This question is from a practice qualifying exam. Here's my attempt (I'm a bit stuck on the continuity part):
Since $f \in L^1(\mathbb{R})$, $f$ is bounded. Then:
$$|g(y)| = |\int_{\mathbb{R}} \sin(y^2x)f(x) dx| \leq \int_{\mathbb{R}} |\sin(y^2x)|f(x)| dx \leq \int_{\mathbb{R}} |f(x)| dx$$
So $g$ is bounded. For continuity, let $y,z \in \mathbb{R}$ and $\epsilon >0$.
$$|g(y)-g(z)| = |\int_{\mathbb{R}} \sin(y^2x)f(x) dx - \int_{\mathbb{R}} \sin(z^2x)f(x) dx| = |\int_{\mathbb{R}} \left[\sin(y^2x)-\sin(z^2x)\right]f(x) dx|$$
Now $f$ is bounded so $|g(y)-g(z)| \leq M\int_{\mathbb{R}} |\sin(y^2x)-\sin(z^2x)|dx$
Since $\sin$ is a continuous function, can I just make $|\sin(y^2x)-\sin(z^2x)| \leq \frac{\epsilon}{M}$, and be done? Or is there a better way to solve this problem?