Isometric embedding of $\ell ^ 1$ in $\ell ^\infty$ in finite dimensions

Question

It is well known that $\mathbb{R}^n$ with $\ell ^1$ norm can be embedded into $\mathbb{R}^k$ with $\ell^\infty$ norm for some $k\in \mathbb{N}.$

But I guess, this is not true in complex case that is $(\mathbb{C}^n, \ell ^1)$ can not be embedded into $(\mathbb{C}^k, \ell ^{\infty})$ for any $k\in \mathbb{N}.$ Can anyone give me some reference of this fact? It would be of great help.

Answer 1

No reference, but here is a proof that $(\mathbb{C}^2, \ell^1)$ does not admit a linear isometric embedding into $(\mathbb{C}^k,\ell^\infty)$ for any $k$. Indeed, such an embedding can be described as a map $$(z,w) \mapsto (\alpha_j z+\beta_j w)_{1\le j\le k}$$ where $|z|+|w| = \max_j|\alpha_j z+\beta_j w|$. Plugging $(1,0)$ and $(0,1)$ for $(z,w)$ shows that $$|\alpha_j|,|\beta_j|\le 1\quad \text{for all }\ j\tag{1}$$

Let $z=e^{it}$ and $w=1$; then $$\max_j|\alpha_j e^{it} + \beta_j| = |z|+|w|= 2 \quad \text{for all }\ t \tag{2}$$ But (1) implies that the function $|\alpha_j e^{it} + \beta_j|$ can attain the value $2$ at most once on the unit circle; hence, (2) can hold for at most $k$ values of $e^{it}$.

Answer 2

Let's show that $\mathbb{C}^2$ with $\ell^1$ norm cannot be embedded in any space $\mathbb{C}^k$ with $\ell^\infty$ norm.

Suppose otherwise, and let $T:(\mathbb{C}^2,\ell^1)\to(\mathbb{C}^k,\ell^\infty)$ be an embedding. We have $\Vert T(1,0)\Vert_\infty=\Vert(1,0)\Vert_1=1$, so by multiplying the coordinates of $T$ by suitable complex numbers of absolute value $1$, and changing the order of the coordinates (both operations being isometric in $(\mathbb{C}^\infty,\ell^\infty)$), we can assume that in fact $$T(1,0)=(\underbrace{1,1,\cdots,1}_{p\text{ times}},t_{p+1},t_{p+2},\ldots,t_k)$$ where $0\leq t_i<1$. Now choose $s_1,\ldots,s_k$ for which $$T(0,1)=(s_1,\ldots,s_k)$$ Then $$2=\Vert(1,1)\Vert_1=\Vert T(1,1)\Vert_\infty=\Vert(1+s_1,\ldots,1+s_p,t_{p+1}+s_{p+1},\ldots,t_k+s_k)\Vert_\infty$$ Now, since $0\leq t_i<1$ and $|s_i|\leq 1$, this can only happen if some of the $s_i$ are equal to $1$. Again changing the order of the coordinates, put these on the beggining. Then, making some abuse of notation, $T$ has the form $$T(x,y)=(x+y,\ldots,x+y,x+s_iy,t_jx+s_iy)$$ where the next-to-last coordinate indicates several coordinates of the form $x+s_iy$, and the last coordinate indicates several coordinates of the form $t_jx+s_iy$, and where all $|s_i|\leq 1$, $s_i\neq 1$, and $|t_j|<1$.

Let's find $y$ for which $\Vert T(1,y)\Vert_\infty\neq\Vert(1,y)\Vert_1$.The absolute value of the coordinates of the form $t_jx+s_iy$ is always $\leq |t_j|+|y|<1+|y|$, so we need to care only about the other coordinates.

Choose a non-real complex number $y$ of absolute value $1$ for which all $s_iy$ are non-real (this can be done since there are only finitely many $s_i$. Then both $|1+y|<1+|y|$ and $|1+s_iy|<1+|s_iy|\leq 1+|y|$. Therefore $\Vert T(1,y)\Vert_\infty<\Vert(1,y)\Vert_1$, a contradiction.