Let the norm of the vector space $\displaystyle C([0,1])$ be $\displaystyle ||f|| = sup \{ | f(x) | : x \in [0,1] \} $, prove that the parallelogram law fails here.

Proof.

Pick $\displaystyle f(x)=x $, and $\displaystyle g(x)=x^2$, then they are both functions of $\displaystyle C([0,1])$.

We have: $\displaystyle ||f|| = 1 \ , \ ||f||=1 \ , \ ||f+g|| = ||x+x^2 ||=2 \ , \ ||f-g||=||x-x^2||= \frac {1}{2} - \frac {1}{4} = \frac {1}{4}$

Then $\displaystyle 2 ||f(x)||^2 + 2 ||g(x)||^2 = 4 \neq ||f+g||^2 + ||f-g||^2 = 4 + \frac {1}{16} $

So the parallelogram law doesn't hold here and the norm does not come from any inner product.

Is this right? Thank you.