Do these properties characterize Hilbert spaces?

Question

Suppose $X$ is a Banach space with the following property: For any $x\in X$ there exists a two dimensional subspace $E$ isometric with $l_2^2$ such that $x\in E$. Does this property characterize a (separable) Hilbert space?

What about the stronger property: For any $x$ and any $n\in\mathbb{N}$ there exists a $n$-dimensional subspace $E$ isometric with $l_2^n$ such that $x\in E$?

Answer 1

For any Banach space $X$ you can consider $X\oplus l^2$, with norm $||(x,y)||:=(||x||^2+||y||^2)^\frac{1}{2}$. Then for each $x\in X$, span$(x)\oplus l^2$ is isometric to $l^2$, so $X\oplus l^2$ is covered by isometric copies of $l^2$

Answer 2

Let $\ B\ $ bee an arbitrary Banach space. If each of its 3-dimensional linear space is isomorphic to $\ \ell_2^3\ $ than the inner product can be introduced there properly, and $\ B\ $ is Hilbert then.

I seem to remember that this works even for $\ \ell_2^2\ $ in place of $\ \ell_2^3.$

Otherwise, $\ B\ $ is not Hilbert, far from this (as the examples above show).

Answer 3

No. That property does not characterize Hilbert spaces. Let $H$ be an infinite dimensional separable Hilbert space. Let $K$ be the space of all bounded sequences $(x_n)_{n=0}^{\infty}$ where $x_n\in H$ for each $n$. Then $K$ is a normed vector space with norm $\|(x_n)_n\|=\sup\{\|x_n\|:n\geq 0\}$, and it is not too hard to see that $K$ is a Banach space.

Suppose now that $(x_n)_{n\in\omega}\in K$. Then let $\alpha_n=\|x_n\|$ whenever $n\geq 0$. Let $(x_{n,m})_{n\geq 0,m\geq 0}$ be a collection such $x_{n,0}=x_n$ for each $n$, for each $n$, the set $(x_{n,m})_{m\geq 0}$ is orthogonal, and where $\|x_{n,m}\|=\alpha_n$ for each $n.$

Let $\mathbf{x}_m=(x_{n,m})_{n\geq 0}$. Let $W$ be the subspace of $K$ spanned by $(\mathbf{x}_m)_{m\geq 0}$. Then whenever $\|\beta_1\mathbf{x}_1+\dots+\beta_r\mathbf{x}_r\|=|\beta_1|^2\cdot\|\mathbf{x}_1\|+\dots+|\beta_r\|^2\cdot\|\mathbf{x}_r\|^2$. In other words, the induced norm on $W$ is induced by an inner product.

On the other hand, $K$ is very far from being an inner product space since $K$ has a copy of $\ell^\infty$.