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$\begingroup$

Does there exist a continuous function $f:[0,1]\rightarrow\mathbb{R}$ such that for any two points P,Q on the curve, there exists a point R on the curve such that PQR is an equilateral triangle? If so, can we find a smooth one?

For bounty, scroll to bottom of page...

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    is $f : [0,1] \to {\mathbb R}^2$?2012-05-04
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    Well you can construct it either way. It's the same thing for the purposes of this question. I guess for a proof it's easier to use f:[0,1]-->R^2 because the geometry is in R^2. Edit: it might actually be easier to work in R. Whatever. It doesn't matter.2012-05-04
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    My only guess for the answer would be to use the Intermediate Value Theorem for a relatively flat (e.g. |gradient| <= 2/3) function to show that it is never true. To this end I'm not making progress, and even so, it says nothing about a function that has |gradient| > 2/3 in places.2012-05-04
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    What curve? The graph of $f$?2012-05-04
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    yes, the graph of f2012-05-04
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    Do you mean to [fill](http://en.wikipedia.org/wiki/Space-filling_curve) an equilaterial triangle?2012-05-04
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    no. Just the vertices to be members of the curve2012-05-04
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    Possibly sdcvvc and yourself might have thought I meant to fill in the equilateral traingles, and then yes, we would have to be in R^2. But I am just thinking of a curve on a sheet of paper such that for any two points P and Q on the curve, then there is a 3rd point on the curve which forms an equilateral triangle PQR2012-05-04
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    Edit: Important question: Are we considering functions f:[0,1] --> R or am I allowing the curve be the (translated and squashed) unit circle (e.g f:[0,1]-->R^2 or f:[0,1]-->C (which for the purposes of this question is the same thing)) ? Well, it doesn't matter for the smooth case because the answer to the original question will always be no is no (see below). For the non-smooth case, I would be interested in a solution f:[0,1]-->C if there are no solutions in f:[0,1]-->R.2012-05-04
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    When you decide what you want, come back and click on the link in the answer.2012-05-04
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    I think I've already made clear what I want. I am looking for a continuous graph with the properties in the OP. This has nothing to do with Peano curves2012-05-04

4 Answers 4

0

Edit: The bounty relates to this post. Basically, fill in all the details. I'm doing this because I am interested in the problem, however I don't have time atm to investigate.

There is no convex subset of R^2 other than:

  1. (The empty set)
  2. One point

which is bounded and has the property that, given any two points in the set, the 3rd point is in the set. This can be proven as such:

  1. Suppose there is such a convex bounded set in R^2 (i.e. we can stick a closed circle (i.e. closed ball) round it)
  2. Stick the smallest possible closed circle/closed ball round it.
  3. Take the two points in the set which have max distance.
  4. If there is a 3rd point in the set then the set must be a convex shaded region between an equilateral triangle and the circle.
  5. An equilateral triangle fails (think about the perpendicular bisector).
  6. Circle fails (diameter)
  7. Anything in between fails (the perpendicular from the equilateral triangle is actually outside the circle by construction (I think)).

Contradiction. Shaded concave sets... part of the bounty.

Now to unbounded sets in R^2...

We can have an unbounded set that works. E.g. take a small straight line and "extend it" by taking two points on the line and filling up the equilateral triangles. I don't know what types of sets we will get but they will be crazy and whether or not they fill up R^2 I don't know...

Still I can't work out if a cts function R-->R exists. Or if such a function R-->R exists at all. if we fill up the equilateral grid and then shift it slightly is that a function? Bounty is in response to this question

