Surjectivity of a class of integrals in dimensions two

Question

Let $\Omega \subset \mathbb{R}^2$ be an open set and $G(x,\theta): \Omega \times [0,2\pi]\rightarrow \mathbb{R}$ be a positive continuous function. Assume $F:\Omega \rightarrow \mathbb{R}^2$ defined as

$$F(x)= \int_0^{2\pi}G(x,\theta) (\cos \theta, \sin \theta) d\theta, $$ is surjective. Does there exists a continuous positive function $g:\mathbb{R} \rightarrow \mathbb{R}$ with $\lim_{t\rightarrow \infty}g(t)=\infty$ such that

$$H(x)= \int_0^{2\pi}g(G(x,\theta)) (\cos \theta, \sin \theta) d\theta $$

is not surjective?

Answer 1

An explicit counterexample

The answer to your question is: in general, no. Here I will show a counterexample.

As in the previous version of the answer, we set $\Omega=\mathbb R^2$, and for the sake of simplicity of notation we identify $\mathbb R^2$ with $\mathbb C$, so that we can write $x=re^{i\varphi}$ with $r\geq 0,\varphi\in[0,2\pi)$. To be clear, we do that for both the domain and the codomain of $F$ and $H$, so $$ F,H\colon\mathbb C\to\mathbb C. $$

The counterexample I found is the function $G$ defined as $$ G(re^{i\varphi},\theta)=G(r,\theta-\varphi), $$ $$ G(r,\theta):=1+r\chi(\cos\theta), $$ where $\chi$ is a non decreasing, smooth function such that $$ \chi(s)=\left\{\begin{aligned} &0& &s\leq 0,\\ &1& &s\geq 1/2. \end{aligned}\right. $$ For this map, the function $H$ will always be surjective under your hypotheses, no matter how you choose $g$. The proof would be similar to the example of the previous version of the answer: the maps $F$ and $H$ are invariant under rotations and $F(0)=H(0)=0$, so by continuity it is enough to show that they are unbounded to prove their surjectivity. The unboundedness of $F$ is not difficult (it is actually a special case of $H$ with $g(t)=t$). For the unboundedness of $H$, you need to work a little more, but the idea is not complicated. First, by rotational symmetry, you check the unboundedness on the subset $\mathbb R^+\subset \mathbb C$, i.e., now the input of $H$ is a real number $r\geq 0$. Since $G$ is even in the variable $\theta$, the imaginary part of $H(r)$ (as I said, I identify $\mathbb R^2$ with $\mathbb C$, so the second component of $H$ is nothing but the imaginary part) is zero, so we have that $H(r)$ is also a real number, defined as $$ H(r)=\int \cos\theta \,g(1+r\chi(\cos\theta))d\theta. $$ The idea is that the contribution when $\cos(\theta)\geq 1/2$ grows arbitrarily as $r\to\infty$, while the contribution when $\cos(\theta)\leq 0$ is fixed and the one from the remaining values of $\theta$ is bounded from below by the minimum of the function $g$. So $r\mapsto H(r)$ is unbounded, hence $H\colon \mathbb C\to\mathbb C$ is surjective.

I hope this explanation is enough. Please let me know if you want more details.