Bochner's theorem. A function (phi: mathbb{R}^{n} ightarrow mathbb{C}) is positive semidefinite, if the matrices (left(phileft(xi_{i}-xi_{k}ight)ight)_{i, k=1}^{m}) are

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Bochner's theorem. A function \(\phi: \mathbb{R}^{n} ightarrow \mathbb{C}\) is positive semidefinite, if the matrices \(\left(\phi\left(\xi_{i}-\xi_{k}ight)ight)_{i, k=1}^{m}\) are positive hermitian for all \(m \geqslant 1\) and \(\xi_{1}, \ldots, \xi_{m} \in \mathbb{R}^{n}\), i.e. \(\sum_{i, k=1}^{m} \phi\left(\xi_{i}-ight.\) \(\left.\xi_{k}ight) \lambda_{i} \bar{\lambda}_{k} \geqslant 0\) for all \(\lambda_{1}, \ldots, \lambda_{m} \in \mathbb{C}\). The steps below sketch the proof of the theorem.

Theorem (Bochner). A continuous function \(\phi: \mathbb{R}^{n} ightarrow \mathbb{C}\) is the Fourier transform of a finite measure \(\mu\) on \(\left(\mathbb{R}^{n}, \mathscr{B}\left(\mathbb{R}^{n}ight)ight)\) if, and only if, \(\phi\) is positive semidefinite.

(i) The Fourier transform \(\widehat{\mu}(\xi)\) of a finite measure \(\mu\) is continuous and positive semidefinite.

(ii) Assume that \(\phi\) is continuous and positive semidefinite. Then \(\phi(0) \geqslant 0, \phi(\xi)=\overline{\phi(-\xi)}\), and \(|\phi(\xi)| \leqslant \phi(0)\).

[use that \((\phi(0))\) and

\[
\left(\begin{array}{cc}
\phi(0) & \phi(-\xi) \\
\phi(\xi) & \phi(0)
\end{array}ight)
\]

are positive semidefinite matrices.]

(iii) Prove that \(u_{\epsilon}(x):=\iint \phi(\xi-\eta)\left(e^{i x \cdot \xi} e^{-2 \epsilon|\xi|^{2}}ight) \overline{\left(e^{i x \cdot \eta} e^{-2 \epsilon|\eta|^{2}}ight)} d \xi d \eta\) is positive and that

\[u_{\epsilon}(x)=\frac{1}{c} \int \phi_{\epsilon}(\eta) e^{i x \cdot \eta} d \eta \quad \text { with } \phi_{\epsilon}(\eta)=e^{-\epsilon|\eta|^{2}} \phi(\eta)\]

(iv) Show that the function \(u_{\epsilon}\) is Lebesgue integrable.

(v) Conclude that \(\phi_{\epsilon}\) is the Fourier transform of \(c u_{\epsilon}\). Apply Lévy's continuity theorem (Problem 21.3 ) to \(\lim _{\epsilon ightarrow 0} \phi_{\epsilon}(\xi)=\phi(\xi)\).

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