Cauchy-Schwarz Inequality

Cauchy–Schwarz Inequality

Algebraic Theory – MIT‑Level Study Notes

Goal – Understand the Cauchy–Schwarz inequality (CS) from first principles, prove it rigorously, explore its deeper algebraic structure, discuss edge cases, and see how it underpins many areas of modern mathematics.

1. Preliminaries

1.1 Vector Spaces and Inner Products

Let (V) be a vector space over (\mathbb{F}\in{\mathbb{R},\mathbb{C}}).

• Inner product: A map (\langle\cdot,\cdot\rangle:V\times V\to\mathbb{F}) satisfying:
  1. Conjugate symmetry
     [ \langle x,y\rangle=\overline{\langle y,x\rangle}\quad(\text{hence } \langle x,x\rangle\in\mathbb{R}_{\ge 0}). ]
  2. Linearity in the first argument (sesquilinear in the complex case): [ \langle ax+by,z\rangle=a\langle x,z\rangle+b\langle y,z\rangle,\qquad a,b\in\mathbb{F}. ]
  3. Positive definiteness
     (\langle x,x\rangle>0) for all (x\neq0).
• Norm induced by the inner product: (|x|:=\sqrt{\langle x,x\rangle}).

Remark: For (\mathbb{F}=\mathbb{C}) the second argument is conjugate‑linear.

2. Statement of Cauchy–Schwarz

Theorem (Cauchy–Schwarz).
For all (x,y\in V), [ |\langle x,y\rangle|\le |x|;|y|. ] Equality holds iff (x) and (y) are linearly dependent: (x=\lambda y) for some (\lambda\in\mathbb{F}).

3. Proofs from First Principles

3.1 Proof via Quadratic Forms

Let (x,y\in V). Consider the real‑valued function [ f(t)=|x-ty|^{2} =\langle x-ty,x-ty\rangle =|x|^{2}-2t,\Re\langle x,y\rangle+t^{2}|y|^{2}\quad(t\in\mathbb{R}). ] Since (|x-ty|^{2}\ge0) for every (t), the quadratic polynomial [ q(t)=|y|^{2}t^{2}-2\Re\langle x,y\rangle,t+|x|^{2} ] has non‑negative discriminant: [ \Delta=4\bigl(\Re\langle x,y\rangle\bigr)^{2}-4|x|^{2}|y|^{2}\le0. ] Thus [ \bigl(\Re\langle x,y\rangle\bigr)^{2}\le |x|^{2}|y|^{2}. ] Repeating the argument with (x+ty) yields [ \bigl(\Im\langle x,y\rangle\bigr)^{2}\le |x|^{2}|y|^{2}. ] Adding the two inequalities gives (|\langle x,y\rangle|^{2}\le|x|^{2}|y|^{2}), i.e. CS.

Edge Case: If (y=0), both sides are zero, so the inequality is trivial.

3.2 Proof via Gram Matrix

Let (x,y\in V). The Gram matrix [ G=\begin{pmatrix} \langle x,x\rangle & \langle x,y\rangle\[4pt] \langle y,x\rangle & \langle y,y\rangle \end{pmatrix} ] is Hermitian and positive semidefinite because for any (\alpha,\beta\in\mathbb{F}), [ \bigl|\alpha x+\beta y\bigr|^{2}= \begin{pmatrix}\overline{\alpha}&\overline{\beta}\end{pmatrix} G\begin{pmatrix}\alpha\\beta\end{pmatrix}\ge0. ] Therefore (\det G\ge0), i.e. [ \langle x,x\rangle\langle y,y\rangle-\langle x,y\rangle\langle y,x\rangle\ge0 ;\Longrightarrow; |\langle x,y\rangle|^{2}\le\langle x,x\rangle\langle y,y\rangle. ]

Equality Condition: (\det G=0) iff (x) and (y) are linearly dependent.

3.3 Proof via Orthogonal Projection

Let (P) be the orthogonal projection onto (\operatorname{span}{y}).
Then (P(x)=\frac{\langle x,y\rangle}{\langle y,y\rangle}y), so [ |P(x)|=\frac{|\l