I've got the following problem:
Proof or disproof: there is a g $\in C(\mathbb{R^2})$ so that a $h(y) := \int_{-\infty}^{\infty}g(x,y)dx$ for every $y \in \mathbb{R}$ exists, but is not continuous.
A hint suggests I should consider that for every $f \in C(\mathbb{R})$ that is improperly integrable and $y>0$ it applies that:
$\int_{-\infty}^{\infty}yf(yx)dx = \int_{-\infty}^{\infty}f(x)dx$
I honestly don't quiet know how to even start solving this problem. I think I don't have the right properties for improper integrals with two variables in mind to work with them.
Does anyone have an Idea?
Following the hint, take $g(x,y) = yf(yx)$ and $f \in C(\mathbb{R})$ which is integrable over $\mathbb{R}$ and such that $\int_{\mathbb{R}} f(x) \, dx \neq 0$. (For example take $f(x) = e^{-x^2}$).
In this case, $h(0) = 0$, but
$$\lim_{y \to 0}h(y) = \lim_{y \to 0}\int_{-\infty}^{\infty}yf(yx) \, dx = \lim_{y \to 0}\int_{-\infty}^{\infty}f(x) \, dx = \int_{-\infty}^{\infty}f(x) \, dx \neq 0$$
Hence, $h$ is not continuous at $y = 0$.