Regarding Kolmogorov's Superposition Theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T20:29:31Z http://mathoverflow.net/feeds/question/47557 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47557/regarding-kolmogorovs-superposition-theorem Regarding Kolmogorov's Superposition Theorem Wang Tao 2010-11-28T05:22:24Z 2013-03-22T23:08:59Z <p>Hi Experts,</p> <p>I have question regarding Kolmogorov's Superposition Theorem:</p> <p>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:</p> <p>${f(x_1,x_2,...,x_m)= \sum_{q=0}^{2m} \Phi_q (\sum_{p=1}^m \phi_{p,q}(x_p))}$</p> <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 ${f(x_1,x_2,...,x_m)}$.</p> <p>Question is: Is it possible to find inner functions ${\phi_p{(x_p)}}$ which is independent of $q$, that satisfies the superposition theorem:</p> <p>${f(x_1,x_2,...,x_m)= \sum_{q=0}^{2m} \Phi_q (\sum_{p=1}^m \phi_p (x_p))}$ </p> <p>Where ${\Phi_q, \phi_p, N}$ can be selected and defined where appropriate.</p> <p>It is critical to our works on nonlinear control, and we look forward to your advises on possible solutions, tips, related documents,etc.</p> <p>Thank You! Wang Tao </p> http://mathoverflow.net/questions/47557/regarding-kolmogorovs-superposition-theorem/47571#47571 Answer by AgCl for Regarding Kolmogorov's Superposition Theorem AgCl 2010-11-28T09:14:43Z 2010-11-28T09:14:43Z <p>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:</p> <blockquote> <p>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 $\mathbb R$.</p> </blockquote> <p>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 <em>one-to-one</em> (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. </p> http://mathoverflow.net/questions/47557/regarding-kolmogorovs-superposition-theorem/47674#47674 Answer by Wang Tao for Regarding Kolmogorov's Superposition Theorem Wang Tao 2010-11-29T13:26:51Z 2010-11-29T13:26:51Z <p>Hi AgCl</p> <p>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?</p> <p>Warm Regards Wang Tao</p> http://mathoverflow.net/questions/47557/regarding-kolmogorovs-superposition-theorem/125331#125331 Answer by xxx for Regarding Kolmogorov's Superposition Theorem xxx 2013-03-22T23:08:59Z 2013-03-22T23:08:59Z <p>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.</p>