Proof. We start by writing the derivative of $\theta _\Lambda (b_\Lambda ;\alpha )$ with respect to x as

$$ \begin{align*} \partial_x \theta_\Lambda(b_\Lambda;\alpha) & = \sum_{l \in \mathbb{Z}} e^{-\pi \alpha y (l+\frac{1}{2}-r)^2} l \vartheta'(\tfrac{1}{2}+xl;\tfrac{\alpha}{y}), \end{align*} $$

where $\vartheta '$ denotes differentiation of $\vartheta $ with respect to the first argument. Due to the asymmetry caused by $r = r(y)$ , we estimate the above expression for $l \geq 1$ and $l \leq - 1$ separately (the case $l = 0$ is trivial and yields 0).

Case A ( $l\geq 1$ ). For $l \geq 1$ and $x \in (0, 1/2)$ , we note that $\vartheta '(1/2+x l)$ is positive for $l=1$ and alternates in sign for $l \geq 2$ . Therefore, we estimate the term for $l=1$ from below by combining Lemma 4.2 and Lemma 4.3. In a similar way, we estimate the absolute values of all other terms from above and assign them a negative sign. In total, we get the following estimate from below:

$$ \begin{align*} & \, \sum_{l \geq 1} e^{-\pi \alpha y (l+\frac{1}{2}-r)^2} l \vartheta'(\tfrac{1}{2}+xl;\tfrac{\alpha}{y})\\ & \geq \, e^{- \pi \alpha y (\frac{3}{2} - r)^2} \underbrace{(-\sin(2 \pi (\tfrac{1}{2}+x))}_{= \sin(2 \pi x)} \sqrt{\tfrac{y}{\alpha}} \, A(\tfrac{y}{\alpha}) - \sum_{l \geq 2} l e^{- \pi \alpha y (l+\frac{1}{2} - r)^2} \sqrt{\tfrac{y}{\alpha}} \, B(\tfrac{y}{\alpha}) \left|\sin(2 \pi (x l+\tfrac{1}{2}))\right|. \end{align*} $$

We wish to show that

$$ \begin{align*} e^{- \pi \alpha y (\frac{3}{2} - r)^2} \sqrt{\tfrac{y}{\alpha}} \, A(\tfrac{y}{\alpha}) \sin(2 \pi x)> \sum_{l \geq 2} l e^{- \pi \alpha y (l+\frac{1}{2} - r)^2} \sqrt{\tfrac{y}{\alpha}} \, B(\tfrac{y}{\alpha}) \left|\sin(2 \pi (x l+\tfrac{1}{2}))\right|, \end{align*} $$

which is equivalent to showing that

$$ \begin{align*} \frac{A(\tfrac{y}{\alpha})}{B(\tfrac{y}{\alpha})}> \sum_{l \geq 2} l \frac{e^{- \pi \alpha y(l+\frac{1}{2}-r)^2}}{e^{-\pi \alpha y (\frac{3}{2}-r)^2}} \frac{\left|\sin(2 \pi x l)\right|}{\sin(2 \pi x)}. \end{align*} $$

Now, by induction and using addition theorems for trigonometric functions, one easily obtains

$$ \begin{align*} \left| \frac{\sin(2 \pi x l)}{\sin(2 \pi x)} \right| \leq l, \quad l \in \mathbb{N}. \end{align*} $$

Furthermore,

$$ \begin{align*} \frac{e^{- \pi \alpha y(l+\frac{1}{2}-r)^2}}{e^{-\pi \alpha y (\frac{3}{2}-r)^2}} = e^{- \pi \alpha y(l^2+l-2-2r(l-1))} \leq e^{- \pi \alpha y (l^2 + \frac{l-3}{2})}, \end{align*} $$

where we used the fact that $r(y) \in [0,\frac {1}{4}]$ for $y \geq 1/\sqrt {2}$ and that the whole expression is maximal for $r = 1/4$ . Hence, we succeed if we can show that

(4.3) $$ \begin{align} \frac{A(\tfrac{y}{\alpha})}{B(\tfrac{y}{\alpha})}> \sum_{l \geq 2} l^2 e^{- \pi \alpha y (l^2 + \frac{l-3}{2})}. \end{align} $$

We need to distinguish the cases $y \leq \alpha $ and $y> \alpha $ .

Case A.1 ( $\alpha \geq y$ ). As we assume $y \geq 1/\sqrt {2}$ , we see that

$$ \begin{align*} \alpha y = \frac{\alpha}{y} y^2 \geq \frac{\alpha}{2 y}. \end{align*} $$

We note that, in the case $y \leq \alpha $ , we have

$$ \begin{align*} \frac{A(\tfrac{y}{\alpha})}{B(\tfrac{y}{\alpha})} = e^{- \pi \frac{\alpha}{4 y}}, \end{align*} $$

by the definitions of the functions in Lemma 4.3. The right-hand side of (4.3) can be estimated by

$$ \begin{align*} \sum_{l \geq 2} l^2 e^{- \pi \alpha y (l^2 + \frac{l-3}{2})} & \leq \sum_{l \geq 2} l^2 e^{- \pi \frac{\alpha}{2 y} (l^2 + \frac{l-3}{2})} = e^{- \pi \frac{\alpha}{4 y}} \sum_{l \geq 2} l^2 e^{- \pi \frac{\alpha}{2 y} (l^2 + \frac{l-4}{2})}\\ & \leq e^{- \pi \frac{\alpha}{4 y}}\sum_{l \geq 2} l^2 e^{- \frac{\pi}{2} (l^2 + \frac{l-4}{2})} \leq 0.0359475 \cdots \times e^{-\pi \frac{\alpha}{4 y}}. \end{align*} $$

Therefore, the strict inequality (4.3) follows in this case.

