Derive Fourier transform from what it should do? I was wondering about the following: 
Imagine you want to figure out whether there is a transform that exchanges differentiation with multiplication and convoution with pairwise transformation for $L^2$ functions. Is there an analytical way to derive from this the Fourier transform?
Why am I asking this? In many practical examples, you want to have a transform that has certain properties like: get rid of differentiation. But in mostly all books, the Fourier transform is just introduced and the properties are derived afterwards. Is it a posteriori possible to derive an integral transform just by looking for something with these properties?
 A: Assuming that we know beforehand that the required transform is an integral transform, I get the following with a bit of handwaving (though I guess this can relatively easily be proved rigorously).
Denote by $\mathcal{T}$ the unknown integral transformation and by $K$ the kernel of that transformation, i.e. let $\mathcal{T}[f](\omega) = \int_{-\infty}^\infty K(x,\omega) f(x)\,d x$.

For the $\delta$ distribution, denote $\delta_a(x) = \delta(x-a)$. We get [using $\delta$ is handwaving #1, but probably we could use limits just everywhere and require some form of continuity]
$$
\mathcal{T}[\delta_a](\omega) = K(a,\omega).
$$
Start from the convolution theorem, i.e.
$$
\mathcal{T}[(f \ast g)](\omega) = \mathcal{T}[(f)](\omega)\mathcal{T}[(g)](\omega).
$$
Plugging in $g=\delta_a$, the LHS becomes
$$
\int_{-\infty}^\infty K(x,\omega) \int_{-\infty}^\infty f(x-s) \delta_a(s)\,ds \,dx = \int_{-\infty}^\infty K(x,\omega) f(x-a)\,dx = \int_{-\infty}^\infty K(x+a,\omega) f(x)\,dx,
$$
and the RHS becomes
$$
\int_{-\infty}^\infty K(x,\omega) f(x)\,dx K(a,\omega) = \int_{-\infty}^\infty K(a,\omega) K(x,\omega) f(x)\,dx .
$$
Because they should be equal for every $f$, we get the equation ($\forall \omega, x, a$)
$$
K(a,\omega)K(x,\omega) = K(a+x,\omega).
$$
With some assumptions [handwawing #2; I think continuity is enough?], we get $K(x,\omega) = e^{\alpha(\omega) x}$ with some unknown function $\alpha$.

Now apply the "differentiation becomes multiplication", i.e.
$$
\mathcal{T}[f'](\omega) = \beta(\omega) \mathcal{T}[f](\omega),
$$
with some unknown function $\beta$. For functions $f$ with compact support, The LHS becomes
$$
\int_{-\infty}^\infty e^{\alpha(\omega) x} f'(x) \, dx = \left[ e^{\alpha(\omega) x} f(x) \right]_{x=-\infty}^\infty - \int_{-\infty}^\infty e^{\alpha(\omega) x} \alpha'(\omega) f(x) \, dx =  - \int_{-\infty}^\infty e^{\alpha(\omega) x} \alpha'(\omega) f(x) \, dx
$$
On the other hand, the RHS is
$$
\int_{-\infty}^\infty e^{\alpha(\omega) x} \beta(\omega) f(x) \, dx,
$$
so we get $\beta(\omega)=-\alpha'(\omega)$.

So from 
$$
\mathcal{T}[(f \ast g)](\omega) = \mathcal{T}[(f)](\omega)\mathcal{T}[(g)](\omega)
$$
and 
$$
\mathcal{T}[f'](\omega) = \beta(\omega) \mathcal{T}[f](\omega),
$$
we get
$$
K(x,\omega) = e^{-\gamma(\omega) x },
$$
where $\gamma$ is an antiderivative of $\beta$. Apparently we need more properties to define the exponent exaclty.
