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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.

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    $\begingroup$ 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.) $\endgroup$
    – MSDG
    Feb 24, 2019 at 12:20
  • $\begingroup$ Aha! I was closer than I realised! Thanks @MisterRiemann! $\endgroup$ Feb 25, 2019 at 6:20

1 Answer 1

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

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