# Infinity-norm and inner-product

• Feb 28th 2010, 12:03 AM
surjective
Infinity-norm and inner-product
Hey,

Could someone assist me in the following math-problem.

I want to show that the norm $\displaystyle \Vert \cdot \Vert_{\infty}$ does not come from an inner product.
Hint: Show that the norm $\displaystyle \Vert \cdot \Vert_{\infty}$ does not satisfy the paralellogram law, e.g., for

$\displaystyle f(x)= \left\lbrace \begin{matrix} x \hspace{0,4cm}, x\in[0,1] \\ 2-x \hspace{0,4cm}, x\in [1,2] \end{matrix} \right.$

$\displaystyle g(x)= \left\lbrace \begin{matrix} 1-x \hspace{0,4cm}, x\in[0,1] \\ x-1 \hspace{0,4cm}, x\in [1,2] \end{matrix} \right.$

The paralellogram law states that if $\displaystyle V$ is a vector space with an inner product $\displaystyle \langle \cdot, \cdot \rangle$ and associated norm $\displaystyle \Vert \cdot \Vert$, then:

$\displaystyle \Vert \textbf{v}+\textbf{w} \Vert^{2} + \Vert \textbf{v}-\textbf{w} \Vert^{2}= 2\left( \Vert \textbf{v} \Vert^{2} + \Vert \textbf{w} \Vert^{2} \right)$

• Feb 28th 2010, 01:05 AM
HallsofIvy
Have you tried to this yourself? What problem did you have? You are told exactly what to do. What are ||f|| and ||g||? What are ||f+ g|| and ||f- g||? What is $\displaystyle 2(||f||^2+ ||g||^2)$?

Surely you see that f+ g= 1 for all x in [0, 2] and that f- g= 2x-1 for x in [0, 1] and 3- 2x for x in [1, 2].
• Feb 28th 2010, 06:14 AM
surjective
Infinity norm and inner-product
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.
• Feb 28th 2010, 08:45 AM
Drexel28
Quote:

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.

You have proved that the parallelogram law need not apply for $\displaystyle \|\cdot\|_{\infty}$ but every norm satisfies this. Thus...
• Feb 28th 2010, 10:26 AM
putnam120
Quote:

Originally Posted by Drexel28
You have proved that the parallelogram law need not apply for $\displaystyle \|\cdot\|_{\infty}$ but every norm satisfies this. Thus...

I think you mean every inner product, not every norm. Since $\displaystyle \|\cdot\|_\infty$ is a norm.
• Feb 28th 2010, 01:54 PM
Drexel28
Quote:

Originally Posted by putnam120
I think you mean every inner product, not every norm. Since $\displaystyle \|\cdot\|_\infty$ is a norm.

Yes, thank you.