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I am trying to show that the sup norm on $\mathbb{R}^2$ is not derived from the inner product on $\mathbb{R}^2$.

What I notice so far is that to suppose $\langle x,y\rangle$ is an inner product, not the dot product, and $|x| = \sqrt{\langle x,x\rangle}$. To compute $\langle x\pm y,x \pm y\rangle = \sqrt{(x \pm y)^2 + (x\pm y)^2}$. Also, what is special about saying $x=e_1$, and $y = e_2$?

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    Yes, but an inner product is a type o$f$ dot product2011-11-05

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Hint: If a norm comes from an inner product, then it satisfies the parallelogram identity.

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    @Buddy: Show the sup norm does *not* satisfy the parallelogram identity by selecting suitable (cleverly chosen) vectors and showing the identity is false. That will prove the sup norm does not derive from an inner product.2011-11-05