I want to apply the Leibniz Integration Rule to a function $f:\mathbb{R}\times \mathbb{R}^n\to \mathbb{R}$ given by (for a compact convex set $K\subseteq \mathbb{R}^n$):
$$f(x,\omega) =\frac{1}{\lVert \omega\rVert_2}\exp\left(-\frac{x^2}{\lVert \omega\rVert_2^2}\right)\chi_K(\omega),$$
where $\chi_K(x)$ is an indicator function.
This is to say that I want to justify the equality
$$\partial_x \int f(x,\omega)d\omega = \int \partial_x f(x,\omega)d\omega.$$
Per Wikipedia, this requires
- $\omega\mapsto f(x,\omega)$ is Lebesgue-integrable for each $x$,
- for almost all $\omega$, $\partial_x f(x,\omega)$ exists for all $x$,
- there is an integrable function $\theta(\omega)$ such that $|\partial_x f(x,\omega)| \leq \theta(\omega)$ for all $x$, for almost all $\omega$.
My issue is that $f(x,\omega)$ is "close" to many standard counterexamples to the Leibniz Integration Rule. While it is different in a way that "helps", its not clear to me if it is enough.
For example, for any $x\neq 0$ it is simple to see that $f(x,\omega)$ is integrable via applying the bound $\exp(x) \geq x^k/k!$ for large enough $k$. This upper bounds $f$ by a polynomial in $\omega$ (it is trivially lower-bounded by 0), and we are integrating over the compact set $K$, so everything is fine. The other properties similarly are straightforward to establish for $x\neq 0$.
There is a potential problem at $x = 0$ though, leading to the question
For $K$ a compact convex subset of $\mathbb{R}^n$, is $f(0,\omega) = \frac{1}{\lVert \omega\rVert_2}\chi_K(\omega)$ lebesgue integrable?
Of course I am also interested if the other conditions hold, but a negative answer to this suffices to answer my question.
The function does satisfy the desired conditions, at least if $n\geq 2$. It is first worth mentioning that condition 2 is easily satisfied, as it must only hold up to a countable set of exceptions (so bad behavior at $\omega = 0$ is fine). This leaves conditions 1 and 3.
Note that for these conditions, we really only need to be concerned with behavior at $x = 0$. For $x\neq 0$, one can apply a bound equivalent to $\exp(x) \geq x^k/k!$ for large-enough $k$ to get that
$$0\leq\int f(x,\omega)d\omega \leq \int \frac{k!}{\lVert\omega\rVert_2}\left(\frac{\lVert \omega\rVert_2^2}{x^2}\right)^k\chi_K(\omega)d\omega \leq \int k! x^{-2k}\lVert\omega\rVert_2^{2k-1}\chi_K(\omega)d\omega,$$
i.e. this is integrable. A similar bound holds for $\partial_x f(x,\omega)$ (and for any $\partial_x^if(x,\omega)$, if one allows $k$ to depend on $i$), i.e. one can easily show integrability for any dimension $n$.
For $x = 0$, this argument breaks down. Fortunately, the barrier to $\int\frac{1}{\lVert\omega\rVert_2}\chi_K(\omega)d\omega < \infty$ (namely the singularity at 0) is not an issue in higher dimensions. This is because one can see by switching to spherical coordinates (where we assume $f$ is radial for simplicity) that $\int f(x,\omega)d\omega \leq O(1)\int_0^\infty f(\lVert \omega\rVert_2)\lVert\omega\rVert_2^{n-1}\chi_K(\omega)d\omega$, where the constant is the surface area of $S^{n-1}$. In particular, provided $n-1+(-1)\geq 0$, $f(x,\omega)$ itself will be integrable. One can check that integrability of $\partial_xf(x,\omega)$ at zero requires no additional assumptions, as it vanishes for $x = 0$, so we solely need $n\geq 2$ overall.