Recall that $\{ x \}$ is a decimal part of a real number $x$. For example, if $x=3.41$, then $\{ x \}=-0.41$.
Part A: Let $f(x)=2x$, with $x \in (-1,1)$ and let $g:R \to R$, $g(x)=2 \{ x \}$. Sketch a graph of $f$ and $g$.
The graph of $f(x)$ is a linear line with a slope of $2$ with $y$-intercept at $0$. The domain is from $(-1,1)$, and the range from $(-2,2)$.
The graph of $g(x)$ are a set of linear lines with just like $f(x)$.
Part B: Compute the Fourier transform of $f$ and the Fourier transform of $g$. How are they related?
$$\hat{f}(y)=\int_R f(x)e^{-ixy}dx=\int_{-1}^1 2xe^{-ixy}dx =2\left.\left(\frac{ix}{y}+\frac{1}{y^2}\right)e^{-ixy}\right|_{-1}^1 =2\left(\frac{i}{y}+\frac{1}{y^2}\right)e^{-iy}-2\left(\frac{-i}{y}+\frac{1}{y^2}\right)e^{iy} =2\frac{i}{y}(e^{-iy}+e^{iy})+2\frac{1}{y^2}(e^{-iy}+e^{iy}) =\frac{4i\cos(y)}{y}+\frac{4i\sin(y)}{y^2} =\frac{4i}{y}\left[\cos(y)+\frac{\sin(y)}{y}\right]$$
$$\hat{g}(y)=\int_R g(x)e^{-ixy}dx =\int_R 2\{ x \}e^{-ixy}dx =\sum_{n=-\infty}^\infty \int_n^{n+1} 2(x-n)e^{-ixy}dx =\sum_{n=-\infty}^\infty 2\left.\left(\frac{i(x-n)}{y}+\frac{1}{y^2}\right)e^{-ixy}\right|_n^{n+1} =\sum_{n=-\infty}^\infty 2\left(\frac{i}{y}+\frac{1}{y^2}\right)e^{-i(n+1)y}-\frac{2}{y^2}e^{iny}$$
$\hat{f}(y)$ and $\hat{g}(y)$ are related because $\hat{f}(y)$ is a special case of $\hat{g}(y)$ where $\sum_{n=-1}^1 \int_n^{n+1} 2(x-n)e^{-ixy}dx$
Part C: In general, let $f \in L^2(-1,1)$ and let $g:R \to R$, $g(x)=f( \{ x \} )$. How are the Fourier transforms of $f$ and $g$ related?
I know that Part B is a specific case of part C. But what can I deduce from it?