# Is there any continuous ternary function which can not be represented by composition of continuous binary functions?

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.

• Related, I think: mathoverflow.net/questions/322184 Mar 15 '19 at 9:55
• @PierrePC Thanks! Kolmogorov superposition theorem was the answer! Mar 15 '19 at 10:09

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.

• Just to clarify, the answer is negative, since OP asked whether there existed $f$ that could not be expressed as $g(h(-,-),-)$. Mar 15 '19 at 13:36
• @NajibIdrissi Corrected. Thank you! 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? Mar 15 '19 at 13:59
• Thanks for the detailed description! However, the notation of index of $\phi$ is wrong at least 1 place. Mar 15 '19 at 14:01
• @SoonwonMoon I modified my answer. I think it is more clear now. Mar 15 '19 at 19:01