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    Step 3 is wrong: If the set is itself a (filled) eqilateral triangle, then two maximally distant points in it do make an equilateral triangle with a third point. (A vertex and the mid-point of the opposite side don't, so your conclusion may be true -- but you haven't proved it here.).2012-05-10
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    Yes. Well I can show it for convex sets of R^2. Concave sets I haven't actually shown yet. Let's change the bounty question...2012-05-10
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    basically, if anyone can fill in all the details and make some concrete proofs or even better some examples of functions, then they get 100 points. I'm going back to revision. Good luck2012-05-10
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    Are you using a definition of convexity that admits isolated points, such as those which are the vertices of an equilateral triangle, as given by your 3rd example? If so, then there are unbounded sets which do not take up all of $\mathbb R ^2$. Consider that, given any two points, we can find exactly two other points which are vertices of an equilateral triangle with the first two. Start with two points, say $(-1,0)$ and $(1,0)$, and repeat this process recursively. The infinite set of points will be unbounded, will satisfy the other requirements, but will not contain the origin.2012-05-10
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    It does not, however, satisfy the typical definition of convexity, and neither does your third example, I believe.2012-05-10
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    convex set means given any two points we can draw a straight line between them and all points on the line will be in the set.2012-05-10
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    @AdamRubinson So then the set of three points which are the vertices of an equilateral triangle (example 3) is not a convex set.2012-05-10
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    And your example of the equilateral triangle lattice is an obvious example of an unbounded set in R^2 satisfying the conditions. But obviously it isn't convex. You can even remove some (in fact, infinite) points in the lattice and that will still be a set satisfying the conditions. What is my third example?2012-05-10
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    Oh yeah. I had added in the "convex" condition afterwards and changed the post, but missed that.2012-05-10
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    Are there, for example, closed, uncountable sets in R^2 with this property? Try to show there are no (nontrivial) closed, countable sets in R^2 other than the 3 points that are vertices of an equilateral triangle2012-05-10
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Not in the smooth case. Here we suppose that $f$ is $C^1$.

Let $s=\frac{\sqrt{3}}3=\tan{\fracπ6}.$

Suppose that for some $x_0$, $|f'(x_0)| then around $x_0$ the graph of $f$ looks like a straight line so horizontal that the third point of a equilateral triangle containing $(a,f(a))$ and $(b,f(b))$ would be of abscissa $x∈[a,b]$.

In the following picture, the case on the left is the one that is impossible.

Slopes

Then $|f'|≥s$ and $f'$ is continuous so $f$ is monotonous so $f$ can't meet your criteria.

I don't know how to generalize.

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    For your line of reasoning, we only need |f'(x_0)| < 2/3, not necc. 1/2. Thanks though, I think I can construct an epsilon-delta proof now for functions that have |f'(x_0)| < 2/3 for some x_0. If f is such that there is no x_0 with |f'(x_0)| < 2/3, then the result is kind of obvious. So that is the smooth case done. The "cts but non-smooth" case still remains to be solved.2012-05-04
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    I made an error, it was not $\frac12=\sin{\fracπ6}$ but $\tan{\fracπ6}$. However $\frac23$ is not small enough.2012-05-05
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    Thanks for the diagram. Just for anyone reading: the shaded-in part of the diagram (equilateral triangles) are not supposed to be part of the function. Only the vertices of the equilateral triangle are meant to be part of the function. The only other option (for the smooth case) is if we have a function which has |f(x)| >= 2/3 everywhere. f must be monotone otherwise there must be a turning point which would mean |f|<=2/3 --><--. But f monotone and gradient >= 2/3 clearly won't work.2012-05-05
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    However if we don't require smoothness then we can try some crazy fractal functions like the Weierstrass function: http://en.wikipedia.org/wiki/Weierstrass_function2012-05-05
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    However if we don't require smoothness then we can try some crazy fractal functions like the Weierstrass function: http://en.wikipedia.org/wiki/Weierstrass_function I guess this is what I am after for the non-smooth case. Also, I don't think we can find a subset of R^2 (e.g. shaded in circle, shaded in square etc.) which works. Suppose there were such a shape, call it a set X in R^2. Just find (z1,z2) in X that maximizes {|zA-zB|: zA,zB in X}}. Then you would need the 3rd vertex (say z3) to be in X and when you colour in the equilateral triangle joining z1, z2 and z3, this shaded triangle2012-05-05
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    @AdamRubinson: indeed, but these $z_i$ are not unique. For example if $X$ is the vertices of an equilateral triangle then it works. However $X$ is not the graph of many continuous functions :-)2012-05-05
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    must lie in X. Now the shaded equilateral triangle doesn't work (just perpendicularly bisect one of the sides to give to two points for which it fails). If any other points in X lie outside the shaded equilateral triangle, then it must lie within the circle of diameter |z2-z1|, otherwise this contradicts the supposed maximal distance of z1 and z2.2012-05-05
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/3341/discussion-between-jmad-and-adam-rubinson)2012-05-05
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    Edit: it doesn't have diameter |z2-z1|, but the circle does have centre point at the centre of equilateral triangle and does have radius from origin to z1. The only thing left to show is that there are no shapes between the equilateral triangle and the circle...2012-05-05
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    Hi @jmad, just wanted to let you know that the ["Italian Language & Usage"](http://area51.stackexchange.com/proposals/42949/italian-language-usage) proposal has restarted. Since you committed to the previous proposal, maybe you are still interested. See you! (I hope this comment is not considered as spam: if so, I beg your pardon)2012-09-11
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Here is the answer to the generalized form of your question in the case of bounded sets. As it turns out, the convexity assumption is unnecessary.

