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Is there any linear map that lowers the number of variables of functions, namely a map that maps a function of several variables to functions of one variable and at the same time the original function can be reconstructed with its image through an explicit form.
For example, $$T: C(\mathbb{R}^n)\rightarrow C(\mathbb{R})$$ $$f(\mathbb {x})\rightarrow F(t)$$ such that $$f(\mathbb {x})=\int_{\mathbb{R}}K(\mathbf{x},t)F(t)dt,\quad \forall f\in C(\mathbb{R}^n)$$ with some kernel $K$?
It has not to be this specific form. My key problem is to find transforms that lowers the number of variables of functions and store $f(x)$ separately in $F(t)$, which preserves the information, and $K(\mathbf{x},t)$, which keeps its "location".

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As your question is not completely precise, this might not really an answer but at least it might be an occasion to modify your question in a more precise way.

If you enlarge the point of view from continuous functions to Borel functions, then the problem is easy: all Polish spaces with no isolated points are Borel-isomorphic, so if for example $x=x(t)$ with inverse $t=t(x)$ is a bijection that preserves Borel sets you can use $F(t)=f(x(t))$ as transformation $T$ (not only linear, but also multiplication and order preserving; it does not preserve the spaces of continuous functions)

Looking only at continuous functions, you can use also a "dual" of the "space filling curve" approach: the Arnold - Kolmogorov theorem about Hilbert 13th problem. There are versions with only one external function (and, as usual, the internal functions are independent from $f$ so that the representation is really due to a continuous embedding of the $n$-dimensional cube in a larger euclidean space in such a way that continuous functions on the $n$-dimensional cube are identified with continuous functions of one real variable, the sum of the coordinates of the embedding). There are also explicitly constructed such embeddings, for example http://wissrech.ins.uni-bonn.de/research/pub/braun/remonkoe.pdf and you can easily search for more.

Edit: I have just seen in Kolmogorov superposition for smooth functions that you already know Vitushkin's theorems that kill the possibility of a general representation with differentiable functions. Even recent activity not cited in that thread confirm the "no-go" status beyond continuity, see for example http://emis.library.cornell.edu/journals/HOA/FPTA/Volume2010/287647.pdf

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  • $\begingroup$ With Kolmogorov superposition theorem, every $f\in C(\mathbb{R}^n)$ has a corresponding $g\in C(\mathbb{R})$, then combine Fourier inverse transform, $F(t)=\hat{g}(t)$. This is something I tried before, but the functions involved in Kolmogorov's theorem are only continuous(or Holder continuous), I wonder if there any other way out? $\endgroup$
    – Lucy
    Apr 10 '14 at 14:02
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Your problem is probably too vaguely formulated for you to expect a precise answer but the following musings might be of interest. In the case of continuous functions, such isomorphisms are quite common. Thus Milyutin's theorem implies that the Banach spaces $C([0,1]^n$ and $C([0,1])$ are isomorphic. However, this is in such way that the explicitness condition fails. On the other hand, if you are happy with a result on smooth functions, specifically the kind of function spaces which abound in distribution theory, then one can get results which do give an explicit form (again, depending on how you interpret the latter phrase). We consider as a generic example the Schwartzian spaces of rapidly decreasing, smooth functions on the plane and the line. Both have natural bases, in the latter case the hermitian functions indexed by the natural numbers, in the former the two dimensional analogue. Then if we choose a bijection $\phi$ between the two index sets, this implements (at the level of the representations of functions as infinite series in terms of the basic sequences) an isomorphism between the two- dimensional and the one dimensional case. Whether this corresponds to your notion of explicit is a question only you can answer.

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  • $\begingroup$ Sorry about the vagueness. I tried my best and modified my question, but I am afraid it is still not very clear. $\endgroup$
    – Lucy
    Apr 7 '14 at 17:49
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Take a space filling continuous curve $c:\mathbb R\to \mathbb R^n$, or at least one with dense image, then $T:f\mapsto f\circ c$ is injective $C(\mathbb R^n)\to C(\mathbb R)$. Now try to describe the image.

If the transform is continuous for the compact open topologies, then $T(f)(t)=\int_{\mathbb R^n} f(x)\mu_t(x)$ for a Radon measure $\mu_t$ on $\mathbb R^n$ of compact support; so you get a continuous map $\mu:\mathbb R\to C(\mathbb R^n)^*$. I guess, that in this situation you cannot get an inversion formula.

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  • $\begingroup$ Thank you, Peter. Using space-filling-curve transform preserves the information of $f$, but loses the "geographical" information of $\mathbf{x}$, in another word, at which point(s) $\mathbf{x}$ does$f$ have a certain value. This causes problems when reconstructing $f$ from $Tf$. $\endgroup$
    – Lucy
    Apr 7 '14 at 17:40

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