# Intersection of a unit sphere of a given norm in finite dimension with an hyperplane.

Let $\|\cdot\|$ be a norm on $\mathbb{R}^n$. Let $C:=\{x\in\mathbb{R}^n\,:\,\|x\| \leq 1\}$, that is to say let $C$ be a convex compact symmetric set of non empty interior. Let $H$ be a linear subspace of $\mathbb{R}^n$ of dimension $n-1$. Is it true that there exists $z\in \mathbb{R}^n$ such as $\|z\|=1$ and \begin{align*} \Big[(H \cap \partial C) + \mathbb{R}z\Big] \cap \mathring{C} = \emptyset \quad\text{ ? } \end{align*} If not, please, give me a counter-example ! It appears to me that $z\in H^\perp$ should be a proper candidate, but though I might see why this is working for the $\| \cdot \|_p$ norms for example in low dimensions, I did not manage to prove it generally.

Thanks in advance.

Xou.

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Did you mean $\leq$ in the definition of $C$? – copper.hat Sep 26 '12 at 15:11
What's for you $\,\Bbb Rz\,$?? – DonAntonio Sep 26 '12 at 15:12
@DonAntonio: Presumably $\{\lambda z \}_{\lambda \in \mathbb{R}}$? – copper.hat Sep 26 '12 at 15:13
what is $\mathring{C}$? – robjohn♦ Sep 26 '12 at 15:16
Most likely the interior. – Stefan Sep 26 '12 at 15:28
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## 2 Answers

If $z \neq 0$ the result is not true in general.

Here is a counterexample.

Let $C = \{ (x_1,x_2) | x_1^2+x_2^2 = 1 \}$. Let $H$ = $\{ x | x_2 = \sqrt{2}\}$. Then $H \cap \partial C = \{(\pm \sqrt{2}, \sqrt{2})^T \}$, and $(H \cap \partial C) + \mathbb{R}z = \{ \lambda z + \sqrt{2} (1,1)^T, \lambda z + \sqrt{2} (-1,1)^T \}_{\lambda \in \mathbb{R}}$.

(Picture inspired by @robjohn.)

Now let $\phi(x) = \frac{1}{2} (x_1^2+x_2^2)$ and let $z \neq 0$. We will show that for any $z$, there exists $x \in H \cap \partial C$ and a $\lambda$ such that $\phi(x+\lambda z) < \frac{1}{2}$, which implies that $x+\lambda z \in C^\circ$.

In particular, all we need to do is to show that $\frac{\partial \phi(x)}{\partial x} z \neq 0$, since if this is the case, it is straightforward to show that there exists a $\lambda$ such that $\phi(x+\lambda z) < \phi(x) = \frac{1}{2}$.

Suppose, to the contrary, that $\frac{\partial \phi(x)}{\partial x}z = x^T z = 0$ for all $x \in H \cap \partial C$. Then this implies that $\sqrt{2} (\pm 1, 1) z = 0$, which implies that $z = 0$ since $(\pm 1, 1)^T$ are linearly independent, which contradicts our assumption that $z \neq 0$.

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By $\frac{\partial \phi(x)}{\partial x}$, do you mean $\nabla\phi$? At least it seems that $\frac{\partial \phi(x)}{\partial x}$ is supposed to be a vector. Assuming so, then this looks good (+1). – robjohn♦ Sep 26 '12 at 17:53
@robjohn: Sort of, $\frac{\partial \phi(x)}{\partial x} h = \langle \nabla\phi(x), h \rangle$. $\frac{\partial \phi(x)}{\partial x}\in (\mathbb{R}^n)^*$, whereas $\nabla\phi \in \mathbb{R}^n$. (And thanks!) – copper.hat Sep 26 '12 at 19:57
I am so sorry, I think my english is not good enough : I meant a linear hyperplan (i.e. containing 0), that is a subspace of dimension $n-1$. In your example, you could only use lines crossing zero, and you may check that the orthogonal direction does the job for $z$. – xounamoun Sep 26 '12 at 21:14

Unless I am misunderstanding something, the following seems to be a counterexample.