Proposition. Let $A\subseteq\Bbb R^2$ be a bounded set, such that for each pair of distinct points $x,y\in A$ there is a point $z\in A$, such that $x,y,z$ are the vertices of an equilateral triangle. Then $A$ is one of the following:

  • empty set,
  • a set containing only one point,
  • the set of vertices of some equilateral triangle.

Proof. Let $A$ be a set satisfying the hypotheses of the proposition. Suppose $A$ contains more than one point. We will show that then $A$ must be the set of vertices of some equilateral triangle.

First, we shall prove the proposition in the case that $A$ is closed. (In the end we shall show that the general case follows easily from this one.) So, let's assume $A$ is closed, i.e. $A$ contains all its limit points. Then $A$ is compact, so there exist points $a,b\in A$ such that $$d(a,b)=\operatorname{diam}A=\sup\lbrace d(x,y)|\text{ }x,y\in A\rbrace,$$ where $d$ is the metric in $\Bbb R^2$ and $\operatorname{diam}$ stands for diameter.

Let $c$ be a point such that $a,b,c$ are the vertices of an equilateral triangle. This exists by the hypotheses. Let $r=d(a,b)$ and let $S=\overline{K}(a,r)\cap\overline{K}(b,r)\cap\overline{K}(c,r)$, where $\overline{K}(x,r)$ denotes the closed ball centered at $x$ with radius $r$. Then $A\subseteq S$, because otherwise we would have $\operatorname{diam} A>r$ which is not the case. By the way, $S$ is a Reuleaux triangle:

enter image description here

We already know that $A$ contains the vertices of this Reuleaux triangle. We shall now show that this is all that $A$ contains. We shall first define some more points. Let $a'$ be the reflection of $a$ across the line through $b$ and $c$. Define $b'$ and $c'$ analogously. (As reflections of $b$ and $c$ across the lines opposite to them.) Let $$S_{a}:=S\setminus(\overline{K}(b',r)\cup\overline{K}(c',r))\\S_{b}:=S\setminus(\overline{K}(c',r)\cup\overline{K}(a',r))\\S_{c}:=S\setminus(\overline{K}(a',r)\cup\overline{K}(b',r))$$ Now, note that every point of $S\setminus\lbrace a,b,c\rbrace$ is contained in at least one of the sets $S_a,S_b,S_c$. So, to complete the proof in the case where $A$ is closed, it suffices to show that $A\cap S_a=\emptyset,A\cap S_b=\emptyset$ and $A\cap S_c=\emptyset$, which we will now do.

Actually, we will only prove the case of $S_a$. The other two can be proved by a completely symmetric argument, so we leave them to the reader. Let $x\in S_a$. Let $u(x)$ be the point that forms an equilateral triangle together with $a$ and $x$, for which the triangle $a,x,u(x)$ is oriented clockwise (i.e. negatively). Let $v(x)$ be the other point that forms an equilateral triangle with $a$ and $x$, i.e. $a,x,v(x)$ is oriented counterclockwise (positively). This defines two functions $u:S_a\to\Bbb R^2$, $v:S_a\to\Bbb R^2$. By definition, $u$ is a rotation around the point $a$ by $-\frac{\pi}{3}$, and $v$ is the rotation around the point $a$ by $\frac{\pi}3$.

