Mathematics Lecture Notes

Theorem6.5Cauchy-Schwarz Inequality

For all vectors $u, v$ in an inner product space $V$ :

$|\langle u, v \rangle|^2 \leq \langle u, u \rangle \cdot \langle v, v \rangle.$

Equality holds if and only if $u$ and $v$ are linearly dependent.

ProofProof via non-negative quadratic form

Case 1: If $v = 0$ , both sides are $0$ and the inequality holds with equality.

Case 2: Assume $v \neq 0$ . For any scalar $t \in \mathbb{R}$ (or $t \in \mathbb{C}$ ), consider the vector $u - tv$ . By positive definiteness:

$0 \leq \langle u - tv, u - tv \rangle = \langle u, u \rangle - t\langle u, v \rangle - \bar{t}\langle v, u \rangle + |t|^2 \langle v, v \rangle.$

Real case: Set $t = \frac{\langle u, v \rangle}{\langle v, v \rangle}$ (the optimal $t$ minimizing the quadratic):

$0 \leq \langle u, u \rangle - 2 \frac{\langle u, v \rangle^2}{\langle v, v \rangle} + \frac{\langle u, v \rangle^2}{\langle v, v \rangle} = \langle u, u \rangle - \frac{\langle u, v \rangle^2}{\langle v, v \rangle}.$

Rearranging: $\langle u, v \rangle^2 \leq \langle u, u \rangle \cdot \langle v, v \rangle$ .

Complex case: Set $t = \frac{\langle u, v \rangle}{\langle v, v \rangle}$ :

$0 \leq \langle u, u \rangle - \frac{\langle u, v \rangle \overline{\langle u, v \rangle}}{\langle v, v \rangle} = \langle u, u \rangle - \frac{|\langle u, v \rangle|^2}{\langle v, v \rangle}.$

Rearranging: $|\langle u, v \rangle|^2 \leq \langle u, u \rangle \cdot \langle v, v \rangle$ .

Equality: We have equality iff $\langle u - tv, u - tv \rangle = 0$ , iff $u - tv = 0$ , iff $u = tv$ . $\blacksquare$

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ExampleVisualizing the quadratic proof in R^2

$u = (3, 1)$ , $v = (1, 2)$ . $\langle v, v \rangle = 5$ , $\langle u, v \rangle = 5$ .

$t = 5/5 = 1$ , $u - tv = (3, 1) - (1, 2) = (2, -1)$ , $\|u - tv\|^2 = 5$ .

$\|u\|^2 - |\langle u, v \rangle|^2 / \|v\|^2 = 10 - 25/5 = 5 \geq 0$ ✓.

If instead $u = (2, 4) = 2v$ : $t = 10/5 = 2$ , $u - tv = 0$ , equality holds.

ExampleComplex case verification

$u = (1, i)$ , $v = (1, 1)$ in $\mathbb{C}^2$ .

$\langle u, v \rangle = 1 \cdot 1 + i \cdot 1 = 1 + i$ , $|\langle u, v \rangle|^2 = 2$ .

$\langle u, u \rangle = 1 + 1 = 2$ , $\langle v, v \rangle = 2$ . Product: $4$ .

$2 \leq 4$ ✓. Strict inequality because $u$ and $v$ are linearly independent (no scalar $t$ with $(1, i) = t(1, 1)$ ).

ProofProof using projection

If $v = 0$ , the result is trivial. Assume $v \neq 0$ .

Decompose $u$ orthogonally with respect to $v$ :

$u = \operatorname{proj}_v(u) + (u - \operatorname{proj}_v(u)), \quad \text{where } \operatorname{proj}_v(u) = \frac{\langle u, v \rangle}{\langle v, v \rangle} v.$

By the Pythagorean theorem (since the two components are orthogonal):

$\|u\|^2 = \|\operatorname{proj}_v(u)\|^2 + \|u - \operatorname{proj}_v(u)\|^2.$

Since $\|u - \operatorname{proj}_v(u)\|^2 \geq 0$ :

$\|u\|^2 \geq \|\operatorname{proj}_v(u)\|^2 = \frac{|\langle u, v \rangle|^2}{\|v\|^2}.$

Multiplying both sides by $\|v\|^2$ : $\|u\|^2 \|v\|^2 \geq |\langle u, v \rangle|^2$ .

