Proof of L²(μ;ℝⁿ) Is a Real Hilbert Space

Throughout, every element of $L^{2}(\mu;\mathbb{R}^{n})$ is $[f]$ for some $f\in\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$, and $[f]=[g]$ if and only if $f\sim g$, in particular whenever $f=g$ (Equivalence Classes Partition the Set: Cover, Disjointness and Representatives; the Quotient Is a Set and the Canonical Projection Is a Surjection (iv), Equivalence Classes Partition the Set: Cover, Disjointness and Representatives; the Quotient Is a Set and the Canonical Projection Is a Surjection (v) and Equivalence Classes Partition the Set: Cover, Disjointness and Representatives; the Quotient Is a Set and the Canonical Projection Is a Surjection (i)); so identities in $L^{2}(\mu;\mathbb{R}^{n})$ may be checked on representatives. Sums, scalar multiples, negatives and differences of representatives lie in $\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$ by Square-Integrable Vector-Valued Functions: Components, Dot Products and Almost-Everywhere Equality (iii).

Vector space. Let $f,g,h\in\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$ and $\lambda,\nu\in\mathbb{R}$. By The Lebesgue Space L²(μ;ℝⁿ) (ii),

$$([f]+[g])+[h]=[(f+g)+h]=[f+(g+h)]=[f]+([g]+[h]),$$

since $(f+g)+h$ and $f+(g+h)$ are the same function by the associativity of the sum of $\mathbb{R}^{n}$ at each point. In the same way, the commutativity of the sum and the laws

$$\lambda(\nu[f])=(\lambda\nu)[f],\qquad 1[f]=[f],\qquad \lambda([f]+[g])=\lambda[f]+\lambda[g],\qquad (\lambda+\nu)[f]=\lambda[f]+\nu[f]$$

follow from the corresponding vector-space laws of $\mathbb{R}^{n}$ at each point. The function $0$ lies in $\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$: it is measurable by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (iii), and $|0|^{2}$ is the zero function by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (iii), with integral $0$ by Basic Properties of the Integral of Nonnegative Measurable Maps (v). As $[f]+[0]=[f+0]=[f]$, the class $[0]$ is a zero as in Vector Space over a Field (iii). If $z\in L^{2}(\mu;\mathbb{R}^{n})$ satisfies $u+z=u$ for every $u$, then $z=z+[0]=[0]+z=[0]$, and $[f]+[-f]=[f+(-f)]=[0]=z$; this is Vector Space over a Field (iv). Hence $L^{2}(\mu;\mathbb{R}^{n})$ is a real vector space (Vector Space over a Field (ix) and Vector Space over a Field (x)), its zero vector is $[0]$ and $-[f]=[-f]$ (The Zero Vector, Negatives and Differences in a Vector Space (i) and The Zero Vector, Negatives and Differences in a Vector Space (ii)). Consequently, by The Zero Vector, Negatives and Differences in a Vector Space (iii),

$$[f]-[g]=[f]+[-g]=[f-g]. \tag{1}$$

Inner product. By The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite (i) and The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite (ii), applied at each point,

$$f\cdot g=g\cdot f,\qquad (f+g)\cdot h=f\cdot h+g\cdot h,\qquad (\lambda f)\cdot g=\lambda\,(f\cdot g)$$

as functions on $X$. These functions lie in $\mathcal{L}^{1}(\mu)$ by Square-Integrable Vector-Valued Functions: Components, Dot Products and Almost-Everywhere Equality (ii), so by The Lebesgue Space L²(μ;ℝⁿ) (iii) and Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii),

$$\langle[f],[g]\rangle=\langle[g],[f]\rangle,\qquad \langle[f]+[g],[h]\rangle=\langle[f],[h]\rangle+\langle[g],[h]\rangle,\qquad \langle\lambda[f],[g]\rangle=\lambda\,\langle[f],[g]\rangle .$$

At each point $f(x)\cdot f(x)=\lVert f(x)\rVert^{2}$ by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (i), so $f\cdot f=|f|^{2}$, and by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi)

$$\langle[f],[f]\rangle=\int f\cdot f\,d\mu=\int|f|^{2}\,d\mu\ge0, \tag{2}$$

the last integral being that of $|f|^{2}$ as an element of $\mathcal{M}^{+}$. If $\langle[f],[f]\rangle=0$, then $\lVert f(x)\rVert^{2}=0$, hence $f(x)=0$, for almost every $x$ (The Lebesgue Integral and Null Sets (iv) and Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (iii)); so $f\sim0$ and $[f]=[0]$. Thus $\langle\cdot,\cdot\rangle$ satisfies Real Inner Product Spaces (i), Real Inner Product Spaces (ii), Real Inner Product Spaces (iii) and Real Inner Product Spaces (iv), so it is an inner product (Real Inner Product Spaces (v)), and $L^{2}(\mu;\mathbb{R}^{n})$ is a real inner product space (Real Inner Product Spaces (vi)).

