 Solution

Choose $\varphi (x)=\int d^{4}p(2\pi {)}^{-4}{e}^{ip\cdot x}\tilde{\varphi }(p)$ in Euclidean signature. The inverse quadratic kernel is the smooth-cutoff scalar propagatorThe subscript records the explicit cutoff dependence implied by the condition $f_{\Lambda }(z)={\Lambda }^{2}z$ at small $z$. For $p^2+m^2\le {\Lambda }^2$, this reduces to $1/(p^2+m^2)$, the usual Euclidean scalar propagator. For $p^2+m^2\gg {\Lambda }^2$, the kernel $f_{\Lambda }$ grows rapidly and its inverse tends to zero.

High-momentum modes are strongly suppressed. For a finite smooth regulator they are not literally identically zero; that statement would require a sharp cutoff. The field variance carried by those Fourier transform modes is correspondingly negligible.