2
$\begingroup$

Consider the set $S_n$ defined as:

$S_n=\lbrace \mathbf{x} \in \mathbb{R}^n: x_i \neq0,\text{} \forall \text{ }i=1,2,..,n\rbrace$,

i.e. $S_n$ is the set of real vectors of dimension $n$ that have all entries different than zero.

Is the function $f_n(r,\theta) = \left(r\cos(\theta),r\cos(2\theta),...,r\cos(n\theta)\right)^T$ a surjection from $\mathbb{R}^2$ to $S_n$ for any natural $n$?

How to prove it?

Thank you in advance!

  • 0
    A plot of $\theta\mapsto (\cos(\theta),\cos(2\theta))$ shows that $f_2\colon \Bbb R\setminus\{0\}\times (\pi/4,3\pi/4]\to \Bbb R^2\setminus\{(0,0)\}$ is bijective. – 2017-01-18
  • 0
    Yes and moreover that is rather easy to check. Say $f_2 = \left(c_1,c_2\right)$. If you set $c_1$ to a fixed value $c_1^*$, you have that $c_2 = c_1^* \frac{cos(\theta)}{cos(2\theta)}$. The range of $\frac{cos(\theta)}{cos(2\theta)}$ is $\mathbb{R}$, so $c_2$ may still be set arbitrarily given $c_1$. – 2017-01-19
  • 0
    I obtain $\frac{c_1}{c_2}=\frac{\cos(2\theta)}{\cos(\theta)}$. To show that $\alpha = g(\theta)$ with $g(\theta)=\frac{\cos(2\theta)}{\cos(\theta)}$ has a solution for every $\alpha$, it is enough to study $\lim_{\theta\to x}g(\theta)$ for $x\in\{\pi/4,\pi/2,3\pi/4\}$ (take left and right limits at $\pi/2$) and use the continuity of $g$ on $[\pi/4,\pi/2)$ and $(\pi/2,3\pi/4]$. This should prove surjectivity for $n=2$. Now, for $n>2$, I still strongly believe that a similar argument as my (wrong) answer will prove that it is not surjective. – 2017-01-19
  • 0
    This was my answer (for $<10k$ users). *(Wrong) Hint for $n\geq 3$:* If $\cos(\theta)=\cos(2\theta)$, then $2\theta = \theta + 2k\pi$ for some $k$, and so $\theta = 2k\pi$ and therefore $\cos(\theta)=\cos(m\theta)$ for every $m=1,\ldots,n$. In particular, any vector of the form $r(1,1,t,\ldots)$ can not be obtained for $0 – 2017-01-19
  • 2
    Are you asking if $S_n \subset f(\mathbb R^2)?$ I don't think surjection is the right term here. – 2017-01-24
  • 0
    @zhw. For the bounty, I'm interested in (dis)proving surjectivity for $f_n\colon \Bbb R^2 \to \Bbb R^n$. Ideally (but not required), you can even fully describe the set $f_n(\Bbb R^2)$. – 2017-01-24
  • 0
    the image of $f_n$ is included in an algebraic surface so it can't be surjective for any $n \ge 3$ – 2017-01-25

2 Answers 2

2

$n=2$: I'll use $t$ for $\theta.$ Let's first take $r=1, t\in [-\pi/4, \pi/4].$ Recall that $\cos 2t = 2\cos^2t - 1.$ So the set of points traced out by $f$ from this subdomain has the form $(\cos t, 2\cos^2t-1), t\in [-\pi/4, \pi/4].$ This is exactly the set $(x,2x^2 -1): x\in [-1/\sqrt 2, 1/\sqrt 2],$ i.e., it's the graph of $y=2x^2-1$ over this interval.

If we look at $r=-1, t\in [-\pi/4, \pi/4],$ we will get the graph of $y=1-2x^2, x\in [-1/\sqrt 2, 1/\sqrt 2].$ In other words, we get the reflection of the first graph with respect to the $x$-axis.

