2
$\begingroup$

I have been given a couple of properties of inner product and a norm in these cambridge notes on metric and topological spaces, page 6.

And then lemma 3.5 (i) talks about cauchy-schwarz inequality. I have looked at their proof and they show that:

$<u,u>-\frac {<u,v>^2}{<v,v>} \geq 0$ and this should be equivalent to $||u||||v|| \geq |<u,v>|$. I fail to see it. Especially, given that they have not shown an explicit form of either the norm or an inner product, rather their properties.

$\endgroup$
  • $\begingroup$ The statement of the lemma starts by defining $\lVert u\rVert \stackrel{\rm def}{=} \langle u,u\rangle$ (Lemma 3.5, p.6). How does that not suffice? $\endgroup$ – Clement C. Dec 29 '16 at 17:30
  • $\begingroup$ $\langle u , u \rangle \langle v, v \rangle \geq \langle u , v \rangle ^2$ can"t you just take the square root of that? $\endgroup$ – Emil Dec 29 '16 at 17:33
1
$\begingroup$

You have noticed something very important:

they have not shown an explicit form of either the norm or an inner product, rather their properties.

This is a level of abstraction and allows flexible adaptation of the inner product. The results shown for such inner products are thus applicable to all mappings satisfying the inner product axioms.

By definition, $\|u\|^2 = \langle u,u \rangle$. Hence $$ \langle u,u \rangle - \frac{\langle u,v \rangle^2}{\langle v,v \rangle} \geq 0 \implies \langle u,u \rangle \geq \frac{\langle u,v \rangle^2}{\langle v,v \rangle} \implies \langle u,u \rangle \langle v,v\rangle \geq \langle u,v\rangle^2 \implies \|u\|^2 \|v\|^2 \geq \langle u,v\rangle^2 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.