I intend to demonstrate Cauchy-Schwarz inequality assuming that Minkovski inequality is true $(i)$. Instead of doing the usual proof I want to apply this second inequality somewhere in the resolution, so that it could get more intuitive.

I thought about this because Minkovski inequality can be easily proven using Cauchy-Schwarz inequality, i.e. Cauchy-Schwarz inequality implies Minkovski inequality $(ii)$. I want to see if this relation is reciprocal:

$$\text{Minkovski inequality} \Leftrightarrow \text{Cauchy-Schwarz inequality}$$

Resuming, I want to prove Cauchy-Schwarz inequality:

For a vectorial space $H$ on the complex space $\mathbb{C}$ with the dot product $<.,.>:H\times H\to \mathbb{R}$ where $f,g\in H$ we have that: $$|<f,g>|\leq \|f\|.\|g\|$$

using Minkovski inequality.

Note:As you can see here the proof (at least in my opinion) implies that you already know how to do it, i.e. the initial procedure isn't intuitive a priori. For example, if I had to demonstrate this on an exame I would need to memorize at least the first managements which is not what I want. So the main point of this question is to seek a more simple demonstration for Cauchy-Schwarz inequality as it worked with the Minkovski inequality demonstration by application of another well knowing theorem. If it not requires the application of Minkovski inequality but it can get more simple than the usual, then you're welcome to post your answer here!

  • $\begingroup$ So you just prove the triangle inequality from Cauchy-Schwarz inequality? $\endgroup$ – raul Oct 16 '15 at 23:44
  • $\begingroup$ There is no question here. $\endgroup$ – Thomas Andrews Oct 17 '15 at 0:04
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    $\begingroup$ The dot product takes negative values, too. $\endgroup$ – Thomas Andrews Oct 17 '15 at 0:06
  • $\begingroup$ The question is if is it possible to prove the converse implication. I agree that it is a bit loosely formulated.(@Elio: IMHO, you should state clearly the theorem that you want to prove. Also, the proof of Minkowski given Cauchy-Schwarz is rather immediate and well-known, so I don't think that reproducing it here adds much to the question). $\endgroup$ – Giuseppe Negro Oct 17 '15 at 0:07
  • $\begingroup$ I suspect you mean "intend" rather than "pretend." But you almost certainly don't mean "pretend." $\endgroup$ – Thomas Andrews Oct 17 '15 at 0:18

Assume that $V$ is a real vector space and that $\langle, \rangle$ is a scalar product on $V$ $^{[1]}$. Suppose you have already proved that $$\tag{1} \|a+b\|\le \|a\|+ \|b\|, \qquad \forall a,b\in V, $$ where $\|a\|^2=\langle a, a\rangle$. Then you can prove that $$\tag{2} \lvert \langle a, b\rangle \rvert \le \|a\|\|b\|, \qquad \forall a, b\in V$$ by using the following algebraic identity, that expresses $\langle,\rangle$ in terms of $\|\cdot\|^2$: $$ \tag{3} 2\langle a,b\rangle = \|a\|^2 + \|b\|^2 -\|a-b\|^2.$$ Indeed, take arbitrary vectors $a, b\in V$. Without loss of generality, suppose that $\langle a, b\rangle \le 0$. (If that is not the case, then substitute $a$ with $-a$. This does not affect the inequality (2) that we are going to prove). This assumption ensures that $$\|a-b\|^2 \ge \|a\|^2 +\|b\|^2, $$ hence, using (3), we have $$\tag{4} 2\lvert \langle a, b\rangle \rvert=\|a-b\|^2-\|a\|^2-\|b\|^2.$$ By the inequality (1), we have $$ \|a-b\|^2\le (\|a\|+\|b\|)^2=\|a\|^2+\|b\|^2 +2\|a\|\|b\|, $$ which inserted into (4) gives (2).

The identity (3) is sometimes known as polarization identity. This proof can probably be adapted to complex vector spaces too, up to a more careful treatment of the polarization identity, which is slightly more complicated in the complex case.

$^{[1]}$ It is enough if $\langle, \rangle$ is a positive semidefinite symmetric bilinear form.


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