# Positivity of $L(1,\chi)$ for real Dirichlet's character

Let $\chi$ be a real nonprincipal Dirichlet's character modulo $m$.

In my answer to the question on $L(1,\chi)$, I explain a trick for showing that $L(1,\chi)>0$ on the simplest examples of the real characters modulo 3 and 4. The proof goes as follows: one takes $$f(x)=\sum_{n=1}^\infty\chi(n)x^n=\frac1{1-x^m}\sum_{j=1}^{m-1}\chi(j)x^j$$ and uses Abel's theorem to write $$L(1,\chi)=\int_0^1f(x)dx;$$ since the corresponding function $f(x)$ is positive on $(0,1)$, the latter integral has to be positive.

Clearly, $1-x^m>0$ on $(0,1)$, so that the required positivity of $f(x)$ reduces to the positivity of the polynomial $$g_\chi(x)=\sum_{n=1}^{m-1}\chi(n)x^n$$ on $(0,1)$. Trying to verify on how generalizable is this method for $m>3$, I was quite surprised to see that it works perfectly further; for example, $$g(x)=x(1-x)(1-x^2)>0 \quad\text{if } m=5$$ or $$g(x)=x(1-x)(1+x^2+2x^3+3x^4+2x^5+x^6+x^8)>0 \quad\text{if } m=11.$$ Honestly saying, the positivity is not so obvious in many other examples (for example, $m=19$) but nevertheless it is always holds for small values $m\le30$.

Question. Given an integer $m>2$ and a real nonprincipal character $\chi$ modulo $m$, is it true that $g_\chi(x)>0$ for $x\in(0,1)$? If not, are there (in)finitely many $m$ for which the positivity does not take place? Is the above strategy for showing $L(1,\chi)\ne0$ discussed in the literature?

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These are called Fekete polynomials, and you can find out a great deal about them here. Unfortunately they tend to have lots of real zeros in $(0,1)$ when $m$ is large.

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Thanks, David! Never heard of them... Feteke or Fekete? And is $L(1,\chi)>0$ for some $m$? – Wadim Zudilin May 26 '10 at 1:00
fekete, sorry! I'm not sure I understand your second question. – David Hansen May 26 '10 at 1:01
Oh, no, the inequality $L(1,\chi)>0$ for $\chi$ real goes back to Dirichlet, and is an immediate consequence of his famous class number formula. – David Hansen May 26 '10 at 1:46
Wadim and David: there's an easier reason why L(1,chi) can't be negative for real nontrivial chi: the function L(s,chi) for s > 0 is obviously real-valued. As s --> infty it tends to 1, which is positive, and we know L(s,chi) is nonzero for s > 1 by the Euler product. Therefore by continuity L(s,chi) > 0 for s > 1, hence by taking a limit from the right L(1,chi) is definitely not negative. It's not 0 either (harder!), so L(1,chi) > 0. – KConrad May 26 '10 at 2:22
It is not the same, but Chowla then Rosser chased around vaguely similar ideas (weighted sums) to show that $L(s,\chi)>0$ for real $s>0$ for various real $\chi$. Chowla: matwbn.icm.edu.pl/ksiazki/aa/aa1/aa119.pdf Rosser: ams.org/journals/bull/1949-55-10/S0002-9904-1949-09306-0/… – Junkie May 26 '10 at 7:17

As David pointed out, it is known that for large m these polynomials have many zeroes. This is very unfortunate for the following reason: If you assume $g_{\chi}$ is non-negative then it follows by Mellin inversion that the L-function $L(s,\chi)$ can not have a Siegel zero (or, more generally, any zero on the positive real axis).

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Thanks, Marc! The heuristics is already indicated in David's reference, but only for the Legendre symbol. I acknowledge your contribution by upvoting. – Wadim Zudilin May 26 '10 at 1:54
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# Positivity of $L(1,\chi)$ for real Dirichlet's character

Let $\chi$ be a real nonprincipal Dirichlet's character modulo $m$.

In my answer to the question on $L(1,\chi)$, I explain a trick for showing that $L(1,\chi)>0$ on the simplest examples of the real characters modulo 3 and 4. The proof goes as follows: one takes $$f(x)=\sum_{n=1}^\infty\chi(n)x^n=\frac1{1-x^m}\sum_{j=1}^{m-1}\chi(j)x^j$$ and uses Abel's theorem to write $$L(1,\chi)=\int_0^1f(x)dx;$$ since the corresponding function $f(x)$ is positive on $(0,1)$, the latter integral has to be positive.

Clearly, $1-x^m>0$ on $(0,1)$, so that the required positivity of $f(x)$ reduces to the positivity of the polynomial $$g_\chi(x)=\sum_{n=1}^{m-1}\chi(n)x^n$$ on $(0,1)$. Trying to verify on how generalizable is this method for $m>3$, I was quite surprised to see that it works perfectly further; for example, $$g(x)=x(1-x)(1-x^2)>0 \quad\text{if } m=5$$ or $$g(x)=x(1-x)(1+x^2+2x^3+3x^4+2x^5+x^6+x^8)>0 \quad\text{if } m=11.$$ Honestly saying, the positivity is not so obvious in many other examples (for example, $m=19$) but nevertheless it is always holds for small values $m\le30$.

Question. Given an integer $m>2$ and a real nonprincipal character $\chi$ modulo $m$, is it true that $g_\chi(x)>0$ for $x\in(0,1)$? If not, are there (in)finitely many $m$ for which the positivity does not take place? Is the above strategy for showing $L(1,\chi)\ne0$ discussed in the literature?

-

These are called Fekete polynomials, and you can find out a great deal about them here. Unfortunately they tend to have lots of real zeros in $(0,1)$ when $m$ is large.

-
Thanks, David! Never heard of them... Feteke or Fekete? And is $L(1,\chi)>0$ for some $m$? – Wadim Zudilin May 26 '10 at 1:00
fekete, sorry! I'm not sure I understand your second question. – David Hansen May 26 '10 at 1:01
Oh, no, the inequality $L(1,\chi)>0$ for $\chi$ real goes back to Dirichlet, and is an immediate consequence of his famous class number formula. – David Hansen May 26 '10 at 1:46
Wadim and David: there's an easier reason why L(1,chi) can't be negative for real nontrivial chi: the function L(s,chi) for s > 0 is obviously real-valued. As s --> infty it tends to 1, which is positive, and we know L(s,chi) is nonzero for s > 1 by the Euler product. Therefore by continuity L(s,chi) > 0 for s > 1, hence by taking a limit from the right L(1,chi) is definitely not negative. It's not 0 either (harder!), so L(1,chi) > 0. – KConrad May 26 '10 at 2:22
It is not the same, but Chowla then Rosser chased around vaguely similar ideas (weighted sums) to show that $L(s,\chi)>0$ for real $s>0$ for various real $\chi$. Chowla: matwbn.icm.edu.pl/ksiazki/aa/aa1/aa119.pdf Rosser: ams.org/journals/bull/1949-55-10/S0002-9904-1949-09306-0/… – Junkie May 26 '10 at 7:17

As David pointed out, it is known that for large m these polynomials have many zeroes. This is very unfortunate for the following reason: If you assume $g_{\chi}$ is non-negative then it follows by Mellin inversion that the L-function $L(s,\chi)$ can not have a Siegel zero (or, more generally, any zero on the positive real axis).

-
Thanks, Marc! The heuristics is already indicated in David's reference, but only for the Legendre symbol. I acknowledge your contribution by upvoting. – Wadim Zudilin May 26 '10 at 1:54
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The text passage contains a mathematical question and provides some discussion and context around it, but does not provide a direct answer to the question.

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