Absolute continuity of $g(x) = \int_{-\infty}^{\infty} \frac{\cos(xy^3)}{1+y^2}dy$.

Question

I'm trying to determine whether the function $$g(x) = \int_{-\infty}^{\infty} \frac{\cos(xy^3)}{1+y^2} dy$$ (a) Is continuous,

(b) Is uniformly continuous,

(c) Is absolutely continuous,

in $(-\infty, \infty)$.

Would $h(x) = \cos(xy^3)$ be absolutely continuous? If so, would it not be enough to write $\cos(xy^3)$ as an indefinite integral since we know that a function is absolutely continuous if and only if it can be written as an indefinite integral? This approach seems to be too easy, which makes me think that there is something that I'm not quite understanding. Any hints would be appreciated.

Answer 1

This only answers (1) and (2) of the OP's questions.

2. $g(x)$ is uniformly continuous in $\mathbb{R}$: $$ \begin{align} |\cos((x+h)y^3)-\cos(xy^3)|&=|\cos(xy^3)\cos(hy^3)-\sin(xy^3)\sin(hy^3)-\cos(xy^3)|\\ &\leq \cos(xy^3)||1-\cos(hy^3)| +|\sin(xy^3)\sin(hy^3)|\\ &\leq |1-\cos(hy^3)|+|\sin(hy^3)| \end{align} $$

Hence $$ |g(x+h)-g(x)|\leq \int^\infty_{-\infty}\frac{|1-\cos(hy^3)|+|\sin(hy^3)|}{1+y^2}\,dy $$ Since $y\mapsto \frac{1}{1+y^2}$ belongs to $L_1(\mathbb{R})$, it follows by dominated convergence that $$ \limsup_{h\rightarrow0}\sup_{x\in\mathbb{R}}|g(x+h)-g(x)|\leq\lim_{h\rightarrow0}\int^\infty_{-\infty}\frac{|1-\cos(hy^3)|+|\sin(hy^3)|}{1+y^2}\,dy=0 $$ This of course implies (1) in the OP.

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3. As for absolute continuity, I would venture to say no. Here I only wave may hands (or rather let my fingers dance on the keyboard). The point is that if $g$ were absolute continuous, then $g'$ would exists almost surely and on any bounded closed interval $[a,b]$ we would have $$g(x)=g(a)+\int^x_a g'(t)\,dt$$ by the (Lebesgue) fundamental theorem of Calculus.

But (and here is my not being completely rigorous) then is tempting to say that $$g'(x)=-\int^\infty_{-\infty}\frac{y^3}{1+y^2}\sin(xy^3)\,dy$$

But $y\mapsto\frac{y^3|\sin(xy^3)|}{1+y^2}$ is not intgrable.

I believe $g$ is differentiable and that $g'$ has an integral form, but not in the sense of Lebesgue but in the sense of the principal value. Here you would have to do some extra work.