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

Question

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?

Answer 1

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.

Answer 2

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).