By exploiting this geometry, one can now easily see that $u$ rotates the set $S_a$ outside of $S$, i.e. $u(S_a)\subseteq \Bbb R^2\setminus S$, and for $v$ the same holds: $v(S_a)\subseteq \Bbb R^2\setminus S$. But this means that for each $x\in S_a$ the only vertices that could form an equilateral triangle with $x$ and $a$ are outside of $S$, hence not in $A$. But then $x$ cannot be an element of $A$. Here's a picture showing $S_a$ (middle), and its two rotations (the Reuleaux triangle around $S_a$ is $S$, and the two rotations lie outside of $S$):

enter image description here

This completes the proof of the closed case.

In the general case of bounded (not necessarily closed) $A$ with at least two points, we proceed as follows. Let $\overline{A}$ be the closure of $A$. Since $\Bbb R^2$ is such a nice (i.e. first-countable) space, this means that $$\overline{A}=\lbrace x\in\Bbb R^2|\text{ there is a sequence } (x_n)_n\text{ with terms in } A\text{, such that } \lim_{n\to\infty}x_n=x\rbrace.$$ Now, clearly $A\subseteq\overline{A}$. Furthermore, $\overline{A}$ also satisfies the hypotheses of the proposition. To see this, let $x,y\in\overline{A}$. Then there are sequences $(x_n)_n$ and $(y_n)_n$ in $A$, such that $\lim_{n\to\infty}x_n=x$, $\lim_{n\to\infty}x_n=x$. But since $x_n,y_n\in A$ for each $n$, we can associate a point $z_n\in A$ to each pair $x_n,y_n\in A$ such that $x_n,y_n,z_n$ are the vertices of an equilateral triangle. Because $\overline{A}$ is compact, we can choose a subsequence of $(z_n)_n$ that converges to a point $z\in\overline{A}$. By continuity of the metric, $x,y,z$ again form an equilateral triangle. So, $\overline{A}$ indeed satisfies the hypotheses of the proposition and is therefore the set of vertices of some equilateral triangle. From this it easily follows that $\overline{A}=A$, which concludes the proof. $\square$

This proves among other things, that there is no continuous function $f:[0,1]\to\Bbb R$, whose graph contains for every pair of points a third point that forms an equilateral triangle with them. It proves even more: there is also no such discontinuous function $f:[0,1]\to\Bbb R$. Why? Suppose there is. Then it must be bounded, since, otherwise its graph would contain two points $x,y$ such that $d(x,y)>100$. There would then have to be a third point on the graph whose first coordinate would lie outside of $[0,1]$, which is not the case. But for bounded sets, the proposition applies. The same applies for any function $f:B\to\Bbb R$, where $B\subseteq\Bbb R$ is a bounded set.

The situation might be more interesting with functions $f:\Bbb R\to\Bbb R$, however. But in this case again, there is no such bounded function. (Same argument as in the previous paragraph.) It might be possible to construct something unbounded and probably horribly discontinuous, though, but I haven't thought much about that.

