Analysis Problem of the Day 46

Today’s problem is a generalization of Problem 5 from the Fall 2001 Analysis Qual:

Problem 5: Let $l^p(n)$ be $\mathbb{R}^n$ endowed with the $l^p$ norm. Find, with proof, all $1 \leq p_1 < p_2 \leq \infty$ such that $l^{p_1}(n)$ is isometrically isomorphic to $l^{p_2}(n).$

Solution: We first consider the case $p'<\infty.$ Suppose $T: l^p(n) \to l^{p'}(n)$ is an isometric isomorphism, i.e. a topological isomorphism. Since duality is functorial, it follows that the adjoint $T^*: l^{q'}(n) \to l^q(n)$ is also an isometric isomorphism. In particular, if $A$ is the matrix of $T$ with respect to the standard basis, $A^T$ is the matrix of $T^*,$ one has $\|Ax\|_{p'} = \|x\|_p, \|A^T x\|_{q} = \|x\|_{q'}.$ Evaluating these expressions at the standard unit basis $e_1,...,e_n$ tells us that the $l^{p'}$ norms of the columns of $A$ and the $l^{q}$ norms of the columns of $A^T$ are all 1. Summing over the rows/columns respectively thus yields

$\sum_{i,j=1}^n |a_{ij}|^{p'} = \sum_{i,j=1}^n |a_{ij}|^{q}=n$

for $A=(a_{ij}).$ Note that all entries are bounded in magnitude in $1$ (by the bound on the rows/columns above), so if $p' \not = q,$ one obtains that $A$ is a generalized permutation matrix (i.e. there is exactly one $\pm 1$ in each row or column). But then, for $x=e_1+e_2, Ax = b_1 e_i + b_2 e_j,$ so

$2^{\frac{1}{p}}=\|x\|_p=\|Ax\|_{p'} = 2^{\frac{1}{p'}},$

so $p=p'.$

We now handle the case $p'=\infty.$ Note that a linear transformation sends straight lines to straight lines. Since the unit ball in $l^\infty(n)$ is a cube, and unit balls aren’t linear for $1<p<\infty,$ it follows that $l^\infty(n)$ and $l^p(n)$ are not isometrically isomorphic for $1<p<\infty.$ For $l^1(n),$ $n \geq 3,$ note that there are $2n$ extreme points of the unit ball in $l^1$ (namely, $\pm e_j$ ), and $2^n$ extreme points of the unit ball in $l^\infty$ (namely, $\sum_{i=1}^n \pm e_i$ ), and since linear isometries preserve extreme points, there is no isometric isomorphism. Finally, for $n=2,$ $l^1(2)$ is indeed isometrically isomorphic to $l^\infty(2)$ according to the map $Tv = \sqrt{2} Av,$ where $A$ is a counterclockwise rotation by $45^\circ.$