Csdrhrt

I would like to prove the following :

There isn't a linear isometric embedding from $(mathbb{R}^2, | cdot |_2)$ to $(l^1, | cdot |_1)$

I don't know how to prove this. So far I am able to prove this result only in the case where the vector $(1,0)$ and $(0,1)$ are sent to sequences that have all positive, or all negative value.

In this case I use the fact that the $2$ norm is not linear whereas the $1$ norm is (ie $| xa + yb | = xa + yb$, $x, y, a, b > 0$).

The problem is that when the sequences have different signs i's hard for me to conclude.

Thank you.

Let the image of $(1,0)$ be the sequence $x=(x_i)_i$ and
let the image of $(0,1)$ be the sequence $y=(y_i)_i$.
Now we define the function
For $alphainmathbb R$ we define
$$
f(alpha) = | x + alpha y| = sum_{i=1}^infty | x_i+alpha y_i |.
$$
Note that because of the isometry we should have
$$
f(alpha) = sqrt{1+alpha^2}.
$$
We cannot say that $f$ is linear, because as $alpha$ changes, there might be sign changes in some of the terms $|x_i+alpha y_i|$.

We pick a fixed $j$ such that $y_jneq0$.
Then we choose $Ninmathbb N$ such that
$sum_{i=N+1}^infty |y_i| < |y_j|/2$.
Clearly, $Ngeq j$ has to be true.
We define
$$
g(alpha):=sum_{i=1}^N | x_i+alpha y_i |
qquad
h(alpha):=sum_{i=N+1}^infty | x_i+alpha y_i |.
$$
Clearly, $f=g+h$.
It can be shown that $h(alpha)$ is Lipschitz continuous with the global Lipschitz constant $|y_j|/2$.

Now we will analyse $g$.
Since there are only finitely many terms, there are only finitely many points where $g$ is not smooth.
In between those points $g$ is affine linear.
If we consider the point $alpha_j:=-x_j/y_j$, then we can find $varepsilon_0>0$ such that
$g$ is affine linear on the intervals
$[alpha_j-varepsilon_0,alpha_j]$ and $[alpha_j,alpha_j+varepsilon_0]$.
Then we have
$$
g(alpha_j+varepsilon)-g(alpha_j) geq varepsilon |y_j|
quadtext{and}quad
g(alpha_j-varepsilon)-g(alpha_j) geq varepsilon |y_j|
$$
for all $varepsilonin (0,varepsilon_0)$.

Combining this with the Lipschitz constant for $h$, we can conclude
$$
f(alpha_j+varepsilon)+f(alpha_j-varepsilon)-2f(alpha_j) geq varepsilon | y_j|.
$$
for all $varepsilonin (0,varepsilon_0)$.
This means that
$$
liminf_{varepsilondownarrow 0}
frac1varepsilon (f(alpha_j+varepsilon)+f(alpha_j-varepsilon)-2f(alpha_j))
geq |y_j| > 0.
$$
holds.
However, this is a contradiction to $ f(alpha) = sqrt{1+alpha^2}$, because here the right-hand term is twice continuously differentiable, which (by using Taylor expansion) implies
$$
lim_{varepsilondownarrow 0}
frac1varepsilon (f(alpha_j+varepsilon)+f(alpha_j-varepsilon)-2f(alpha_j))
=0
$$

The following construction gives an approximate such imbedding. Choose an $Ngg1$, and put
$$f(1,0)=(x_k)_{0leq kleq N-1},qquad f(0,1)=(y_k)_{0leq kleq N-1}$$
with
$$x_k:={piover 2N}cos{2pi kover N},quad y_k:={piover 2N}sin{2pi kover N}qquad(0leq kleq N-1) .$$
Then, by linearity,
$$eqalign{|f(cosphi,sinphi)|&={piover 2N}sum_{k=0}^{N-1}left|cosphicos{2kpiover N}+sinphisin{2kpiover N}right| cr &=
{1over 4}sum_{k=0}^{N-1}{2piover N}left|cosleft(phi-{2pi kover N}right)right|cr &approx{1over4}int_{-pi}^pi|cos t|>dt=1 ,cr}$$
independently of $phi$.

I'm not so sure that an exact isometric imbedding is impossible.