{\partial k}\left[\int_{\Omega} \big[ v\left( f( k, z ) \right) \big]^\alpha \mu \left(\mathrm{d}z\right)\right] = \int_{\Omega} \frac{\partial }{\partial k}\big[ v\left( f( k, z ) \right) \big]^\alpha \mu\left(\mathrm{d}z\right) = \int_{\Omega} \alpha v^{\alpha-1}\left( f( k, z ) \right) v'\left( f( k, z )\right) \frac{\partial }{\partial k}\big[ f( k, z ) \big] \mu (\mathrm{d}z) = \alpha\int_{\Omega} v^{\alpha-1}\left( f \right) v'\left( f \right) f_k\left( k, z \right) \mu \left(\mathrm{d}z\right) $$

Substituting the last expression into previous formula for $g'(k)$ we get

$$ g'\left(k\right) = \frac{1}{\alpha} \left(\int_\Omega \nu^\alpha\left(f\right)\mu\left(\mathrm{d}z\right)\right)^{\frac{1}{\alpha}-1} \cdot \alpha\int_{\Omega} v^{\alpha-1}\left( f \right) v'\left( f \right) f_k\left( k, z \right) \mu \left(\mathrm{d}z\right) $$ $$ \boxed{g'\left(k\right) = \left(\int_\Omega \nu^\alpha\left(f\right)\mu\left(\mathrm{d}z\right)\right)^{\frac{1}{\alpha}-1} \int_{\Omega} v^{\alpha-1}\left( f \right) v'\left( f \right) f_k\left( k, z \right) \mu \left(\mathrm{d}z\right)} $$

where $f=f\left(k,z\right)$ and $f_k\left(k,z\right)=\dfrac{\partial f\left(k,z\right)}{\partial k}$.

Justification for Using Leibniz's rule

As pointed out in comments by the OP, I have not provided justification for using Leibniz's rule in measure-theoretical form:

Let $X$ is an open subset of $ \mathbb{R} $, and $\Omega$ is a measure space. Suppose $ F\colon X\times \Omega \rightarrow \mathbb {R}$ satisfies the following conditions:

  • $ F(x,\omega )$ is a Lebesgue-integrable function of $\omega$ for each $ x\in X$.

  • For almost all $\omega \in \Omega$, the derivative $ F_{x} $ exists for all $ x\in X$.

  • There is an integrable function $\displaystyle \theta \colon \Omega \rightarrow \mathbf {R}$ such that $$\left\lvert F_{x}(x,\omega )\right\rvert\leq \theta (\omega )$$ for all $ x\in X$ and almost every $\omega \in \Omega $.

Then, for all $x\in X$, $$ {\frac {d}{dx}}\int _{\Omega }F(x,\omega )\,d\omega =\int _{\Omega }F_{x}(x,\omega )\,d\omega . $$

Let us verify the last condition for the integrand

$$ F\left(k,z\right)=\left[ v\big( f( k, z ) \big) \right]^\alpha , \qquad 0<\alpha<1. $$

Observe that $F\left(k,z\right)$ is superposition of three concave increasing functions (treating $k\mapsto f\left(k,z\right)$ as function of single variable $k$)

$$ F\left(\cdot\right) = F_1\circ F_2\circ F_3 = F_3\left(F_2\left(F_1\left(\cdot\right)\right)\right), $$


$$ \begin{aligned} F_1\left(\tau\right) &= f\left(\tau,z\right), &&\text{where $z$ is fixed} \\ F_2\left(\tau\right) &= v\left(\tau\right), &&\\ F_3\left(\tau\right) &= \tau^{\alpha}, && 0<\alpha<1 \end{aligned} $$

with $\tau$ being dummy variable. As an exercise, I propose you to try to show explicitly that superposition of $F\left(\cdot\right) = F_1\circ F_2\circ F_3 = v^\alpha\left( f\left( k, z \right) \right)$ is convex in $k$.

Recall that concave function of single variable has monotonically decreasing derivative, which means that $F'\left(\tau\right)\leq F'\left(\tau_0\right)$ whenever $\tau\leq\tau_0$. For example, we can get bound

$$ \begin{aligned} F'\left(\tau\right)&<F'\left(0\right)& \text{since}& &f \colon \mathbb{R}_+ \times \Omega \to \mathbb{R}_+ \implies \tau>0 \end{aligned} $$

where $F'\left(0\right)$ does not depend on $\tau$ (and possibly depends on $\omega$).

Thus, choosing nonnegative constant $k_0$ we can bound derivative as $$ \frac{\partial}{\partial k} \Big[v^\alpha\big( f\left( k, z \right) \big)\Big] \leq \left.\frac{\partial}{\partial k} \Big[v^\alpha\big( f\left( k, z \right) \big)\Big]\right\rvert_{k_0} \qquad\text{ whenever } k\geq k_0. $$

Thus the third condition is satisfied.

  • $\begingroup$ Thanks for your reply @Vlad . It seems to be working fine now. When I asked you in previous comment, my brain was dead and I thought the bound derivative might not be integrable. Please give me more time to think about it more carefully. Your kind help is much appreciated! :) $\endgroup$ – Paradiesvogel May 30 '18 at 7:59
  • $\begingroup$ But why is $\left.\frac{\partial}{\partial k} \Big[v^\alpha\big( f\left( k, z \right) \big)\Big]\right\rvert_{k_0}$ in $L^1(\mu)?$ $\endgroup$ – zhw. Jun 1 '18 at 17:51
  • $\begingroup$ @zhw. My reasoning goes as following: I. Since $f\left(k,z\right)$ is measurable in $z$ and concave, increasing, and continuously differentiable in $k$, the derivative $\partial_x f\left(k,z\right)$ exists, is continuous, and monotone decreasing. II. Since composition of measurable and continuous functions is measurable, the derivative $\partial_k\left[v^\alpha\big( f\left( k, z \right) \big)\right]$ Lebesgue–integrable for any finite number $k_0\in\mathbb{R}^+$. $\endgroup$ – Vlad Jun 2 '18 at 2:52
  • $\begingroup$ @Vlad I don't understand why you say $\int_\Omega \partial_k v^\alpha(f(k,z)) \,d\mu(z)<\infty.$ Have I missed something? I would think the boundedness of $v$ has to play a role, but I didn't see you use it. $\endgroup$ – zhw. Jun 4 '18 at 22:07
  • $\begingroup$ WHen you conclude "the third condition is satisfied", isn't there an absolute value missing? How do you justify that $|\partial_k\ldots|$ is bounded? $\endgroup$ – Arnaud Mortier Jun 5 '18 at 23:57

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