# Thread: Norm versus inner product

1. ## Norm versus inner product

Hey,

For any $\bf{v} \in V$ i would like to show the following:

$
\Vert \bf{v} \Vert = \sup \{ | \langle \bf{v},\bf{w} \rangle| \big| \bf{w}\in V, \Vert w\Vert=1\}
$

I have done the following: To show an equality one can show that the case is true for both $\geq$ and $\leq$. For the $\geq$-case I have used the Cauchy-Schwarz inequality:

$|\langle \bf{v},\bf{w} \rangle| \leq \Vert \bf{v} \Vert \Vert \bf{w} \Vert$

Now since $\Vert \bf{w}\Vert=1$ we have that:

$|\langle \bf{v},\bf{w} \rangle| \leq \Vert \bf{v} \Vert$

Taking the $\sup$ on both sides of the inequality yields:

$\sup|\langle \bf{v},\bf{w} \rangle| \leq \sup\Vert \bf{v} \Vert$

If I could somehow get rid of the $\sup$ on the right side I would be home free. I could be tempted to reason that $\sup\Vert\bf{v}\Vert=\Vert \bf{v}\Vert$, but somehow that does not sound correct since $\bf{v}\in \mathcal{H}$.

2. "||v||= sup ||v||" is trivial because ||v|| is a single number amd the "sup" of a single number is that number. Also, although you have been talking about vectors in V you say " $\bf{v}\in \mathcal{H}$. Where did $\mathcal{H}$ come from?

I am concerned about the whole thing. Since, in an inner product space, $\bf{||v||}$ is defined as $\sqrt{}$, shouldn't there be square roots throughout that?

3. Originally Posted by surjective
Hey,

For any $\bf{v} \in V$ i would like to show the following:

$
\Vert \bf{v} \Vert = \sup \{ | \langle \bf{v},\bf{w} \rangle| \big| \bf{w}\in V, \Vert w\Vert=1\}
$

I have done the following: To show an equality one can show that the case is true for both $\geq$ and $\leq$. For the $\geq$-case I have used the Cauchy-Schwarz inequality:

$|\langle \bf{v},\bf{w} \rangle| \leq \Vert \bf{v} \Vert \Vert \bf{w} \Vert$

Now since $\Vert \bf{w}\Vert=1$ we have that:

$|\langle \bf{v},\bf{w} \rangle| \leq \Vert \bf{v} \Vert$

Taking the $\sup$ on both sides of the inequality yields:

$\sup|\langle \bf{v},\bf{w} \rangle| \leq \sup\Vert \bf{v} \Vert$ That should say $\color{red}\sup|\langle \bf{v},\bf{w} \rangle| \leq \| \bf{v} \|$ – see HallsofIvy's comment.
What you should now do is to take $\bf{w} = \frac{\bf{v}}{\|\bf{v}\|}$. Then $\|\bf{w}\| = 1$ and $\langle \bf{v},\bf{w}\rangle = \frac{\|\bf{v}\|^2}{\|\bf{v}\|} = \|\bf{v}\|$.

(Of course, you can't divide by 0 so that argument won't deal with the case when $\bf{v}=\bf{0}$. So you have to deal with that case separately.)