# Hilbert-irreducible Banach space

A Banach space $X$ is called Hilbert-irreducible if it satisfies the following condition:

If a subspace $Y\subset X$ satisfies the parallelogram equality, then $Y$ is necessarilly a one dimensional space.

Does $M_{n}(\mathbb{R})$ with operator norm satisfy this property? What is an example of an infinite dimensional Banach space with this property?

• Note that $M_n \cong \ell_2^n \otimes_\varepsilon \ell_2^n$ isometrically. Oct 30, 2016 at 19:48

In other words, a (real) Banach space $X$ is Hilbert irreducible iff it has no $2$-dimensional subspace isometric to $\mathbb R^2$ with the Euclidean norm.
In $M_n(\mathbb R)$, the subspace $Y$ consisting of matrices whose entries below the first row are $0$ satisfies the parallelogram law.
The space $\mathbb c$ of real sequences converging to $0$ with supremum norm is Hilbert irreducible. To prove this, consider two linearly independent members $x$ and $y$ of $c$. It is easy to show that there is $\epsilon > 0$ such that $\|x + t y\|$ is an affine function of $t$ for $0 < t < \epsilon$. On the other hand, in $\mathbb R^2$ with Euclidean norm $\|(1,t)\|$ is strictly convex.
• Is there a finite dimensional banach space which can not be embedded in $M_{n}(\mathbb{R})$ isometrically? Nov 3, 2016 at 16:16
• The norm in $M_n(\mathbb R)$ is an algebraic function: $\|A\|^2 = \|A^T A\|$ is an eigenvalue of $A^T A$, and thus satisfies the characteristic polynomial of $A^T A$. So take any finite-dimensional Banach space whose norm is a transcendental function of the coordinates in some basis. Nov 4, 2016 at 15:00
• In $\mathbb R^2$, for $|y| \le |x|$, $\|(x,y)\| = |x| \sqrt{1 + 4 \sin^2(x/(2y))}$; for $|x| \le |y|$, $|y| \sqrt{1 + 4 \sin^2(y/(2x))}$. Nov 4, 2016 at 16:58