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    I did suspect that convexity was unnecessary for proofs (especially general ones). And for geometric purposes, it seems as though you "need" convexity, even though you have shown you don't if you use compactness of R^n. In fact, the compactness method you use adds to the beauty of the problem in my opinion. HOWEVER, in my question I require for you to also show the equilateral triangle property holds/does not hold for cts, unbounded f:R-->R functions. If you can do this you will achieve the bounty... I don't think this is unreasonable as it clearly states it in my question at the bottom.2012-05-12
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    I can think of ideas that might work for "horrible discontinuous unbounded functions f:R-->R with the property" I am thinking: take the "equilateral triangle grid": http://exchangedownloads.smarttech.com/public/content/29/29b3fb04-3db1-43b3-8e3a-36864768e8d7/previews/medium/0001.png considering only the vertices, not the lines joining the vertices. Rotate the whole graph by pi/4 radians anti-clockwise (for example). Then I think every point is no on top any other point. Maybe it doesn't fill up R in the domain but maybe we can add another point somewhere and "expand out" so that it does.2012-05-12
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    @AdamRubinson: If I think of something, I'll let you know. I won't be around for two days, though, since I have some stuff to do ...2012-05-12
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    thanks v. much. To try to be fair, if someone else finds the answer before you, then if possible I'll see if I can share the bounty between you two. I'll ask a mod if I can do that if that situation arises...2012-05-12
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Suppose that the set $S$ is bounded, and suppose that it has more than one point. Let $A$ and $B$ be two points in $S$. Then it must have a third point $C$ which is (another) vertex of the equilateral triangle with one side as $AB$, and by our condition of convexity the entire triangle $ABC$ must be in $S$.

Now let $D$ be the point of intersection of the perpendicular bisector of $\angle BAC$ and ${BC}$. Then there must be a point $E$ which is (another) vertex of the equilateral triangle with one side as $AD$. Let the distance of $AB$ be equal to $d$. Then, by several applications of the pythagorean theorem, we have that either the length of $CE=\frac {\sqrt 7} 2 d$ or $BE=\frac{\sqrt 7} 2 d$. Either way, this shows that the distance between points is unbounded, which is a contradiction with our hypothesis. Hence $S$ may not have more than one point.


If the sets are unbounded, it need not be all of $\mathbb R ^2$. For instance, any half-plane should do.


Here's a nice simple proof for why a continuous function is not possible. I've tried to be more laconic, and just give the idea of why it will not work, as I believe is in line with your preference (from the comments). Please let me know if you would like more details.

enter image description here

For any point (in the picture I place the point at the origin, but it's no matter) we cannot have any point of our graph on the lines emanating from this point at a $\pi/6$ angle with the horizontal, for if we did it would quickly follow that the only way to create an equilateral triangle would be to place a point above/below one of these points, and thus $f$ would not be a function. By the intermediate value theorem, this means that we cannot place any points in the shaded area, for in order to create an equilateral triangle with the point at the origin and a point in the shaded area we would need to place a point outside the shaded area, and at some point $f$ would have to cross the line. Now place a point anywhere in the non-shaded area. Immediately, from our discussion earlier, we know that $f$ cannot cross the dotted lines emanating from this point either, and furthermore the point needed to create an equilateral triangle will lie in between these lines, which we know (by the intermediate value theorem) is not a point which $f$ can reach. Hence, such a continuous function is impossible.


Also, my proof at the beginning, that a bounded set $S$ which is convex is not possible, does not really deal with showing that $S$ cannot be between a circle and a triangle, it is much simpler than that. Read it again, and note that all it shows is that, given two points $A, B$ with distance $d$, we are guaranteed that there are points $C, D$ which have distance $\frac {\sqrt 7} 2 d$. The reason this shows that the set $S$ cannot be bounded is that if it were bounded by some ball with diameter $h$, we are guaranteed that there exist points in $S$ with a distance greater than $h$. We simply pick any two points $A, B$ with distance $d$, and then repeat the process above $k$ times, where $\left(\frac {\sqrt 7} 2\right) ^k d>h$. Therefore $S$ cannot be bounded. Note that nowhere do we assume anything about bounding $S$ by the smallest ball possible, or choosing points yielding the maximum distance because (in particular, if $S$ were open) such points may not exist.

