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Under $H_0$, the Hilbert-space central limit theorem giveswhere $G$ is a centered Gaussian random element with covariance $C_X$. Distinct leading eigenvalues imply consistency of the empirical eigenvalues and eigenfunctions, up to irrelevant signs. The first $K$ standardized functional principal component scores of $G$ are independent $N(0,1)$ variables, so Slutsky theorem yieldsAn asymptotic level-$\alpha$ test rejects above the $(1-\alpha )$ quantile of the chi-squared distribution.

Under a fixed alternative, put $\delta =\mu -{\mu }_0$. The weak law of large numbers and eigenpair consistency giveThus the statistic diverges and the test is consistent whenever at least one retained projection is nonzero. A fixed alternative orthogonal to the first $K$ eigenfunctions is invisible to this fixed-$K$ statistic.

Letand define the functional principal component scoresUse empirical centered scores ${\hat{\xi }}_{ik},{\hat{\eta }}_{il}$ from the sample covariance eigenfunctions and formThe finite-dimensional statistic is

Under $H_0$, $Y=\epsilon$ is independent of $X$. The population cross-covariances vanish, andThe multivariate central limit theorem, consistency of the empirical eigenpairs, and Slutsky theorem therefore giveRejecting above the $(1-\alpha )$ quantile gives an asymptotic level-$\alpha$ test.

Under a fixed alternative, $\mathbb{E}[{\eta }_l{\xi }_k]$ is the $(l,k)$ coordinate of the cross-covariance operator $C_{YX}=B{C}_X$, where $B$ is the Hilbert-Schmidt operator with kernel $\beta$. ThusThe test is consistent whenever this retained block contains a nonzero cross-covariance; fixed truncation can miss alternatives outside the selected principal-component subspaces.

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

If the $B_k$ form the eigenbasis of $C_X$, thenThe retained functional principal component scores are therefore uncorrelated with every omitted score, so $g=0$. The least-squares estimates of $c_1,\dots ,c_K$ are consistent despite truncation. Fixed $K$ still estimates only the projection ${\beta }_K$; consistency for the complete function requires the truncation level to increase and the tail error to vanish.

Let $B$ be the Hilbert-Schmidt operator with kernel $\beta$, so the function-on-function linear model is $Y=BX+\epsilon$. Independence and centering give $\mathbb{E}[Y\mid X]=BX$. Letand define the functional principal component scoresThe cross-covariance operator identity $C_{YX}=B{C}_X$ gives, for every ${\lambda }_j>0$,

Since $BX$ is centered, the requested integrated variance is $\mathbb{E}\Vert BX{\Vert }^2$. Applying the Karhunen–Loève expansion to $X$ and the Parseval identity in the ${\psi }_k$ basis yieldsEquivalently, if ${\rho }_{jk}=\mathrm{Corr}({\xi }_j,{\eta }_k)$, the expression is ${\sum }_k{\gamma }_k{\sum }_{j:{\lambda }_j>0}{\rho }_{jk}^2$.