Let $\|\cdot\|$ be a norm on $\mathbb R^2$. Show that the perimeter of the closed unit ball is between $4$ and $12$.

Let $u$ be a point on the boundary $\partial B$ of the unit ball $B$ of a norm on $\mathbb R^2$. Then $\partial B$ and the boundary $u+\partial B$ of the unit ball with center $u$ intersects (for instance by the Intermediate Value Theorem). Let $v$ be a point in the intersection. Then $u$, $v$, $v-u$, $-u$, $-v$, $u-v$ are 6 points in this order on the boundary, with successive points at distance $1$ apart. They subdivide $\partial B$ into 6 arcs, each of length (in the norm) of at least $1$. This shows that the length of $\partial B$ is at least $6$.

Next, among all $u,v\in\partial B$, choose two for which the area $\det(u,v)$ of the parallelogram with vertex set $\{o, u, v, u+v\}$ is maximized. Then the line through $u$ parallel to $v$ supports $B$ at $u$ (otherwise $\det(u,v)$ could be increased) and the line through $v$ parallel to $u$ supports $B$ at $v$. It follows that the parallelogram $P$ with vertices $\pm u\pm v$ contains $B$ and subdivides $B$ into $4$ parts. One of the parts is the arc of $\partial B$ from $u$ to $v$, which is contained in the convex hull of $o, u, u+v, v$. It can be shown (using the triangle inequality) that the length of the arc of $\partial B$ from $u$ to $v$ is at most the length of the corresponding arc of $\partial P$, which consists of the two segments $[u,u+v]$ and $[u+v,v]$, and thus has length $2$. It follows that the length of $\partial B$ is at most $8$.

Remarks

1. The above two bounds were first proved by Stanislaw Golab in 1932. It can also be found in Chapter 4 of Thompson's book Minkowski Geometry.
2. The result can also be shown by choosing $P$ to be a parallelogram of minimum area containing $B$.
3. There is, up to isometry, only one norm where the length of the unit ball is $6$, namely when $B$ is an affine regular hexagon.
4. There is, up to isometry, only one norm where the length of the unit ball is $8$, namely when $B$ is a parallelogram.
5. See my survey with Martini and Weiß for an elementary proof from the triangle inequality that if a convex arc $\gamma$ from $a$ to $b$ is contained in the triangle $\triangle abc$, then the length of $\gamma$ is at most $\|a-c\|+\|b-c\|$. Thompson proves it indirectly using an area argument.

References

S. Golab, Some metric problems in the geometry of Minkowski (Polish. French summary), Prace Akademii Górniczej w Krakowie 6 (1932), 1-79.

H. Martini, K. J. Swanepoel, and G. Weiß, The geometry of Minkowski spaces — a survey. Part I, Expositiones Mathematicae 19 (2001), 97-142.

A. C. Thompson, Minkowski Geometry, Cambridge University Press, Cambridge, 1996.

w.l.g. normalize the norm such that $||(1,0)|| = 1$, and take $a,b>0$ such that $||(1/2,a)|| = 1$ and $||(0,b)||=1$, then we can consider the following paralelograms (with sides paralels to the axes):

Polygon $P_1$: such that its corners are $\{(1/2,a), (-1/2,a), (-1/2,-a), (1/2,-a)\}$.

Polygon $P_2$: such that its corners are $\{(1,b), (-1,b), (-1,-b), (1,-b)\}$.

Then notice that the ball of radius one $\mathbb{B} \subset \mathbb{R}^2$ satisfies $$ P_1 \subset \mathbb{B} \subset P_2$$

To prove this, just use the convexity of the sets.

Notice that since $||(1/2,a)|| = 1$ and $||(1/2,0)|| = 1/2$, employing the triangular inequality we can show that $1/2 \leq ||(0,a)||$. Therefore the perimeter of $P_1$, that is composed of 4 strokes, can be estimated as: $$\text{Perim} (P_1) = 4 ||(1/2,0)|| + 4||(0,a)|| \geq 4\cdot 1/2 + 4\cdot 1/2 = 4$$

Similarly with $P_2$, its perimeter can be computed as $$\text{Perim}(P_2) = 4||(1,0)|| + 4||(0,b)|| = 8$$.

So we proved that $$ 4 \leq \text{Perim}(\mathbb{B}) \leq 8 \leq 12 $$