I am trying to understand the Wiener-Ikehara Tauberian theorem which can be a step to understanding the prime number theorem. Let

$$ \hat{a}(s) = \int_0^\infty e^{-us}\, da(u) $$

with $a(u)$ some non-negative and increasing function on $\mathbb{R}$, and this Laplace transform converges for $s = \sigma + i \tau$ with $\sigma > 1$. $$ \hat{a}(s) - \frac{1}{s-1} $$ should be continuous on $\sigma \geq 1$ half-plane. Then $$ \int_0^x 1 \, da(u) = c e^x + o(e^x) $$

Mathworld slams us with this rather abstract characterization:

If $f \in L^1(\mathbb{R})$ then the translates of f span a dense subspace iff the Fourier transform is nonzero everywhere.

The statement (taken from Montgomery-Vaughan) is actually really elaborate and I have many questions, but to fit the MO format, I'll ask just one. In my mind, I am linking these results to the Tauberian Theorems one finds in a good Complex Analysis textbook.

The proof of this theorem seems to rest on two lemmas. First, a Laplace transform identity: $$ \mathcal{L}: e^u \to \frac{1}{s-1} $$ I think that just says: $\int_0^\infty e^{-u(1-s)} \, du = \frac{1}{s-1}$ but the main point is when we have $$\sum \Lambda(n) n^{-s} = \frac{1}{s-1} + O(...)$$ that pole gets turned into an exponential main term, and we are left bounding the noise term - potentially larger than the main term itself!

We estimate something smaller than $\int 1 \, da(u)$:

$$ \int_0^x e^{-\delta u} \, da(u) < e^x \big[ o(1) + (1 + \epsilon) \big]$$

In order to do as much Montgomery and Vaughan use a single "trigonometric" lemma. There exist two functions $f\pm (x) \in L^1(\mathbb{R})$

$$ f_-(x) \leq e^{-x} \mathbf{1}_{\mathbb{R}^-} \leq f_+(x) $$

where $f_\pm (x)$ are made of only specific wavelengths, $ \hat{f}(t)= 0$ unless $|t| < T$. Finally the total mass of $f$ is somewhat near $1$: $$ 1 - \epsilon < \int_\mathbb{R} f_-(x) dx < \int_\mathbb{R} f_+(x) dx < 1 + \epsilon $$ How is such a construction achieved: This looks somewhat like a low pass filter:

The book uses a mix of the Fejer Kernel and Jackson Kernel. Can anyone comment on why this proof looks more and more like signal processing ?

  • $\begingroup$ my apologies in advance I'm trying here $\endgroup$ – john mangual Aug 3 '16 at 17:04

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