# Norms and Parallelogram Identity

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

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 $$p \neq 2$$.