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Choose $C_{\mathbf{p}}=a_{\mathbf{p}}^0-a_{\mathbf{p}}^3$ with the standard temporal and longitudinal polarization vectors. The physical pre-space is the common kernel of all $C_{\mathbf{p}}$. In each regulated mode, $a^0$ acts on creator polynomials as $-{\partial }_{a^{0†}}$ and $a^3$ as ${\partial }_{a^{3†}}$, so the constraint kernel consists of transverse creator polynomials and polynomials in $C^{†}=a^{0†}-a^{3†}$. Since $[C,C^{†}]=0$, these latter excitations remain constrained. They are orthogonal to every constrained state because $⟨\Phi |C^{†}=⟨C\Phi |=0$. Quotienting this radical of a Hermitian form leaves only the transverse bosonic Fock space with a positive inner product. The constraint alone gives a positive semidefinite Hermitian form; quotienting removes its null directions. If starting from finite-particle creator polynomials, take the Hilbert space completion of this positive quotient to obtain the physical Hilbert space.

For every nonempty closed convex set $C$ in a Hilbert space and every $x$ in that space, there is a unique $p\in C$ satisfyingClosedness and completeness let a limiting approximation attain the infimum; convexity prevents two different minimizers. The proof idea is to choose a sequence approaching the infimum. The parallelogram identity and the admissible midpoint make it a Cauchy sequence; applying the same identity to two putative minimizers proves uniqueness.

For a closed subspace of a Hilbert space, this nearest-point map is an orthogonal projection and is linear. For a general closed convex set it need not be linear: projecting the real line onto $[0,1]$ clips values below zero to zero and values above one to one. The theorem is therefore useful for both linear approximation and constrained nearest-point problems.^[ref]

Closedness is essential even in familiar examples. For the open interval $(0,1)\subseteq \mathbb{R}$ and $x=0$, the distance infimum is zero but no nearest point exists. The same obstruction appears when approximating $(1/k{)}_{k\ge 1}$ by finitely supported sequences: the infimum is zero, but the limit lies outside that subspace until we take its Hilbert space completion.

Start with finite real linear combinations of $1$, $\cos (\lambda x)$ and $\sin (\lambda x)$ with arbitrary positive real frequencies, and use the inner product $⟨f,g{⟩}_M={\lim }_{R\to \infty }{R}^{-1}{\int }_{-R}^{R}fg$. Product-to-sum identities give existence of these averages and mutual orthogonality of distinct frequencies. The squared norm is $2a_0^2+{\sum }_{\lambda }(a_{\lambda }^2+b_{\lambda }^2)$. Its Hilbert space completion contains an uncountable orthonormal set, so is not a separable Hilbert space. This construction does not equip all locally square-integrable functions with an inner product: compactly supported nonzero functions have zero mean square, and other functions have divergent or nonexistent averages.

A separable Hilbert space has a countable subset dense in its norm topology.

The printed assertion about all locally square-integrable functions is false. The proposed average is not even finite for every such function: for $f(x)=x$,It also fails positive definiteness. The nonzero function $f=1_{[0,1]}$ hasConsequently this formula cannot define an inner product, much less a Hilbert space, on $L_{\mathrm{loc}}^2(\mathbb{R})$.

A precise version of the intended nonseparability argument uses the mean-square completion of trigonometric polynomials. Start with the real vector space $V$ of finite linear combinations of $1$, $\cos (\lambda x)$, and $\sin (\lambda x)$, with arbitrary $\lambda >0$. Product-to-sum identities show that all the proposed cross averages exist. Distinct frequencies are orthogonal, each sine and cosine has squared norm one, and the constant function has squared norm two. Thus, after collecting equal frequencies,This is positive definite on $V$. Its Hilbert space completion $H$ contains the uncountable orthonormal set $\{\cos (\lambda x):\lambda >0\}$. The distance between two distinct members is $\sqrt{2}$. Their open balls of radius $1/2$ are pairwise disjoint, and a dense subset must meet each one. A countable dense subset is therefore impossible: this corrected completed space is nonseparable. Completion is an essential additional construction; it does not validate the printed claim about all of $L_{\mathrm{loc}}^2$.

Equip $C_0^{\infty }(D)$ with the Dirichlet inner productThe zero-boundary Sobolev space $H_0^1(D)$ is its Hilbert space completion in the norm $\Vert f{\Vert }_{\nabla }=(f,f{)}_{\nabla }^{1/2}$.

A Hermitian form is positive semidefinite when $h(v,v)\ge 0$ for every vector. It need not define an inner product, because a nonzero vector may have zero squared norm. The Cauchy-Schwarz inequality still holds: applying nonnegativity to $v+zw$ and minimizing the quadratic expression in $z$ gives $|h(v,w){|}^2\le h(v,v)h(w,w)$ when $h(w,w)>0$; when $h(w,w)=0$, varying $z$ forces $h(v,w)=0$. Thus its zero-norm vectors are exactly the radical of a Hermitian form. Quotienting this radical of a Hermitian form gives a positive inner product; taking its Hilbert space completion then gives a Hilbert space. This is the final positivity step in the Gupta-Bleuler null-state quotient.