If every two-dimensional (vector) subspace of a normed space is an inner product space, then so is that normed space

Question

Let $\big( X, \lVert \cdot \rVert \big)$ be a (real or complex) normed space. Suppose that, for every two-dimensional (vector) subspace $Y$ of $X$, the norm on $Y$ (i.e. restriction of the norm of $X$ to $Y$) satisfies the parallelogram identity and thus is induced by a certain inner product on $Y$. Can we conclude from this that the norm of $X$ is also induced from an inner product on $X$?

That is, if every two-dimensional (vector) subspace of a normed space is an inner product space, then can we prove that that normed space is itself an inner product space?

Answer 1

Since $|a+b|^2+|a-b|^2=2(|a|^2+|b|^2)$ is trivial when $a\parallel b$, and otherwise $a,\,b$ span a $2$-dimensional subspace of $X$, the identity is true in general, and the inner product on $X$ is then defined by a suitable polarization identity.