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A smooth uniformly elliptic differential expression with symmetric Dirichlet bilinear form has an unbounded self-adjoint operator realization $A_D$ on $L^2(U)$, whose domain is $H^2(U)\cap H_0^1(U)$ on a smooth bounded domain. If the form is strictly positive, the Lax-Milgram theorem and elliptic regularity make $A_D^{-1}$ a bounded map from $L^2$ to $H^2\cap H_0^1$. Its action on $L^2$ is compact by the Rellich-Kondrachov compactness theorem and self-adjoint by symmetry of the form. The spectral theorem for compact Hermitian operators therefore gives an orthonormal basis of smooth Dirichlet eigenfunctions with positive eigenvalues tending to infinity.

Apply the spectral theorem for compact Hermitian operators to positive $A=T^{\ast }T$. Its positive eigenvalues ${\mu }_i$ have orthonormal eigenvectors $u_i$ spanning $\ker A^{\perp }$, with a finite or countable list. In the infinite case ${\mu }_i\to 0$. DefineThen $(v_i,v_j)=(u_i,T^{\ast }T{u}_j)/({\lambda }_i{\lambda }_j)={\delta }_{ij}$, so the $v_i$ are orthonormal. Also $\ker A=\ker T$, because $(x,Ax)=\Vert Tx{\Vert }^2$. Decompose $x=x_0+{\sum }_i(u_i,x)u_i$ with $x_0\in \ker T$ and apply bounded $T$ to the norm-convergent expansion. This gives the singular value decompositionThe series converges in norm; its squared tail norm is ${\sum }_{i>m}{\mu }_i|(u_i,x){|}^2\to 0$. No separability of the whole Hilbert space is necessary: the nonzero spectral subspace of a compact operator is separable. For $T=0$ use the empty list $N=0$.

First construct the Dirichlet eigensystem. By part (i) and the Lax-Milgram theorem, the inverse $S=A_D^{-1}$ exists from $L^2$ into $H_0^1$. Elliptic regularity gives $Sf\in H^2\cap H_0^1$, and the Rellich-Kondrachov compactness theorem makes $S$ a compact operator on $L^2$. Symmetry givesso $S$ is a self-adjoint operator. It is injective, and $(Sf,f)=B[Sf,Sf]>0$ for $f\ne 0$. The spectral theorem for compact Hermitian operators supplies an orthonormal basis $(w_m)$ with $S{w}_m={\mu }_m{w}_m$, ${\mu }_m>0$, and ${\mu }_m\to 0$. Put ${\lambda }_m={\mu }_m^{-1}$. ThenRepeated elliptic regularity makes each $w_m$ smooth up to the boundary.

As with part (ii), the stated characterization omits boundary compatibility. At $k=0$, its right-hand side is finite for every $u\in L^2$ by Parseval identity, whereas its left-hand side would require $u\in H_0^1$. At higher orders, the preceding polynomial example is also decisive for the Dirichlet eigensystem: on $(0,\pi )$,Although this function belongs to every ordinary $H^k$, the weighted sum diverges at $k=3$.

The correct spectral characterization of elliptic Dirichlet domains iswith $X_k$ defined in part (ii). For finite eigenfunction sums the squared norm $((u,u){)}_k$ is exactly the weighted sum: even orders follow from $L^l{w}_m={\lambda }_m^l{w}_m$, and odd orders also use $B[w_m,w_j]={\lambda }_m{\delta }_{mj}$. These sums are dense in $X_1$: if $v$ is $B$-orthogonal to every $w_m$, then $0=B[v,w_m]={\lambda }_m(v,w_m)$, so $v=0$. They are dense in $X_0$ by construction, and the isomorphisms $A_D:X_{k+2}\to X_k$ propagate density to every $X_k$. Completion in the equivalent norms from part (ii) proves both directions of the corrected equivalence. In particular, the printed characterization is correct for $k=1,2$.