Case A.2 ( $\alpha \leq y$ ). Now, by the definitions in Lemma 4.3, the left-hand side of (4.3) becomes

$$ \begin{align*} \frac{A(\tfrac{y}{\alpha})}{B(\tfrac{y}{\alpha})} = \frac{1-\frac{1}{3000}}{1+\frac{1}{3000}} = \frac{2999}{3001}. \end{align*} $$

By assumption $\alpha y \geq \alpha ^2 \geq 1$ . So, the right-hand side of (4.3) can be estimated by

$$ \begin{align*} \sum_{l \geq 2} l^2 e^{- \pi \alpha y (l^2 + \frac{l-3}{2})} \leq \sum_{l \geq 2} l^2 e^{- \pi (l^2 + \frac{l-3}{2})} \leq 0.0000671031 \cdots \leq \frac{2999}{3001}. \end{align*} $$

This settles Case A.

Case B ( $l \leq -1$ ). For $l \leq -1$ and $x \in (0, \tfrac {1}{2})$ , we estimate

$$ \begin{align*} \sum_{l \leq -1} e^{-\pi \alpha y (l+\frac{1}{2}-r)^2} l \vartheta'(\tfrac{1}{2}+xl;\tfrac{\alpha}{y}) = \sum_{l \geq 1} e^{- \pi \alpha y (l-\tfrac{1}{2}+r)^2} l \vartheta'(\tfrac{1}{2} + x l; \tfrac{\alpha}{y}), \end{align*} $$

where we used the fact that $\vartheta '(1/2 + x l;\alpha /y) = - \vartheta '(1/2 - x l;\alpha /y)$ , which can be deduced from Lemma 4.1. We see that the expression above is certainly no less than

$$ \begin{align*} e^{- \pi \alpha y (\frac{1}{2} + r)^2} \underbrace{(-\sin(2 \pi (\tfrac{1}{2}+x))}_{= \sin(2 \pi x)} \sqrt{\tfrac{y}{\alpha}} \, A(\tfrac{y}{\alpha}) - \sum_{l \geq 2} l e^{- \pi \alpha y (l-\frac{1}{2} + r)^2} \sqrt{\tfrac{y}{\alpha}} \, B(\tfrac{y}{\alpha}) \left|\sin(2 \pi (x l+\tfrac{1}{2}))\right|. \end{align*} $$

We wish to show that

$$ \begin{align*} e^{- \pi \alpha y (\frac{1}{2} + r)^2} \sin(2 \pi x) \sqrt{\tfrac{y}{\alpha}} \, A(\tfrac{y}{\alpha})> \sum_{l \geq 2} l e^{- \pi \alpha y (l-\frac{1}{2} + r)^2} \sqrt{\tfrac{y}{\alpha}} \, B(\tfrac{y}{\alpha}) \left|\sin(2 \pi x l)\right|, \end{align*} $$

or, equivalently,

(4.4) $$ \begin{align} \frac{A(\tfrac{y}{\alpha})}{B(\tfrac{y}{\alpha})}> \sum_{l \geq 2} l e^{-\pi \alpha y (l^2-l+2r(l-1))} \frac{\left| \sin(2 \pi l x)\right|}{\sin(2 \pi x)}. \end{align} $$

We argue again in two cases.

Case B.1 ( $\alpha \geq y$ ). The left-hand side of (4.4) is simply given by

$$ \begin{align*} \frac{A(\tfrac{y}{\alpha})}{B(\tfrac{y}{\alpha})} = e^{- \pi \frac{\alpha}{4 y}}. \end{align*} $$

For the right-hand side of (4.4), we observe that it is maximal if $r=0$ , so

$$ \begin{align*} \sum_{l \geq 2} l e^{-\pi \alpha y (l^2-l+2r(l-1))} \underbrace{\frac{\left| \sin(2 \pi l x)\right|}{\sin(2 \pi x)}}_{\leq l} & \leq \sum_{l \geq 2} l^2 e^{-\pi \frac{\alpha}{2 y} (l^2-l)} = e^{-\pi \frac{\alpha}{4 y}} \sum_{l \geq 2} l^2 e^{-\pi \frac{\alpha}{2 y} (l^2-l-\frac{1}{2})}\\ & \leq e^{-\pi \frac{\alpha}{4 y}} \sum_{l \geq 2} l^2 e^{-\frac{\pi}{2} (l^2-l-\frac{1}{2})} \leq 0.380714 \cdots \times e^{-\pi \frac{\alpha}{4 y}}. \end{align*} $$

This shows (4.4) in this case.

Case B.2 ( $\alpha \leq y$ ). The left-hand side of (4.4) is now

$$ \begin{align*} \frac{A(\tfrac{y}{\alpha})}{B(\tfrac{y}{\alpha})} = \frac{2999}{3001}. \end{align*} $$

As we now have $\alpha y \geq \alpha ^2 \geq 1$ , the right-hand side can be estimated by

$$ \begin{align*} \sum_{l \geq 2} l e^{-\pi \alpha y (l^2-l+2r(l-1))} \frac{\left| \sin(2 \pi l x)\right|}{\sin(2 \pi x)} \leq \sum_{l \geq 2} l^2 e^{-\pi (l^2-l)} \leq 0.00746983 \ldots \,. \end{align*} $$

Thus (4.4) is true in this case as well.

