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Perron´s formula is in some sense just Fourier inversion, but I have never seen proven it that way in a textbook. I take this must be because the conditions for the Fourier inversion formula to hold may be difficult to verify in this case. Or are they? Is it feasible to prove Perron´s formula using mainly just the fact that the Fourier transform is self-dual?

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3 Answers 3

Harald,

My personal stance on this is that I like to try and avoid using Perron's formula in the "traditional" form. Instead, I like to see the Prime Number Theorem (say) as a statement about $\sum \Lambda(n) \phi(n)$, where $\phi$ is a $C^{\infty}_0$ cutoff function approximating the interval $[1,X]$. To relate this to $\zeta'/\zeta$, you need the Mellin inversion formula for $\phi$ on the vertical line $\Re s = \sigma$, and this really is precisely the same thing as the Fourier inversion formula for the function $e^{\sigma u}\phi(e^u)$. Since everything is a compactly supported smooth function, and in particular a Schwartz function, the analytic issues involved with inverting the Fourier transform are as mild as they can be.

My point of view on this is elaborated upon in in chapter 1 of this course http://www.dpmms.cam.ac.uk/~bjg23/ANT.html.

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I might be missing something, but a small question: For $\sigma>\max(\sigma_c,0)$, we would want to prove that $$\sum_n a_n \phi(n)=\int_{(\sigma)} \alpha(s)\mathcal{M}(\phi)(s) ds.$$ If $\sigma>\sigma_a$, the abscissa of absolute convergence, then we can switch orders, but as I remember the analytic details in the proof of Perrons formula come from the subtle problem of justifying the switching of the order in the case where we do not have absolute convergence. How do you get around this easily? (The case where $a_n=\Lambda(n)$ has $\sigma_c=\sigma_a=1$, so it works out nicely) –  Eric Naslund Mar 28 '12 at 19:29
    
Allow me to second Eric's question (can one really "second" a question?). –  H A Helfgott Mar 29 '12 at 12:17
    
Also - can this approach (which I agree is natural) be used to simplify the complex Tauberian proof of PNT? There, one usually smooths things by convolving with a very specific function (often the continuous Fejer kernel), so as to make the Fourier transform (read: Mellin/Laplace transform on a vertical line) have compact support; then, after proving things (using Fourier inversion, Riemann-Lebesgue, uniform convergence) one undoes the convolution by a Tauberian lemma. Would any $C_0^\infty$ smoothing be enough (so that the convolution and its opposite step become redundant)? –  H A Helfgott Mar 29 '12 at 12:17
    
The problem mentioned by E. Naslund does occur in many applications where $a_n$ is oscillating, say if $a_n$ is the $n$-th Hecke eigenvalue of some cusp form. In many cases, the question becomes: is one really ultimately interested in (1) proving a bound for $\sum_{n\leq x}{a_n}$; or (2) using such a bound to prove, e.g., a subconvexity bound for an $L$-value. In case (1), quoting some form of Perron's formula might be the easiest way to do things quickly, but in case (2), you would only require estimates for smoothed sums anyway. –  Denis Chaperon de Lauzières Mar 31 '12 at 17:26
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I think these answers may be elaborated upon together. Assuming $\sum |a_n| n^{-c}$ converges, the formula to which BR refers explains the sense in which the "traditional" form of Perron's formula is a Fourier integral. By a change of variables, you have

$$\frac{1}{2\pi}\int_{-T}^{T}\left(\sum_{1}^{\infty}\frac{a_n}{n^{c+iy}}\right)\frac{e^{ixy}dy}{c+iy}=e^{-cx}\sum_{n< e^x}a_n+O\left(\frac{1}{T}\sum_{1}^{\infty}\frac{|a_n|}{n^c|x-\log n|}\right)$$

(see Titchmarsch for a generalization that leads to PNT, p61-63). As $T\rightarrow\infty$, the (sharp) estimate on the r.h.s. shows that the convergence is not uniform, even though the interchange of limits on the l.h.s. is justified because the Dirichlet series converges absolutely (hence uniformly).

This is obviously where the Schwartz function $\phi$ to which Ben refers justifies appeal to Fourier duality, instead of having to justify interchanging limits beyond the scope of dominated convergence. It also gives you more parameters to tweak for other purposes, if required. Yet, without introducing a further limit, the resulting formula is not Perron's so that doesn't resolve your question (but perhaps Perron's formula is obsolete?).

As I understand it, this was the precisely the controversy with Riemann's statement of the Fourier expansion of the prime counting function $J(x)$ - he just appealed to Fourier duality and left it there, but that was ultimately verified by Von-Mangoldt, so perhaps there is a proof. Certainly Fourier-Stieltjes applies when the coefficients are positive, so maybe that can be deployed by someone who knows more about measure than me.

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Generally, Perron's formula is the calculation of the inverse Mellin (or Laplace or Fourier) transform of a particular function. When the function's representation as a Mellin transform is known, this is simple. Otherwise, some work in necessary. To a certain extent, I guess it also depends on what you mean by "Perron's formula".

One version of Perron's forumla calculates the inverse Mellin transform of $I_T(s)/s$, where $I_T$ is the indicator function of the strip $|{\rm Im}(s)|< T$, a statement being (from Patterson's book on the zeta function), for $c>0$, $${1\over 2\pi i}\int_{c-iT}^{c+iT}{x^s\over s}ds=\cases{O\big(x^c/T\log(x)\big)& $0< x< 1$\cr 1/2+O(T^{-1}) & $x=1$\cr 1+O\big(x^c/T\log(x)\big)& $x>1$}$$ Since it is a statement about a particular function, you can't really get a general proof. On the other hand, I bet you could extract a certain amount of information as you do in the Paley-Wiener theorem (and similar results).

Wikipedia's version of Perron's formula, which is also an application of the above formula, is amenable to "proof via inversion". Let $g(s)=\sum_{n\ge 1} a_n/ n^s$ be a Dirichlet series converging absolutely for ${\rm Re}(s)>\sigma$. Re-write this as $g(s)=s\int_0^\infty A(x)x^{-s-1}\ dx$, where $A(x)=\sum_{n\le x}a_n$ (with some complication when $x$ is an integer).

If we set $B(x)=A(1/x)$, after changing variables $x\rightarrow x^{-1}$, this becomes $g(s)=s\int_0^\infty B(x)x^{s-1}\ dx=s{\cal M}B(s)$, where $\cal M$ denotes "Mellin transform" (otherwise $g(s)={\cal M}A(-s)$). Divide by $s$ and apply Mellin inversion to both sides (this requires some bound on the decay of $g(s)$ in the region of convergence, which isn't difficult), with $c\gg\sigma$, $${1\over 2\pi i}\int_{c-i\infty}^{c+i\infty}g(s){x^{-s}\over s}\ ds=B(x)$$ Send $x\rightarrow x^{-1}$ to get $${1\over 2\pi i}\int_{c-i\infty}^{c+i\infty}g(s){x^{s}\over s}\ ds=A(x)$$

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So, the first formula above seems to be a justification in the limit that the inverse Mellin transform of $1/s$ is the shifted Heaviside step function H(x-1) with the derivative of both sides giving the inverse Mellin transform of 1 as the shifted Dirac delta function $\delta(x-1)$. –  Tom Copeland Apr 21 '12 at 2:20
    
I think that's a reasonable interpretation. –  B R Apr 21 '12 at 13:36
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