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I was looking at a number of different proofs of the Cauchy-Schwarz inequality in an inner product space ($\mathbb{R}^n$ or $\mathbb{C}^n$).

All of them used the idea of $\lVert x-sy\rVert$ where $s$ was selected in particular fashion which in the real case, $s$ would be the value that minimized the function $f(s) = \lVert x-sy\rVert$

The thing I am confused on is that the books say "$s$ was selected so that $\lVert x-sy\rVert$ would be minimized". I don't understand how beforehand minimizing $\lVert x-sy\rVert$ would be known to be relevant to the inequality.

What is the intuitive link between the minimum of $f(s) = \lVert x-sy\rVert$ and the Cauchy-Schwarz inequality?

Xander Henderson
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Mike
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3 Answers3

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Intuitively perhaps the Cauchy-Schwarz theorem wants to say that there exists an angle $\gamma$ for any pair $x,y$ of vectors, such that $\langle x,y\rangle =|x|\cdot|y|\cdot \cos\gamma$.

Let $e$ denote the line $(x+sy)_{s\in\mathbb R}$, then minimalizing the norm among them means exactly orthogonal projection of the origo to $e$, making there a nice triangle with a right angle and with $\gamma$ and $\cos\gamma$.

Berci
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You want to bound $x \cdot y$ in terms of $x \cdot x$ and $y \cdot y$. The only inequality you know about $\cdot$ is that $z \cdot z \ge 0$ for all $z$. So it is reasonable to look at $z \cdot z$ where $z$ is a linear combination of $x$ and $y$, say $z = r x + s y$. By homogeneity, we might as well take $r = 1$. Now expand: $$0 \le (x + s y) \cdot (x + s y) = x \cdot x + 2 s x \cdot y + s^2 y \cdot y$$ For this to give us an upper bound on $x \cdot y$, we need $s$ to be negative. For convenience, write $s = -t$. Thus for all $t > 0$, $$ x \cdot y \le \frac{x \cdot x + t^2 y \cdot y}{2 t}$$ Each positive number $t$ gives us an upper bound on $x \cdot y$. We want the best possible upper bound, so we minimize the right side.

Robert Israel
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The basic idea is that the map $s \mapsto \|x-sy\|$ (or, equivalently, its square) has a minimising value $\hat{s}$ corresponding to the closest point of the line $s \mapsto x-sy$ to the origin, and, of course, $\|x-\hat{s}y\| \ge 0$.

Let $s\in\mathbb{R}$, then expanding gives $\|x-sy\|^2=\|x\|^2+s^2\|y\|^2-2s\text{Re} \langle x,y\rangle$, which is a quadratic in $s$.

Note that the Cauchy-Schwarz inequality is trivially true if $y=0$, so assume $y\neq 0$. Then minimizing $\|x-sy\|^2$ with respect to $s$ gives $\hat{s}=\frac{\text{Re}\langle x,y\rangle}{\|y\|^2}$, and substituting this value into the above gives $\|x-\hat{s}y\|^2=\|x\|^2-\frac{(\text{Re}\langle x,y\rangle)^2}{\|y\|^2}\geq 0$, from which we obtain $|\text{Re}\langle x,y\rangle|\leq\|x\|\|y\|$.

To finish, note that if $\theta\in\mathbb{C}$ with $|\theta|=1$, we have $|\text{Re}\langle\theta x,y\rangle|\leq\|x\|\|y\|$, and we can choose $\theta$ such that $\text{Re}\langle\theta x,y\rangle=\langle x,y\rangle$, which gives the desired result.

However, the essence of the proof is that $\|x-sy\|\geq 0$ for all $s$.

copper.hat
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