# Proof of Cauchy-Schwarz inequality from Terry Tao's notes, meaning of “cancelling the phase”

I was reading Tao's proof of the Cauchy-Schwarz inequality by exploiting certain inherent symmetries and making some transformations. He says that first we use the fact that the norm is positive, i.e.

$$\|u-v\|^2>0$$ to conclude that $\operatorname{Re} (\langle u,v\rangle) \leq \frac{1}{2}\|u\|^2+\frac{1}{2}\|v\|^2$, and then he claims that we make a transformation $$v \mapsto ve^{\iota \theta}$$ and the RHS is clearly preserved under the transformation whereas the LHS changes. Now he says that we choose a $\theta$ to make the LHS as high as possible such that it cancels the phase of $\langle u,v\rangle$.

The rest of the proof is clear. But I don't get what he means by cancelling the phase. I mean the transformation is such that $\operatorname{Re} (\langle u,v\rangle)e^{\iota \theta} \mapsto |\langle u,v\rangle|$ for the theta which cancels the phase and for this particular theta we are kind of recovering the imaginary part of the complex scalar and then take its length. But how does it work? I don't understand it.

Cancelling the phase means to choose $\theta$ so that $e^{i\theta}\langle u,v\rangle$ is real. A priori all we know about $\langle u,v\rangle$ is that it is some complex number $\langle u,v\rangle = re^{i\varphi}$, so choosing $\theta = -\varphi$ (i.e. "cancelling the phase" on this complex number) will make $e^{i\theta}\langle u,v\rangle$ a (nonnegative) real number. This optimizes the inequality because $|\text{Re}(e^{i\theta}\langle u,v\rangle)| \leq |\langle u,v\rangle|$, and equality holds precisely when $e^{i\theta}\langle u,v\rangle$ is real.
• Thanks . The explanation was super clear . One further question how does the mapping work on the RHS. I thought the map was $v \mapsto ve^{\iota \theta}$ and $u \mapsto ue^{\iota \theta}$ but on the LHS the map is something like $<u,v> \mapsto e^{\iota \theta}<u,v>$. I mean if the map was what I think it is then $<u,v> \mapsto <ue^{ \iota \theta} ,ve^{ \iota \theta}>$ which would yield $<u,v>$ So probably I was under the illusion that I understood it. Do you get my question? Thanks – user3503589 May 15 '15 at 23:12
• The map is only $v\mapsto ve^{i\theta}$. If you apply this rotation to both $u$ and $v$, then you preserve $\langle u,v\rangle$, which is not what you want. Professor Tao's point is that in the inequality $|\text{Re}\langle u,v\rangle| \leq \frac{1}{2}(\|u\|^2 + \|v\|^2)$, there is an imbalance of rotation symmetry, and consequently there is a little wiggle room on the non-rotation-invariant LHS. Therefore the idea is to find a rotation that does not preserve the LHS, and $v\mapsto ve^{i\theta}$ is enough. – Gyu Eun Lee May 17 '15 at 5:09