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If the norm on a real or complex normed vector space is defined by $\|x\| = \sqrt{\langle x,x\rangle}$ where $\langle\cdot,\cdot\rangle$ is an inner product, then the norm satisfies the parallelogram law $$ \|x+y\|^2 + \|x-y\|^2 = 2\|x\|^2 + 2\|y\|^2. $$ That's easy to show, and it's a bit more work, but not a lot more, to show that "only if " also holds. (That is done by showing how to define the inner product as a function of the norm, and then showing that that inner product gives you back the same norm.)

If I wanted to prove that the normed space $L^1$ is not an inner product space, I would find a counterexample to the parallelogram law in that space.

So my question is: Is there some reasonable way to write such a proof other than that or things that are in some sense trivially equivalent to showing that the parallelogram law fails?

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Norms that arise from a scalar product are called "Euclidean". A gazillion very different characterizations of Euclidean norms appeared in the literature since this question first began to be studied early in the 20th century. See my modest contribution to this subject in https://arxiv.org/abs/0910.0608 (later published in the AMM) and (more importantly) look into Dan Amir's compendious book on the subject referenced therein.

So, for example, Aronszajn's criterion for a norm to be Euclidean is that the length of two sides and one diagonal of a parallelogram should determine the length of the other diagonal. This is easy to refute in the taxi-cab norm on $\Bbb{R}^2$ without doing the calculations involved in refuting the parallelogram law.

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