Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi Experts,

I have question regarding Kolmogorov's Superposition Theorem:

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

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

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

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

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

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

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

Thank You! Wang Tao

share|improve this question
add comment

3 Answers

up vote 3 down vote accepted

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

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

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

share|improve this answer
add comment

Hi AgCl

Thank you very much for the reply. It helps to save my time and effort. As this topic is important to me, I want to think about my answer to you carefully. Allow me some days to draft a proper reply to you?

Warm Regards Wang Tao

share|improve this answer
1  
@Wang Tao: usually you can comment on answers submitted to questions that you asked. In this case you apparently have two separate accounts, which is why you posted this as an answer. I've requested your original account be merged into your new account. –  Willie Wong Nov 30 '10 at 12:08
add comment

The inner functions can be constructed indepent of q, which means only one outer function phi is needed. however phi is discontinuous in this case.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.