Given the standard Euclidean norm in $\mathbb{R}^3$ and the plane $x_1=.6$, $H\cap\partial C$ is the circle $$\{x:x_1=.6\text{ and }x_2^2+x_3^2=.64\}$$ No matter what $z$ is (other than $0$), the elliptic cylinder $(H \cap \partial C) + \mathbb{R}z$ will intersect $\mathring{C}$.

$\hspace{3.3cm}$

Given any $z$, if there is an $h\in H\cap\partial C$ so that $h+\mathbb{R}z$ does not intersect $\mathring{C}$, then $z$ must be perpendicular to the normal to $\partial C$ at $h$. In order that $h+\mathbb{R}z$ does not intersect $\mathring{C}$ for all $h\in H\cap\partial C$, $z$ would have to be perpendicular to each normal to $\partial C$ at a point of $H\cap\partial C$.

Therefore, $z$ would need to be perpendicular to $(.6,.8,0)$ and $(.6,0,.8)$, thus parallel to their cross product $(.64,-.48,-.48)$. Furthermore, $z$ would need to be perpendicular to $(.6,-.8,0)$ and $(.6,0,.8)$, thus parallel to their cross product $(-.64,-.48,.48)$. However, since $(.64,-.48,-.48)$ and $(-.64,-.48,.48)$ are not parallel, $z$ cannot be parallel to them both, unless it is $0$.

Thus, there is no $z$ that so that $\left[(H\cap\partial C)+\mathbb{R}z\right]\cap\mathring{C}=\emptyset$.

Revised Question

Consider the maximum norm on $\mathbb{R}^3$ given by $$\|x\|=\max(|x_1|,|x_2|,|x_3|)$$ Under this norm, $C$ is a cube of side $2$.

Let $H$ be the plane $x_1+x_2+x_3=0$. This plane intersects all $6$ sides of $\partial C$. Each of the following points is on $H$ and a different side of $\partial C$: $$\small\left\{\left(1,-\frac12,-\frac12\right),\left(-\frac12,1,-\frac12\right),\left(-\frac12,-\frac12,1\right),\left(-1,\frac12,\frac12\right),\left(\frac12,-1,\frac12\right),\left(\frac12,\frac12,-1\right)\right\}$$

As argued in the answer to the original question, $z$ must be perpendicular to the normal to $C$ at all points of $H\cap\partial C$. However, the normals to the cube span $\mathbb{R}^3$, so there can be no such $z$ other than $z=0$.

True in Two

Due to the fact that $\|-x\|=\|x\|$, the normal to $\partial C=\{x:\|x\|=1\}$ at any $x$ is the opposite of the normal at $-x$. In $\mathbb{R}^2$, $H\cap\partial C$ consists of two points, $h$ and $-h$. Thus, if $z$ is perpendicular to the normal to $\partial C$ at $h$, then neither $h+\mathbb{R}z$ nor $-h+\mathbb{R}z$ intersect $\mathring{C}$.

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Same remark as above. I am so sorry because you both gave good, precise and furthermore drawn answers ! Since the plane has to pass through zero, again the orthogonal direction seems to work (the intersection is made through one of the "big circles" of the sphere). – xounamoun Sep 26 '12 at 21:17
– robjohn♦ Sep 27 '12 at 14:57
Ok, I agree ! Thanks ! – xounamoun Sep 27 '12 at 16:02
@xounamoun: However, it appears to be true in $\mathbb{R}^2$. – robjohn♦ Sep 27 '12 at 16:53
Yeah, I noticed this before, this is probably because the intersection seems to be always reduced to 2 points. Actually, this question is related to a lemma used in the study of Schauder bases. I was quite sure that the result may not be true generally in finite dimension because it would simplify considerably some classical proof ! Ok I just read the proof " true in two ", I agree also ! – xounamoun Sep 28 '12 at 7:54
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