Proofs of the Cauchy-Schwarz Inequality?

Question

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

Answer 1

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: $$ |\lambda|^2 |\langle x, x \rangle| = |\lambda|^2 \|x\|\|x\| $$

Answer 2

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. $x\in\mathbb{F}y$.

Answer 3

Here is a more general and natural version of Cauchy-Schwarz inequality, called Gram's inequality.

Let $ V $ be a real vector space, with a positive definite symmetric bilinear function $ (x,y) \rightarrow \langle x, y \rangle $.
Examples : $ V = \mathbb{R}^n $ with $ \langle x, y \rangle = x^{T} y $ ; $ V = \{ $ all continuous functions $ [a,b] \rightarrow \mathbb{R} \, \} $ with $ \langle f, g \rangle = \int_{a}^{b} f(t) g(t) dt $.

Key Lemma : Let $ v_1, \ldots, v_k \in V $ be linearly independent, and $ U = \text{span}\{v_1, \ldots, v_k \} $. Then $ U $ has an orthonormal basis (that is, there exists a basis $ e_1, \ldots, e_k $ of $ U $, such that $ \langle e_i, e_j \rangle $ is $ 1 $ if $ i=j $ and $ 0 $ otherwise).
Proof : See Gram-Schmidt process.

Theorem : Let $ v_1, \ldots, v_k \in V $. We can form a $ k \times k $ matrix $ G $ with $ (i,j)^{\text{th}} $ entry $ \langle v_i, v_j \rangle $. Then $ \det(G) $ is $ 0 $ if $ v_1, \ldots, v_k $ are linearly dependent, and $ > 0 $ if $ v_1, \ldots, v_k $ are linearly independent.
Proof : Suppose $ v_1, \ldots, v_k $ are linearly dependent. So $ v_1 \lambda_1 + \ldots + v_k \lambda_k = 0 $ for some $ \lambda_1, \ldots, \lambda_k $ not all $ 0 $. Applying $ \langle v_j, - \rangle $ on this equation for each $ j $, we get $ G \lambda = 0 $ where $ \lambda := (\lambda_1, \ldots, \lambda_k)^{T} $. Since $ \lambda \neq 0 $, we have $ \det(G) = 0 $, as needed.
Now suppose $ v_1, \ldots, v_k $ are linearly independent. By the lemma, $ U := \text{span}\{v_1, \ldots, v_k\} $ has an orthonormal basis $ e_1, \ldots, e_k $. Writing $ v_j $s w.r.t this new basis, we get $ (v_1, \ldots, v_k) = (e_1, \ldots, e_k) P $ for an invertible $ k \times k $ matrix $ P = (p_{ij}) $. Now notice $ \langle v_i, v_j \rangle $ $ = \langle p_{1i} e_1 + \ldots + p_{ki} e_k, p_{1j} e_1 + \ldots + p_{kj} e_j \rangle $ $ = p_{1i} p_{1j} + \ldots + p_{ki} p_{kj} $, which is $ P_i ^{T} P_j $ (where $ P_1, \ldots, P_k $ are columns of $ P $). Hence $ G = P^{T} P $, and taking $ \det $ gives $ \det(G) = \det(P) ^{2} > 0 $, as needed.

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[Note $V=\mathbb{R}^{n}$, $ \langle x, y \rangle = x^{T} y $, $ k = 2 $ gives the usual Cauchy-Schwarz inequality for vectors. Also $ V = \{ $continuous functions $ [a,b] \rightarrow \mathbb{R}\} $, $ \langle f, g \rangle = \int f(t)g(t)dt$, $ k = 2 $ gives the usual Cauchy-Schwarz inequality for continuous functions].

Answer 4

Comment on a previous answer by Pauly B., whose method can be applied to complex vectors with minor modifications.

