Let $f\in L^1(\mathbb{R})$. Show that the function $g$ defined on $\mathbb{R}$ by $$ g(x) = \int_{\mathbb{R}} \sin(xy)f(y)dy$$ is well defined and continuous on the real line.
So I want to prove that if $x_n \rightarrow x$ then $g(x_n) \rightarrow g(x)$. Expanding this $$ \left|\lim_n \int_{\mathbb{R}} \sin(x_ny)f(y)dy - \int_{\mathbb{R}} \sin(xy)f(y)dy\right| = \lim_n\left|\int_{\mathbb{R}} (\sin(x_ny)- \sin(xy))f(y)dy\right|.$$ Can I move the limes inside the integral because of the continuity of continuous of integrals. Or could I use the dominated convergence somehow? Thanks
You have the trivial estimate $|\sin (x y) f(y)| \le |f(y)|$ so that the integral defining $g$ is defined and finite. Moreover, $|f|$ is an $L^1$ majorant of the sequence $f_n(y) = \sin(x_ny) f(y)$. Since $\sin (x_n y) f(y) \to \sin(xy) f(y)$ for almost every $y$ (in fact every $y$ where $f(y)$ is defined and finite) you have $$ \lim_{n \to \infty} \int \sin(x_n y) f(y) \, dy = \int \sin (xy) f(y) \, dy $$ by the dominated convergence theorem.