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

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?

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.