Equality iff $u = \operatorname{proj}_v(u)$ , iff $u$ is a scalar multiple of $v$ . $\blacksquare$

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ExampleProjection proof visualized

$u = (4, 3)$ , $v = (1, 0)$ .

$\operatorname{proj}_v(u) = 4 \cdot (1, 0) = (4, 0)$ . Residual: $(0, 3)$ .

$\|u\|^2 = 25 = 16 + 9 = \|\operatorname{proj}\|^2 + \|\text{residual}\|^2$ .

Cauchy--Schwarz: $|\langle u, v \rangle|^2 = 16 \leq 25 \cdot 1 = 25$ ✓. The gap is $\|\text{residual}\|^2 = 9$ .

ExampleProjection proof for functions

$f(x) = x$ , $g(x) = 1$ on $[0, 1]$ .

$\operatorname{proj}_g(f) = \frac{\langle f, g \rangle}{\|g\|^2} g = \frac{1/2}{1} \cdot 1 = \frac{1}{2}$ .

$\|f\|^2 = 1/3 \geq |\langle f, g \rangle|^2 / \|g\|^2 = 1/4$ ✓.

The residual is $f - 1/2 = x - 1/2$ , with $\|x - 1/2\|^2 = 1/3 - 1/4 = 1/12$ .

ProofProof via Lagrange identity (real case, R^n)

For $u, v \in \mathbb{R}^n$ , the Lagrange identity states:

$\|u\|^2 \|v\|^2 - (u \cdot v)^2 = \sum_{1 \leq i < j \leq n} (u_i v_j - u_j v_i)^2.$

The right side is a sum of squares, hence $\geq 0$ . This gives $\|u\|^2 \|v\|^2 \geq (u \cdot v)^2$ .

Equality holds iff all $u_i v_j - u_j v_i = 0$ for $i < j$ , iff all $2 \times 2$ minors of $\begin{pmatrix} u_1 & \cdots & u_n \\ v_1 & \cdots & v_n \end{pmatrix}$ vanish, iff $u$ and $v$ are proportional. $\blacksquare$

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ExampleLagrange identity in R^2

$u = (a, b)$ , $v = (c, d)$ :

$\|u\|^2\|v\|^2 - (u \cdot v)^2 = (a^2 + b^2)(c^2 + d^2) - (ac + bd)^2 = (ad - bc)^2$ .

For $u = (1, 2)$ , $v = (3, 4)$ : $(1 + 4)(9 + 16) - (3 + 8)^2 = 125 - 121 = 4 = (4 - 6)^2 = 4$ ✓.

ExampleLagrange identity in R^3

$u = (1, 0, 0)$ , $v = (0, 1, 0)$ :

$\|u\|^2\|v\|^2 - (u \cdot v)^2 = 1 \cdot 1 - 0 = 1$ .

$(u_1 v_2 - u_2 v_1)^2 + (u_1 v_3 - u_3 v_1)^2 + (u_2 v_3 - u_3 v_2)^2 = 1 + 0 + 0 = 1$ ✓.

Note: in $\mathbb{R}^3$ , $\|u\|^2\|v\|^2 - (u \cdot v)^2 = \|u \times v\|^2$ , where $u \times v$ is the cross product.

ExampleDeriving the triangle inequality

$\|u + v\|^2 = \|u\|^2 + 2\operatorname{Re}\langle u, v \rangle + \|v\|^2 \leq \|u\|^2 + 2\|u\|\|v\| + \|v\|^2 = (\|u\| + \|v\|)^2$ .

The key step uses $\operatorname{Re}\langle u, v \rangle \leq |\langle u, v \rangle| \leq \|u\|\|v\|$ (Cauchy--Schwarz).

ExampleDeriving Bessel's inequality

For an orthonormal set $\{e_1, \ldots, e_k\}$ and any $v$ , let $w = v - \sum_i \langle v, e_i \rangle e_i$ . Then $w \perp e_j$ for all $j$ , and:

$0 \leq \|w\|^2 = \|v\|^2 - \sum_i |\langle v, e_i \rangle|^2$ .

This is Bessel's inequality, which is a multi-dimensional Cauchy--Schwarz.

