Let $$L(C,s)=\sum_{n=1}^\infty \frac{a_n}{n^s}$$ be the Dirichlet series of the Hasse--Weil L-function of an elliptic curve $C$ over $ℚ$. The modularity theorem implies that $L(C,s)$ is the $L$-function of a holomorphic cusp form for a congruence subgroup and it is entire function and have a holomorphic continuation. Also there is a rapidly-converging series $f(s)$ expression $L(C,s)$ for any complex number $s$ given in http://modular.math.washington.edu/books/bsd/ on page 9.

I am interested on the real points $a∈ℝ$ such that the equation $$f^{(k)}(s)=a$$ have a finite number of real solutions $s$ for some $k$. Unfortunately, I have no idea to start.


I think that for any real number $a$ and any $k \geq 0$ there should be infinitely many solutions to $f^{(k)}(s) = a$ (where $f(s) = L(C,s)$). The idea is that (as you observed in your previous question), we have $f(m) = 0$ for any integer $m \leq 0$ and that $f(m+1/2)$ tends to infinity (in absolute value) and alternates sign. (This should be easy to justify using that $f(s) \to 1$ as $s \to \infty$ and the functional equation.)

Now, it suffices to prove the following lemma:

Suppose that $g(x)$ is an infinitely differentiable function and $a_{1}, a_{2}, \ldots$ is a sequence of real numbers with $\cdots < a_{k} < a_{k-1} < \cdots < a_{1}$ so that

$\bullet$ $g(a_{i}) = 0$ for all $i$,

$\bullet$ There is a constant $C$ so that $a_{i} - a_{i+1} \leq C$ for all $i$, and

$\bullet$ $g(x)$ is neither bounded from above nor bounded from below on $(-\infty,a_{1})$.

Then the same is true of $g'(x)$. That is there is a sequence $\cdots < b_{k} < b_{k-1} < \cdots < b_{1}$ so that $g'(b_{i}) = 0$, there is a constant $D$ so that $b_{i} - b_{i+1} \leq D$ for all $i$, and $g'(x)$ is neither bounded from above nor bounded from below on $(-\infty,b_{1})$.

This lemma should be easy to prove using the mean value theorem.

  • $\begingroup$ @ Jeremy Rouse: It is not clear how to prove the second and the third items for $g'$. $\endgroup$ – China-Hong Kong May 9 '14 at 10:27
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    $\begingroup$ For the second item, note that there must be a zero of $g'$ between $a_{i}$ and $a_{i+1}$ for all $i$. For the third, note that if $g(x) > A$ for some $x < a_{1}$, there must be an $a_{i} < x$ with $x - a_{i} < C$. Now apply the mean value theorem on the interval $[a_{i},x]$. (The same argument works to construct negative values of $g'(x)$.) $\endgroup$ – Jeremy Rouse May 9 '14 at 12:05
  • $\begingroup$ @ Jeremy Rouse: Ok and thank you very much. $\endgroup$ – China-Hong Kong May 9 '14 at 13:19

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