The Cameron-Martin space of a Gaussian measure $\mu =\mathcal{N}(0,\Sigma )$ isThe inverse is defined on the range of ${\Sigma }^{1/2}$. In this infinite-dimensional setting “positive definite” must mean $⟨\Sigma h,h⟩>0$ for every nonzero $h$, rather than a uniform lower bound: a trace-class operator cannot be uniformly positive on an infinite-dimensional Hilbert space.

By the spectral theorem for compact Hermitian operators, choose an orthonormal basis $(e_j)$ with $\Sigma e_j={\lambda }_j{e}_j$, ${\lambda }_j>0$, and ${\sum }_j{\lambda }_j<\infty$. (A strictly positive trace-class operator also forces the ambient Hilbert space to be separable.) In coordinates $h_j=⟨h,e_j⟩$,This is a Hilbert space with its indicated norm, even though its range need not be closed in the ambient norm.

The Cameron-Martin theorem for a Gaussian measure states that the translated law ${\mu }_h(B)=\mu (B-h)=\mathcal{N}(h,\Sigma )(B)$ is an equivalent probability measure to $\mu$ exactly when $h\in E$. For such $h$, its Radon-Nikodym derivative isThe series has mean-square convergence and converges almost surely under $\mu$, since its independent summands have total variance ${\sum }_j{h}_j^2/{\lambda }_j$. Its law is $\mathcal{N}(0,\Vert h{\Vert }_E^2)$, so the density has expected value one. It is sometimes formally written ${\ell }_h(x)=⟨h,x{⟩}_E$, but a typical infinite-dimensional Gaussian sample is not in $E$; the series interpretation is essential. If $h\notin E$, the two laws are mutually singular measures.

The Karhunen–Loève expansion has the following Hilbert-space form. Let $X$ be a square-integrable Random element of a Hilbert space $H$, with mean $\mu$ and covariance operator $C$. There are nonnegative eigenvalues ${\lambda }_1\ge {\lambda }_2\ge \cdots$ and orthonormal eigenvectors ${\varphi }_k$ spanning the closure of the range of $C$ such thatin $L^2(\Omega ;H)$. The functional principal component scoressatisfyIf $X$ is a Gaussian random element, the scores are independent normal random variables.

For the proof, $C$ is positive, self-adjoint, and trace-class, withIt is therefore compact, so the spectral theorem for compact Hermitian operators supplies the eigenpairs. Their score covariance isFor the residual $R_m=X-\mu -{\sum }_{k=1}^m{\xi }_k{\varphi }_k$, Parseval identity and the trace formula giveThe centered variable's projection onto $\ker C$ has zero second moment and is therefore zero almost surely, which completes the mean-square expansion. When $C$ has a continuous covariance kernel, Mercer's theorem additionally expands that kernel as $c(s,t)={\sum }_k{\lambda }_k{\varphi }_k(s){\varphi }_k(t)$.

The operator $A^{\ast }A$ is positive, self-adjoint, and compact. Because $\mathrm{ran}A$ is infinite dimensional, the spectral theorem for compact Hermitian operators gives positive spectral values tending to zero. Even if $\ker A=\{0\}$, zero remains in the spectrum as a limit point. Hence for every $\tau$,The strict inequality $\Vert I-\tau K\Vert <1$ needed for the operator-norm Neumann series is impossible, so formula (3) cannot be applied directly to invert $A^{\ast }A$.

The spectral theorem for compact Hermitian operators says that the nonzero spectrum of a compact Hermitian $T$ consists of real eigenvalues of finite multiplicity, with zero as the only possible accumulation point. Eigenvectors for distinct eigenvalues are orthogonal, and an orthonormal basis of ${\ell }^2$ can be chosen from eigenvectors together with a basis of $\ker T$. Thuswhere the nonzero eigenvalues are repeated according to multiplicity and ${\lambda }_n\to 0$ if there are infinitely many.

DefineEach $T_N$ is finite-rank and Hermitian, andHence every compact Hermitian operator is a norm limit of finite-rank Hermitian operators, by finite-rank truncation of a compact Hermitian operator.