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.



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...)

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.