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.


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 $g(h(-,-),-)$. $\endgroup$ Mar 15 '19 at 13:36
  • $\begingroup$ @NajibIdrissi Corrected. Thank you! $\endgroup$ Mar 15 '19 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 '19 at 13:59
  • $\begingroup$ Thanks for the detailed description! However, the notation of index of $\phi$ is wrong at least 1 place. $\endgroup$
    – damhiya
    Mar 15 '19 at 14:01
  • $\begingroup$ @SoonwonMoon I modified my answer. I think it is more clear now. $\endgroup$ Mar 15 '19 at 19:01

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