# 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.

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What do you mean with the dual of a function? –  Jonas Teuwen Jan 20 '11 at 16:48
Hmm, if you just take the inverse Fourier transform of both equations and you divide them then you get $\beta/\alpha = G(x)$ for some $G(x)$, but that $G(x)$ is uniquely determined because the Fourier transform is. Do I miss something? In this case it is sufficient for $h_i g$ to be a tempered distribution. –  Jonas Teuwen Jan 20 '11 at 19:21
@JonasT: Thank you for your thoughts! By the dual I mean it in the sense as in the Dirac notation, en.wikipedia.org/wiki/Bra-ket_notation . What you meantion in your second comment is exactly the problem. If $\alpha$ and $\beta$ are just constants (as I might not have pointed out very direct), taking the Fourier transform and dividing, they will be functions... This is one thing I am puzzled with :) Greets –  Robert Filter Jan 21 '11 at 9:12
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.