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?

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

1 Answer 1


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.

  • $\begingroup$ Is there a finite dimensional banach space which can not be embedded in $M_{n}(\mathbb{R})$ isometrically? $\endgroup$ Nov 3, 2016 at 16:16
  • $\begingroup$ 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. $\endgroup$ Nov 4, 2016 at 15:00
  • $\begingroup$ Thank you for your comment. What is an explicit example of such transcendental norm? $\endgroup$ Nov 4, 2016 at 15:11
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    $\begingroup$ In $\mathbb R^2$, for $|y| \le |x|$, $\|(x,y)\| = |x| \sqrt{1 + 4 \sin^2(x/(2y))}$; for $|x| \le |y|$, $|y| \sqrt{1 + 4 \sin^2(y/(2x))}$. $\endgroup$ Nov 4, 2016 at 16:58

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