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Use the Fourier transform convention $\hat{\nu }(\xi )=\int e^{-ix\cdot \xi }d\nu (x)$ throughout this question. The decay assumption implies square integrability, since polar coordinates giveNear zero the integrand is bounded by $r$, and at infinity it is bounded by $r^{-1-2ϵ}$. The positive $ϵ$ is what makes the latter integrable.

By the Plancherel theorem, there is a function $g\in L^2({\mathbb{R}}^2)$ whose Fourier transform is $\hat{\mu }$. It is the density of $\mu$ with respect to Lebesgue measure. To justify this step rather than assume a density, for every Schwartz function $\phi$, Fourier inversion givesThe finite measure and the locally integrable function thus define the same tempered distribution, so they agree as measures: $d\mu =gdx$. In particular $g\ge 0$ almost everywhere and $\int g=\mu ({\mathbb{R}}^2)<\infty$. This is the L2 density from a square-integrable Fourier transform principle.

For $f\in L^{\infty }(\mu )$, the density of $fd\mu$ is $fg$. The inequality $|f|\le \Vert f{\Vert }_{L^{\infty }(\mu )}$ holds wherever $g>0$, apart from a Lebesgue measure zero set, soA second application of the Plancherel theorem yields the explicit boundThe implicit constant in the requested estimate may depend on the measure and its Fourier-decay bound, but is independent of $f$.

First take a Schwartz function, so that Fourier inversion and the exchange of integrals are justified. Inserting the inverse transform of the complex conjugate givesFor $f\in L^1\cap L^2$, approximate it by Schwartz functions in both norms. The identity just proved makes their transforms a Cauchy sequence in L2 space, while convergence in L1 space makes the transforms converge uniformly to the defining transform of $f$. Thus the two transform limits agree, and passing to the $L^2$ norms proves the finite Parseval identity. This is the normalization of the Plancherel theorem appropriate to the paper.

There is a qualification to the printed hypothesis: absolute integrability alone does not guarantee a finite square integral. For instance $f(x)=x^{-3/4}$ on $(0,1)$ and zero elsewhere belongs to $L^1$ but not $L^2$. For an $L^1$ function of this kind the transform is not in $L^2$ either, by L2 density from a square-integrable Fourier transform, so both sides are infinite if the identity is interpreted in that extended sense. The finite equality requires square integrability as well.