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How many proofs of the Cauchy-Schwarz inequality are there? Is there some kind of reference that lists all of these proofs?

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4  
It's Cauchy-Schwarz (no t, please!). –  t.b. Feb 24 '11 at 12:58
10  
I would try the book "The Cauchy-Schwarz Masterclass". –  user3533 Feb 24 '11 at 13:03
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Your first question is essentially unanswerable, except, maybe, by "many"... –  Mariano Suárez-Alvarez Feb 24 '11 at 22:52
12  
There are exactly 42 proofs. No more. No less. –  Eric Naslund Feb 25 '11 at 15:58
3  
There HAS TO BE 42 proofs. Or the Universe will collapse. –  Did Jul 2 '12 at 9:58

3 Answers 3

Here is one:

Claim: $|\langle x,y \rangle| \leq \|x\|\|y\| $

Proof: If one of the two vectors is zero then both sides are zero so we may assume that both $x,y$ are non-zero. Let $t \in \mathbb C$. Then

$$ \begin{align} 0 \leq \|x + ty \|^2 &= \langle x + ty, x + ty\rangle \\ &= \langle x,x\rangle + \langle x,t y\rangle + \langle yt, x\rangle + \langle ty,ty\rangle \\ &= \langle x,x\rangle + \bar{t} \langle x,y\rangle + t \overline{\langle x,y\rangle} + |t|^2 \langle y,y\rangle \\ &= \langle x,x\rangle + 2 \Re(t \overline{\langle x,y\rangle}) + |t|^2 \langle y,y\rangle \end{align}$$

Now choose $t := -\frac{\langle x, y \rangle}{\langle y, y \rangle}$. Then we get $$ 0 \leq \langle x,x\rangle + 2 \Re(- \frac{|\langle x,y\rangle|^2}{\langle y, y \rangle}) + \frac{|\langle x,y\rangle|^2}{\langle y, y \rangle} = \langle x, x \rangle - \frac{|\langle x,y\rangle|^2}{\langle y, y \rangle}$$

And hence $|\langle x,y \rangle| \leq \|x\|\|y\| $.

Note that if $y = \lambda x$ for $\lambda \in \mathbb C$ then equality holds: $$ |\lambda|^2 |\langle x, x \rangle| = |\lambda|^2 \|x\|\|x\| $$

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I think that a proof of the Cauchy-Schwarz inequality should also include a discussion of the equality case (which is also straightforward from this argument). –  t.b. Jul 2 '12 at 9:55
    
@t.b. Like this? –  Matt N. Jul 2 '12 at 10:22
    
This is half of what I had in mind. More interesting is the fact that if equality $\lvert\langle x,y\rangle\rvert = \lVert x\rVert \lVert y \rVert$ holds then $y = \lambda x$ or $x = 0$. –  t.b. Jul 2 '12 at 10:26
    
In other words: equality holds if and only if $x$ and $y$ are linearly dependent. –  t.b. Jul 2 '12 at 10:33
    
@t.b. There's nothing I'd want to do less than disappoint you. Can we use $\|x\| \|y\| \cos \theta = \langle x,y \rangle$ to get the other direction? Or did you have something else in mind? –  Matt N. Jul 2 '12 at 10:37

Without loss of generality, assume $\|y\|=1$. Write $x=\left<x,y\right>y+z$. Then $z$ is orthogonal to $y$, because $$\left<x,y\right>=\left<(\left<x,y\right>y+z),y\right>=\left<x,y\right>\left<y,y\right>+\left<z,y\right>,$$ indeed yields $\left<z,y\right>=0$. Hence $$\|x\|^2=\left<x,x\right>=|\left<x,y\right>|^2+\left<z,z\right>\geq |\left<x,y\right>|^2,$$ with equality iff $z= 0$, i.e. $x\in\mathbb{F}y$.

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I like this proof for real vectors a lot. Recall that an inner product for real vectors has the following properties:

$\langle x,y\rangle=\langle y,x\rangle$

$\langle ax+y,z\rangle=a\langle x,z\rangle+\langle y,z\rangle$

$\langle x,x\rangle\geq0$

Then $0\leq\langle lx+y,lx+y\rangle=l^2\langle x,x\rangle+l\langle x,y\rangle+l\langle y,x\rangle+\langle y,y\rangle=l^2\langle x,x\rangle+2l\langle x,y\rangle+\langle y,y\rangle$

$Let\:a=\langle x,x\rangle, b=\langle x,y\rangle,c=\langle y,y\rangle$, then the equation becomes

$al^2+bl+c\geq0$

This is a quadratic equation in $l$ with at most 1 real root. Therefore

$b^2-4ac\leq 0$

$\implies4{\langle x,y\rangle}^2-4\langle x,x\rangle\langle y,y\rangle\leq 0$

$\implies{\langle x,y\rangle}^2\leq\langle x,x\rangle\langle y,y\rangle$

Not bad huh? Sadly it doesn't work out so nicely with complex vectors $:($

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