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The induced length is a norm

Statement

Let $V$ be a real or complex inner-product space with induced length $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ . Then $∥ v ∥ \geq 0$ with $∥ v ∥ = 0$ exactly for $v = 0$ , $∥ λ v ∥ = ∣ λ ∣ ∥ v ∥$ for every scalar $λ$ , and $∥ u + v ∥ \leq ∥ u ∥ + ∥ v ∥$ ; hence $∥ \cdot ∥$ is a norm on $V$ , read over $R$ by A norm on a real vector space, the induced metric, and the dictionary with the metric axioms and over $C$ by Real and complex scalar conventions for normed spaces.

Facts & Assumptions

[A1]

The induced length is the unique nonnegative square root of the diagonal pairing, with $∥ v ∥ \geq 0$ and $∥ v ∥ = 0$ exactly for $v = 0$ (The norm $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ induced by a real or complex inner product).

[A2]

Cauchy–Schwarz gives $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[A3]

A norm on a real vector space satisfies separation, absolute homogeneity and the triangle inequality, and a complex normed space is defined by the same clauses with the complex modulus (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, Real and complex scalar conventions for normed spaces).

[A4]

For complex scalars $z \overline{z} = ∣ z ∣^{2}$ and $λ \overline{λ} = ∣ λ ∣^{2}$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive), and for real scalars $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ with $∣ x ∣ \geq 0$ (Basic properties of the absolute value).

[A5]

For nonnegative reals $a \leq b$ if and only if $a^{2} \leq b^{2}$ (Squaring is monotone on the nonnegatives), and each nonnegative real has a unique nonnegative square root (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ).

[A6]

If $z = a + b i$ then $Re z = a$ and $∣ z ∣ = \sqrt{a^{2} + b^{2}}$ , so $Re z \leq ∣ z ∣$ (Real and imaginary parts, complex conjugation, and modulus).

Proof

technique · direct

Given: A real or complex inner-product space $V$ , vectors $u, v \in V$ and a scalar $λ$ ; in the real case conjugation is the identity and $∣ \cdot ∣$ is the absolute value of $R$ , so the second clause of [A4] is read in the real field.

1.1

Nonnegativity and separation are [A1]: $∥ v ∥ \geq 0$ , and $∥ v ∥ = 0$ exactly when $⟨ v, v ⟩ = 0$ , that is exactly when $v = 0$ .

A1A3

1.2

Homogeneity: $⟨ λ v, λ v ⟩ = λ \overline{λ} ⟨ v, v ⟩ = ∣ λ ∣^{2} ∥ v ∥^{2}$ by sesquilinearity and [A4], and both $∥ λ v ∥$ and $∣ λ ∣ ∥ v ∥$ are nonnegative with equal squares, so $∥ λ v ∥ = ∣ λ ∣ ∥ v ∥$ by uniqueness of the nonnegative square root.

A1A4A5algebra

1.3

Triangle inequality: expanding and using conjugate symmetry gives $∥ u + v ∥^{2} = ∥ u ∥^{2} + 2 Re ⟨ u, v ⟩ + ∥ v ∥^{2}$ , and $Re z \leq ∣ z ∣$ for every scalar $z$ , since either $Re z < 0 \leq ∣ z ∣$ or $Re z = a \geq 0$ with $a^{2} \leq a^{2} + b^{2} = ∣ z ∣^{2}$ by [A6]; with [A2] this gives $∥ u + v ∥^{2} \leq ∥ u ∥^{2} + 2 ∥ u ∥ ∥ v ∥ + ∥ v ∥^{2} = (∥ u ∥ + ∥ v ∥)^{2}$ .

A1A2A4A6algebra

2.1

Both sides of the inequality in step 1.3 are nonnegative, so monotonicity of squaring on the nonnegatives turns it into $∥ u + v ∥ \leq ∥ u ∥ + ∥ v ∥$ .

A5step 1.3

3.1

Steps 1.1, 1.2 and 2.1 are exactly the clauses of [A3] over $R$ , and over $C$ they are the same clauses with the complex modulus, so $∥ \cdot ∥$ is a norm on $V$ in either scalar field.

A3step 1.1step 1.2step 2.1∎

Depends on

Used by

Hilbert space Definition
Orthogonality and the orthogonal complement Definition
The standard inner products make K n, ell two and quotient L two Hilbert spaces Example
Orthogonal complements are closed Lemma
The inner product is jointly continuous Lemma
Two dimensional numerical range is convex Lemma
A Hilbert space with a dense sequence has a finite or countable orthonormal basis Theorem
The double orthogonal complement of a subspace is its closure Theorem
The norm completion of an inner-product space is a Hilbert space Theorem
The parallelogram law Theorem

Dependency tree · two levels

43 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, §1.3.3, pp.38–39 (standard reference, not scraped)
Andrew Lin and Casey Rodriguez, MIT 18.102 Introduction to Functional Analysis, Lectures 15–16 (standard reference, not scraped)