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Distinct squared traction-free eigenfrequencies in a bounded lossless elastic body have eigenfunctions orthogonal in $L^2(\mathcal{D},\rho dV{)}^3$, by the Betti identity for elastodynamic fields. Normalize one field from each eigenspace. This is an orthonormal set in a separable Hilbert space, so it is countable: disjoint balls of radius less than $1/\sqrt{2}$ around its unit vectors must contain distinct points of a countable dense set. Positive and negative frequency partners add at most a factor of two. This argument proves countability without requiring a separate compact-resolvent proof.

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.

Let $K=\sigma (T)$ for a bounded normal operator on a nonzero complex Hilbert space. The zero Hilbert space has the trivial calculus. Take the inner product linear in its first argument. The continuous functional calculus is an isometric unital star-representation $\pi :C(K)\to \mathcal{B}(H)$ with $\pi (z)=T$. The Borel calculus will act on $B_b(K)$, the algebra of actual bounded Borel measurable functions. A quotient by null functions is introduced only after specifying the spectral measure; there is no preferred scalar measure such as plane area on $K$.

Here is a construction that also covers a nonseparable Hilbert space. For $v\in H$, the positive functional $f\mapsto ⟨\pi (f)v,v⟩$ has a finite regular Borel measure ${\mu }_v$ by the Riesz-Markov-Kakutani representation theorem. On continuous functions,Continuous functions are dense in $L^2(K,{\mu }_v)$. Hence $f\mapsto \pi (f)v$ extends to a unitary map from that space onto $H_v=\overline{\pi (C(K))v}$. This subspace reduces $T$ and $T^{\ast }$, which correspond to multiplication by $z$ and $\overline{z}$. Decompose $H$ as an orthogonal direct sum of such cyclic reducing subspaces, using maximality: a nonzero vector in the remaining orthogonal complement would generate another summand. The family need not be countable. This is the cyclic multiplication model for a normal operator.

On each summand define $f(T)$ to be multiplication by $f$, for $f\in B_b(K)$, and take their bounded direct sum. It defines a unital star-homomorphismThus $(fg)(T)=f(T)g(T)$, $\overline{f}(T)=f(T{)}^{\ast }$, and positivity of $f$ gives positivity of $f(T)$. The continuous functional calculus is recovered on $C(K)$.

Set $E(S)=\Phi (1_S)$ for a Borel subset $S$ of $K$. Then $E(S)$ is an orthogonal projection, $E(K)=I$, $E(S)E(R)=E(S\cap R)$, and disjoint countable unions add in the strong operator topology. Indeed on every multiplication summand the indicators add pointwise, and the tail squared norm is the measure of a decreasing sequence of sets tending to the empty set. Thus $E$ is a projection-valued measure, andFor $v,w\in H$, polarization gives scalar measures ${\mu }_{v,w}(S)=⟨E(S)v,w⟩$ satisfying $⟨f(T)v,w⟩=\int fd{\mu }_{v,w}$. Regularity of these finite scalar measures and uniqueness in the Riesz-Markov-Kakutani representation theorem make the projection-valued measure unique: its values on continuous functions determine all scalar measures and hence every projection. This also makes the calculus independent of the chosen cyclic decomposition.

The correct convergence assertion is bounded pointwise convergence in the strong operator topology. If $f_n\to f$ pointwise and ${\sup }_n\Vert f_n{\Vert }_{\infty }<\infty$, thenby the dominated convergence theorem. This need not be convergence in operator norm. For multiplication by $t$ on $L^2[0,1]$, the functions $1_{[0,1/n]}$ converge pointwise to $1_{\{0\}}$, their operators converge strongly to zero, but their operator norms are all one.

A Borel measurable function can vanish spectrally without vanishing pointwise. Precisely, $f(T)=0$ if and only if $E(\{|f|>\epsilon \})=0$ for every $\epsilon >0$, andThe upper norm bound follows by integrating $|f{|}^2$ against vector measures; a unit vector in any nonzero projection $E(\{|f|>C\})$ gives $\Vert f(T)\Vert \ge C$. The induced map on the quotient by these spectral-null Borel measurable functions is isometric. Every nonempty relatively open subset of $K$ has nonzero spectral projection, since otherwise a nonzero continuous function supported there would contradict faithfulness of the continuous functional calculus. This full support gives the usual continuous spectral mapping theorem, but for a Borel measurable function the exact statement is insteadIf an indicated neighborhood has zero projection, define a bounded reciprocal of $f-\lambda$ off it and zero on it; the calculus supplies an inverse. Conversely unit vectors in the nonzero projections for arbitrarily small neighborhoods give $\Vert (f(T)-\lambda )v\Vert$ arbitrarily small, contradicting invertibility. This is spectral essential range in Borel functional calculus. For example $1_{\{0\}}(T)=0$ in the multiplication model on $[0,1]$, although the pointwise range of that function is $\{0,1\}$.

