The comparison constants between $\parallel \cdot {\parallel }_1$, $\parallel \cdot {\parallel }_2$ and $\parallel \cdot {\parallel }_{\mathrm{\infty }}$ on ${\mathbb{R}}^2$, and vectors attaining each

On ${\mathbb{R}}^2$ with the norms of The $p$-norms $\parallel x{\parallel }_p$ for rational $p\ge 1$, and $\parallel x{\parallel }_{\mathrm{\infty }}$, the comparison chain of The finite and reverse triangle inequalities for a norm; and for $n\ge 1$ every norm $N$ on ${\mathbb{R}}^n$ satisfies $N(x)\le C\parallel x{\parallel }_1$ and is Lipschitz, hence continuous, for $d_2$ clause 3 reads

$\parallel x{\parallel }_{\mathrm{\infty }}\le \parallel x{\parallel }_2\le \parallel x{\parallel }_1\le \iota (2)\parallel x{\parallel }_{\mathrm{\infty }},\parallel x{\parallel }_1\le \sqrt{\iota (2)}\parallel x{\parallel }_2\mathrm{.}$

Each of these four constants is attained, so none can be improved:

• $e_0=(1,0)$ has $\parallel e_0{\parallel }_{\mathrm{\infty }}=\parallel e_0{\parallel }_2=\parallel e_0{\parallel }_1=1$, so the first and second inequalities are equalities there;
• $(1,1)$ has $\parallel (1,1){\parallel }_{\mathrm{\infty }}=1$, $\parallel (1,1){\parallel }_2=\sqrt{\iota (2)}$ and $\parallel (1,1){\parallel }_1=\iota (2)$, so the third and fourth inequalities are equalities there.

The general theorem For $n\ge 1$ all norms on ${\mathbb{R}}^n$ are equivalent supplies constants but no attaining vectors; that is what this computation adds.

Unit balls. Writing $B_p:=\{x\in {\mathbb{R}}^2:\parallel x{\parallel }_p\le 1\}$ for $p\in \{1,2,\mathrm{\infty }\}$, the chain gives $B_1\subseteq B_2\subseteq B_{\mathrm{\infty }}$, and both inclusions are strict: $(1,1)$ lies in $B_{\mathrm{\infty }}$ and not in $B_2$, and $(3\mathrm{/}\iota (5))(1,1)$ lies in $B_2$ and not in $B_1$. The scalar has to be chosen strictly between $1\mathrm{/}\iota (2)$ and $1\mathrm{/}\sqrt{\iota (2)}$: at the endpoint $1\mathrm{/}\iota (2)$ the vector $(1,1)\mathrm{/}\iota (2)$ has $\parallel \cdot {\parallel }_1=1$ and so still lies in $B_1$.

• Sharpness is not the same as equivalence. For $n\ge 1$ all norms on ${\mathbb{R}}^n$ are equivalent asserts that constants exist and produces some; nothing in it says which are smallest. The computation above supplies attaining vectors, and those are what make the constants of the chain best possible on ${\mathbb{R}}^2$.

• Both attaining vectors are extreme in the expected way. A vector with a single nonzero coordinate makes all three norms agree; a vector whose two coordinates have equal absolute value spreads the mass as evenly as possible and is where $\parallel \cdot {\parallel }_1$ is largest relative to the other two. On ${\mathbb{R}}^n$ the same two vectors give equality with $\iota (n)$ and $\sqrt{\iota (n)}$ in place of $\iota (2)$ and $\sqrt{\iota (2)}$; only the case $n=2$ is verified here.

• The strictness computation in step 2.3 is arithmetic, not geometry. The scalar $3\mathrm{/}\iota (5)$ was chosen to lie strictly between $1\mathrm{/}\iota (2)$ and $1\mathrm{/}\sqrt{\iota (2)}$; any scalar in that open interval would serve, and the interval is nonempty exactly because $\sqrt{\iota (2)}<\iota (2)$.