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Can there exist two functions $f,g: \mathbb R \to \mathbb R$ so that $f$ is continuous, $g$ is discontinuous, and their graphs $\Gamma_f, \Gamma_g \subseteq \mathbb R^2$ are related by an isometry? (I think you can assume the isometry is a rotation.)

The graph of a continuous function must be path connected, so a natural intermediate question is, "can a discontinuous function have path connected graph?"

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  • $\begingroup$ Are you sure that claim is valid for all (graphs of) continuous functions? How about a constant function $\mathbb{R} \to \mathbb{R}$, say $x \mapsto 0$? Unless you rotate it by 90 degrees the graph will be a straight line and hence continuous, whereas if you rotate by 90 degrees it is not the graph of a function at all anymore. I think more details of the question are in order. $\endgroup$
    – Chris Heunen
    Dec 11, 2010 at 7:41
  • $\begingroup$ This sounds similar to taking the inverse of an injective bounded operator with dense range - this is usually not continuous and only densely defined. The graphs are rotations of one another, but it is not classical analysis.. $\endgroup$
    – Ollie
    Dec 11, 2010 at 8:25
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    $\begingroup$ Even now the question is poor: take $f(x)=x$ on $[0,1)$ and $x+1$ on $[1,2]$. Then $f$ is discontinuous and $f^{-1}$ is continuous (on their domains, of course). If we assume the set $S$ to be compact, then every function arising this way is continuous and the question is vacuous again. We can, probably, try to salvage it in some way but I prefer to leave that to OP. $\endgroup$
    – fedja
    Dec 11, 2010 at 15:26
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    $\begingroup$ tea.mathoverflow.net/discussion/828/… $\endgroup$
    – Anton Geraschenko
    Dec 11, 2010 at 20:55
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    $\begingroup$ As stated the answer is no. The proof is an exercise in the intermediate value theorem. $\endgroup$
    – Ryan Budney
    Dec 12, 2010 at 2:06

2 Answers 2

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I think the question has an easy answer, and it's essentially a 1st course in analysis type question that reduces pretty quickly to an intermediate value theorem application. Here's a more general statement. Let $f : \mathbb R \to \mathbb R$ be continuous and $h : \mathbb R^2 \to \mathbb R^2$ a homeomorphism. Then if $h(graph(f))$ is the graph of a function $g : \mathbb R \to \mathbb R$, then $g$ is continuous.

Sketch: consider the function $P : \mathbb R \to \mathbb R$ given by $P(x) = \pi_1 \circ h(x,f(x))$, where $\pi_1(x,y)=x$. This is a continuous monotone function by the assumptions, so it is an open map by the intermediate value theorem, so it's inverse exists and is continuous. The claimed function $g$ is then $g(x)=\pi_2 h(P^{-1}(x), f(P^{-1}(x))$ where $\pi_2(x,y)=y$.

The question gives me the feeling it's a homework problem, but I've never seen it before. This was solved in meta, with a correction by Anton. I wish I had known about this question yesterday -- I would have put it on my analysis class final exam!

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    $\begingroup$ I hadn't realized it earlier, but this argument (with slight modification) actually proves the stronger statement that the graph of a function is path connected if and only if the function is continuous. Very nice. $\endgroup$
    – Anton Geraschenko
    Dec 12, 2010 at 6:13
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The only meaning of the statement I can see is, the (quite obvious) fact that a 90 degrees rotation of the graph of a continuous, increasing function $f$ gives the graph of a function, plus the jumps.

Talking of rotating graphs, a more interesting fact is that a 45 degree rotation transforms 1-Lipschitz graphs into monotone increasing graphs, and conversely. As a consequence, the theorems of a.e. differentiability for Lipschitz functions, and for monotone functions, can be deduced from each other.

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    $\begingroup$ Isn't f(x)=x 1-Lipschitz? $\endgroup$
    – TonyK
    Dec 11, 2010 at 9:46
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    $\begingroup$ Isn't g(x)=0 increasing? $\endgroup$
    – Pietro Majer
    Dec 11, 2010 at 15:38
  • $\begingroup$ @Pietro: do you mean"nondecreasing"? $\endgroup$
    – Thierry Zell
    Dec 12, 2010 at 1:32
  • $\begingroup$ @Pietro: if I'm interpreting your answer correctly, rotating the graph of f(x)=x gives a vertical line, not a horizontal one. If you meant the other rotation, then TonyK could ask "Isn't f(x)=-x 1-Lipschitz?" I think you need the function to be (1-ε)-Lipschitz, or to define the inequality in the definition of Lipschitz to be strict. Then you get a strictly increasing function. $\endgroup$
    – Anton Geraschenko
    Dec 12, 2010 at 1:33
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    $\begingroup$ Thierry: yes, of course (to me "increasing" means "f(x)≤f(x) whenever x≤y"; for the strict variant I use "strictly increasing"). @Anton: yes, of course. If we want to include 1-Lip, extended real-valued functions have to be considered. Given the nature of the question as given in the first version, it didn't seem to me the case of proposining more structured statements. $\endgroup$
    – Pietro Majer
    Dec 12, 2010 at 13:31

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