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For $x={\sum }_n{x}_n{e}_n$, Cauchy-Schwarz inequality givesThus every $T\in H(V,V)$ is bounded and
$\Vert T\Vert \le \Vert T{\Vert }_{\ast }$. The truncationshave finite rank and satisfyHence every such Hilbert-Schmidt operator is compact.

DefineThis is an inner product inducing $\Vert \cdot {\Vert }_{\ast }$. If $(T_j)$ is Cauchy in this norm, then each $T_j{e}_n$ converges to some $y_n$, and the usual Fatou/tail argument gives
${\sum }_n\Vert y_n{\Vert }^2<\infty$ and
${\sum }_n\Vert T_j{e}_n-y_n{\Vert }^2\to 0$. Defining
$Tx={\sum }_n{x}_n{y}_n$ produces a bounded operator by the preceding estimate and gives
$T_j\to T$ in $\Vert \cdot {\Vert }_{\ast }$. Thus $(H(V,V),\Vert \cdot {\Vert }_{\ast })$ is a Hilbert space.

The norms are not equivalent in infinite dimension. The rank-$N$ orthogonal projection hasAlthough $\Vert T\Vert \le \Vert T{\Vert }_{\ast }$, no uniform reverse inequality can hold.

Write $x\otimes y$ for the rank-one operator $h\mapsto ⟨h,y⟩x$. The estimator is the kernel representation of the empirical covariance operatorSince $\mathbb{E}X=0$,Thus the estimator has the fixed bias of an estimator ${\mu }_0\otimes {\mu }_0$ for $C_X$.

The fourth-moment assumption makes $(X_i-{\mu }_0)\otimes (X_i-{\mu }_0)$ square-integrable in the Hilbert space of Hilbert-Schmidt operators. The weak law of large numbers therefore givesConsequently consistency for $C_X$ holds exactly when ${\mu }_0=0$. More precisely, if $Y=(X-{\mu }_0)\otimes (X-{\mu }_0)$, thenThe Hilbert-space central limit theorem also yieldswhere $G$ is a centered Gaussian random element in the Hilbert-Schmidt operator space with covariance determined by $Y$. Relative to $C_X$, the same fluctuation is displaced by $\sqrt{n}({\mu }_0\otimes {\mu }_0)$ and hence does not have a finite centered limit when ${\mu }_0\ne 0$.

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 ${\sum }_k{\gamma }_k{\sum }_{j:{\lambda }_j>0}{\rho }_{jk}^2$.

Let $T=AB$ with $A,B$ Hilbert--Schmidt, and write $T=U|T|$. Since $|T|=U^{\ast }AB$, part iii and the Cauchy-Schwarz inequality for series giveThus $T$ is trace class with the required bound.

The kernel $k(x,y)=e^{-|x-y|}$ is real and symmetric. For $f,g\in L^2[0,1]$, Fubini's theorem givesso $K$ is self-adjoint. It is linear, and because $k\in L^2([0,1{]}^2)$ it is a Hilbert-Schmidt operator, hence bounded, with

The Schatten norm Hölder inequality gives $\Vert AB{\Vert }_r\le \Vert A{\Vert }_p\Vert B{\Vert }_q$ when $1/r=1/p+1/q$. In particular, the product of two Hilbert-Schmidt operators is trace class and

For $(\Delta +k^2)\psi =Q$, the outgoing Green function for the three-dimensional Helmholtz equation yields the far-field patternThe sign corresponds to the Green function satisfying $(\Delta +k^2)G=-\delta$. This Hilbert-Schmidt operator samples the source Fourier transform on the sphere $|\mathbf{\xi }|=k$; for a ball of radius $r_0$, its squared Hilbert-Schmidt norm is $r_0^3/3$.