Tao's proof on the Cauchy-Schwarz inequality

Question

The Cauchy-Schwarz inequality says that

$$|\langle v, w \rangle| \leq \|v\| \|w\|\tag{1}$$ for vectors $v, w$ in a complex Hilbert space.

By exploiting the obvious inequality

$$\|v-w\|^2 \geq 0\tag{2}$$

and expanding everything out, one gets the weaker inequality $$ \hbox{Re} \langle v, w \rangle \leq \frac{1}{2} \|v\|^2 + \frac{1}{2} \|w\|^2. \tag{3} $$ Now (3) is weaker than (1). Firstly, observe that the phase rotation symmetry $v \mapsto e^{i\theta} v$ preserves the RHS of (3) but not the LHS. We exploit this by replacing $v$ by $e^{i\theta} v$ in (3) for some phase $\theta$ to be chosen later, to obtain $$ \hbox{Re} e^{i\theta} \langle v, w \rangle \leq \frac{1}{2} \|v\|^2 + \frac{1}{2} \|w\|^2. $$ By choosing $e^{i\theta}$ to cancel the phase of $\langle v, w \rangle$, we obtain $$ |\langle v, w \rangle| \leq \frac{1}{2} \|v\|^2 + \frac{1}{2} \|w\|^2 \tag{4} $$ This is closer to (1); we have fixed the left-hand side, but the right-hand side is still too weak. But we can amplify further, by exploiting an imbalance in a different symmetry, namely the homogenisation symmetry $(v,w) \mapsto (\lambda v, \frac{1}{\lambda} w)$ for a scalar $\lambda > 0$, which preserves the left-hand side but not the right. Inserting this transform into (4) we conclude that $$ |\langle v, w \rangle| \leq \frac{\lambda^2}{2} \|v\|^2 + \frac{1}{2\lambda^2} \|w\|^2 $$ where $\lambda > 0$ is at our disposal to choose. We can optimise in $\lambda$ by minimising the right-hand side, and indeed one easily sees that the minimum (or infimum, if one of $v$ and $w$ vanishes) is $\|v\| \|w\|$ (which is achieved when $\lambda = \sqrt{\|w\|/\|v\|}$ when $v,w$ are non-zero, or in an asymptotic limit $\lambda \to 0$ or $\lambda \to \infty$ in the degenerate cases), and so we have amplified our way to the Cauchy-Schwarz inequality (1).

The above is a proof of the Cauchy-Schwarz inequality from one of Terry Tao's expository articles.

In Folland's Real Analysis, a similar construction is given, which also proves the condition when the equality in (1) is true:

Here is my question:

following Tao's argument, can one prove that the equality in (1) is true only when $v,w$ are linearly dependent?

Answer 1

It is not entirely clear what you mean by 'following Tao's argument'.

Suppose $|\langle v,w\rangle| = \|v\| \|w\|$.

Let $w=\alpha v +h$, with $h \bot v$. Then $\|w\|^2 = |\alpha|^2 \|v\|^2 + \|h\|^2$ and $|\langle v,w\rangle| = |\alpha| \|v\|^2$.

Then the first squared line gives: $|\alpha|^2 \|v\|^4 = \|v\|^2 (|\alpha|^2 \|v\|^2 + \|h\|^2)$.

Hence either $v=0$ and we have $v = 0 \cdot w$, or $h=0$ in which case $w = \alpha v$.

Alternative approach:

If $w=0$ or $v=0$ the result is true so suppose both are not zero.

Suppose $|\langle v,w\rangle| = \|v\| \|w\|$. Replacing $v$ by $\theta v$, with $|\theta| = 1$ does not change the formula, so we can assume that $\langle v,w\rangle = \|v\| \|w\|$.

Note that replacing $(v,w)$ by $(tv, {1 \over t} w)$, with $t>0$ does not change the formula.

Now note that $(t\|v\|-{1 \over t}\|w\|)^2 = \|tv-{1 \over t}w\|^2$.

Now choose $t=\sqrt{\|w\| \over \|v\|}$ to get $w = {\|w\| \over \|v\|} v$.

Answer 2

I may not ask the queation in a clear good way. I was looking for an argument similar to Tao's one, which is conceptually easy and natural to remember. Here is one that I find natural.

Suppose $|\langle x,y\rangle|=\|x\|\|y\|\neq 0$. It suffices to show that there exists $t\in\mathbb{C}$ such that $$ \|x-ty\|^2=0.\tag{1} $$ Expanding everything out, one has $$ \|x-ty\|^2=\|x\|^2+|t|^2\|y\|^2-2\hbox{Re}(t\langle x,y\rangle)\tag{2} $$ Our goal is to choose $t$ so that the right hand side is $0$. As a complex number, we can write $t=re^{-\theta}$ where $r=|t|$.

Let $\theta=-\hbox{sgn}(\langle x,y\rangle)$ where $\hbox{sgn}(z):=z/|z|$ for $z\neq 0$. Then (2) becomes $$ \|x-ty\|^2=\|x\|^2+r^2\|y\|^2-2r\cdot|\langle x,y\rangle|=\|x\|^2+r^2\|y\|^2-2r\cdot\|x\|\|y\|.\tag{3} $$ In other word we can choose the phase angle of $t$ to "cancel" out that of $\langle x,y\rangle$.

Now, choosing $r=\|x\|/\|y\|$, we are done: $$ \|x\|^2+r^2\|y\|^2-2r\cdot\|x\|\|y\|=0. $$

Answer 3

Our papers below offer some new, elementary approaches to Schwarz inequalities:

1. Z. Boros and A. Szaz, Generalized Schwarz inequalities for .... Novi Sad J. Math. 47(2017), 177--188.

2. Z. Boros and A. Szaz, A weak Schwarz inequality for .... Rostock Math. Koll. 71(2016), 28--40.