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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.

damhiya
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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.

Piotr Hajlasz
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    Just to clarify, the answer is negative, since OP asked whether there existed $f$ that could not be expressed as $g(h(-,-),-)$. – Najib Idrissi Mar 15 '19 at 13:36
  • @NajibIdrissi Corrected. Thank you! – Piotr Hajlasz Mar 15 '19 at 13:38
  • 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?

     – Skeeve Mar 15 '19 at 13:59
  • Thanks for the detailed description! However, the notation of index of $\phi$ is wrong at least 1 place. – damhiya Mar 15 '19 at 14:01
  • @SoonwonMoon I modified my answer. I think it is more clear now. – Piotr Hajlasz Mar 15 '19 at 19:01
  • @Skeeve I modified my answer. I think it is more clear now. – Piotr Hajlasz Mar 15 '19 at 19:01
  • @PiotrHajlasz thanks a lot for the reference! So the functions $\phi_q$ are even Lipschitz... very interesting! But I'm still a bit mislead by the sentence the answer follows from the Kolmogorov theorem. – Skeeve Mar 15 '19 at 19:30
  • @Skeeve "represented by composition of continuous functions" is a vague statement since it does not specify what sort of representation we have in mind. But the author of the question said "Kolmogorov superposition theorem was the answer!" so the answer indeed follows from the Kolmogorov theorem. – Piotr Hajlasz Mar 15 '19 at 19:40
  • @PiotrHajlasz now I see why you chose this interpretation. Instead I was looking at the equation $f(x,y,z) = g(h(x,y),z)$ in the OP. Anyway, the picture is clear now! – Skeeve Mar 15 '19 at 20:13