Regarding Kolmogorov's Superposition Theorem

Printable View

November 26th 2010, 08:17 PM
WangTaoSG

1 Attachment(s)

Regarding Kolmogorov's Superposition Theorem

Hi Experts,

We have a question regarding Kolmogorov's Superposition Theorem. Would you please look at the attachment for details? It is better formated for ease of reading.

It is critical to our research on nonlinear control, and we look forward to your advises on possible solutions, tips, related documents,etc.

Thank You!
Wang Tao
November 27th 2010, 02:23 AM
Defunkt

A few pointers:

1) Not many people are willing to open .pdf files - you would be better off writing your questions in TeX.

2) Since this is a research question, I doubt you will have many answers here. I think you'll have a greater chance of getting an answer if you post this in Math Overflow as well.

Good luck.
November 27th 2010, 05:59 AM
WangTaoSG

Hi Defunkt,

Good tips, will post there as suggested.

Thank you!
November 27th 2010, 07:57 PM
WangTaoSG

As suggested by Defunt, repost in LaTex:

It is known that:
Let $\displaystyle{\[f(x_1,x_2,...,x_m): \Re^m :=[0,1]^m \to \Re }$ be an arbitrary multivariate continuous function. From Kolmogorov’s Superposition Theorem we have the following representation:

$\displaystyle{\[f(x_1,x_2,...,x_m)= \sum_{q=0}^{2m} \Phi_q (\sum_{p=1}^m \phi_{p,q}(x_p))}$

With continuous one-dimensional outer functions $\displaystyle{\Phi_q}$ and inner functions $\displaystyle{\phi_{p,q}}$ . All these functions are defined on real line. The inner functions are independent of function $\displaystyle{\[f(x_1,x_2,...,x_m)}$

Question is:
Is it possible to find inner functions $\displaystyle{\phi_p{(x_p)}}$ which is independent of q, that satisfies the superposition theorem:

$\displaystyle{\[f(x_1,x_2,...,x_m)= \sum_{q=0}^{2m} \Phi_q (\sum_{p=1}^m \phi_p(x_p))}$

Where $\displaystyle{\Phi_q, \phi_p, N}$ can be selected and defined where appropriate.

Thank you!
November 28th 2010, 10:18 PM
WangTaoSG

Hi Defunkt,

As you expected I have reply from Math Overflow and the answer is NO. Good that we don't have to waste more time on this. Thank you very much!

Attached the answer from Math Overflow before from AgCl:

If I understand correctly, that doesn't seem possible. If the inner functions are independent of $\displaystyle{q}$ , then the sum of outer functions collapses to a single function $\displaystyle{\Phi}$ with $\displaystyle{\Phi(\cdot) = \sum_{q=0}^{2m} \Phi_q(\cdot)}$ . So the stronger form of the theorem that you are looking for would be equivalent to:

For every dimension $\displaystyle{m}$ , there exists continuous functions $\displaystyle{\phi_1,\phi_2,\cdots ,\phi_m}$ from $\displaystyle{[0,1]}$ to $\displaystyle{\mathbb R}$ such that, any continuous function $\displaystyle{f\colon [0,1]^m \rightarrow \mathbb R}$ can be written as $\displaystyle{f(x_1,x_2,...,x_m)= \Phi_f \left(\sum_{p=1}^m \phi_p (x_p)\right) }$ for some continuous function $\displaystyle{\Phi_f}$ from $\displaystyle{\mathbb R}$ to $\displaystyle{\mathbb R}$ .

But by taking $\displaystyle{f_i}$ with $\displaystyle{f_i(x_1,x_2,...,x_m)=x_i}$ respectively, it can be shown that, that would imply the existence of a continuous map $\displaystyle{F\colon [0,1]^m\rightarrow \mathbb R}$ which is also one-to-one (since all the coordinates can be recovered from it). But a continuous map from an open set in $\displaystyle{\mathbb R^m}$ to $\displaystyle{\mathbb R}$ cannot be one-to-one for $\displaystyle{m>1}$ , so we get a contradiction.