Putting together the results from Case A and Case B finishes the proof.

Proof. We start by computing the derivative with respect to x. We get

$$ \begin{align*} \partial_x \widehat{\theta}_\Lambda(b_\Lambda;\alpha) & = \sqrt{\tfrac{y}{\alpha}} \, \sum_{l \in \mathbb{Z}} (-1)^l e^{-\pi \alpha y l^2} \sum_{k \in \mathbb{Z}} e^{-\pi \frac{y}{\alpha} (k+b_2)^2} (-2 \pi k l) \sin(2 \pi k l x)\\ & = -4 \pi \sqrt{\tfrac{y}{\alpha}} \, \sum_{l \geq 1} (-1)^l l e^{-\pi \alpha y l^2} \sum_{k \in \mathbb{Z}} k e^{-\pi \frac{y}{\alpha} (k+b_2)^2} \sin(2 \pi k l x). \end{align*} $$

We see immediately that the expression vanishes for $x \in \frac {1}{2} \mathbb {Z}$ . Also, the first terms in k and l should dominate the rest, and the zeros should only come from the sine function. This needs to be proved, of course. This time, we have a symmetry in l and only need to distinguish the cases $y \geq \alpha $ and $y < \alpha $ .

First case ( $\alpha \leq y$ ). The term $l = 1$ (ignoring a factor of $+4 \pi \sqrt {y/\alpha }$ in front of the series) contributes

$$ \begin{align*} & e^{-\pi \alpha y} \sum_{k \geq 1} k \left( e^{-\pi \frac{y}{\alpha} (k+b_2)^2} + e^{-\pi \frac{y}{\alpha} (k-b_2)^2} \right) \sin(2 \pi k x)\\ & \geq 2 e^{-\pi \alpha y} \Bigg( e^{-\pi \frac{y}{\alpha}(1+b_2^2)} \cosh(2 \pi \tfrac{y}{\alpha} b_2) \sin(2 \pi x) - \sum_{k \geq 2} k e^{-\pi \frac{y}{\alpha} (k^2+b_2^2)} \cosh(2 \pi k \tfrac{y}{\alpha} b_2) |\sin(2 \pi k x)|\Bigg). \end{align*} $$

Dropping the factor 2, we wish to show that, for $x \in (0, \frac {1}{2})$ ,

$$ \begin{align*} e^{-\pi \alpha y} e^{-\pi \frac{y}{\alpha} (1+b_2^2)} \cosh(2 \pi \tfrac{y}{\alpha} b_2) \sin(2 \pi x) \geq e^{-\pi \alpha y} \sum_{k \geq 2} k e^{-\pi \frac{y}{\alpha} (k^2+b_2^2)} \cosh(2 \pi k \tfrac{y}{\alpha} b_2) |\sin(2 \pi k x)|\\ + \sum_{l \geq 2} l e^{-\pi \alpha y l^2} \sum_{k \geq 1} k e^{-\pi \frac{y}{\alpha} (k^2+b_2^2)} \cosh(2 \pi k \tfrac{y}{\alpha} b_2) |\sin(2 \pi k l x)|. \end{align*} $$

Again, we face the problem that we do not have room for estimates near $x=0$ and $x=\frac {1}{2}$ , since these values yield critical points where the derivative vanishes. In order to overcome this problem, we may divide by the sine function, which is justified by l’Hospital’s rule. This gives us the chance to actually estimate the second derivative of $\widehat {\vartheta }_\Lambda (b_\Lambda ;\alpha )$ close to the critical points (up to the sign at $x=1/2$ ) and $\partial _x \widehat {\vartheta }_\Lambda (b;\alpha )$ away from the critical points at the same time. Also, since the leading term should dominate everything, we see that, as a function of x, $\partial _x \widehat {\vartheta }_\Lambda (b_\Lambda ;\alpha )$ should actually behave like a sine function with respect to x. So, dividing by $\sin (2 \pi x)$ leaves us to prove that

$$ \begin{align*} e^{-\pi \alpha y} e^{-\pi \frac{y}{\alpha} (1+b_2^2)} \cosh(2 \pi \tfrac{y}{\alpha} b_2) \geq & e^{-\pi \alpha y} \sum_{k \geq 2} k e^{-\pi \frac{y}{\alpha} (k^2+b_2^2)} \cosh(2 \pi k \tfrac{y}{\alpha} b_2) \frac{|\sin(2 \pi k x)|}{\sin(2 \pi x)}\\ + & \sum_{l \geq 2} l e^{-\pi \alpha y l^2} \sum_{k \geq 1} k e^{-\pi \frac{y}{\alpha} (k^2+b_2^2)} \cosh(2\pi k \tfrac{y}{\alpha} b_2) \frac{|\sin(2 \pi k l x)|}{\sin(2 \pi x)}. \end{align*} $$

Again, we use the formula

$$ \begin{align*} \left| \frac{\sin(2 \pi N x)}{\sin(2 \pi x)}\right| \leq N, \quad N \in \mathbb{N}. \end{align*} $$

Hence, as $\sin (2 \pi x) = |\sin (2\pi x)|$ for $x \in [0, 1/2]$ , we succeed if we can show the stronger inequality

