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


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

For every $ m \in \mathbb{N} $, there exist continuous functions $ \phi_{1},\phi_{2},\ldots,\phi_{m}: [0,1] \to \mathbb{R} $ such that any continuous function $ f: [0,1]^{m} \to \mathbb{R} $ can be written as $$ f(x_{1},x_{2},\ldots,x_{m}) = {\Phi_{f}} \left( \sum_{p = 1}^{m} {\phi_{p}}(x_{p}) \right) $$ for some continuous function $ \Phi_{f}: \mathbb{R} \to \mathbb{R} $.

For each $ i \in \{ 1,\ldots,m \} $, take $ f_{i} $ to be the $ i $-th projection function (i.e., $ {f_{i}}(x_{1},x_{2},\ldots,x_{m}) = x_{i} $). It can be shown that this would imply the existence of a continuous function $ F: [0,1]^{m} \to \mathbb{R} $ that is also one-to-one (as all the coordinates can be recovered from it). However, a continuous function from an open set in $ \mathbb{R}^{m} $ to $ \mathbb{R} $ cannot be one-to-one for $ m > 1 $ (by Invariance of Domain), so we obtain a contradiction.


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

  • 4
    $\begingroup$ @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. $\endgroup$ Nov 30 '10 at 12:08

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.


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