Fourier Integral funny notation

Question

If $f$ is differentiable and $\int\limits_{-\infty}^\infty|f(x)|dx<\infty$ show that $\widehat{(f'_c)}(\omega)=\omega\hat{f_s}(w)$ and $\widehat{(f'_s)}(\omega)=-\omega\hat{f_c}(w)$

$f(x)=\int\limits_0^\infty A(\omega)\cos(\omega x)+B(\omega)\sin(\omega x)d\omega$

$A(\omega):=\int\limits_{-\infty}^\infty f(v)\cos(\omega v)dv$

$B(\omega):=\int\limits_{-\infty}^\infty f(v)\sin(\omega v)dv$

$\hat{f_c}(\omega)=\sqrt{\frac{2}{\pi}}\int\limits_0^\infty f(v)\cos(\omega v)dv$

$\hat{f_s}(\omega)=\sqrt{\frac{2}{\pi}}\int\limits_0^\infty f(v)\sin(\omega v)dv$

All these definitions to give context. Obviously this is a homework question, but I am not seeking the answer [unless you can't help it ;) ]. Really what I am not understanding is this notation $\widehat{(f'_c)}(\omega)$, what does this mean? I'm sure I can prove this once I know that.

Answer 1

Your definitions actually do contain the explanation for $\hat{\circ}$; as for $\widehat{\circ\circ}$, it is just another way to write the same symbol. – Ivan Neretin

You have: $\quad\widehat {f_c}(\omega)\mathop{:=}\sqrt{\tfrac 2\pi\;}\int\limits_{0}^\infty f(\nu)\cos(\nu\omega)\operatorname d \nu$

Then: $\qquad$ $ \widehat{f'_c}(\omega) = \sqrt{\tfrac 2 \pi \;} \int\limits_{0}^\infty f'(\nu) \cos(\nu\omega)\operatorname d \nu \\[3ex] \widehat{f_c}'(\omega) = \dfrac{\operatorname d}{\operatorname d \omega} \left(\sqrt{\tfrac 2 \pi \;} \int\limits_{0}^\infty f(\nu) \cos(\nu\omega)\operatorname d \nu\right)$