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For a nearest-neighbor auxiliary-field representation of an Ising model, sharing each local potential between adjacent bonds gives the displayed symmetric integral operator. Its periodic partition function is the trace of its $N$th power, up to Gaussian normalization constants. In the slowly varying continuum limit, a slice of length $ϵ$ has kernel $[M/(2\pi ϵ){]}^{1/2}\exp [-M(m^{\prime }-m{)}^2/(2ϵ)-ϵ\{U(m^{\prime })+U(m)\}/2]$. Expanding its action on a smooth function gives $1-ϵ\hat{H}+O({ϵ}^2)$, with $\hat{H}=-(2M{)}^{-1}{\partial }_m^2+U$. Thus fixed endpoints give an imaginary-time transition kernel and periodic endpoints give a Euclidean path integral trace. The finite-spacing integral operator and the continuum differential operator need not have identical high-energy spectra.

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.

An integral kernel is a function or distribution specifying an integral operator through $(Tf)(x)=\int K(x,y)f(y)dy$. A Green function is an integral kernel that inverts a differential operator under specified boundary conditions.

For $\alpha \ne 0$, the Fredholm alternative gives a unique solution exactly when $\alpha \ne -{\lambda }_n$ for all $n\ge 0$, unless a resonant problem could exceptionally be compatible. Here none is: with ${\kappa }_n=n+1/2$,At every resonance the compatibility condition therefore fails.

The case $\alpha =0$ must be checked separately because $L$ is a compact first-kind integral operator. For every admissible $y$, $Ly$ is continuous and $(Ly)(\pi )=0$, whereas $\cos \pi =-1$. Hence $Ly=\cos x$ has no solution. Combining these cases,In the nonresonant nonzero case the shifted eigenvalues stay away from zero, since ${\lambda }_n\to 0$ and $\alpha \ne 0$, so inversion is bounded on the Hilbert space.

Write $v=Ly$ asDifferentiating twice, the moving-limit terms cancel at first order and yield $\boxed{v^{\prime \prime }+v=y}$ at second order. Also $v^{\prime }(0)=0$ and $v(\pi )=0$. Thus this integral operator is the inverse of $D^2+1$ with mixed Neumann boundary condition and Dirichlet boundary condition.

For ${\kappa }_n=n+1/2$, $y_n(x)=\cos ({\kappa }_{n}x)$ satisfies those conditions and $(D^2+1)y_n=(1-{\kappa }_n^2)y_n$. The corresponding mixed homogeneous equation has no nonzero solution, so uniqueness givesThis verifies the asserted eigenvalues and eigenfunctions for every $n\ge 0$.

Write $u(t)=e^t$ and $v(t)=e^{-t}$. Both lie in the Hilbert space $L^2(-\pi ,\pi )$, and the integral operator factors as the rank-one operatorThe Cauchy-Schwarz inequality gives $\Vert Tf\Vert \le \Vert u\Vert \Vert v\Vert \Vert f\Vert$, and equality is attained at $f=v/\Vert v\Vert$. Hence the operator norm is $\Vert T\Vert =\Vert u\Vert \Vert v\Vert$. Direct integration givesFor any Hilbertian basis $(h_k)$, the Parseval identity for a Hilbertian basis givesThis is the general formula for the norms of a rank-one operator. In particular $T$ is a Hilbert-Schmidt operator, andThis computation needs no explicit Fourier basis. The printed optional basis formula omits the variable $t$: the intended Hilbertian basis is $e_n(t)=(2\pi {)}^{-1/2}{e}^{int}$, $n\in \mathbb{Z}$. Taken literally, the printed expressions are constant functions and cannot form an orthonormal basis of $L^2(-\pi ,\pi )$.

Let $\Vert K{\Vert }_{\infty }={\max }_{[0,1{]}^2}|K|$. For $f,g\in C([0,1])$,Taking the supremum over $x$ proves that the integral operator $T$ is Lipschitz continuous, hence continuous, in the uniform norm.

It remains to see that $Tf$ is continuous. Since $K$ is uniformly continuous on the compact square and $f$ is bounded,as $x^{\prime }\to x$. Therefore $T:C([0,1])\to C([0,1])$ is well defined and continuous.

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.