(4.5) $$ \begin{align} e^{-\pi \alpha y} e^{-\pi \frac{y}{\alpha} (1+b_2^2)} \cosh(2 \pi \tfrac{y}{\alpha} b_2) \geq & e^{-\pi \alpha y} \sum_{k \geq 2} k^2 e^{-\pi \frac{y}{\alpha} (k^2+b_2^2)} \cosh(2 \pi k \tfrac{y}{\alpha} b_2) \nonumber \\ + & \sum_{l \geq 2} l^2 e^{-\pi \alpha y l^2} \sum_{k \geq 1} k^2 e^{-\pi \frac{y}{\alpha} (k^2+b_2^2)} \cosh(2\pi k \tfrac{y}{\alpha} b_2), \end{align} $$

or, equivalently,

$$ \begin{align*} \cosh(2 \pi \tfrac{y}{\alpha} b_2) \geq & \sum_{k \geq 2} k^2 e^{-\pi \frac{y}{\alpha} (k^2-1)} \cosh(2 \pi k \tfrac{y}{\alpha} b_2)\\ & + \sum_{l \geq 2} l^2 e^{-\pi \alpha y (l^2-1)} \sum_{k \geq 1} k^2 e^{-\pi \frac{y}{\alpha} (k^2-1)} \cosh(2\pi k \tfrac{y}{\alpha} b_2). \end{align*} $$

By assumption, $\alpha \geq 1$ and $\alpha \leq y$ ; therefore,

$$ \begin{align*} \alpha y \geq \alpha^2 \geq 1. \end{align*} $$

Also, we have $b_2 \in [1/4, 1/2]$ (as $y \geq 1/\sqrt {2}$ ). By underestimating the left-hand side and overestimating the right-hand side, we obtain the stronger inequality

$$ \begin{align*} \cosh(\tfrac{\pi}{2}) & \geq \sum_{k \geq 2} k^2 e^{-\pi \frac{y}{\alpha} (k^2-1)} e^{\pi k \tfrac{y}{\alpha}} + \sum_{l \geq 2} l^2 e^{-\pi \alpha y (l^2-1)} \sum_{k \geq 1} k^2 e^{-\pi \frac{y}{\alpha} (k^2-1)} e^{\pi k \tfrac{y}{\alpha}}\\ & = \sum_{k \geq 2} k^2 e^{-\pi \frac{y}{\alpha} (k^2-k-1)} + \sum_{l \geq 2} l^2 e^{-\pi \alpha y (l^2-1)} \sum_{k \geq 1} k^2 e^{-\pi \frac{y}{\alpha} (k^2-k-1)}. \end{align*} $$

The right-hand side is at most

$$ \begin{align*} \sum_{k \geq 2} k^2 e^{-\pi (k^2-k-1)} + \sum_{l \geq 2} l^2 e^{-\pi (l^2-1)} \sum_{k \geq 1} k^2 e^{-\pi (k^2-k-1)} = 0.180383 \cdots < 1 < \cosh(\tfrac{\pi}{2}) \end{align*} $$

and so this case is settled.

Second case ( $\alpha \geq y$ ). Now, as we assume $\alpha \geq 1$ and $y \geq 1/\sqrt {2}$ , we have

$$ \begin{align*} \alpha y = \frac{\alpha}{y} y^2 \geq \frac{\alpha}{2 y}. \end{align*} $$

Interestingly, the proof of this case now differs somewhat from that of the previous cases (at least the proof we could find), but the overall idea remains the same. We reenter the first case at the inequality (4.5):

$$ \begin{align*} e^{-\pi \alpha y} e^{-\pi \frac{y}{\alpha} (1+b_2^2)} \cosh(2 \pi \tfrac{y}{\alpha} b_2) \geq & e^{-\pi \alpha y} \sum_{k \geq 2} k^2 e^{-\pi \frac{y}{\alpha} (k^2+b_2^2)} \cosh(2 \pi k \tfrac{y}{\alpha} b_2)\\ + & \sum_{l \geq 2} l^2 e^{-\pi \alpha y l^2} \sum_{k \geq 1} k^2 e^{-\pi \frac{y}{\alpha} (k^2+b_2^2)} \cosh(2\pi k \tfrac{y}{\alpha} b_2). \end{align*} $$

The problem is now that $\frac {y}{\alpha }$ is potentially small, and so a lot of terms may contribute to the series in k. The trick is, of course, to use the Poisson summation formula again, to invert the parameters in the exponentials. In order to be able to use Poisson summation for the right-hand side, we wish to sum over all integers again. So, we start by adding the left-hand side, which corresponds to the case $(k,l)=(1,1)$ , on both sides of the inequality:

$$ \begin{align*} 2 e^{-\pi \alpha y} e^{-\pi \frac{y}{\alpha} (1+b_2^2)} \cosh(2 \pi \tfrac{y}{\alpha} b_2) \geq & \sum_{l \geq 1} l^2 e^{-\pi \alpha y l^2} \sum_{k \geq 1} k^2 e^{-\pi \frac{y}{\alpha} (k^2+b_2^2)} \cosh(2\pi k \tfrac{y}{\alpha} b_2). \end{align*} $$

Showing that this is true is equivalent to showing that

$$ \begin{align*} 2 e^{-\pi \frac{y}{\alpha}} \geq \sum_{l \geq 1} l^2 e^{-\pi \alpha y (l^2-1)} \sum_{k \geq 1} k^2 e^{-\pi \frac{y}{\alpha} k^2} \tfrac{\cosh(2 \pi \frac{y}{\alpha} k b_2)}{\cosh(2 \pi \frac{y}{\alpha} b_2)}. \end{align*} $$

