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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?

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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.

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  • $\begingroup$ thanks very much! $\endgroup$ Aug 6, 2020 at 11:58

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