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    Assuming what you wrote is correct... I still want to know if there is a $cts$ function f:R-->R with the equilateral triangle properties (I think this is equivalent to asking the same question for f:[0,infinity)-->R )2012-05-11
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    @AdamRubinson Do you mean that the image of $f$ is a set $S$ which has these properties?2012-05-12
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    @AdamRubinson It doesn't make sense to ask for a continuous function $f:\mathbb R \to \mathbb R$ such that the image set satisfies these properties because the image of $f$ will be one dimensional. What I think you mean is to ask for a function $f:\mathbb R \to \mathbb R ^2$. A simple answer to your question is that a [Peano curve](http://www.cut-the-knot.org/do_you_know/hilbert.shtml) can be made with domain $\mathbb R$, whose range is all of $\mathbb R ^2$. Such a function would be continuous (read the linked page on Peano curves to see why)...2012-05-12
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    @AdamRubinson ... and it would clearly map $R$ to a set which satisfies the properties you have described (except for boundedness of course - the proof above shows that a bounded set doesn't exist).2012-05-12
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    It should be obvious what I am asking, given what I have found out and what you have found... it is obvious a function R-->R^2 exists... just start with a small bounded line of some sort in R^2 and extend it with 3rd-equilateral-triangle points. Whether or not all these sets necc. fill up R^2 can be added to the bounty if you wish... but I am not so interested in that...2012-05-12
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    I guess I want a function from R-->R, with the triangle properties. TO be absolutely clear (I hope): So think about R^2. I want a line such that there are no two points (x1,y1) and (x2,y2) on the line s.t. x1=x2.2012-05-12
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    @AdamRubinson So then you want the *graph* of $f:\mathbb R \to \mathbb R$ to satisfy those properties. For obvious reasons, if the desired graph is to be convex it must be a straight line, but since a straight line does not meet the equilateral triangle criteria we have that such a function does not exist. If we drop convexity, something interesting might happen.2012-05-12
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    What is your definition of convex here?2012-05-12
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    @AdamRubinson Same as before, treating the graph of $f$ as a subset of $\mathbb R ^2$. If we ask about a [convex function](http://en.wikipedia.org/wiki/Convex_function) instead of a convex set, that's different.2012-05-12
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    @AdamRubinson It's no matter, because I believe there's a simple proof that a continuous function doesn't exist anyway. I'll try to write it up tomorrow, but I believe I have the general idea.2012-05-12
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    In that case, you say "For obvious reasons, if the desired graph is to be convex it must be a straight line". Why is this the case?2012-05-12
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/3411/discussion-between-michael-boratko-and-adam-rubinson)2012-05-12
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    for some reason it won't let me log into chat...2012-05-12
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    hmm, ok, well the basic idea is to consider two points, $(0,f(0))$ and $(1,f(1))$2012-05-12
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    What about those points?2012-05-12
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    @AdamRubinson Sorry, the site is being temperamental. So consider the set represented by the graph of $f$. Now since this set is convex, the line connecting $(0,f(0))$ and $(1,f(1))$ must be in the set and thus the values for $(x,f(x))$ where $x\in [0,1]$ are entirely determined by this line. But now consider some point $(x_0,f(x_0))$ which does not lie on the (extension of the) line. Then the line defined by the points $(1/2,f(1/2))$ and $(x_0,f(x_0))$ must be in the graph of $f$. But this is a contradiction, because $f$ is now no longer a function since some points $x \in [0,1]$ have ...2012-05-12
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    @AdamRubinson ... two corresponding values of $f$. For instance, if $x_0>1$, then the point $f(3/4)$ will need to correspond to two points, one on the first line and one on the second, and these two points cannot be the same because the lines are not coincident.2012-05-12
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    @AdamRubinson Thus the proposed point $(x_0,f(x_0))$ cannot exist, and hence the graph of $f$ is a straight line.2012-05-12
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    ... Okay, but I don't know why you wanted to use the definition of convexity for the graph of f. I only wanted to use the idea of convexity because I thought it might be useful to prove the original question for shaded regions (in R^2). You have shown that the graph of f must be convex, but this is "trivial", and I don't see how it helps us. If you could give your proof that the continuity function doesn't exist, that would be awesome (and I will try to share the bounty points between you and Dejan Govc. Can't promise anything though, but I will message a mod now).2012-05-12
