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I am studying Fourier analysis and finding two Bochner's theorem:

Bochner's theorem

VERSION $1$: In order that a function $f:\mathbb{R}^d\rightarrow \mathbb{R}$ be positive definite and continuous, it is necessary and sufficient that it be the Fourier transform of a nonnegative finite-valued Borel measure on $\mathbb{R}^d$. [In Cheney's book---A course in approximation theory]

VERSION $2$: Let $f:\mathbb{R}^d\to \mathbb{R}$ be a bounded and continuous function on $\mathbb{R}^d$. Then $f$ is a positive semi-definite function on $\mathbb{R}^d$ if and only if there is a probability measure $\mu$ on $\mathbb{R}^d$ such that $$ f(x) =f(0) \int_{\mathbb{R}^d} e^{2\pi i x\cdot \xi}\, d\mu(\xi),\ \ \forall x\in \mathbb{R}^d.$$ [In my course note]

My question is in the necessary parts of version 1 and 2. I am wondering whether they are same? or is there a relationship between them?

In my understanding, VERSION $1$ says if $f$ is a positive definite then it be the Fourier transform of a Borel measure. But in VERSION $2$, it says if $f$ is a positive semi-definite then it be the inverse Fourier transform of a Borel measure(i.e. the probability measure in version 2).

Thanks in advance!

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    $\begingroup$ But both versions are incorrect: $f$ should be complex-valued. Once this is fixed, the two versions are equivalent, since $\xi\mapsto-\xi$ makes swaps Fourier transform and its inverse. $\endgroup$ Aug 21 '18 at 23:25
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    $\begingroup$ Apart from the $\mathbb{C}$-valuedness issue mentioned by @CaveJohnson, version 1 should also say positive semidefinite. Sometimes people are sloppy and say "postive definite" when they mean "positive semidefinite". $\endgroup$ Aug 22 '18 at 13:32
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To be hugely explicit about why the two are clearly equivalent:

If $\mu$ is a measure define a new measure $\nu$ by $$\nu(E)=\mu(-E).$$Then for any $\phi$ such that one of the integrals is defined, $$\int\phi(t)\,d\nu(t) =\int\phi(-t)\,d\mu(t).$$(Proof: True for $\phi=\chi_E$ by definition. Hence true if $\phi$ is simple...)

In particular the Fourier transform of $\mu$ is the inverse Fourier transform of $\nu$.

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    $\begingroup$ Does the downvoter care to explain? This is a well-written answer that offers great clarification in case needed. I can't believe this was downvoted. $\endgroup$ Aug 23 '18 at 18:55

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