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More precisely, can we build a norm $N$ on $\mathbb{R}^2$, such that the ratio circumference / diameter (computed with norm $N$) of a standard circle is $42$?

(By standard circle, I mean a circle defined with the Euclidean norm, i.e. $x^2+y^2 = r^2$, and not a circle defined with the norm $N$).

Thus, this question is different to this one and this one.

Community
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Basj
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    Possible duplicate of $\pi$ in arbitrary metric spaces –  Dec 21 '16 at 08:34
  • @Rohan it's a totally different question. – Crostul Dec 21 '16 at 08:36
  • @Crostul See Fixed Point's answer and also read the last part of the question. –  Dec 21 '16 at 08:37
  • @Rohan i don't mean a circle built with the new norm, like what is suggested on your linked question, so it's not a duplicate. – Basj Dec 21 '16 at 08:42
  • @Rohan can you revert the duplicate notice, as it's not. Once again I don't mean the circle built with the new norm but I want to measure the ratio $circumference / diameter$ (computed with the new norm) of the circle built with the traditional norm $x^2+y^2=r$. – Basj Dec 21 '16 at 08:58
  • @Basj: in the "traditional" norm, $x^2+y^2=r\color{red}{^2}$. –  Dec 21 '16 at 09:07
  • You're right @YvesDaoust, thanks! – Basj Dec 21 '16 at 09:25
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    The radius of the circle must be the same in all directions ? – Gribouillis Dec 21 '16 at 09:26
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    A norm is entirely defined by its unit ball. If the radius of the circle is 1 in all directions, the circle is the unit ball and the norm is euclidian. – Gribouillis Dec 21 '16 at 09:28
  • What Gribouillis says. If the metric comes from a norm, and if all the points $(x,y)$ such that $x^2+y^2=r^2$ have distance $r$ from the origin, then it follows that we have the usual metric and the circumference/diameter ratio is the usual one. Basically this is because every vector in $\Bbb{R}^2$ is a scalar multiple of a vector from the origin to a point on $x^2+y^2=r^2$. – Jyrki Lahtonen Dec 21 '16 at 11:07
  • In other words, with this formulation of the question there is even less wiggle room than what we had in the related questions. – Jyrki Lahtonen Dec 21 '16 at 11:08
  • Are you asking for a norm which measures the length of curves differently than the length of straight lines? – MNKY Dec 21 '16 at 11:49
  • @gribouillis It seams that we are speaking about two different things. I'm measuring the circumference / diameter ratio (with a specific non euclidean norm) of the euclidean unit circle $ x^2 + y^2 = 1$. – Basj Dec 21 '16 at 11:52
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    Then the diameter of the circle is not constant for all straight lines passing through the center. Or perhaps you mean the largest diameter (that's the definition of diameter in metric spaces) ? – Gribouillis Dec 21 '16 at 11:59
  • yes you're right, it's non constant, and I mean this definition of diameter $\max_{x^2+y^2=1} ||x-y||_N$ – Basj Dec 21 '16 at 12:07
  • @Basj: Do never ever again vandalize other people's answers! If there is something wrong with them, say it so in comments: it is their choice to edit their posts according to your observations or not. When you saw that several of your edit attempts got rejected, you should have stopped trying to perform those vanadalizing edits! – Alex M. Dec 22 '16 at 10:21
  • @AlexM.: only one edit was rejected, that's when I realized the little mistakes were still there. I didn't do anything in the goal of vandalizing, I wanted to add constructive notes. But you're right, yes I'll do another way next time. I agree. – Basj Dec 22 '16 at 10:41

2 Answers2

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It's hopeless to get anything greater than $\pi$ with your definition:

Let $N$ be a norm on $\mathbb{R}^2$. By the positive homogeneity of the norm, it's enough to address only the case of the Euclidean circle $C$ centered at $(0,0)$ of radius $1$.