ExampleAngle is well-defined

Cauchy--Schwarz guarantees $-1 \leq \frac{\langle u, v \rangle}{\|u\|\|v\|} \leq 1$ for real inner product spaces, so $\theta = \arccos\left(\frac{\langle u, v \rangle}{\|u\|\|v\|}\right) \in [0, \pi]$ is well-defined.

Without Cauchy--Schwarz, the argument of $\arccos$ could exceed $[-1, 1]$ , making the angle undefined.

RemarkThree perspectives on Cauchy-Schwarz

Quadratic form proof: The most general, works in any inner product space (real or complex, finite or infinite-dimensional). Shows that the inequality is equivalent to non-negativity of a specific scalar quadratic in $t$ .
Projection proof: Geometrically intuitive -- the inequality says the projection of $u$ onto $v$ is at most as long as $u$ itself. The "gap" $\|u - \operatorname{proj}_v(u)\|^2$ measures how far $u$ is from being a multiple of $v$ .
Lagrange identity proof: Algebraically explicit (for $\mathbb{R}^n$ only), showing the "defect" $\|u\|^2\|v\|^2 - (u \cdot v)^2 = \sum(u_iv_j - u_jv_i)^2$ as a sum of squares. In $\mathbb{R}^3$ , this equals $\|u \times v\|^2$ , connecting to the cross product.

RemarkHistory of the inequality

The inequality is named after Augustin-Louis Cauchy (who proved the finite sum version in 1821), Viktor Bunyakovsky (who proved the integral version in 1859), and Hermann Amandus Schwarz (who independently proved the integral version in 1884). In some traditions it is called the Cauchy--Bunyakovsky--Schwarz inequality or CBS inequality.

The inequality appears in every branch of mathematics:

Analysis: the foundation of $L^p$ space theory.
Probability: bounds on covariance and correlation.
Physics: uncertainty principles in quantum mechanics.
Geometry: the angle between vectors, curvature bounds.
Combinatorics: counting arguments via the second moment method.

ExampleUncertainty principle from Cauchy-Schwarz

In quantum mechanics, observables $A, B$ act on a Hilbert space $\mathcal{H}$ . For a state $|\psi\rangle$ , the uncertainty principle:

$\Delta A \cdot \Delta B \geq \frac{1}{2}|\langle [A, B] \rangle|$

follows from Cauchy--Schwarz applied to $|u\rangle = (A - \langle A \rangle)|\psi\rangle$ and $|v\rangle = (B - \langle B \rangle)|\psi\rangle$ .

For position $X$ and momentum $P$ with $[X, P] = i\hbar$ : $\Delta X \cdot \Delta P \geq \hbar/2$ .

ExampleCauchy-Schwarz in number theory

For real sequences $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ :

$\left(\sum a_i b_i\right)^2 \leq \left(\sum a_i^2\right)\left(\sum b_i^2\right).$

Setting $a_i = 1$ and $b_i = d(i)$ (number of divisors of $i$ ): $\left(\sum_{i=1}^n d(i)\right)^2 \leq n \sum_{i=1}^n d(i)^2$ . Since $\sum d(i) \sim n \log n$ , this gives $\sum d(i)^2 \gg n(\log n)^2$ .

ExampleGeneralization: Holder's inequality

Cauchy--Schwarz is the $p = q = 2$ case of Holder's inequality: for $\frac{1}{p} + \frac{1}{q} = 1$ ,

$\sum |a_i b_i| \leq \left(\sum |a_i|^p\right)^{1/p} \left(\sum |b_i|^q\right)^{1/q}.$

The $p = q = 2$ case is exactly Cauchy--Schwarz.

RemarkThe cornerstone inequality

The Cauchy--Schwarz inequality is the single most consequential inequality in inner product space theory. From it flow:

The triangle inequality (hence metric space structure).
Bessel's inequality (finite energy of Fourier coefficients).
The well-definedness of angles.
The optimality of orthogonal projections.
Uncertainty principles in physics.

Its proof, via the non-negativity of $\|u - tv\|^2$ , is a masterclass in the power of positive definiteness.

Math Notes

Proof of Cauchy-Schwarz Inequality

Statement

Proof 1: The quadratic form argument

Proof 2: Via orthogonal projection

Proof 3: Lagrange identity (for R^n)

Consequences derived

Comparison of proofs

Historical note

Extended examples

Summary