17
$\begingroup$

Kolmogorov superposition theorem states that a continuous function $f(x_1,\ldots,x_n)$ can be written as

$$f(x_1,\ldots,x_n)=\sum_{q=0}^{2n}\Phi_q\left(\sum_{p=1}^{n}\phi_{q,p}(x_p)\right)$$

for certain continuous functions $\Phi_q$ and $\phi_{q,p}$.

For smooth, $f\in C^{\infty}$, can we obtain such a representation with the $\Phi_q$ and $\phi_{q,p}$ smooth too? Moreover, I am particularly interested in the case in which more than smooth, $f$ belongs to some Denjoy-Carleman quasianalytic class, but the question about smoothness would be a point to start.

I don't assume independence of $\Phi_q$ and $\phi_{q,p}$ of their parameters or of $f$.

$\endgroup$

2 Answers 2

21
$\begingroup$

The answer is no. There exist analytic functions $f$ of 3 variables such that they cannot be represented as a composition of continuously differentiable functions of two variables. This is old result of Vitushkin. You can find nice story of Hilbert's thirteen problem in Vitushkin, A. G. Hilbert's thirteenth problem and related questions. Russian Math. Surveys 59 (2004), no. 1, 11–25). Edit: I mixed two different results of Vitushkin. In 1964 Vitushkin shown that not every analytic function of three variables can be written in the desired form with continuously differentiable functions $\phi_{p,q}$. That's answer ABC's question. In 1954 Vitushkin proved that not all continuously differentiable functions of three variables can be represented as superposition of continuously differentiable functions of two variables.

$\endgroup$
2
  • $\begingroup$ This result of Vitushkin is about linear superpositions of analytic functions by differentiable functions. The general problem in the class $C^\infty$ is open as far as I know. $\endgroup$
    – Andrew
    Commented Aug 31, 2013 at 16:37
  • $\begingroup$ @Andrew Thank's for correction. I edited my answer. $\endgroup$ Commented Aug 31, 2013 at 17:44
6
$\begingroup$

The inner functions $\phi_{qp}$ are independent in Kolmogorov's superposition theorem and they are very "bad" functions like cantor function. The smoother $f$ is, the more singular $\Phi_q$ has to be to cancel out the singularities of $\phi_{pq}$ except in some special cases, such as $f$ is a constant. There is no linear relation between the smoothness of $f$ and it representing function $\Phi_q$. The smoothness of $f$ mainly depends on the behaviour of $\Phi_q$ on the image of singular points of $\sum_p \phi_{qp}$ .

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .