I was wondering if it possible to find a measure $\eta$ on $\mathbb{R}$ such that

$$ L(x):=\log\frac{1}{|x|}=\int e^{itx}\;d\eta(t) $$ for every $x\in [0,1/2]$.

On a structural ground, this question looks to be related to the Bochner Theorem which states that a continuous function $Q:\mathbb{R}\rightarrow\mathbb{C}$ is the Fourier transform of some $finite$ measure iff $Q$ is positive definite (that is for all $n\geq 0$ and all $x_1,\ldots,x_n\in\mathbb{R}$ we have $\det \big[L(x_i-x_j)\big]_{i,j=1}^n\geq 0$).

Note that here $L(0)=+\infty$, and thus doesn't fit with the setting of Bochner Theorem. Nevertheless, one could allow $\eta$ to be an infinite Radon measure, so that $$ \int e^{it 0}d\eta(t)=\eta(\mathbb{R})=+\infty. $$

An other question is : Is $L$ positive definite on [0,1/2] in some sense ? Note that by restricting $L$ to $[0,1/2]$ we have $L(x-y)\geq 0$ for any $x,y\in[0,1/2]$.

Do you know any generalizations of the Bochner theorem which deal with functions which may take the value $+\infty$ ?

Thanks for your help in advance !

  • 1
    $\begingroup$ Basically you can (inverse) Fourier transform a lot of things and since your $L$ is locally integrable there is distribution $\eta$. By the way: wolframalpha gives a Fourier transform of $L$... $\endgroup$
    – Dirk
    Commented Nov 10, 2011 at 7:35
  • 2
    $\begingroup$ The Fourier transform of any even positive integrable function that is convex on $[0,+\infty)$ is non-negative. It isn't hard to expand your restriction of logarithm to such a function. $\endgroup$
    – fedja
    Commented Nov 10, 2011 at 18:28
  • 1
    $\begingroup$ @fedja: so it is not necessary for the function to be monotonically decreasing in $[0,\infty)$ in addition to the requirements that you placed? $\endgroup$
    – Suvrit
    Commented Nov 14, 2011 at 13:20


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