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Let $f:\mathbb R\to\mathbb R$ be a monotone function. Let $\gamma=\{(x,y)\ |\ y=f(x)\}$ is a curve in $\mathbb R^2$.

Does there exists a $f$ such that $\gamma\cap L\neq \emptyset \ \forall L\subset\mathbb R^2$; that is, $\gamma$ intersects with all lines?


Intuitively there must exist some $f$; imaging a "ladder" bouncing between $y=x$ and $y=1.1x$. But I failed to rigoroze this.

What is the sufficient and necessary condition for a curve that does not intersect with at least one line (which means: $\exists L\subset\mathbb R^2$ such that $\gamma\cap L=\emptyset$)?

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  • $\begingroup$ As long as your curve has two distinct points it will intersect at least one line. $\endgroup$ – John Douma Mar 14 at 22:32
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    $\begingroup$ What about $y=x^3$? $\endgroup$ – Fabio Lucchini Mar 14 at 22:36
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    $\begingroup$ @JohnDouma: I think the last line was asked in the sense that some line does not intersect with a curve, i.e. the opposite of intersecting with every line. I would begin by observing that a monotone increasing function with the given property (intersects every straight line in $\mathbb R^2$) is easily converted into a monotone decreasing function with the same property. Given a monotone increasing function that is unbounded in both directions, it must intersect not only every vertical line (by def. of function) but every horizontal line and every line with negative slope. $\endgroup$ – hardmath Mar 14 at 22:38
  • $\begingroup$ @FabioLucchini I was tooo slow in math I have to admit that $\endgroup$ – High GPA Mar 14 at 22:38
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    $\begingroup$ @tangentbundle That's not a monotone function. $\endgroup$ – Mars Plastic Mar 14 at 23:46
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Let $f:\Bbb R\to\Bbb R$ be a continuous function. If $f (x)/x\to +\infty$ as $x\to\pm\infty $, then the graph $\gamma $ of $f $ meets all lines of the plane.

Clearly $\gamma $ meets all vertical ones. If $f (x)/x\to+\infty$ as $x\to\pm\infty $ then $$f (x)-mx-q\sim f (x)= \begin{cases} +\infty&x\to+\infty\\ -\infty&x\to-\infty \end{cases}$$ as $x\to\pm\infty $, hence $f (x)-mx-q$ has at least a root by intermediate values theorem. Example: $f (x)=x^3$

Let $f:\Bbb R\to\Bbb R$ be a continuous function. If $f $ is (upper and lower) unbounded and $f (x)/x\to 0$ as $x\to\pm\infty $, then the graph $\gamma $ of $f $ meets all lines of the plane.

Since $f $ is unbounded, its graph meets all orizontal lines. On the other hand, for $m\neq 0$ $$f (x)-mx-q\sim -mx$$ as $x\to\pm\infty $, hence $f (x)-mx-q$ as at least a root by intermediate value theorem. Example: $f (x)=\sqrt [3]x $.

Let $f:\Bbb R\to\Bbb R$ be continuous. If there exists a line which doesn't meets the graph $\gamma$ of $f$, then there exists infinitely many lines which doesn't meet the graph $\gamma$ of $f$.

For if $f(x)-mx-q\neq 0$ for every $x\in\Bbb R$, then $f(x)-mx-q>0$ or $f(x)-mx-q<0$ for every $x\in\Bbb R$. If wlog, $f(x)-mx-q>0$ for every $x\in\Bbb R$, then $f(x)-mx-q'>0$ for every $x\in\Bbb R$ and $q'<q$.

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Concerning your first question: What you are looking for is a monotone function $f\colon\Bbb R \to \Bbb R$ such that for any $a,b\in\Bbb R$ the equation $$ f(x)-(ax+b)=0$$ has at least one solution $x\in\Bbb R$. As Fabio Lucchini has already pointed out in the comments, $f(x)=x^3$ will do the trick, just like any other monotone polynomial function with uneven degree greater than 1.

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    $\begingroup$ Note that if function $f$ satisfies the conditions (of monotonicity and intersection with every line in the plane), then $f^{-1}$ exists (as a function) and also satisfies the conditions. $\endgroup$ – hardmath Mar 15 at 2:35
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    $\begingroup$ $f(x)=x$ is a monotone polynomial with uneven degree. $\endgroup$ – TonyK Mar 15 at 12:04
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If we restrict the problem for only "nice" monotonic functions $f$, which have the derivative in $\mathbb R$, then such $f$ must fulfill two conditions:

  1. It has neither upper, no lower bound.

enter image description here

  1. Its derivative $f'$ has no upper bound if $f$ is increasing.
    (No lower bound for $f$ decreasing.)

enter image description here

The pictures show simple counter-examples with disjunct line $a$, when the function $f$ don't fulfill the corresponding condition.

In other words, we have to get rid of disjunct lines $y = kx+q$ - both for $k=0$ and $k \ne 0$.
(Al lines $x=c$ intersect your curve - the graph of $f$.)

What is the sufficient and necessary condition for a curve that does not intersect with at least one line (which means: $\exists L\subset\mathbb R^2$ such that $\gamma\cap L=\emptyset$)?

For "nice" functions which have derivative in $\mathbb R$, the opposite of conditions "1. and 2.", i.e. "not 1 or not 2".

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  • $\begingroup$ You seem to limit consideration to monotone increasing functions, which is not an intrinsic limitation (see my Comment on Question). However an upper bound on derivative is not necessary, e.g. $y = \sqrt[3]{x}$, which has unbounded slope at the origin. $\endgroup$ – hardmath Mar 15 at 0:32
  • $\begingroup$ @hardmath: $\sqrt[3]{x}$ does not have a derivative at the origin, so it does not satisfy the assumption. $\endgroup$ – tomasz Mar 15 at 18:35
  • $\begingroup$ Arcus tangent is not a very good example: it has a bounded derivative. A better example would be the exponential function: it has unbounded derivative, but is bounded from below. $\endgroup$ – tomasz Mar 15 at 18:37

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