Let $f : X^3 \rightarrow X$.

If $X$ is $\mathbb Z$, then there will be a couple of functions $g,h$ from $\mathbb Z^2$ to $\mathbb Z$ that satisfies $f(x,y,z) = g(h(x,y),z)$ since there is a bijection $h :\mathbb Z^2 \rightarrow \mathbb Z$.

However, If $X$ is $\mathbb R$, then there are no continuous bijection from $\mathbb R^2$ to $\mathbb R$.

My question is : Is there any continuous function $f : \mathbb R^3 \rightarrow \mathbb R$ that can't be represented by composition of continuous functions $g_i = \mathbb R^2 \rightarrow \mathbb R$? And similar questions for $f : \mathbb R^n \rightarrow \mathbb R$.

Sorry, I'm not sure about my tags.

New contributor
Soonwon Moon is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.

As pointed out by Pierre PC in his comment, the answer follows from the Kolmogorov theorem. Just for a record let us state one of the version of the theorem due to Lorentz (there are many other more refined versions; the reader will not have difficulties to find the references).

Theorem. There exist constants $0<\lambda_p\leq 1$, $1\leq p\leq n$ and strictly increasing functions $\phi_q:[0,1]\to [0,1]$, $0\leq q\leq 2n$ such that if $f:[0,1]^n\to\mathbb{R}$ is continuous, then there is another continuous function $g:[0,n]\to\mathbb{R}$ such that $$ f(x_1,\ldots,x_n)=\sum_{q=0}^{2n} g\left(\lambda_1\phi_q(x_1)+\ldots+\lambda_n\phi_q(x_n)\right). $$

It is very surprising that neither the functions $\phi_q$ nor the constants $\lambda_p$ depend on $f$.

For a proof see pages 168-174 in

G. G. Lorentz, Approximation of functions, 1966.

  • 1
    $\begingroup$ Just to clarify, the answer is negative, since OP asked whether there existed $f$ that could not be expressed as $g(h(-,-),-)$. $\endgroup$ – Najib Idrissi Mar 15 at 13:36
  • $\begingroup$ @NajibIdrissi Corrected. Thank you! $\endgroup$ – Piotr Hajlasz Mar 15 at 13:38
  • $\begingroup$ Would you please add some details on how exactly this theorem leads to the negative answer to the OP? I have an impression that the answer is negative just because you have constructed (in the linked post) an explicit counterexample. And one more question: this variant of the Kolmogorov theorem looks a bit different from the one on Wikipedia (where $\phi_q$ depends also on $q$ and are not claimed to be increasing). Could you add some reference to the variant you provided? $\endgroup$ – Skeeve Mar 15 at 13:59
  • $\begingroup$ Thanks for the detailed description! However, the notation of index of $\phi$ is wrong at least 1 place. $\endgroup$ – Soonwon Moon Mar 15 at 14:01
  • $\begingroup$ @SoonwonMoon I modified my answer. I think it is more clear now. $\endgroup$ – Piotr Hajlasz Mar 15 at 19:01

Your Answer

Soonwon Moon is a new contributor. Be nice, and check out our Code of Conduct.

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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