Norm. By Real Inner Product Spaces (vii) and The k-th Root and the Square Root of a Nonnegative Real Number (ii), $\|[f]\|^{2}=\langle[f],[f]\rangle$, and (ii) follows from (2) and Square-Integrable Vector-Valued Functions: Components, Dot Products and Almost-Everywhere Equality (i). With (1) and $(f-g)_{k}=f_{k}-g_{k}$ (Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (iv)), this gives, for $f,g\in\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$,

$$\|[f]-[g]\|^{2}=\sum_{k=1}^{n}\|f_{k}-g_{k}\|_{2}^{2}. \tag{3}$$

Completeness. Assume the axiom of dependent choice, and let $(u_{m})$ be a Cauchy sequence in $L^{2}(\mu;\mathbb{R}^{n})$ for the norm metric $d(u,v)=\|u-v\|$ of its induced norm. Each $u_{m}$ is a subset of $\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$ with at least one element (Equivalence Classes Partition the Set: Cover, Disjointness and Representatives; the Quotient Is a Set and the Canonical Projection Is a Surjection (iii) and Equivalence Classes Partition the Set: Cover, Disjointness and Representatives; the Quotient Is a Set and the Canonical Projection Is a Surjection (iv)), so by Countable Choice Follows from Dependent Choice (i) there is a sequence $(f_{m})$ in $\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$ with $f_{m}\in u_{m}$, that is, $u_{m}=[f_{m}]$ (Equivalence Classes Partition the Set: Cover, Disjointness and Representatives; the Quotient Is a Set and the Canonical Projection Is a Surjection (v)). From now on everything is at the level of functions.

Fix $k\in[n]$. By (3) and Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (vi),

$$\|(f_{m})_{k}-(f_{\ell})_{k}\|_{2}\le\|u_{m}-u_{\ell}\|\qquad(m,\ell\in\mathbb{N}),$$

so for every $\varepsilon>0$ there is $N\in\mathbb{N}$ with $\|(f_{m})_{k}-(f_{\ell})_{k}\|_{2}<\varepsilon$ for all $m,\ell\ge N$; the components $(f_{m})_{k}$ lie in $\mathcal{L}^{2}(\mu)$ by Square-Integrable Vector-Valued Functions: Components, Dot Products and Almost-Everywhere Equality (i). By The Riesz–Fischer Theorem (ii) with $p=2$, the set $C_{k}$ of $g\in\mathcal{L}^{2}(\mu)$ with $\|(f_{m})_{k}-g\|_{2}\to0$ has an element. As $[n]$ is finite (Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (vii)), Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (xi) gives a map $k\mapsto g_{k}$ on $[n]$ with $g_{k}\in C_{k}$ for every $k$. Let $f:X\to\mathbb{R}^{n}$ be the function with components $f_{k}=g_{k}$ (Component Maps of a Map into $\mathbb{R}^{q}$, Entry Maps of a Map into Matrices, and Maps Assembled from Them (iii)); then $f\in\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$ by Square-Integrable Vector-Valued Functions: Components, Dot Products and Almost-Everywhere Equality (i). By (3), Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (vi) and Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (ix),

$$\|u_{m}-[f]\|^{2}=\sum_{k=1}^{n}\|(f_{m})_{k}-g_{k}\|_{2}^{2}\longrightarrow0\qquad(m\to\infty).$$

Given $\varepsilon>0$, there is therefore $N\in\mathbb{N}$ with $\|u_{m}-[f]\|^{2}<\varepsilon^{2}$, hence $d(u_{m},[f])<\varepsilon$, for every $m\ge N$; so $u_{m}\to[f]$ (Convergent Sequence in a Metric Space (i)). Every Cauchy sequence converges, so the norm metric space of $L^{2}(\mu;\mathbb{R}^{n})$ is complete (Complete Metric Space (i)), $L^{2}(\mu;\mathbb{R}^{n})$ is a real Banach space, and hence a real Hilbert space.