Now, we make the simple observation that $\frac {\cosh (R x)}{\cosh (x)}$ is growing for $x> 0$ and any fixed $R> 1$ . This is easily shown by finding all $x> 0$ such that

$$ \begin{align*} \dfrac{d}{dx} \frac{\cosh(R x)}{\cosh(x)} = \frac{R \ \sinh(R x)\cosh(x)-\cosh(R x) \sinh(x)}{\cosh(x)^2}>0. \end{align*} $$

Rearranging this inequality leads to

$$ \begin{align*} R \ \tanh(R x)> \tanh(x), \end{align*} $$

which is true for all $x> 0$ and $R> 1$ . So, by replacing $b_2$ by $\frac {1}{2}$ , we succeed if we can show the resulting stronger inequality

$$ \begin{align*} 2 e^{-\pi \frac{y}{\alpha}} \geq \sum_{l \geq 1} l^2 e^{-\pi \alpha y (l^2-1)} \sum_{k \geq 1} k^2 e^{-\pi \frac{y}{\alpha} k^2} \tfrac{\cosh(2 \pi \frac{y}{\alpha} k \frac{1}{2})}{\cosh(2 \pi \frac{y}{\alpha} \frac{1}{2})}, \end{align*} $$

as $b_2 \in [1/4, 1/2]$ . Another simple computation shows that, for any $k \geq 1$ ,

$$ \begin{align*} \frac{\cosh(\pi \frac{y}{\alpha} k)}{\cosh(\pi \frac{y}{\alpha})} \leq e^{\pi \frac{y}{\alpha} (k-1)}, \end{align*} $$

which leads us to the new and stronger inequality

(4.6) $$ \begin{align} 2 e^{-\pi \frac{y}{\alpha}} & \geq \sum_{l \geq 1} l^2 e^{-\pi \alpha y (l^2-1)} \sum_{k \geq 1} k^2 e^{-\pi \frac{y}{\alpha} k^2} e^{\pi \frac{y}{\alpha}(k-1)} \nonumber \\ & = \sum_{l \geq 1} l^2 e^{-\pi \alpha y (l^2-1)} \sum_{k \geq 1} k^2 e^{-\pi \frac{y}{\alpha} (k-\frac{1}{2})^2} e^{-\pi \frac{3y}{4\alpha}}. \end{align} $$

Next, we use the simple estimate (note the range of the index)

$$ \begin{align*} \sum_{k \geq 1} k^2 e^{-\pi \frac{y}{\alpha} (k-\frac{1}{2})^2} e^{-\pi \frac{3y}{4\alpha}} \leq \sum_{k \in \mathbb{Z}} k^2 e^{-\pi \frac{y}{\alpha} (k+\frac{1}{2})^2} e^{-\pi \frac{3y}{4\alpha}}. \end{align*} $$

As k is now running through the integers again, we may apply the Poisson summation formula. We start by rewriting the sum and get

$$ \begin{align*} \sum_{k \in \mathbb{Z}} k^2 e^{-\pi \frac{y}{\alpha} (k+\frac{1}{2})^2} & =\sum_{k \in \mathbb{Z}} \left[\left(k+\tfrac{1}{2}\right)^2 -\left(k+\tfrac{1}{2}\right)+\tfrac{1}{4}\right] e^{-\pi \frac{y}{\alpha} (k+\frac{1}{2})^2}\\ & = \sum_{k \in \mathbb{Z}} \left(k+\tfrac{1}{2} \right)^2 e^{-\pi \frac{y}{\alpha} (k+\frac{1}{2})^2} -\underbrace{\sum_{k \in \mathbb{Z}} \left(k+\tfrac{1}{2} \right) e^{-\pi \frac{y}{\alpha} (k+\frac{1}{2})^2}}_{=0} +\tfrac{1}{4}\sum_{k \in \mathbb{Z}} e^{-\pi \frac{y}{\alpha} (k+\frac{1}{2})^2}\\ & = \sum_{k \in \mathbb{Z}} \left(k+\tfrac{1}{2} \right)^2 e^{-\pi \frac{y}{\alpha} (k+\frac{1}{2})^2} + \tfrac{1}{4}\left(\tfrac{\alpha}{y} \right)^{\frac{1}{2}}\sum_{k \in \mathbb{Z}} (-1)^k e^{-\frac{\pi \alpha}{y}k^2}. \end{align*} $$

Note that the negative part canceled due to symmetry (we thank the referee for pointing this out, thus simplifying our argument). Similar results may be found in [Reference Bochner16]. We next compute

$$ \begin{align*} \sum_{k \in \mathbb{Z}} \left(k+\tfrac{1}{2} \right)^2 e^{-\pi \frac{y}{\alpha} (k+\frac{1}{2})^2} & = -\tfrac{1}{\pi}\partial_t\left[\sum_{k \in \mathbb{Z}} e^{-\pi t (k+\frac{1}{2})^2} \right]_{t=\frac{y}{\alpha}} = -\tfrac{1}{\pi}\partial_t\left[\tfrac{1}{\sqrt{t}}\sum_{k \in \mathbb{Z}} (-1)^k e^{-\frac{\pi}{t}k^2}\right]_{t=\frac{y}{\alpha}}\\ & = -\tfrac{1}{\pi}\sum_{k \in \mathbb{Z}} (-1)^k e^{-\frac{\pi}{t}k^2}\left[-\tfrac{1}{2t^{\frac{3}{2}}}+\tfrac{\pi k^2}{t^{\frac{5}{2}}} \right]_{t=\frac{y}{\alpha}}\\ & = \sum_{k \in \mathbb{Z}} (-1)^k e^{-\frac{\pi \alpha}{y}k^2} \left( \tfrac{1}{2\pi}\left( \tfrac{\alpha}{y} \right)^{\frac{3}{2}} - \left( \tfrac{\alpha}{y} \right)^{\frac{5}{2}}k^2 \right). \end{align*} $$

The interchange of summation and differentiation is justified due to the rapid decay of the series. It follows that

$$ \begin{align*} \sum_{k \in \mathbb{Z}} k^2 e^{-\pi \frac{y}{\alpha} (k+\frac{1}{2})^2}=\left(\tfrac{\alpha}{y}\right)^{3/2} \sum_{k \in \mathbb{Z}} (-1)^k e^{-\pi \frac{\alpha}{y} k^2} \left(\tfrac{1}{2 \pi} +\tfrac{y}{4 \alpha} - \tfrac{\alpha}{y} k^2\right). \end{align*} $$

As $\alpha /y$ becomes large, the exponential terms in k become small rather quickly. However, the factor in front of the series becomes huge, but it cannot beat the exponential decay. In fact, we have the following inequality, which can be checked by very elementary means: for any $t> 1$ , we have

(4.7) $$ \begin{align} t^{3/2} e^{-\pi \frac{t}{4}} < e^{-\pi \frac{1}{4 t}}. \end{align} $$

The aim of inequality (4.7) is to compensate for the inversion of the exponent and the appearance of the monomial, due to the Poisson summation formula, in (4.6). By using the substitution $t \mapsto \frac {1}{t}$ , we see that we have the equivalent inequality

$$ \begin{align*} e^{-\pi \frac{t}{4}}> \left(\tfrac{1}{t}\right)^{3/2} e^{-\frac{\pi}{4 t}} \quad \text{ for } t \in (0,1). \end{align*} $$

As $\alpha> y$ , we also have

$$ \begin{align*} 2 e^{-\pi \frac{y}{4\alpha}}> 2 \left(\tfrac{\alpha}{y}\right)^{3/2} e^{-\pi \frac{\alpha}{4y}}, \end{align*} $$

which now leads us to showing the stronger inequality

$$ \begin{align*} 2 \left(\tfrac{\alpha}{y}\right)^{3/2} e^{-\pi \frac{\alpha}{4y}} \geq \left(\tfrac{\alpha}{y}\right)^{3/2} \sum_{l \geq 1} l^2 e^{-\pi \alpha y (l^2-1)} \sum_{k \in \mathbb{Z}} (-1)^k e^{-\pi \frac{\alpha}{y} k^2} (\tfrac{1}{2 \pi} + \tfrac{y}{4\alpha} - \tfrac{\alpha}{y} k^2). \end{align*} $$

Now, we succeed if we can show the even stronger inequality

$$ \begin{align*} 2 \geq \sum_{l \geq 1} l^2 e^{-\pi \alpha y (l^2-1)} \left( (\tfrac{1}{2\pi} + \tfrac{y}{4 \alpha})e^{\pi\frac{\alpha}{4y}} + 2 \sum_{k \geq 1} e^{-\pi \frac{\alpha}{y} (k^2-\frac{1}{4})} ( \tfrac{\alpha}{y} k^2 + \tfrac{1}{2 \pi} + \tfrac{y}{4 \alpha}) \right). \end{align*} $$

As preparation for estimating the right-hand side, we observe that, for $t> 0$ ,

$$ \begin{align*} t \mapsto t k^2 e^{-\pi t (k^2 - \frac{1}{4})} \end{align*} $$

has a maximum at $t = \frac {4}{(4k^2-1)\pi }$ and is decreasing afterwards. In particular, this implies that $t \mapsto t k^2 e^{-\pi t (k^2 - \frac {1}{4})}$ is maximal for $t = 1$ on $[1,\infty )$ and any fixed $1 \leq k \in \mathbb {N}$ . Furthermore, as $y^2 \geq \frac {1}{2}$ , we have

$$ \begin{align*} e^{-\pi \alpha y(l^2-1)} e^{\pi \frac{\alpha}{4 y}} = e^{- \pi \alpha y (l^2-1-\frac{1}{4y^2})} \leq e^{-\pi \alpha y (l^2 - 1 - \frac{1}{2})}. \end{align*} $$

Therefore, all that is left to observe is the fact that

$$ \begin{align*}\hspace{-12pt} 2> \left( \tfrac{1}{2\pi} + \tfrac{1}{4} \right) \sum_{l \geq 1} l^2 e^{-\frac{\pi}{2} (l^2-\frac{3}{2})} + 2 \sum_{l \geq 1} l^2 e^{-\frac{\pi}{2} (l^2-1)} \sum_{k \geq 1} e^{-\pi (k^2-\frac{1}{4})} (k^2 + \tfrac{1}{2 \pi} + \tfrac{1}{4}) = 1.20646 \ldots.\\[-36pt] \end{align*} $$

Proof. This is a simple consequence of Lemma 3.1.

