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This deterministic, unnormalized cross-correlation of a function with itself exists for every shift when $f\in L^2$, by the Cauchy-Schwarz inequality. It satisfies $h(0)=\Vert f{\Vert }_2^2$ and $h(-x)=\overline{h(x)}$. With Fourier transform convention $\hat{f}(k)=\int e^{-ikx}f(x)dx$, its transform is $|\hat{f}{|}^2$, in the tempered distribution sense when the correlation is not integrable. Fourier inversion at zero gives the Parseval identity, first for Schwartz functions and then by approximation in L2 space. Unlike the normalized statistical autocorrelation, its value at zero is not generally one.

With the stated Fourier transform convention, the inverse Fourier transform and the deterministic autocorrelation of a square-integrable function areAssume initially sufficient decay and smoothness, for example a Schwartz function, to justify interchanging integrations and Fourier inversion. With $s=x+y$, the cross-correlation identity isEvaluating $h$ directly and by its inverse transform at zero proves the Parseval identity:Approximation extends the identity to L2 space, as the Plancherel theorem.

For the compact cosine profile, the Fourier transform of a truncated cosine follows from $\cos x=(e^{ix}+e^{-ix})/2$:At $k=\pm 1$ the apparent singularities are removable, with value $\pi /2$. The direct squared norm is ${\int }_{-\pi /2}^{\pi /2}{\cos }^{2}xdx=\pi /2$. Hence Parseval identity and evenness givePutting $t=\pi k/2$ makes the left side ${\pi }^3/8$ times the requested integral. ThereforeAt $t=\pi /2$ the integrand's continuous extension is $1/{\pi }^2$, so no principal-value prescription is required.

Set $g=f$. The cross-correlation at zero is $h(0)={\int }_{\mathbb{R}}|f(x){|}^{2}dx$, while part (i) gives $\hat{h}=|\hat{f}{|}^2$. Evaluating the inverse Fourier transform at zero yields the Parseval identityThis derivation is immediate for Schwartz functions, for which all integrations and Fourier inversion are justified. Approximation by such functions and completeness then extend the identity to L2 space, defining its Fourier transform by the corresponding norm limit. This is the Plancherel theorem extension; no absolute integrability of an arbitrary square-integrable function is presumed.

First take $f,g\in L^1(\mathbb{R})$, so Fubini's theorem applies to the cross-correlation double integral. With $s=x+y$, the Fourier transform isThe complex conjugation and the sign in $x+y$ are both essential to this convention.