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I have recently started learning about Norms and Inner Products. I have came across the idea that for an inner product to introduce a norm the parallelogram Identity must be true.

The proof my books gives is just that it is apparently easy to see that the parallelogram identity fails when the norm, p is not = 2. Also that you can easily find a counterexample in R^2

I am quite confused here any help would be great.

Thanks

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In $\mathbb R$, the parallelogram identity will always hold for any $p$-norm, as I am about to show:

Let $p \geq 1$, and $\Vert\cdot\Vert_p$ denote the $p$-norm. Let $x,y \in \mathbb R$. Then $$2\Vert x\Vert_p^2 + 2\Vert y\Vert_p^2 = 2x^2 + 2y^2 = (x+y)^2 + (x-y)^2 = \Vert x+y\Vert_p^2 + \Vert x-y\Vert_p^2.$$ However, in general, the parallelogram identity will fail for $p \neq 2$. Let $n > 1$. Then $\mathbb R^2$ sits in $\mathbb R^n$. Therefore, by finding a counter-example in $\mathbb R^2$, we prove that the identity fails for $\mathbb R^n$.

Let $x = (1,0), y = (0,1)$. Let $p > 1, p \neq 2$. Then $$2\Vert x\Vert_p^2 + 2\Vert y\Vert_p^2 = 2(0^p+1^p)^{2/p} + 2(1^p + 0^p)^{2/p} = 4.$$ However, $$\Vert x+ y\Vert_p^2 + \Vert x - y\Vert_p^2 = (1^p+1^p)^{2/p} + (1^p + 1^p)^{2/p} = 2\times4^{1/p}.$$ Clearly $2\times4^{1/p} = 4$ only if $1/p = 1/2$. But this is not the case, so we have found a counterexample to the parallelogram identity for $p \neq 2$.

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