Convolution, Fourier transforms, and area preservation Consider the convolution of two functions, f * g. And let us assume, for practicality, some example case where an integral of f or g can be interpreted as the "area under the curve" (or the mass from an integrated density or...).
I've learned from this answer that the integral of a convolution f * g (i.e. the area of the convolution) yields the product of the integrals of f and g (i.e. a product of areas):

Af *g = Af * Ag

But from this answer I've learned that convolution corresponds to polynomial multiplication in one space and term-wise multiplication in the dual space.
Now I'm trying to bring together three seemingly different types of products:

*

*Multiplication of "areas" (i.e. of sums or integrals)

*Multiplication of polynomials (i.e. of corresponding monoms + summation)

*Term-by-term multiplication

I'm lacking the intuition how this is all supposedly the same, i.e. with respect to the multiplication of areas, and how the Fourier transformation comes into all this.
Example: Consider two unit rectangle functions (unit area). If we slice both of them in halfs, multiply the areas piece-wise, and take the sum, we get a different result than when multiplying the areas directly:

1 * 1 = 1

but

1/2 * 1/2 + 1/2 * 1/2 = 1/4 + 1/4 = 1/2

...so term-wise multiplication obviously does not result in a product of areas. But convolution does!
Consider another simple, but more instructive example from this answer:

x[n] = {2, 4, 1} and p[n] = {5, 1, 8}
with areas Ax = 2+4+1 = 7 and Ap = 5+1+8 = 14
Product of areas (= "multiplication in real space"):
=> Ax * Ap = 7 * 14 = 98
Term-wise multiplication ("convolution in real space"):
termw[n] = {2⋅5, 4⋅1, 1⋅8} = {10, 4, 8}
=> Atermw = 10 + 4 + 8 = 22
Convolution ("polynomial multiplication in dual space"):
conv[n] = {5⋅2, 5⋅4+1⋅2, 5⋅1+1⋅4+8⋅2, 1⋅1+8⋅4, 8⋅1} = {10, 22, 25, 33, 8}
=> Aconv = 10 + 22 + 25 + 33 + 8 = 98

How is this all the same, yet not the same? Am I misinterpreting the concept of duality?
To make the confusion complete, the Fourier transformation is obviously not area-preserving as can be easily seen from these graphs:
Fourier Transforms
But it does seem to preserve the "multiplicativity of areas" when it comes to convolutions when transforming convolution integrals (areas) between real and dual space (i.e. convolution/deconvolution).
There must be some underlying concept that unifies all this clearly. I'm not a mathematician, but I've read that the mechanics behind convolution (crosscorrelation, autocorrelation) and Fourier transformation is intimately linked to the concept of products (Plancherel's theorem, square-integrable functions etc.). I remember from basic functional analysis the concept of complete orthonormal bases (ONBs) and that a Fourier series "is" a projection of functions onto an ONB of harmonic functions. And somehow, this translates into two different ways of multiplication (term-wise vs. polynomial). From this, convolution appears to be some kind of "generalized product" defined on functions and if we represent functions by harmonic series, Fourier transformation somehow transforms the "mechanics" of the multiplication (term-wise vs. polynomial).
I have a hunch to how Fourier transformation might serve to "simplify" a polynomial multiplication (which scales with the square of the number of monoms) into a term-wise multiplication (which scales linearly with the number of monoms): In a Fourier transformation, a function gets "projected" onto a complete ONB of harmonic functions (represented by complex exponential functions) and integrated. So any harmonic component in "disharmony" with an exponential's frequency will get averaged out. So for the polynomial multiplication, I imagine that all mixed terms are being averaged out leaving only monomic terms.
In other words, we are plugging prior information about the posited "harmonicity" of our functions (i.e. they can be represented by Fourier series) into our multiplicative operation in order to "save some work".
Of course, this is a very imprecise way of describing things - we are not losing any information by applying the Fourier transform as the original functions can be perfectly retrieved via the reverse transform, but somehow we have a representation of functions that lends itself "inherently" for doing multiplicative calculations (products). And it is the nature of this "inherent structure" that I'm asking about here.
I'm sorry if I cannot get any clearer here - but the fact that I cannot is precisely the issue I'm trying to clear up.
 A: You may find it helpful to think of the area integral $\int f(x)dx $ as the zero-frequency component $\hat{f}(0)$ of the Fourier transform $\hat{f}(\omega)$. The statement "the area of a convolution equals the product of areas" is then the special $\omega\rightarrow 0$ case of the more general statement that the frequency-$\omega$ component of a convolution is the product of the frequency-$\omega$ components of the Fourier transforms.
This way of thinking may help to resolve the "misinterpretation of the concept of duality" and the confusion about "area preservation". The zero-frequency component of a function is defined in $\omega$-space, so it is not subject to duality. And the answer to the question "does Fourier transformation preserve a certain frequency component" is "yes" by definition.
The dual in $x$-space of the $\omega\rightarrow 0$ component in $\omega$-space is $x\rightarrow 0$ limit of the function $f(x)$. And indeed, we have the dual statement that the position-$x$ component of a convolution in $\omega$-space is the product of the position-$x$ components of the inverse Fourier transforms.