Spectral projections provide reducing subspaces, and $E(\{\lambda \})H=\ker (T-\lambda I)$: the squared norm of $(T-\lambda )v$ is $\int |z-\lambda {|}^{2}d{\mu }_v$. In finite dimension the calculus is simply $f(T)={\sum }_{j}f({\lambda }_j)P_j$, the sum over eigenspaces. In general it also permits discontinuous cuts and Borel choices of square roots, such as $s(z)=|z{|}^{1/2}{e}^{i\mathrm{Arg}(z)/2}$ with $s(0)=0$, giving a normal operator $s(T)$ with $s(T{)}^2=T$ even when a continuous scalar branch is unavailable.

Any bounded operator commuting with both $T$ and $T^{\ast }$ commutes with their continuous functional calculus and with all $E(S)$. To justify the latter, regularity approximates $1_S$ by uniformly bounded continuous functions in the $L^2$ norms of any finite list of vector measures; the resulting net converges strongly to $E(S)$. Thus the Borel calculus lies in the Von Neumann algebra generated by $T$. On a separable Hilbert space, a countable dense list of vectors gives a controlling measure $\mu ={\sum }_j{2}^{-j}{\mu }_{v_j}/(1+\Vert v_j{\Vert }^2)$; spectral-null sets are exactly its null sets, and the range is the represented $L^{\infty }(K,\mu )$, equal to that Von Neumann algebra. Without separability this last equality is not automatic: for the diagonal operator with one eigenvector for each $t\in [0,1]$, the generated Von Neumann algebra has arbitrary bounded diagonal functions, while the original Borel calculus has Borel diagonal functions. The spectral-measure construction and the strong convergence theorem themselves require no separability. This distinguishes the Borel functional calculus for a normal operator from its continuous predecessor without confusing pointwise functions, spectral equivalence classes and topological closure.

Choose field $a$ to be $({\hat{\mathbf{u}}}^a{)}^{\ast }$ at frequency $-{\omega }_a$, and field $b$ to be ${\hat{\mathbf{u}}}^b$ at ${\omega }_b$. Zero body forces and zero boundary tractions make all source and surface terms vanish. ThereforeFor distinct squared frequencies this proves weighted orthogonality,Assume positive mass density bounded above and below and finite-energy fields in the ordinary bounded domain. The weighted space $H=L^2(\mathcal{D},\rho dV{)}^3$ is a separable Hilbert space. Select one unit eigenfunction for each distinct squared frequency. They form an orthonormal set. Let $\{d_j\}$ be a countable dense subset of $H$; assign to each eigenfunction a point $d_j$ within distance $1/3$. Two orthogonal unit eigenfunctions have distance $\sqrt{2}$, so the same $d_j$ cannot be assigned to both. This gives an injection into a countable set.

Hence there are at most countably many squared eigenfrequencies, and at most twice that many signed frequencies. Multiplicity at one frequency does not change the argument. A zero frequency associated with rigid translations or rotations is included. This proves the countability of elastic eigenfrequencies directly; proving discreteness or finite multiplicity would require additional spectral regularity arguments.

Choose an orthonormal basis $(e_j)$ of the separable Hilbert space and independent standard normal random variables $(Z_j)$ on a common probability space. In finite dimension the sums below are finite; in dimension zero take $X(0)=0$. In infinite dimension defineIndependence, centring and unit variance giveBy the Parseval identity for a Hilbertian basis, the coefficient sequence is square-summable. Completeness of $L^2(\Omega )$ therefore supplies a limit, and we defineEvery partial-sum map is linear, and passage to the $L^2$ limit preserves this identity. Thus for each fixed $a,b,g,h$, $X(ag+bh)=aX(g)+bX(h)$ as an $L^2$ identity, hence an almost sure equality. This constructs an isonormal Gaussian process.

The usual meaning is almost-sure linearity for each fixed choice of arguments. A version linear simultaneously for every argument can also be chosen: select a Hamel basis of $H$, choose a measurable representative of $X$ on each basis vector, and extend each sample algebraically by finite sums. For every fixed $h$, this extension equals the constructed $L^2$ variable almost surely, so all its required distributions are unchanged.

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 $L_{\mathrm{loc}}^2$.

A projection-valued measure on the Borel sets of $\mathbb{C}$ is a map $E$ into the orthogonal projections on a separable Hilbert space $\mathcal{H}$ such thatand for pairwise disjoint $B_j$,for every $v\in \mathcal{H}$, with convergence in norm.

The spectral theorem for normal operators on a separable Hilbert space states that a bounded normal operator $A$ has a unique projection-valued measure supported on $\mathrm{Sp}(A)$ for whichMore generally, the Borel functional calculus for a normal operator is

For $v,w\in \mathcal{H}$, the scalar spectral measures areIf $A$ is self-adjoint, its spectrum and hence the support of $E$ lie in $\mathbb{R}$. Moreover,so ${\mu }_v$ is a positive measure, andThe paper prints total mass $\Vert v\Vert$; with the standard definition it is $\Vert v{\Vert }^2$, so the unsquared norm is a typographical error.