3
$\begingroup$

I am new to this site so I am not sure if this is the right place to be asking this question but I will try anyway. I am reading an economics paper for my research and the author does the following: $\frac{\partial}{\partial C_t(j)} \int_0^1 P_t(j) C_t(j) dj = P_t(j)$

I feel that this derivative is not properly defined but I am probably missing something obvious because the author knew what he was doing. Could someone please tell me if this is a legitimate derivative?

Thanks,

2 Answers 2

4

There is a mathematically serious notion of a functional derivative, but it requires distributions in order to obtain a rigorous foundation. I'll shoot for a hand-wavy view of the situation.

We define the Dirac delta function $\delta$ to be such that $\displaystyle \int_{-\infty}^{+\infty}\delta(x)f(x)dx=f(0)$. Such a thing doesn't technically exist, but if you play by certain rules you can "act" like it does and remain consistent; you'll have to dive more into the theory of distributions if you want a more satisfying explanation.

Additionally, while a function $f$ maps numbers to numbers typically (for example $\mathbb{R}\to\mathbb{R}$), we introduce what's called a functional, which maps functions to numbers. Now we define a derivative

$\frac{\delta \mathcal{F}}{\delta \phi}=\lim_{\epsilon\to0}\frac{\mathcal{F}(\phi(\cdot)+\epsilon\,\delta(\;\cdot\;-y))-\mathcal{F}(\phi(\cdot))}{\epsilon}$

Here $\phi(\cdot)$ is a function, $\mathcal{F}(\;)$ is a functional, $\epsilon$ is restricted to actual numbers, and the derivative here is actually a function of $y$. So if we have a linear functional

$\mathcal{F}(f)=\int_a^b K(x)f(x)dx$

then our derivative is given by

$\lim_{\epsilon\to0}\frac{\int_a^b K(x)[f(x)+\epsilon\,\delta(x-y)]dx-\int_a^b K(x)f(x)dx}{\epsilon}=\lim_{\epsilon\to0}\frac{\int_a^b \epsilon\, K(x)\delta(x-y)dx}{\epsilon}=K(y).$

(Actually it would evaluate to $0$ if $y$ is outside of $[a,b]$.) This is consistent with what the author wrote, with the caveat that there should be a non-$j$ variable on the right-hand side. But as I said, you will need to go at least knee-deep into the theory of distributions in order to fully make sense of these concepts and see how they are rigorously defined. Hope that helps.

  • 0
    Thanks, that was very helpful.2012-01-29
0

That derivative can be properly defined if and only if


there exists an appropriate pair of values $t$ and $j$ such that $\: C_t(j) = 0 \:$

and

all appropriate values of $t$ such that
[there exists an appropriate value of $j$ such that $\: C_t(j) = 0 \:$]
give the same value of the integral

and

$0$ is a limit point of the set of values taken by $C_t(j)$ for appropriate values of $t$ and $j$


("appropriate values" are those associated with the definition of $C_t(j)$, and even if the derivative "can be properly defined", it may still fail to exist, as is the case with non-differentiable functions).


If either of the first two fail there is not a unique "base value" to plug into the difference quotient.
If the last one fails, it is not clear what the definition of "limit" would be.