Q:

A Projection Problem in Functional Analysis - Uniqueness of a Solution

I have the following system of equations:
$$\alpha f_1(x) = \int_\mathbb{R} g(k)  h_1(k) e^{\mathrm{i}kx} dk$$
$$\beta f_2(x) = \int_\mathbb{R} g(k)  h_2(k) e^{\mathrm{i}kx} dk$$
with known functions $f_i(x)$ and $h_i(x)$ and unknown $\alpha$, $\beta$ and $g(k)$.
Before outlining my ansatz, here directly the question:

Under which conditions on $f_1(x)$ and $f_2(x)$ can we calculate the ratio $\frac\beta\alpha$ uniquely?

My approach
The first step was to integrate one of the equations,
$$\alpha \int_\mathbb{R} f_1(x) e^{-\mathrm{i}\lambda x}dx = \int_{\mathbb{R}^2} g(k)  h_1(k) e^{\mathrm{i}(k-\lambda)x} dxdk$$
resulting in
$$\alpha F_1(k) = 2\pi g(k) h_1(k)$$
because of
$$\delta\left(k-\lambda\right) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{\mathrm{i}\left(k-\lambda\right)x}dx$$
hence $g(k)$ can be seen as calculated. Plugging this into the second equation, one gets
$$2\pi\frac\beta\alpha f_2(x) = \int_\mathbb{R} F_1(k)  \frac{h_2(k)}{h_1(k)} e^{\mathrm{i}kx} dk\, .$$
Now I think the answer to the question relates to a further integration of this equation via some kernel $K(x)$. I would think that this is somehow the dual of the function $f_2(x)$ like in a Hilbert space. But I don't know if this really gives me a well defined unique result, and, how to calculate the dual of $f_2(x)$.
I would be thankful for any hint
Sincerely
Robert
A little excuse
As some might have noticed, I am at the moment occupied with some questions arising in, say, applied functional analysis. As a physicist I am often not very sure about the justification of what I am doing so I hope it is ok to ask these questions here even though the answers might be obvious to most of the people in this community.

A:

Fold each integral with $\int_{\mathbb{R}} d x\, e^{-i\lambda x}$, 
$$\begin{eqnarray*}
\alpha F_1(k) &=& 2\pi g(k) h_1(k) \\
\beta F_2(k) &=& 2\pi g(k) h_2(k). 
\end{eqnarray*}$$
Therefore, 
$$\alpha h_2(k) F_1(k) = \beta h_1(k) F_2(k),$$
or 
$$\alpha \mathcal{F}\{ \eta_2 * f_1\} 
= \beta \mathcal{F}\{ \eta_1 * f_2\},$$
where $*$ is the convolution and $\eta_i = \mathcal{F}^{-1}\{h_i\}$.  Thus, if
$$\eta_2 * f_1 \propto \eta_1 * f_2$$
and
$$\eta_1 * f_2 \ne 0$$
the ratio $\beta/\alpha$ can be found uniquely, it is 
$(\eta_2 * f_1)/(\eta_1 * f_2)$. 
As @JonasTeuwen mentioned in the comments, a sufficient condition for this all to work is $f_i$, $h_i$, and $g$ are tempered distributions.