Proof. We write

$$ \begin{align*} k = r \cos{\theta} \qquad \mbox{and} \qquad l = r \sin{\theta}. \end{align*} $$

Then, after some elementary algebra, we have

$$ \begin{align*} Q_1(k,l) = \frac{2}{3\sqrt{3}} + \frac{6 r^2}{3 \sqrt{3}} + \frac{2 r (r \sin (t) \cos (t)+\sin (t)+\cos (t))}{\sqrt{3}}. \end{align*} $$

Using $\sin (t) \cos (t) \geq -1/2$ and $\cos (t) + \sin (t) \geq -\sqrt {2}$ , we see that for $r \geq 8.3 = \sqrt {68.89}$ ,

$$ \begin{align*} Q_1(k,l) & \geq \frac{2}{3\sqrt{3}} + \frac{3 r^2}{3 \sqrt{3}} + \frac{2 r (\sin (t)+\cos (t))}{\sqrt{3}} \\ & \geq \frac{2}{3\sqrt{3}} + \frac{3 r^2}{3 \sqrt{3}} - \frac{\sqrt{8} r }{\sqrt{3}} \geq \frac{2r^2}{3\sqrt{3}}. \end{align*} $$

It thus suffices to check all lattice points $(k,l) \in \mathbb {Z}^2$ for which $k^2 + l^2 \leq 69$ and we see that the inequality holds for these.

Proof. We know (Lemma 5.4) that there is a critical point at $y = \sqrt {3}/2$ . It thus suffices to show that the second derivative is negative. This ends in explicit computations in the region

$$ \begin{align*} (\alpha, y) \in (1,6) \times (\sqrt{3}/2, 1) = \Omega. \end{align*} $$

The function $f_{\alpha }$ can be differentiated in closed form and the derivatives inherit the exponential decay (see Lemma 5.13). The rapid exponential decay allows us to observe that

$$ \begin{align*} \sup_{(\alpha, y) \in \Omega} \quad \frac{d^2 }{d y^2} f_{\alpha}\left( \frac{1}{2}, y \right) \leq -0.01. \end{align*} $$

More precisely, we see that in the closure of that region, the maximum of the second derivative is attained for $y=1$ and $\alpha = 6$ and evaluates roughly to $-0.0115$ . We will not use this fact (which would require more extensive numerics) and use it for illustrative purposes: since we do not use the precise result $-0.0115$ and are content with a much weaker estimate (even $-0.001$ would suffice), the continuity and exponential decay of the functions involved reduce the verification to a finite number of computations (see Lemma 5.13).

Proof. We first establish the statement by hand for $|k|, |l| \leq 20$ . In that region, the minimal ratio is at least $0.62$ and attained only for $(k,l)$ being one of $(0,0), (-1,0), (0,-1)$ (these are the three dominant terms). We aim to prove a lower bound on

$$ \begin{align*}\phi_{k,l}(y) = \frac{(2 k+l+1)^2}{4 y}+y \left(l+\frac{1}{2} -\frac{1}{8 y^2}\right)^2.\end{align*} $$

We note that for $y \geq \sqrt {3}/2$ , we have

$$ \begin{align*}\frac{1}{3} \leq \frac{1}{2} -\frac{1}{8 y^2} \leq \frac{1}{2}.\end{align*} $$

Therefore, we have, for $\sqrt {3}/2 \leq y$ ,

$$ \begin{align*} \left(l+\frac{1}{2} -\frac{1}{8 y^2}\right)^2 \geq \begin{cases} l^2 \quad &\mbox{if}~l \geq 0 \\ \left(|l| - \frac{1}{2}\right)^2 \quad &\mbox{if}~l < 0 \end{cases} \geq \left(|l| - \frac{1}{2}\right)^2 \end{align*} $$

and thus

$$ \begin{align*}\phi_{k,l}(y) \geq \frac{(2 k+l+1)^2}{4 y} + y \left(|l| - \frac{1}{2}\right)^2.\end{align*} $$

This leads us to consider general functions of the type

$$ \begin{align*}h(y) = \frac{A}{y} + B y,\end{align*} $$

where $A,B> 0$ . It is easy to see that h assumes its global minimum at $y = (A/B)^{1/2}$ (with value $2(AB)^{1/2}$ ) and is then monotonically increasing. Since we only care about $y \geq \sqrt {3}/2$ , this leads to two different cases: if $A/B \leq 3/4$ , then the minimum is assumed for a value of $y \leq \sqrt {3}/2$ and, by convexity of the function h, the minimal value in the region of interest is thus assumed at $y = \sqrt {3}/2$ . Conversely, if $A/B \geq 3/4$ , then the minimal value is $2 (AB)^{1/2}$ . The argument then follows from comparing quadratic forms (as above). We quickly note the precise results that result from this approach: the approach does not imply the result for small values of $(k,l)$ (here: $(0,-1)$ and $(-1,1)$ ) where the bound is too lossy; however, these cases have been excluded by hand before. More computations would imply that the constant $1/2$ could be increased to $4/5$ (except for the three dominant terms), but this will not be necessary.

