The Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality is one of the most important inequalities in mathematics, relating inner products to norms. It underlies the triangle inequality and angle definition.

TheoremCauchy-Schwarz Inequality

For any vectors $\mathbf{u}, \mathbf{v}$ in an inner product space: $|\langle \mathbf{u}, \mathbf{v} \rangle| \leq \|\mathbf{u}\| \|\mathbf{v}\|$

Equality holds if and only if $\mathbf{u}$ and $\mathbf{v}$ are linearly dependent (one is a scalar multiple of the other).

This inequality ensures that $-1 \leq \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{u}\|\|\mathbf{v}\|} \leq 1$ , validating the cosine formula for angles. It appears in countless contexts from analysis to probability theory.

ExampleCauchy-Schwarz in $\mathbb{R}^n$

For $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$ , Cauchy-Schwarz states: $\left|\sum_{i=1}^n x_iy_i\right| \leq \sqrt{\sum_{i=1}^n x_i^2} \cdot \sqrt{\sum_{i=1}^n y_i^2}$

Example: $\mathbf{x} = (1,2,3)$ , $\mathbf{y} = (4,5,6)$ : $|1 \cdot 4 + 2 \cdot 5 + 3 \cdot 6| = 32$ $\sqrt{1^2+2^2+3^2} \cdot \sqrt{4^2+5^2+6^2} = \sqrt{14} \cdot \sqrt{77} \approx 32.9$

Indeed, $32 \leq 32.9$ . ✓

TheoremTriangle Inequality

For any vectors $\mathbf{u}, \mathbf{v}$ in an inner product space: $\|\mathbf{u} + \mathbf{v}\| \leq \|\mathbf{u}\| + \|\mathbf{v}\|$

Proof: Expand $\|\mathbf{u} + \mathbf{v}\|^2$ : $\|\mathbf{u} + \mathbf{v}\|^2 = \langle \mathbf{u} + \mathbf{v}, \mathbf{u} + \mathbf{v} \rangle = \|\mathbf{u}\|^2 + 2\langle \mathbf{u}, \mathbf{v} \rangle + \|\mathbf{v}\|^2$

By Cauchy-Schwarz: $\langle \mathbf{u}, \mathbf{v} \rangle \leq |\langle \mathbf{u}, \mathbf{v} \rangle| \leq \|\mathbf{u}\|\|\mathbf{v}\|$

Therefore: $\|\mathbf{u} + \mathbf{v}\|^2 \leq \|\mathbf{u}\|^2 + 2\|\mathbf{u}\|\|\mathbf{v}\| + \|\mathbf{v}\|^2 = (\|\mathbf{u}\| + \|\mathbf{v}\|)^2$

Taking square roots gives the result. ∎

TheoremBessel's Inequality

Let $\{\mathbf{e}_1, \ldots, \mathbf{e}_k\}$ be an orthonormal set in inner product space $V$ . For any $\mathbf{v} \in V$ : $\sum_{i=1}^k |\langle \mathbf{v}, \mathbf{e}_i \rangle|^2 \leq \|\mathbf{v}\|^2$

If the orthonormal set is a basis, this becomes Parseval's identity (equality holds).

TheoremPythagorean Theorem

If $\mathbf{u} \perp \mathbf{v}$ (orthogonal), then: $\|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2$

Proof: $\|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + 2\langle \mathbf{u}, \mathbf{v} \rangle + \|\mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2$ since $\langle \mathbf{u}, \mathbf{v} \rangle = 0$ . ∎

Remark

The Cauchy-Schwarz inequality is fundamental to analysis: it ensures that the norm induced by an inner product satisfies the triangle inequality, making the space a metric space. In probability, it becomes the correlation coefficient bound $|\text{Corr}(X,Y)| \leq 1$ . In quantum mechanics, it underlies the uncertainty principle. Its ubiquity reflects its role as a bridge between algebra (inner products) and geometry (lengths and angles).