The union of the above two graphs, let's call it $G,$ has the property that for every $a\in [0,2\pi],$ there exists $s(a) >0$ such that $s(a)e^{ia} \in G.$ Now $G\subset f(\mathbb R^2).$ Since $uf(\mathbb R^2)\subset f(\mathbb R^2)$ for any real $u,$ the line thru the origin determined by $e^{ia}$ is contained in $f(\mathbb R^2)$ for every $a\in [0,2\pi].$ Thus $f(\mathbb R^2)=\mathbb R^2.$

$n>2:$ Here $f$ is nonsurjective to say the least. Suppose $n=3.$ Let $$\gamma = \{(\cos t, \cos 2t, \cos 3t):t\in [0,2\pi]\}.$$ Then $f(\mathbb R^2) = \mathbb R\cdot \gamma.$ But $\gamma$ only contains one point in the $y-z$ plane, namely, $(0,-1,0).$ Thus the intersection of $\mathbb R\cdot \gamma $ with the $y-z$ plane is just the line $(0,y,0): y \in \mathbb R.$ The result for $n>3$ follows.

Another way to proceed for $n>2$ is through measure theory: If $f:\mathbb R^j\to \mathbb R^k$ is a smooth map and $j

  • 0
    Thank you very much for your answer, this is pretty much exactly what I expected. For completeness I would add the property on $G$ follows from the fact that $2x^2-1$ is a convex parabola taking its minimum at $-1$ with two distinct zeros implying the existence of $s(a)>0$ for $a\in [\pi,2\pi]$ (the case $a\in [0,\pi]$ follows from the same argument on the other parabola which is concave). I'm also very happy about the other to proceed which was my intuition from the beginning but I did not know how to formulate it rigorously. Do you have a reference for this measure theory result? – 2017-01-24
  • 1
    Convexity is a quick way to see this property, but it's not necessary. Suppose $g,h$ are continuous on $[-a,a],$ $g>0$ on $(-a,a),$ $h<0$ on $(-a,a),$ and both equal $0$ at the endpoints. Then the union of the graphs of $g,h$ over $[-a,a]$ has this property. This follows from the connectedness of these graphs. I'll get back to you on the measure theory. – 2017-01-25
  • 0
    This makes fully sense, at the end you just need a curve whose graph is the boundary of a set which contains an open neighborhood of the origin. I should just admit that with these kind of arguments, it is not always clear to me where to stop with *intuitive arguments*. "I'll get back to you on the measure theory": ok. I will leave the question a few days more in the featured list and then (highly probably) give you the bounty. – 2017-01-25
  • 1
    Let me illustrate the measure theory approach in a simpler setting. Suppose $f:[0,1]\to \mathbb R^2$ and that $|f(y)-f(x)| \le C|y-x|.$ Let $n\in \mathbb N,$ and define $I_k = [(k-1)/n,k/n], k = 1,\dots , n.$ Then $f([0,1]) = \cup_k f(I_k).$ Verify $f(I_k) \subset D(f(k/n),C/n),k=1,\dots, n.$ Thus $$f([0,1]) \subset \cup_k D(f(k/n),C/n),$$ which implies $$m_2(f([0,1])) \le \sum_k \pi(C/n)^2 = n\cdot \pi(C/n)^2 = \pi C^2/n.$$ – 2017-01-25
  • 1
    . . . The above is true for any $n,$ and this shows $m_2(f([0,1]))=0.$ Now try to make this work for $f:[0,1]^2 \to \mathbb R^3$ with the same kind of Lipschitz condition on $f.$ The proof here is very similar to the above. – 2017-01-25
1

No. For example $(1,0, \dots, 0)$ is not in the image.

  • 1
    What about $n=2$, $r=\sqrt 2$ and $\theta = \frac{1}{4}\pi$? – 2017-01-18
  • 0
    The question is asking if for *any* $n$ this map is surjective. – 2017-01-18
  • 0
    Yes, you are correct. Actually, I am not interested in vectors with entries equal to zero. I have edited the post. Thank you! – 2017-01-18