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Is there a non-decreasing function $f:\mathbb{R}\to\mathbb{R}$ such that the convex hull of its graph covers $\mathbb{R}^2$?

Recall that $Graph(f):=\{(x,f(x))\mid x\in\mathbb{R}\}$ and $conv(A):=\{\sum_{i=1}^n t_i a_i\mid n\in\mathbb{N}, \forall i=\overline{1,n},\ t_i\ge0, a_i\in A,\ \sum_{i=1}^n t_i=1\}$ is the smallest convex set that contains $A$.

zhw.
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Neutral Element
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3 Answers3

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The convex hull of the graph of $f(x) = x^3$ covers $\mathbb R^2$.

For a point $(x,y)$ with $y > x^3$, take the line through $(x,y)$ and $(x-1, (x-1)^3)$. This is a line with positive slope that's above the graph of $f$ at $(x,y)$; however, $f$ grows faster than any linear function, so this line eventually hits the graph of $f$ again at $(x', x'^3)$ for some $x'>x$.

Therefore $(x,y)$ lies on the line segment connecting $(x-1, (x-1)^3)$ and $(x', x'^3)$, so it's in the convex hull of the graph of $f$.

For a point $(x,y)$ with $y < x^3$, the same argument applies, by symmetry. (Then $-y > -x^3$, so $(-x,-y)$ is on a line segment with endpoints $(a,a^3)$ and $(b,b^3)$, and $(x,y)$ is therefore on a line segment with endpoints $(-a,-a^3)$ and $(-b,-b^3)$.)

Misha Lavrov
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  • +1: Nice. If $y \neq x^3$ you can always find a line through $(x,y)$ with large enough slope so that the resulting cubic has two (and hence three) distinct real roots. – copper.hat Nov 17 '17 at 21:52
  • @Misha Lavrov The first part is ok. The part with $y<x^3$ is incomplete. I do not know what you meant by symmetry but the slope through $(x,y)$ and $((x-1),(x-1)^3)$ remains positive only if $y\ge(x-1)^3$. I am not sure this part can be corrected. – Neutral Element Nov 17 '17 at 21:52
  • @copper.hat please write what you mean by large enough slope. Is that slope non-negative? – Neutral Element Nov 17 '17 at 21:54
  • @NeutralElement: If $y>x^3$ then yes, otherwise it is negative. Draw a picture. All you need to do is to find one crossing, then the resulting cubic equation must have another real root (on the 'other side'). – copper.hat Nov 17 '17 at 21:55
  • @copper.hat Please write your comments as an answer so I can give you the points – Neutral Element Nov 17 '17 at 22:07
  • Well, this is Misha's answer :-). – copper.hat Nov 17 '17 at 22:08
  • Misha's answer is not ok. Given a point $(x,y)$ and a line through it with non-negative slope that cuts the graph of $y=x^3$ it can cut it only once and that is not ok. You mentioned that if we increase the slope then we may cut twice. That is your idea and if you justify that then you should get the points – Neutral Element Nov 17 '17 at 22:08
  • I guess one needs $\forall x_0,\ y_0\ge0,\ \exists m\ge0$ such that $x^3-m(x-x_0)-y_0=0$ has three real roots and $x_0$ should be greater than or equal to the smallest root and smallest than or equal to the greatest root. – Neutral Element Nov 17 '17 at 22:26
  • I have clarified what I meant by symmetry. – Misha Lavrov Nov 17 '17 at 23:09
  • @MishaLavrov So by symmetry you mean now that if $y<x^3$ then you consider the point$((x+1),(x+1)^3)$ – Neutral Element Nov 17 '17 at 23:31
  • That's what it comes to, though I didn't explicitly think about it; the regions ${(x,y) : y < x^3}$ and ${(x,y) : y > x^3}$ are congruent, so the argument should translate, even before we figure out how. – Misha Lavrov Nov 17 '17 at 23:33
  • It's not. Can you give me a point, then, for which this argument does not produce a line segment whose endpoints lie on the graph of $y = x^3$? – Misha Lavrov Nov 17 '17 at 23:36
  • It is OK now I think – Neutral Element Nov 17 '17 at 23:55
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I agree that the convex hull of $f(x)=x^3$ covers $\mathbb{R}^2$. First note that in the first quadrant the set:

Above$=\{(x,y)\mid x>0,\ y\ge x^3\}$ is included in the convex hull because every line with positive slope through the origin intersects $x^3$

By symmetry -Above which is Below $x^3$ in the third quadrant is in the convex hull

The convex hull of Above$\cup$-Above covers the positive $y-$axis and below $x^3$ in the first quadrant. Therefore the first (and by symmetry) the third quadrants are in the convex hull and that is enough.

Neutral Element
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Define the rays $R_1 = \{(t,t):t\le 0\}, R_2 = \{(t,2t):t\ge 0\}, R_3 = \{(t,t/2):t\ge0\}.$ Note that the convex hull of $R_1\cup R_2 \cup R_3$ is all of $\mathbb R^2.$

Now we can choose sequences $P_n \in R_2, Q_n \in R_3,$ both sequences $\to \infty,$ such that the polygonal path

$$\tag 1 R_1 \cup[(0,0),P_1]\cup [P_1,Q_1]\cup [Q_1,P_2]\cup [P_2,Q_2] \cup [Q_2,P_3] \cup \cdots$$

is the graph $G_f$ an increasing continuous function $f$ on $(-\infty,\infty).$

Because $(0,0),P_1,P_2,\dots \in R_2$ and $(0,0),Q_1,Q_2,\dots \in R_3,$ both $R_2,R_3$ are subsets of the convex hull of $G_f.$ So is $R_3.$ Since the convext hull of $R_1\cup R_2 \cup R_3$ is all of $\mathbb R^2,$ the same is true for the convex hull of $G_f.$

zhw.
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