Regarding Kolmogorov's Superposition Theorem - MathOverflow

Regarding Kolmogorov's Superposition Theorem

2010-11-28T05:22:24Z

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

Answer by AgCl for Regarding Kolmogorov's Superposition Theorem

2010-11-28T09:14:43Z

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.

Answer by Wang Tao for Regarding Kolmogorov's Superposition Theorem

2010-11-29T13:26:51Z

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

Answer by xxx for Regarding Kolmogorov's Superposition Theorem

2013-03-22T23:08:59Z

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.