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The defining integral is well-defined and bounded for every dimension: $|\mathcal{F}f|\le \Vert f{\Vert }_1$ and the multiplier is supported on a finite-volume ball. However the asserted general-dimensional $L^1$ conclusion is false for the Euclidean radial cutoff printed in the original PDF. The nonintegrability of a radial triangular Fourier cutoff in three dimensions is a source issue, not an omitted conjugate or OCR substitution.

For a direct counterexample in $n=3$, take the Schwartz function $f(x)=e^{-\pi |x{|}^2}$, whose Fourier transform is the same Gaussian. With $r=|x|$ and $a=2\pi r$, radial integration givesThe amplitude $b(\rho )=\rho (1-\rho /R)e^{-\pi {\rho }^2}$ vanishes at both endpoints, and $b^{\prime }(R)=-e^{-\pi R^2}$. Two integrations by parts, with a further one to bound the remainder, giveThe integral ${\int }_1^{\infty }{r}^2|F_R(r)|dr$ diverges logarithmically for every fixed $R>0$. For example, restrict to the periodic subintervals on which $|\sin (2\pi Rr)|\ge 1/2$; the error is smaller than half the leading term at sufficiently large $r$. Thus $F_R\notin L^1({\mathbb{R}}^3)$, even for this smooth integrable $f$.

The familiar intended argument is valid in one dimension. There the inverse Fourier kernel isand $F_R=K_R\ast f$ by Fubini's theorem. Its mass outside any fixed neighbourhood of zero tends to zero by scaling, soby continuity of translations in $L^1$. For arbitrary dimension, replacing the radial cutoff by the product ${\prod }_{j=1}^n(1-|{\xi }_j|/R{)}_{+}$ gives a tensor product of these integrable kernels and the same approximate identity proof. These are explicit valid repairs, not claims that the PDF specified a product cutoff.