6
$\begingroup$

To prove the differentiability of a function of the form $$g(x) = \int f(t,x)dt$$ where $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is $C^1$ in the second variable I sometimes see applications of the mean-value theorem as follows:

Let $h\in \mathbb{R}$ and define $\varphi_t(s) = f(t,x+sh)$. Then by the mean-value theorem $\varphi_t(1)-\varphi_t(0) = \varphi_t'(\xi(t)) = \partial_xf(t,x+\xi(t) h)h$ where $\xi(t)$ lies between $0$ and 1. Therefore

\begin{align*} \frac{g(x+h)-g(x)}{h} = \int \frac{f(t,x+h)-f(t,x)}{h}dt = \int \partial_xf(t,x+\xi(t)h)dt. \end{align*}

So for instance if we assume som integrability condition on $\partial_x f(t,x)$ (for instance suppose we can find an integrable majorant) then by Lebesgue's dominated convergence theorem $g$ is differentiable with derivative

$$g'(x) = \int \partial_x f(t,x)dt.$$

Now what I wonder is if the function $\xi(t)$ appearing in the application of the mean-value theorem is actually measurable? As far as I understand its existence is only a consequence of the fact that a continuous function on a compact set has a maximum and a minimum, since this is how one shows Rolle's theorem which one can then use to prove the mean-value theorem.

If $\xi$ is not uniquely defined one must apply the axiom of choice when defining the function right?


The measurability of $\xi(t)$ is not vital for the differentiation: Since $$\partial_xf(t,x+\xi(t)h) = \frac{f(t,x+h)-f(t,x)}{h}$$ and since the right-hand side is measurable we find that $t\mapsto \partial_xf(t,x+\xi(t)h)$ is measurable. Supposing there exists some integrable function $F(t)\geq 0$ such that $\partial_xf(t,x)\leq F(t)$ for all $x$ then we can take the limit.

$\endgroup$

1 Answer 1

3
$\begingroup$

Two very vague comments:

  • If you assumed that $f \in C^1(\mathbb{R}^2; \mathbb{R})$ and if you have sufficient regularity assumptions, you could use the implicit function theorem to show the existence of $\xi$ and also its continuity. If $\xi$ is continuous, it will also be measurable.

  • Proving this theorem of differentiation under an integral really does not need the mean value theorem in my opinion. Personally, I find the proof even more difficult to understand with the mean value theorem, instead of just writing down the limit of the finite difference $$\lim_{h \rightarrow 0} \int\frac{f(t,x+h) - f(t,x)}{h} \mathrm{d}t$$ and employing dominated convergence to show that $\lim$ and $\int$ may be interchanged.

$\endgroup$
3
  • $\begingroup$ Ok thanks for the answer. In the book I used for measure theory I saw what you use under your second point as a proof of derivative under the integral, however I tend to see applications of the mean-value theorem from time to time and this spurred my interest concerning its measurability. I see how one would be able to use the implicit function but do we really need $f\in C^1(\mathbb{R}^2,\mathbb{R})$ for measurability to hold? $\endgroup$
    – OgvRubin
    Feb 2, 2020 at 15:01
  • 1
    $\begingroup$ I would be surprised, if $f \in C^1(\mathbb{R}^2;\mathbb{R})$ were necessary to show measurability of $\xi$. This seemed just like one way one could go. $\endgroup$
    – Jonas
    Feb 2, 2020 at 15:11
  • $\begingroup$ Yes I see what you mean it is a nice way to go, also most examples I encounter allow for this. Also I think one doesn't need measurability of $\xi(t)$ to be able to use the mean-value method. Since it equals the difference quotient $\psi_h(t) \equiv \partial_x f(t,x+\xi(t) h)$ will at least be measurable so the proof would still go through by application of Lebesgue $\endgroup$
    – OgvRubin
    Feb 2, 2020 at 15:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .