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My question was, in an uncountably infinite-dimensional vector spaces, how to represent a vector by a list of parameters, as we do in finite-dimensional spaces? I was assuming that if we can not express a vector as a list of discrete parameters, we have a big issue...but during the writting up of this question, it seems that there is no big issue, the parameters just change to a function and the sum changes to integration.

But I am not sure if my reasoning below is correct, so I still post it below, please correct me if there are something wrong:

In a finite-dimensional vector space $\Omega$, each vector (or point) $\mathbf{v}$ is represented as a list of numbers $c_i, i\in\mathbb{N}$, which can be seens as the parameters or coefficients of the vector, in the sense that the vector is the sum of the products of the parameters with the respective base $e_i$:

$$\mathbf{v}=\sum_{i=1}^{n} c_i e_i$$

When $e_i=x^i$, then $\Omega$ is a finite-dimensional polynomial function space.

Now, if the dimension is not finite, it seems there are two possibilities:

  1. countably infinite dimensional
  2. uncountably infinite dimensional, e.g. $e_i=x^i, i\in \mathbb{R}$

For the first case, the vector $\mathbf{v}$ in the polynomial function space can be expressed similarly:

$$\mathbf{v}=\sum_{i=1}^{\infty} c_i e_i$$ and $\mathbf{v}$ can also be represented by its parameters, i.e., a list of infinite members, as $c_i, i\in\mathbb{N}$.

But for the 2nd case, it seems it's not possible to represent $\mathbf{v}$ by a list of discrete parameters any more...the parameters is itself a continuous function $c:\mathbb{R}\rightarrow\mathbb{R}$. In this case, the vector $\mathbf{v}$ should be expressed as an integration in terms of its parameters and related base:

$$\mathbf{v}=\int_{-\infty}^{\infty} c(x) e(x)\mathrm{d}x$$

My question was from Fourier transform. Now with the understanding above, I have the following: loosely speaking, all functions (who's Fourier Fransform exist) form an uncountably-infinite-dimensional vector space, with uncountably-infinite number of basis $e^{j\omega t}, \omega\in\mathbb{R}$. Each function's parameter in this space is a function $X(\omega)$, which is defined by the Fourier Transform formula:

$$X(\omega)=\int_{-\infty}^{\infty} f(t) e^{-j\omega t}\mathrm{d}t$$ The inverse fourier transform formular is just the way to express the vector $f(t)$ in terms of its parameters and related base:

$$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} X(\omega) e^{j\omega t}\mathrm{d}\omega$$

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  • $\begingroup$ Your intuition seems correct to me. $\endgroup$ – Atul Ingle Aug 16 '17 at 20:06
  • $\begingroup$ Your idea is a very useful heuristic: it can be "helpful" to think of the Fourier transform as giving the "coefficients" in a "continuous linear combination." But you should be aware that these are only metaphors and that your idea requires a great deal of technical care to make precise and rigorous. $\endgroup$ – symplectomorphic Aug 17 '17 at 3:03
  • $\begingroup$ For example, if $f$ is an integrable complex-valued function on the real line (meaning $\int |f|$ converges), then your integral $X(\omega)$ converges. But it is not necessarily true that the next integral you say is equal to $f(t)$ in fact converges to $f(t)$. The content of that assertion is called the "inversion theorem," and you need to impose more constraints on $f$ in order to guarantee it holds. $\endgroup$ – symplectomorphic Aug 17 '17 at 3:13
  • $\begingroup$ Thanks for your comments, Atul and symplectomorphic. Btw, from another book I read that a tuple (or a list of numbers, finite or countably-infinite) can also be seen as a function , e.g., $c:\mathbb{Z}^+\rightarrow\mathbb{R}$. With this concept, it seems that the understanding of the representation of an element in vector space can be unified in finite-dimensional and infinite-dimensional cases, i.e., an element in vector space is represented as a function. $\endgroup$ – bruin Aug 19 '17 at 10:04
  • $\begingroup$ It also seems that, in some cases the representation of a element is treated as the element itself, if the mapping from representation to element is implicitly clear...but I guess this case can also be abstracted as an identical mapping from the representation to element. $\endgroup$ – bruin Aug 19 '17 at 10:05

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