Hilbert transform linearity Explain why the Hilbert transform of $f(t)=\operatorname{sinc}(at) \cos(2 \pi \nu_c t)$ is

$$\hat{f} (t) = \operatorname{sinc} (at) \sin(2 \pi \nu_c t),$$

where $0&lt;a&lt;\nu_c.$

Attempt:

I have already found the Hilbert transform of $\operatorname{sinc}(at)$ here. Now I want to use linearity property of the Hilbert transformation. So writing $f(t)$ as:

$$f(t) = \frac{1}{2} \left( \operatorname{sinc}(at) e^{j 2 \pi \nu_c t} + \operatorname{sinc}(at) e^{-j 2 \pi \nu_c t} \right) \tag{i}$$

Applying linearity:

$$\hat{f}(t) = \frac{1}{2} \left( \frac{\sin^2(at/2)}{at/2} e^{j 2 \pi \nu_c t} + \frac{\sin^2(at/2)}{at/2} e^{-j 2 \pi \nu_c t} \right) \tag{ii}$$

But this doesn't look right. So why does the linearity not work? How can I arrive at the correct Hilbert transform expression given above?

P. S. 

I know that this result follows from Bedrosian's theorem but I am not allowed to use that for this problem. I know the Hilbert transform of $f(t)$ has to be $\operatorname{sinc} (at) \sin(2 \pi \nu_c t)$ since I've already plotted its envelope and it looks correct:
Transform the signal into frequency domain

$$F(f) = \frac{1}{\left|a\right|} \left[~\frac{\Pi(f-\nu_{c})}{2} + \frac{\Pi(f+\nu_{c})|}{2} ~\right]$$

As you know the fourier domain representation of the hilbert operator is

$$H(f) = -j\text{sgn}(f)$$

Applying this transform means you just multiply $H(f)$ to $F(f)$. That gets you

$$\hat{F}(f) = -j\text{sgn}(f) \cdot \frac{1}{\left|a\right|} \left[~\frac{\Pi(f-\nu_{c}}{2} + \frac{\Pi(f+\nu_{c})|}{2} ~\right] $$

$$ \hat{F}(f) = \frac{1}{\left|a\right|} \left[~\frac{\Pi(f-\nu_{c})}{2j} + \frac{-\Pi(f+\nu_{c})|}{2j} ~\right] ~~ \text{Remember} -j = \frac{1}{j}$$

which is same as 

$$ \hat{F}(f) = \frac{\Pi(f)}{\left|a\right|} \ast  \left[~\frac{1}{2j} \delta(f-\nu) + \frac{-1}{2j} \delta(f+\nu) ~\right] $$

Inverse Fourier Transformation of this gives you

$$\hat{f}(t) = \text{sinc}(at)\sin{\left(2\pi\nu_{c}t\right)}$$

Hope this helps!