2
$\begingroup$

I'm trying to work through Exercise 3 from this blog post, which is essentially a proof of the validity of the $l^2$ norm:

Exercise 3: Let $(\mathcal{V},\left<\cdot,\cdot\right>)$ be an inner product space. Show, that $||x|| = \sqrt{\left<x,x\right>}\ \forall x \in \mathcal{V}$ is a normed vector space.

So far my working is as follows (I apologise for the hand-written working, I can convert it to MathJax if required by the site standards);

Hand working out

Which is where I get stuck. Can anyone advise how to finish showing the triangle inequality holds for the $l^2$ norm? I think I may need to use the Schwartz inequality, but I'm not sure how to apply it in the context.

$\endgroup$
  • 2
    $\begingroup$ Note that the LHS in your last step is $$ \langle x,y \rangle + \langle y,x \rangle = \langle x,y \rangle + \overline{\langle x,y \rangle} = 2 \Re \langle x,y \rangle, $$ and conclude using the Schwartz inequality. (Recall also that $\Re z \leq |z|$ for any complex number.) $\endgroup$ – MisterRiemann Feb 24 at 12:20
  • $\begingroup$ Aha! I was closer than I realised! Thanks @MisterRiemann! $\endgroup$ – aaronsnoswell Feb 25 at 6:20
2
$\begingroup$

Any inner product satisfies the C-S inequality $ |\langle x, y \rangle| \leq \|x\| \|y\|$ from which you can complete your argument immediately.

$\endgroup$

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.