Originally Posted by

**surjective** Hey,

Yes indeed I have tried myself. The following is what I have so far:

By definition I know that:

$\displaystyle \Vert h \Vert_{\infty}=\displaystyle\smash{\max_{x \in [a,b]}}\vert h(x) \vert$

It is clear that:

$\displaystyle \displaystyle\smash{\max_{x \in [0,2]}}\vert f(x)+g(x)\vert=1$

$\displaystyle \displaystyle\smash{\max_{x \in [0,1]}}\vert f(x)-g(x)\vert=\displaystyle\smash{\max_{x \in [0,1]}}\vert 2x-1\vert=1$

$\displaystyle \displaystyle\smash{\max_{x \in [1,2]}}\vert f(x)-g(x)\vert=\displaystyle\smash{\max_{x \in [1,2]}}\vert -2x+3\vert=1$

$\displaystyle \displaystyle\smash{\max_{x \in [0,2]}}\vert f(x)\vert=1$

$\displaystyle \displaystyle\smash{\max_{x \in [0,2]}}\vert g(x)\vert=1$

Applying the definition in the beginning we have:

$\displaystyle \Vert f+g \Vert_{\infty}^{2}+\Vert f-g \Vert_{\infty}^{2}=2\left( \Vert f \Vert_{\infty}^{2}+\Vert g \Vert_{\infty}^{2} \right) \Leftrightarrow $

$\displaystyle 1+1=2(1+1) \Leftrightarrow$

$\displaystyle 2=4$

We conclude that the norm $\displaystyle \Vert \cdot \Vert_{\infty}$ does not satisfy the parallelogram law. My question is : "How does this show that the infinity-norm does not come from an inner-product? Is is simply because the above shows that the infinity-norm does not fulfill the parallelogram law which is one of the requirements for a vectorspace with an inner-product and associated norm?"

Thanks.