For complex vectors $x$ and $y$, $$ 0 \le \langle lx + y, lx+y \rangle = |l|^2 \langle x,x \rangle + 2 \text{Re}\, (l \langle x,y \rangle) + \langle y,y \rangle $$ Take $l = r\,\overline{\langle x,y\rangle}$ where $r \in \mathbb R$. Then the above inequality reduces to $$ 0 \le r^2 |\langle x,y \rangle|^2 |\langle x,x\rangle| + 2 r |\langle x,y \rangle|^2 + \langle y,y \rangle, $$ where the right side is also a quadratic polynomial in $r$ with at most 1 real root.

The proof above applies to a general vector space with complex inner products.

Answer 5

Proof by Jensen's inequality

Jensen's inequality for concave functions $f(\cdot)$ states that $\sum_i w_i f(a_i) \le f (\sum_i w_i a_i)$ with positive weights $w_i$ which sum to one: $\sum_i w_i = 1$. Equality holds iff all $a_i$ are equal.

We want to prove the discrete version of Cauchy-Schwarz, $\sum_i x_i y_i \le \sqrt{\sum_i x_i^2}\sqrt{\sum_i y_i^2}$, rewriting Cauchy-Schwarz in the following form: $$ \frac{\sum_i x_i y_i }{{\sum_i x_i^2}} \le \frac{1}{\sqrt{\sum_i x_i^2}}\sqrt{\sum_i y_i^2} $$

To bring this into Jensen's form, rewrite this again as $$ \sum_i \frac{x_i^2}{\sum_j x_j^2} \sqrt{\frac{y_i^2}{x_i^2}} \le \sqrt{\sum_i \frac{x_i^2}{\sum_j x_j^2} {\frac{y_i^2}{x_i^2}} } $$
But this is exactly Jensen's inequality, with $f(\cdot) = \sqrt{(.)}$ and weights summing to one, as indicated here: $$ \sum_i {\underbrace{\frac{x_i^2}{\sum_j x_j^2}}_{w_i}} \sqrt{\underbrace{\frac{y_i^2}{x_i^2}}_{a_i}} \le \sqrt{\sum_i \underbrace{ \frac{x_i^2}{\sum_j x_j^2}}_{w_i} \underbrace{\frac{y_i^2}{x_i^2}}_{a_i} } $$ The equality conditions also translate: equality holds due to Jensen iff all $a_i$ are equal, which means here that all $y_i/x_i$ are equal, which means that the $y$-vector is a scalar multiple of the $x$-vector, which is the well-known equality condition in Cauchy-Schwarz.

Answer 6

In Euclidean space, the Cauchy-Schwarz inequality is equivalent to the assertion that for all $\mathbf u, \mathbf v\in\mathbb R^n$, $$ \left(\sum_{i=1}^{n}u_iv_i\right)^2\le\left(\sum_{i=1}^{n}u_i^2\right)\left(\sum_{i=1}^{n}v_i^2\right) $$ A particularly elegant proof is given on the Wikipedia page of the inequality, which I will reproduce here.

Let $\mathbf u, \mathbf v\in\mathbb R^n$ be given. Assume that $\mathbf{u}\neq\mathbf{0}$, for otherwise the inequality is trivial. Consider the following polynomial in $x$: $$ (u_1x+v_1)^2+\dots+(u_nx+v_n)^2=\left(\sum_{i=1}^{n}u_i^2\right)x^2+\left(2\sum_{i=1}^{n}u_iv_i\right)x+\left(\sum_{i=1}^{n}v_i^2\right) \, . $$ Since $\mathbf u\neq\mathbf 0$, the above polynomial is a quadratic in $x$. By considering the left hand side, we see that that this quadratic is nonnegative and therefore either has no real roots or has one repeated root. Hence, the discriminant must be negative or zero, i.e. $$ 4\left(\sum_{i=1}^{n}u_iv_i\right)^2-4\left(\sum_{i=1}^{n}u_i^2\right)\left(\sum_{i=1}^{n}v_i^2\right) \, \le0 \, . $$ The desired inequality follows at once.