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Could we always locally represent a continuous function $F(x,y,z)$ in the form of $g\left(f(x,y),z\right)$ for suitable continuous functions $f$, $g$ of two variables? I am aware of Vladimir Arnold's work on this problem, but it seems that in that context $F(x,y,z)$ is written as a sum of several expressions of this form. Can one reduce it to just a single superposition $g\left(f(x,y),z\right)$; or does anyone know a counter example?

For related posts see:

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

Kolmogorov superposition for smooth functions

Kolmogorov-Arnold theorem for (just-)functions

KhashF
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1 Answers1

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Proposition. The function $F(x,y,z)=x(1-z)+yz$ cannot be represented as $F(x,y,z)=g(f(x,y),z)$, where $f,g:\mathbb{R}^2\to\mathbb{R}$ are continuous.

Proof. Suppose to the contrary that we have such a representation. Let $g_1(t)=g(t,0)$. Then $$ g_1(f(x,y))=g(f(x,y),0)=F(x,y,0)=x. $$ Let $g_2(t)=g(t,1)$. Then $$ g_2(f(x,y))=g(f(x,y),1)=F(x,y,1)=y. $$ Let $h=(g_1,g_2):\mathbb{R}\to\mathbb{R}^2$, $h(t)=(g_1(t),g_2(t))$. Then $$ (h\circ f)(x,y)=(g_1(f(x,y)),g_2(f(x,y)))=(x,y). $$ In particular, it implies that $f:\mathbb{R}^2\to\mathbb{R}$ is one-to-one.

It remains to show there there are no one-to-one continuous functions from $\mathbb{R}^2$ to $\mathbb{R}$. Suppose to the contrary that such a function $f$ exists. Let $\mathbb{S}^1\subset\mathbb{R}^2$ be the unit circle. Then $$ f|_{\mathbb{S}^1}:\mathbb{S}^1\to\mathbb{R} $$ is continuous and one-to-one. Since $\mathbb{S}^1$ is compact, $f|_{\mathbb{S}^1}$ is a homeomorphism of $\mathbb{S}^1$ onto $f(\mathbb{S}^1)$.

Since $\mathbb{S}^1$ is connected and compact $f(\mathbb{S}^1)\subset\mathbb{R}$ is connected and compact. Therefore it is a closed interval, $f(\mathbb{S}^1)=[a,b]$ and we arrive to a contradiction, because $[a,b]$ is not homeomorphic to $\mathbb{S}^1$ (to see this observe that removing an interior point from $[a,b]$ makes the space disconnected while if we remove a point form $\mathbb{S}^1$, the space will remain connected).

Remark. For a statement and a proof of the Kolmogorov theorem, see for example:

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

Edit. I modified my original answer that was not fully correct. A mistake was pointed out by Aleksei Kulikov in his comment. The correction is related to the arguments in the comments of KhashF and user44191.

