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I'm studying Bochner's theorem: If $\phi:\mathbb{R}\to\mathbb{C}$ is a Hermitian, positive definite, uniformly continuous function such that $|\phi(x)|\leq \phi(0)=1$ for all $x\in\mathbb{R}$, then $\phi$ is the Fourier transform $\hat{\mu}$ of some probability measure $\mu$ on $\mathbb{R}$.

I was wondering: Can one give a simpler, or more direct proof of Bochner's theorem if one assumes, in addition, that $\phi$ is integrable. I was hoping this would be a more accessible case since, then, there is a natural candidate for $\mu$, namely, the measure associated to the inverse Fourier transform of $\phi$: Let $$ f_\phi(t) := \int e^{-iut}\phi(u)du $$ be the inverse Fourier transform of $\phi$. It's well defined because $\phi$ is assumed to be integrable.

Question: Does it follow that $$ \phi(u) = \int e^{itu}f_\phi(t)dt? $$

This doesn't follow directly from Fourier inversion because $f_\phi$ isn't known to be $L^1$.

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    $\begingroup$ What does it mean to say $\phi$ is Hermitian? (In any case, that's not a hypothesis in the standard version: en.wikipedia.org/wiki/Bochner%27s_theorem) $\endgroup$ – David C. Ullrich May 24 '18 at 15:17
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    $\begingroup$ Your last statement is incorrect. If $f \in L^{1}(\mathbb{R})$ and $\hat{f} \in L^{1}(\mathbb{R})$, then $f(x) = \int_{\mathbb{R}} \hat{f}(\xi) e^{i 2 \pi \xi x} \, d\xi$. That is, Fourier inversion does hold for such $f$. $\endgroup$ – fourierwho May 24 '18 at 15:23
  • $\begingroup$ @fourierwho It's not quite that simple. If $f$ and $\hat f$ are both integrable then $f$ equals that integral almost everywhere; strictly speaking if you want to just say $f(x)=$ etc you need to assume $f$ is continuous. $\endgroup$ – David C. Ullrich May 24 '18 at 15:32
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    $\begingroup$ You're right, but that's a bit pedantic considering the question. The OP wants to know if $\mu$ is absolutely continuous with the derivative given by inverting the characteristic function, in which case the almost everywhere behavior of the derivative is all that matters. $\endgroup$ – fourierwho May 24 '18 at 15:43
  • $\begingroup$ @DavidC.Ullrich Hermitian means that $\phi(-x)=\overline{\phi(x)}$. I think it's corresponds to the associated $\mu$ being real-valued. $\endgroup$ – fmg May 25 '18 at 7:14
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Suppose $\mu$ is a finite signed Borel measure on $\mathbb{R}$. (Let's say it's real for convenience.) Suppose in addition that $\hat{\mu}$ (its Fourier transform) is an $L^{1}(\mathbb{R})$ function. I claim that if $E$ is a Borel set, then $\mu(E) = \int_{E} f(x) \, dx$, where $f(x) = \int_{\mathbb{R}} \hat{\mu}(\xi) e^{i 2 \pi \xi} \, d\xi$. I'll show this by proving that \begin{equation*} \forall \varphi \in \mathcal{S}(\mathbb{R}) \quad \int_{\mathbb{R}} \varphi(x) \, \mu(dx) = \int_{\mathbb{R}} \varphi(x) f(x) \, dx, \end{equation*} where $\mathcal{S}(\mathbb{R})$ is the Schwarz space.

The first thing to ascertain is whether or not $\hat{f}(\xi) = \hat{\mu}(\xi)$ on $\mathbb{R}$. However, this is clearly true in the sense of distributions so I won't attend to it further. (Note that $f \in L^{\infty}(\mathbb{R})$ is clear, but it's not immediately obvious to me at this point that $f \in L^{1}(\mathbb{R})$.)

