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I am reading a paper and the following came up:

Given a probability density function, $\rho(x)$, such that for $\epsilon > 0$ $$ \int_{-\infty}^{\infty} |\rho(x)|^{1+\epsilon}dx < \infty $$ $$ \int_{-\infty}^{\infty} x^2 \rho(x) dx = \tau^2 < \infty $$ and $$ \rho(-x) = \rho(x) $$ the following hold:

(i) $ \forall \mu_1 > 0$, $ \exists c_1 > 0$, possibly depending on $\mu_1$, s.t. when $|f| \ge \mu_1$ , we have $|\hat{\rho}(f)| < e^{-c_1} $

(ii) For any $n>n_0$, where $n_0$ is the solution to $\frac{1}{1 + \epsilon} + \frac{1}{n_0} = 1$, then $$ \int_{-\infty}^{\infty} |\hat{\rho}(f)|^n \le \int_{-\infty}^{\infty} |\hat{\rho}(f)|^{n_0} = C_0 < \infty$$

(iii) $\hat{\rho}(f) \to 0$ as $|f|\to \infty$

(iv) $ \exists c_2 > 0 $ s.t. for $\mu_1> 0$ small enough, whenever $|f| \le \mu_1$ we have $|\hat{\rho}(f)| \le e^{-c_2 f^2}$

I don't really know how to even approach these (standard? obvious?) properties. All the properties I know about (scaling, linearity, translation, etc.) don't seem to apply here. I got as far as the (trivial) step of noticing that the Fourier transform is real since the distribution is symmetric, but I couldn't make any further progress.

Any hints would be appreciated. More generally, references to literature that covers material useful for this type of analysis would be appreciated.

I am also confused as to the $\int_{-\infty}^{\infty}|\rho(x)|^{1+\epsilon}dx < \infty$ criterea...can anyone shed some intuition on the motivation for this?

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The key is arguably (ii). The discussion relates "size" of pdf to "size" of the transform via 1+e. Functional analysis texts with chapters on Fourier analysis, generally. It's not about the standard "engineering" properties. –  daniel Nov 15 '11 at 11:06

1 Answer 1

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As stated the statements are not necessarily true. You need additional that $\epsilon \leq 1$ and that $\tau \neq 0$.


Since a probability density function is necessarily (Lebesgue) integrable, statement (iii) is just the Riemann-Lebesgue Lemma.

Statement (i) as stated is just a corollary of the Riemann-Lebesgue Lemma. Maybe your copied your statement incorrectly or possibly omitted some quantifiers?


Statement (ii) follows from the Hausdorff-Young inequality together with real interpolation, if you include the additional assumption that $\epsilon \leq 1$:

For $n > n_0$, you have, by a use of Hoelder's inequality that

$$ \int |\hat{\rho}(f)|^n = \int |\hat{\rho}(f)|^{n_0} |\hat{\rho}(f)|^{n-n_0} \leq \left(\int |\hat{\rho}(f)|^{n_0}\right) (\sup |\hat{\rho}|) $$

The second term in the far right, as a fundamental property of the Fourier transform, is bounded by the total mass of $\rho$, which as a probability density function, is 1. The first term in the far right is bounded by $C_0 < \infty$, if $1+\epsilon \leq 2$, as a consequence of the Hausdorff-Young inequality. If $\epsilon > 1$, this inequality may fail.


Lastly we come to Statement (iv). This is the only one of the four statements using the second moment condition and the symmetry condition. Firstly, for a probability distribution $\hat{\rho}(0) = 1$. Using that $\rho$ is a even function, you have that $\hat{\rho}'(0) = \int i x\rho(x) dx = 0$. And that the second derivative

$$ \hat{\rho}''(0) = - \int x^2 \rho(x) dx = - \tau^2 $$

Now, if $\tau$ is assumed to be not equal to 0, by choosing $c_2 = \tau/2$ and using an application of Taylor's theorem with remainders you see that Statement (iv) is true. On the other hand, were $\int x^2 \rho(x) dx = 0$, statement (iv) is actually false (which you can see by considering, again, Taylor's theorem.

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From the paper I was reading it is clear that the second moment is not zero. I would have added that extra condition had I known. No mention (as far as I can see) was made on $\epsilon$ aside from it being larger than 0 so perhaps they implicitly assumed it to be less than 1. For the curious, paper is here –  user4143 Nov 15 '11 at 17:50
    
Well, from the notation, I would say that it is implicit. I am thinking that $n$ and $n_0$ must be positive integers. So if $\epsilon$ is a finite, positive, real number such that $(1+\epsilon)^{-1} + (n_0)^{-1} = 1$, one must necessarily have that $n_0 \geq 2$ and $\epsilon \leq 1$. –  Willie Wong Nov 16 '11 at 9:25

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