linear isometric embedding from $(\mathbb{R}^2, \| \|_2)$ to $(l^1, \| \|_1)$ 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.
 A: 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 $\alpha\in\mathbb R$ 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 $y_j\neq0$.
Then we choose $N\in\mathbb N$ such that
$\sum_{i=N+1}^\infty |y_i| < |y_j|/2$.
Clearly, $N\geq j$ 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
$$
g(\alpha_j+\varepsilon)-g(\alpha_j) \geq \varepsilon |y_j|
\quad\text{and}\quad
g(\alpha_j-\varepsilon)-g(\alpha_j) \geq \varepsilon |y_j|
$$
for all $\varepsilon\in (0,\varepsilon_0)$.
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 $\varepsilon\in (0,\varepsilon_0)$.
This means that
$$
\liminf_{\varepsilon\downarrow 0}
\frac1\varepsilon (f(\alpha_j+\varepsilon)+f(\alpha_j-\varepsilon)-2f(\alpha_j))
\geq |y_j| > 0.
$$
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
$$
\lim_{\varepsilon\downarrow 0}
\frac1\varepsilon (f(\alpha_j+\varepsilon)+f(\alpha_j-\varepsilon)-2f(\alpha_j))
=0
$$
A: The following construction gives an approximate such imbedding. Choose an $N\gg1$, and put
$$f(1,0)=(x_k)_{0\leq k\leq N-1},\qquad f(0,1)=(y_k)_{0\leq k\leq N-1}$$
with
$$x_k:={\pi\over 2N}\cos{2\pi k\over N},\quad y_k:={\pi\over 2N}\sin{2\pi k\over N}\qquad(0\leq k\leq N-1)\ .$$
Then, by linearity,
$$\eqalign{\|f(\cos\phi,\sin\phi)\|&={\pi\over 2N}\sum_{k=0}^{N-1}\left|\cos\phi\cos{2k\pi\over N}+\sin\phi\sin{2k\pi\over N}\right| \cr  &=
{1\over 4}\sum_{k=0}^{N-1}{2\pi\over N}\left|\cos\left(\phi-{2\pi k\over N}\right)\right|\cr  &\approx{1\over4}\int_{-\pi}^\pi|\cos t|\>dt=1\ ,\cr}$$
independently of $\phi$.
I'm not so sure that an exact isometric imbedding is impossible.
