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When a covariance operator on a function space is an integral operator, its covariance kernel satisfies $(C_{X}f)(s)=\int c_X(s,t)f(t)dt$. For a centered process, $c_X(s,t)=\mathbb{E}[X(s)X(t)]$ whenever point evaluation is meaningful.

Functional data analysis treats each observation as a function or another infinite-dimensional object. Means, covariance operators, spectral coordinates, and regression operators replace their finite-dimensional vector and matrix counterparts.

Functional principal component analysis diagonalizes a compact covariance operator. Its eigenfunctions are principal component functions, and projecting a centered observation onto them gives uncorrelated functional principal component scores.

A random element $G$ of a Hilbert space is Gaussian when every continuous linear functional of $G$ has a normal distribution. Its mean and covariance operator determine its distribution.

For independent identically distributed centered random variables in a separable Hilbert space with finite second moment, $n^{-1/2}{\sum }_{i=1}^n{X}_i$ converges in distribution to a centered Gaussian random element with the same covariance operator.

Write $Z=X-\mu$, so the covariance operator isFor $g,h\in L^2[0,1]$,so $C_X$ is self-adjoint. Moreover,so it is a positive operator.

Let $(e_j)$ be any orthonormal basis. Tonelli theorem and Parseval identity giveA positive operator with finite trace is a trace-class operator, completing the proof.

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 $c(s,t)={\sum }_k{\lambda }_k{\varphi }_k(s){\varphi }_k(t)$.

Let $({\lambda }_j,{\varphi }_j)$ be the ordered eigenpairs of the common covariance operator $C_X$, and write $\delta =\mu -{\mu }^{\ast }$. Estimate $C_X$ by the pooled covariance operatorand denote its first $K$ eigenpairs by $({\hat{\lambda }}_j,{\hat{\varphi }}_j)$. The sign ambiguity of each eigenfunction disappears after squaring. Consider the Two-sample FPCA mean statistic

Under $H_0$, the Hilbert-space central limit theorem giveswhere $G$ is a centered Gaussian random element with covariance $C_X$. The assumed eigenvalue gaps give consistency of the estimated eigenvalues and eigenfunctions, so Slutsky's theorem yieldsAn asymptotic level-$\alpha$ test therefore rejects when $T_{n,K}$ exceeds the $(1-\alpha )$-quantile of the chi-squared distribution with $K$ degrees of freedom.

Under a fixed alternative,The test is consequently consistent whenever the mean difference has a nonzero projection onto one of the retained principal components. A difference orthogonal to their span is invisible to this fixed-$K$ test, so $\mu \ne {\mu }^{\ast }$ alone does not guarantee consistency. Increasing $K$ with $n$ can recover such alternatives, but requires additional eigenvalue and approximation control.

This symmetric finite-rank integral operator is not positive. For $f(t)=1-\frac{3}{2}t$, set $A_j={\int }_0^1{t}^{j}f(t)dt$. Directly,The associated quadratic form isEvery covariance operator is a positive operator, so this operator is not a covariance operator.

The identity operator is positive and self-adjoint, but on the infinite-dimensional Hilbert space $L^2[0,1]$ it has eigenvalue one with infinite multiplicity. ConsequentlyA square-integrable Hilbert-space random variable has a trace-class covariance operator with trace $\mathbb{E}\Vert X{\Vert }^2<\infty$. The identity therefore cannot be such a covariance operator.

This is the Brownian covariance kernel, so its integral operator is a covariance operator. To obtain its eigendecomposition, suppose $Uf=\lambda f$ with $\lambda \ne 0$. Splitting the integral at $s$ givesDifferentiation yields $\lambda f^{\prime }(s)={\int }_s^{1}f(t)dt$ and $\lambda f^{\prime \prime }(s)=-f(s)$, with boundary conditions $f(0)=0$ and $f^{\prime }(1)=0$. Hence the normalized eigenpairs areThe eigenvalues are positive and summable, consistently with positivity and the trace-class operator property of a covariance operator.

Let $({\hat{\lambda }}_k,{\hat{\varphi }}_k)$ be the leading eigenpairs of the sample covariance operator. For fixed $K$ with ${\lambda }_1>\cdots >{\lambda }_K>{\lambda }_{K+1}$ and ${\lambda }_K>0$, the FPCA mean test usesUnder the null, consistency of the empirical eigenpairs and the multivariate central limit theorem implyThe level-$\alpha$ test therefore rejects above the $(1-\alpha )$ quantile of the chi-squared distribution with $K$ degrees of freedom.

For a fixed mean $\mu$, if at least one leading coordinate $⟨\mu ,{\varphi }_k⟩$, $k\le K$, is nonzero, then $T_{K,n}\to \infty$ in probability and the test is consistent. It has only null-level asymptotic power against means orthogonal to the first $K$ principal component functions. Under ${\mu }_n=n^{-1/2}\delta$, the limit is noncentral chi-squared with noncentrality

The minimizer is the arithmetic meanIt remains a positive self-adjoint trace-class operator and hence a covariance operator. The Hilbert-Schmidt inner product gives, for every Hilbert-Schmidt operator $C$,because ${\sum }_k(C_k-\overline{C})=0$. Thus $\overline{C}$ is the unique minimizer, including when the minimization is restricted to covariance operators.

A principal component function is a normalized eigenfunction ${\varphi }_k$ of a covariance operator, ordered by decreasing eigenvalue ${\lambda }_k$.