Piotr Hajlasz
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  • Thanks Piotr! But how did you deduce that $f$ has derivative? Does your example work in the continuous setting as well? – KhashF Feb 01 '19 at 02:15
  • Partial derivatives aren't necessary for this proof; if you let $c = f(0, 0)$, you get that $f(x, y) = g_1^{-1}(x) = f(x, 0) = g_2^{-1}(0) = f(0, 0) = c$. – user44191 Feb 01 '19 at 02:29
  • @user44191 Thank you. I modified my answer. – Piotr Hajlasz Feb 01 '19 at 02:38
  • @KhashF I modified my answer. – Piotr Hajlasz Feb 01 '19 at 02:38
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    I’m sorry but where you used continuity of $f$ and $g$? Because without it answer to question in OP is that every function can be represented in such a manner (via bijection between $\mathbb{R}$ and $\mathbb{R}^2$). And also what is $g_1^{-1}(x)$ really? I see no reasons for $g_1$ to be injective. – Aleksei Kulikov Feb 01 '19 at 02:51
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    @AlekseiKulikov Very good point. However, under the continuity assumption my proof works. The continuity assumption is used implicitly and I will explain how. I will add details tomorrow. – Piotr Hajlasz Feb 01 '19 at 04:44
  • If I may; $f(\cdot,y):x\mapsto f(x,y)$ is continuous injective, so it must be open. Note also that if f(0,y)=f(x,0), then in fact x=0. By considering the clopen set $f(0,\cdot)^{-1}({f(0,0)})={y:f(0,y)=f(0,0)}={y:\exists x,f(0,y)=f(x,0)}=f(0,\cdot)^{-1}(f(\mathbb R,0))$, we see that f is constant with respect to $y$, and the symmetrical argument shows that it is constant with respect to $x$. – Pierre PC Feb 01 '19 at 17:47
  • (Sorry about the LaTeX on two lines...) – Pierre PC Feb 01 '19 at 17:56
  • I think the implicit point in Piotr's solution where the continuity indeed plays a rule is how to get from $g_1(f(x,y))=x$ or $g_2(f(x,y))=y$ to $g_1^{-1}(x)=f(x,y)$ or $g_2^{-1}(y)=f(x,y)$: If you want to use these identities to argue that $g_1$ (resp. $g_2$) is injective, you have to show that for a fixed $y$ (resp. x) every number can be written as $f(x,y)$. This of course fails for (trivially non-continuous) bijections $f:\Bbb{R}^2\rightarrow\Bbb{R}$. Therefore, you need an argument like Pierre's. – KhashF Feb 01 '19 at 20:53
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    P.S. I think even the following works: $g_1(f(x,y))=x$, $g_2(f(x,y))=y$ imply that the continuous $f:\Bbb{R}^2\rightarrow\Bbb{R}$ is injective, a contradiction. – KhashF Feb 01 '19 at 20:53
  • @KhashF I am busy now, but I will add details to my solution later so it will be clear and correct. – Piotr Hajlasz Feb 01 '19 at 20:57
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    As an alternative approach, $h(t) := (g_1(t), g_2(t))$ is a continuous inverse to the continuous function $f(x, y)$, and therefore a homeomorphism. This approach should work category-theoretically: you can construct such functions $f, g$ on $X^2$ only if $X \simeq X^2$ in whichever category you've chosen - and since $\mathbb{R} \not \simeq \mathbb{R}^2$ topologically, we're done. – user44191 Feb 01 '19 at 21:08
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    It should be noted that there are non continuous functions f and g for this example, which depend on f "encoding" two reals into one and g using the decoding inverses to f to represent F. Gerhard "Enhancing The Continuity Of Explication" Paseman, 2019.02.13. – Gerhard Paseman Feb 13 '19 at 18:10
  • @GerhardPaseman It has already been pointed out by Aleksei Kulikov in his comment. I need to edit my answer and I will do it, but I am too busy in the next two weeks. – Piotr Hajlasz Feb 13 '19 at 18:27
  • @PiotrHajlasz Not to pressure you, but two weeks have passed, and the argument is still missing something; I think my latest comment above provides a solution (and allows an answer to a partial generalization of the question). Are you planning to finish the answer, and if not, would you mind if I edited my idea into your answer? – user44191 Mar 20 '19 at 15:03
  • @user44191 I modified my answer. Thank you for reminding me of my unfinished chore. – Piotr Hajlasz Mar 20 '19 at 18:59
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    It is maybe a bit confusing that $g$ is used for two different functions... – Pietro Majer Jul 11 '19 at 06:06
  • Notice that the OP asked about local representation. But this proof still works with minor changes: replace $0$ and $1$ with $z_0\ne z_1$ and still get a bijection between $\mathbb{R}$ and $\mathbb{R}^2$, then use any small circle in place of $S^1$. – Yaakov Baruch Feb 18 '20 at 09:37
  • I agree with Pietor Majer about the two $g$s being confusing. Also it seems that user44191 proposed the notation $h$ for the second $g$ in the comments. Can you please use that notation in the answer as well? – Vincent Mar 03 '20 at 15:32
  • @Vincent Corrected. Is it okay now? – Piotr Hajlasz Mar 03 '20 at 15:54
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    Yes, thank you! BTW I really like this answer. – Vincent Mar 03 '20 at 16:02