Suppose $\varphi \in \mathcal{S}(\mathbb{R})$. Then $\varphi(x) = \int_{\mathbb{R}} \hat{\varphi}(\xi) e^{i 2 \pi \xi x} \, d\xi$ and, thus, \begin{align*} \int_{\mathbb{R}} \varphi(x) \, \mu(dx) &= \int_{\mathbb{R}} \int_{\mathbb{R}} \hat{\varphi}(\xi) e^{i 2 \pi \xi x} \, d \xi \mu(dx) \\ &= \int_{\mathbb{R}} \hat{\varphi}(\xi) \int_{\mathbb{R}} e^{i 2 \pi \xi} \mu(dx) d \xi \\ &= \int_{\mathbb{R}} \hat{\varphi}(\xi) \hat{\mu}(-\xi) \, d \xi \\ &= \int_{\mathbb{R}} f(x) \varphi(x) \, dx. \end{align*} Above I used the fact that $\hat{f} = \hat{\mu}$ in the sense of distributions. Since $\mathcal{S}(\mathbb{R})$ is dense in $C_{0}(\mathbb{R})$ --- indeed, the former contains $C_{c}^{\infty}(\mathbb{R})$---, it follows that $\mu = f \, dx$.

EDIT: I never completely addressed the question. Suppose now that $\varphi \in BUC(\mathbb{R}) \cap L^{1}(\mathbb{R})$ (BUC = bounded uniformly continuous functions) is Hermetian (meaning $\varphi(-\xi) = \overline{\varphi(\xi)}$), positive definite, and $|\varphi(x)| \leq \varphi(0) = 1$. The arguments presented above show that if I define $f \in L^{\infty}(\mathbb{R}; \mathbb{C})$ by $f(x) = \int_{\mathbb{R}} \varphi(\xi) e^{i 2 \pi \xi} \, d \xi$, then $\hat{f} = \varphi$ provided I can prove $f \in L^{1}(\mathbb{R})$. I claim that $f \geq 0$ and $\int_{\mathbb{R}} f(x) = 1$.

First, observe that $f(x) \in \mathbb{R}$ a.e. This follows from the Hermetian symmetry of $\varphi$. Also observe that $1 = \varphi(0) = \int_{\mathbb{R}} f(x) \, dx$. It remains to show $f \geq 0$. We do this by showing that if $\psi \in \mathcal{S}(\mathbb{R})$, then $\int_{\mathbb{R}} \psi(x)^{2} f(x) \, dx \geq 0$. Indeed, given such a $\psi$, we can write $\psi(x) = \int_{\mathbb{R}} \hat{\psi}(\xi) e^{i 2 \pi \xi x} \, d\xi$ and the integral converges uniformly (w.r.t. $x$). Therefore, if we write $\psi_{n}(x) = N^{-1} \sum_{j = -N^{2}}^{N^{2}} \hat{\psi}(jN^{-1}) e^{i 2 \pi jN^{-1} x}$, then $\psi_{n} \to \psi$ uniformly on $\mathbb{R}$. Integrating, we appeal to positive definiteness to find \begin{align*} \int_{\mathbb{R}} |\psi_{n}(x)|^{2} f(x) \, dx &= N^{-2} \sum_{i,j = -N^{2}}^{N^{2}} \hat{\psi}(iN^{-1}) \overline{\hat{\psi}(jN^{-1})} \int_{\mathbb{R}} e^{-i 2 \pi (i - j)N^{-1} x} f(x) \, dx \\ &= N^{-2} \sum_{i, j = -N^{2}}^{N^{2}} \hat{\psi}(iN^{-1}) \overline{\hat{\psi}(jN^{-1})} \varphi(iN^{-1} - jN^{-1}) \geq 0. \end{align*} Sending $n \to \infty$, we find \begin{align*} \int_{\mathbb{R}} \psi(x)^{2} f(x) \, dx \geq 0. \end{align*} Since this is true when $\psi \in \mathcal{S}(\mathbb{R})$, it follows that it's true when $\psi \in L^{1}(\mathbb{R})$ by density (and boundedness of $f$). Finally, observe that if $g \in C_{c}(\mathbb{R})$ and $g \geq 0$, then $\sqrt{g} \in C_{c}(\mathbb{R})$ and, thus, $\int_{\mathbb{R}} g(x) f(x) \, dx \geq 0$. Since this is true independently of $g$, it follows that $f \geq 0$. This completes the proof of Bochner's Theorem.

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  • $\begingroup$ Ok, I see. $\int f(x)dx=1$ and $f\geq 0$ imply that $f$ is absolutely integrable, hence $L^1$, so you can apply Fourier inversion. And, to establish these facts about $f$, we need all the hypotheses of Bochner's theorem. Nice!! $\endgroup$ – fmg May 25 '18 at 7:20

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