  • On the one hand, the $N$-perimeter of $C$ is: $$P_N=\int_0^{2\pi}N(-\sin t,\cos t)\,\mathrm{d}t=\int_0^{2\pi}N(\cos t,\sin t)\,\mathrm{d}t.$$ For the last equality we only used the $2\pi$-periodicity of $t\mapsto N(\cos t,\sin t)$.
  • On the other hand, the $N$-diameter of $C$ is: $$d_N=2\max_{t\in[0,2\pi]}N(\cos t,\sin t).$$ To see this: the definition of the diameter is: $$d_N=\max_{u,v\in D}N(u-v),$$ where $D$ denotes the set $$D=\bigl\{(x,y)\in\mathbb{R}^2\bigm\vert x^2+y^2\leq1\bigr\}.$$ Note that it's a maximum since $D$ is closed and bounded with respect to the $\lVert\cdot\rVert_2$ (hence compact) and the norm $N$ is continuous with respect to the topology induced by $\lVert\cdot\rVert_2$ in virtue of the equivalence of all the norms on $\mathbb{R}^2$ (since it's a finite-dimensional vector space). Now, if the max defining $d_N$ is attained for $u,v\in D$ such that $u\neq-v$ (and of course $u\neq v$) then $\lVert u-v\rVert_2<2$ hence taking $u'=(u-v)/\lVert u-v\rVert_2$ and $v'=(v-u)/\lVert u-v\rVert_2$ yields $u',v'\in D$ and $$N(u'-v')=N\bigl(2(u-v)/\lVert u-v\rVert_2\bigr)=N(u-v)\frac2{\lVert u-v\rVert_2}>N(u-v).$$ Hence the maximum for $d_N$ is attained for opposite vectors. Hence $$d_N=2\max_{u\in D}N(u).$$ By a similar argument, it is easy to show that the max is attained for a vector $u$ on $C=\partial D$, hence the result.

Now we clearly have $$P_N\leq\pi d_N,$$ hence $$\pi_N=\frac{P_N}{d_N}\leq\pi.$$

The mistake in Yves Daoust's answer is in point 2 when he says by symmetry we can assume $x=y$. This is wrong for $p>2$: the vectors on $C$ that lie on the line $x=y$ will have a minimal norm, so the diameter is not attained there. It's attained on the axes, so that $r_p=1$ when $p\geq2$.

gniourf_gniourf
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  • In fact, once you have $\forall u\in \mathbb{R}^2,\ N(u)\le \frac{d_N}{2} |u|_2$, the $N$-length of any curve is smaller that $\frac{d_N}{2}$ times its euclidian length. – Gribouillis Dec 21 '16 at 13:59
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A few notes on @YvesDaoust's answer (my edit was not accepted, so I post the notes as an additional answer).

Let $\mathcal{C}$ be the euclidean unit circle in $\mathbb{R}^2$ defined by $x^2 + y^2 = 1$.

1) Using a $p$-norm $||\cdot||_p$, the circumference of $\mathcal{C}$ will be $$h_p=2\int_0^{\pi}\sqrt[p]{|\cos t |^p+|\sin t|^p}\,dt,$$ which grows to infinity for $p\to0$. But for it to be a norm, one must have $p\ge1$.

From this integral we have $$h_{10} \approx 5.68...$$ $$h_2=2 \pi \approx 6.28...$$ $$h_{1.5} \approx 6.74...$$ $$h_1=8$$ and can't achieve higher value than $h_1$ with a $p$-norm.

2) Let the diameter and radius be defined by $$d_p = \max\limits_{a, b \in \mathcal{C}} ||a-b||_{p}$$ $$r_p = \max\limits_{x^2+y^2=1} \{ (|x|^p + |y|^p)^{1/p} \}$$

When $1 \leq p \leq 2$, by symmetry, we can find that $$r_p=\sqrt[p]{\frac1{\sqrt2^p}+\frac1{\sqrt2^p}}=\frac{\sqrt[p]2}{\sqrt2}=2^{1/p - 1/2}.$$

As noted by user @gniourf_gniourf, this is no more true when $p \geq 2$.

When $p \geq 2$, $r_p = 1$.

If we evaluate $d_p$ in a similar fashion, we see that $d_p = 2\ r_p = 2^{1/p + 1/2}$ is always true for $p \geq 1$.

3) Then the circumference divided by the diameter

$$\pi_p = \frac{h_p}{2\ r_p}= \frac{h_p}{d_p}$$

satisfies (according to numerical integration)

$$\pi_1=\frac{8}{2^{3/2}} = 2\sqrt2 \approx 2.828... $$ $$\pi_{1.5} \approx 3.00... $$ $$\pi_{2} = \frac{2 \pi}{2} = \pi \approx 3.14... $$ $$\pi_{3} \approx 2.96968... $$ $$\pi_{10} \approx 2.84... $$

Deeper analysis is required to find the maximum value, which seems to be reached for $p = 2$.

This is in agreement with @gniourf_gniourf's answer.

Basj
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