# Surjectivity of a class of integrals in dimensions two

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?

• I am not sure about the quantifiers here. Is the statement in question of the form "For each $\Omega$ and each $G$ such that ... there is some $g$ such that ..."? Or is the statement in question of the form "there are $\Omega$, $G$, and $g$ such that ... "? Or is the statement in question something else? Commented Sep 24, 2023 at 1:35
• I meant the form "For each Ω and each 𝐺 such that ... there is some 𝑔 such that ..."? Commented Sep 24, 2023 at 3:45

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.