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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.

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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 $$ \lim_{\varepsilon\downarrow 0} \frac1\varepsilon (f(\alpha_j+\varepsilon)+f(\alpha_j-\varepsilon)-2f(\alpha_j)) =0 $$

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  • $\begingroup$ That is a good point. I think i found a fix and will update soon $\endgroup$ – supinf Nov 29 '18 at 11:15
  • $\begingroup$ @supinf To get a contradiction you can simply use the fact that : $\sum_{i = 1}^{\infty} \mid x_i + x \cdot y_i \mid$ is convex whereas $\sqrt{x^2+1}-\mid x_0 + x \cdot y_0 \mid$ is not convex ? $\endgroup$ – Thinking Nov 29 '18 at 14:05
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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.

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  • $\begingroup$ It is impossible by a convexity argument (see my comment under @supinf answer). Nevertheless this a nice approximation how did you get that idea ? $\endgroup$ – Thinking Nov 29 '18 at 15:06

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