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Does a bounded function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $$ f(x) = 1 - c \, \, |x|^2 \, \, \, \mbox{ if $|x| \in [0,1]$} $$ and such that $f$ is the Fourier transform of a finite non-negative measure on $\mathbb{R}$ exist for a small enough positive constant $c$?

Note that by Bochner's theorem such a last requirement is equivalent to asking that $f$ is a continuous positive-definite function.

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No.

Lemma. Suppose $\mu\ge0$ is a finite measure on $\Bbb R$ and $f=\hat\mu$. Then $\int\xi^2\,d\mu(\xi)\le\limsup_{h\to0}\frac{2f(0)-(f(h)+f(-h))}{h^2}$

Proof: Write $$\Delta(h,\xi)=2\frac{1-\cos(h\xi)}{h^2}$$and $$\delta(h)=\frac{2f(0)-(f(h)+f(-h))}{h^2}.$$For every $A>0$ we have $$\int_{-A}^A\Delta(h,\xi)\,d\mu(\xi)\le\int_{-\infty}^\infty\Delta(h,\xi)\,d\mu(\xi)=\delta(h).$$ Dominated convergence shows that $$\int_{-A}^A\xi^2\,d\mu(\xi)\le\limsup_{h\to0}\delta(h).$$Let $A\to\infty$.

Now suppose that $f$ is as in your question and $f=\hat\mu$ for some finite measure $\mu\ge0$. The lemma implies that $\int\xi^2\,d\mu(\xi)<\infty$. So we can define a finite measure $\nu\ge0$ by $$d\nu(\xi)=\xi^2d\mu(\xi).$$Now $$\hat\nu(\xi)=-f''(\xi)=2c,\quad(|\xi|<1),$$so $\nu$ is supported at the origin, contradiction. (If it's not clear that this implies that $\nu$ is supported at the origin, note that the lemma implies that $\int\xi^2\,d\nu(\xi)=0$.)

Ah. The lemma implies the result I conjectured below, not that we needed it:

Corollary Suppose $\mu\ge0$ is a finite measure on $\Bbb R$, $f=\hat\mu$, and $f''(0)$ exists. Then $\int\xi^2\,d\mu(\xi)=-f''(0)$.

Proof: The lemma implies that $\int\xi^2\,d\mu<\infty$, so the result follows by dominated convergence.

Note A similar result follows for fourth derivatives and fourth moments, by applying the corollary to $\nu$ defined by $d\nu=\xi^2d\mu(\xi)$, as above. Similarly for $2k$-th derivatives and moments.

Original: I strongly suspect the answer is no. A not-quite-proof:

Suppose that $\mu\ge0$ and $f=\hat\mu$. If $\int\xi^4\,d\mu<\infty$ then $f^{(4)}(0)=0$ implies that $\int\xi^4\,d\mu=0$, so that $\mu$ is supported at the origin.

Seems likely to me that the finiteness of $f^{(4)}(0)$ actually implies that $\mu$ has a finite fourth moment - no proof of that. (But it would be enough to prove the same thing for the second derivative and second moment...)

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  • $\begingroup$ Thanks a lot! If the function was defined as $$ 1 + 2c - c |x|^2 $$ would the answer to the question be positive? $\endgroup$ Feb 7 '18 at 21:23
  • $\begingroup$ Does the same argument apply? $\endgroup$ Feb 7 '18 at 21:31

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