# 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.

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

• That is a good point. I think i found a fix and will update soon – supinf Nov 29 '18 at 11:15
• @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 ? – Thinking Nov 29 '18 at 14:05

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.

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