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    Also, everything you write after "but now consider some point (x0,f(x0))..." is redundant, because you have just shown that: "If the graph is convex then it must be a straight line", which is what you set out to prove...2012-05-12
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    @AdamRubinson It wasn't clear to me from your question that you wanted to drop the convexity criterion when considering the graph of a function $f$, that's why I was trying to show that if the graph of $f$ is convex then it must be a straight line.2012-05-12
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    @AdamRubinson All I had shown up to that point is that if the graph of $f$ is convex then the values of $f$ on $[0,1]$ are a straight line. Granted, one could simply provide a constructive proof, and use arbitrary values instead of particular ones.2012-05-12
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    Okay, well... actually to me, it sounds like you are "being caught up in definitions", which is something every mathematician experiences, but I think is bad (no offence intended). Of course rigor is important, but these sets of questions I have posed is really an investigation into the qualitative nature of such functions in R and R^2. Thinking about convexity when considering the shaded regions is an obvious qualitative way to start investigation for shaded regions in R^2. But the question about continuous graphs in R has nothing to do with convexity. So maybe, even if you find the answer2012-05-12
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    you might miss the "big picture" here, and also the beauty and "niceness" of the problem (well, in my opinion anyway). Basically rigor is not so important here.... Maybe I am wrong and you do understand the nature of the problem though... oh well, I dunno2012-05-12
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    @AdamRubinson I found a question interesting, but apparently the question I found interesting wasn't the one you were asking.2012-05-12
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    what was the question you found interesting?2012-05-12
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    @AdamRubinson The one I answered above :)2012-05-12
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    I agree that your conclusion is nice but tbh I saw that argument close to the beginning of attempting to solve the problem. I couldn't quite write down concretely why we can't have regions in between the equilateral triangle and the circle, and you have done that concretely to some extent (I think). I must admit I didn't think of using Pythagoras...2012-05-12
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    @AdamRubinson I've tried to write a proof of why a continuous function is impossible in line with your preferences.2012-05-12
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    Yes. That is actually an extremely nice proof in my opinion. I'll read the second half tomorrow. Also, are there any moderators on this site? I had a quick glance at all users yesterday and it was not obvious that there were...2012-05-13
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    @AdamRubinson There are moderators, but I'm not so concerned with receiving the bounty. Glad to know the proof was clear.2012-05-13
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    Well, thanks for making this easy for me. I will copy and paste your answer into Dejan Govc's so at least we have a complete set of answers in one post.2012-05-13
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    @AdamRubinson: Michael and Dejan are the respective owners of their answers, even though this is your question. I think it's inappropriate to combine their answers without asking both users first, as well as unnecessarily redundant. There are four mods (Qiaochu, Mariano, Willy, Zev) and a couple more on the way, but IIRC moderators do not have the power to split bounties, and the only way to award a second bounty is to increase the amount on it. ([Related](http://meta.math.stackexchange.com/questions/4105/why-cant-i-offer-a-bounty-of-less-than-500-reputation).)2012-05-13
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    Michael Bortako said he didin't mind about the bounty. Whatever. It's not so important as the maths... and I think it reads better to have the whole answer in one place.2012-05-13
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    @AdamRubinson (&anon) - There's a lot of unconventionality going on with this question. For one thing, Adam added an answer as a way of clarifying his question, as opposed to editing his original post. In general I try not to get hung up on "meta" issues like this and focus on the math side of things. That being said, I'm more interested in what part of your question my answer did not cover which Dejan's did. I believe mine proved that there was no convex subset of $\mathbb R ^2$, that an unbounded set doesn't necessarily have to be all of $\mathbb R ^2$, and that there are no cnts. functions.2012-05-13
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    Michael, you showed that there were no bounded convex subsets of R^2 satisfy my relation. However, Dejan showed that ALL BOUNDED (not just convex) subsets of R^2 satisfy my relation. You filled in the gaps by showing there is no f:R-->R. There maybe is one more small gap to fill (if the answer is simple. Otherwise enough work has been done for the moment...).2012-05-13
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    Actually, I can think of lots of such shaded regions, e.g. shade in all of R^2 and take away the unit disc. Also, we can take away as many discs as we like... they don't even have to be circles...so yeah let's forget that...2012-05-13