Proof. We start by noting that, as above, there is some room to maneuver since

$$ \begin{align*} f_{\alpha}\left(\tfrac{1}{2}, 1\right) \leq 0.99 \cdot f_{\alpha}\left(\tfrac{1}{2}, \tfrac{\sqrt{3}}{2} \right). \end{align*} $$

The precise constant is $0.9875\dots $ , which is only assumed for $\alpha =1$ ; then, the ratio is steadily decreasing. For $\alpha = 6$ , the ratio is approximately $0.8977$ . Moreover, we claim that, for $1 \leq \alpha \leq 6$ and $y \geq 1$ ,

$$ \begin{align*} \sum_{\substack{k,l \in \mathbb{Z}\\|k|, |l| \leq 5}} e^{- \pi \alpha \, \phi_{k,l}(y)} - \sum_{\substack{k,l \in \mathbb{Z} \\ |k|, |l| \leq 2}} e^{- \pi \alpha \, \phi_{k,l}(y)} \leq \tfrac{1}{500} \cdot f_{\alpha}\left(\tfrac{1}{2}, \tfrac{\sqrt{3}}{2} \right). \end{align*} $$

This inequality is again far from sharp: the constant could be replaced by $1/1500$ for $\alpha = 1$ and by much smaller terms when $\alpha $ is even slightly bigger (the terms that do not cancel have faster exponential decay in $\alpha $ ). Using Lemma 5.7, we see that the remaining terms are actually quite small: we have

$$ \begin{align*} \sum_{\substack{(k,l) \in \mathbb{Z}^2 \\ \max \left\{ |k|, |l| \right\}> 5}} e^{- \pi \alpha \, \phi_{k,l}(y)} \leq \sum_{\substack{(k,l) \in \mathbb{Z}^2 \\ \max \left\{ |k|, |l| \right\} > 5}} e^{- \pi \alpha \, \left(\min_{y \geq \sqrt{3}/2}\phi_{k,l}(y)\right)} \leq \sum_{\substack{(k,l) \in \mathbb{Z}^2 \\ \max \left\{ |k|, |l| \right\} > 5}} e^{- \pi \alpha \, \frac{1}{2} \sqrt{Q_1(k,l)}}. \end{align*} $$

A short computation shows that if $ \max \left \{ |k|, |l| \right \}> 5$ , then

$$ \begin{align*} Q_1(k,l) \geq \tfrac{146}{3 \sqrt{3}} \sim 28.09 \end{align*} $$

and thus

$$ \begin{align*} \frac{1}{2} \sqrt{Q_1(k,l)} \geq 2.65. \end{align*} $$

We note that these terms therefore undergo exponential decay many orders faster than the leading order terms, which are of size

$$ \begin{align*} e^{-\frac{2\pi \alpha}{3 \sqrt{3}}} \sim e^{- 0.38 \, \pi \alpha}. \end{align*} $$

Indeed, we have, for example, for $\alpha \geq \alpha _0 \sim 0.95$ , that

$$ \begin{align*} \sum_{\substack{(k,l) \in \mathbb{Z}^2 \\ \max \left\{ |k|, |l| \right\}> 5}} e^{- \pi \alpha \cdot \frac{1}{2} \sqrt{Q_1(k,l)}} < \tfrac{1}{200} e^{-\frac{2\pi \alpha}{3 \sqrt{3}}}.\\[-30pt] \end{align*} $$

Proof. Since all three functions are identical, it suffices to consider $(k, l) = (0,0)$ . A short computation shows that

$$ \begin{align*} \frac{d^2}{dy^2} e^{-\pi \alpha \, \phi_{0,0}(y)} =\frac{\pi \alpha \, e^{-\frac{\pi \alpha \left(4 y^2+1\right)^2}{64 y^3}} \left(\pi \alpha \left(-16 y^4+8 y^2+3\right)^2-256 y^3 \left(4 y^2+3\right)\right)}{4096 y^8}. \end{align*} $$

We note that in the range

$$ \begin{align*} \frac{\sqrt{3}}{2} \leq y \leq \frac{\sqrt{3}}{2} + \frac{1}{3\sqrt{\alpha}} \leq \frac{\sqrt{3}}{2} + \frac{1}{3\sqrt{6}}, \end{align*} $$

we have the inequalities

$$ \begin{align*} & & \pi \alpha \left(-16 y^4+8 y^2+3\right)^2 &\leq 1450 \, \pi \alpha \left( y - \frac{\sqrt{3}}{2} \right)^2 \\ & \text{ and } & - 256 y^3 \left(4 y^2+3\right) &\leq - \!990 - \!4500 \left( y - \frac{\sqrt{3}}{2} \right) \ll 0. \end{align*} $$

Using these two inequalities, we can bound the second derivative from above:

$$ \begin{align*} \frac{\partial^2}{\partial y^2} e^{-\pi \alpha \phi_{0,0}(y)} \leq \frac{\pi \alpha e^{-\frac{\pi \alpha \left(4 y^2+1\right)^2}{64 y^3}}}{4096 y^8} \left(1450 \, \pi \alpha \left( y - \frac{\sqrt{3}}{2} \right)^2 -\!\! 990 -\!\! 4500 \left( y - \frac{\sqrt{3}}{2} \right) \right). \end{align*} $$

A quick computation shows that in the range $\sqrt {3}/2 \leq y \leq \sqrt {3}/2 + 1/(3\sqrt {\alpha })$ , we have (uniformly for $\alpha \geq 6$ ) that

$$ \begin{align*} 1450 \, \pi \alpha \left( y - \frac{\sqrt{3}}{2} \right)^2 - \!990 - \!4500 \left( y - \frac{\sqrt{3}}{2} \right) \leq -480. \end{align*} $$

The function $\left (4 y^2+1\right )^2/(64 y^3)$ is monotonically increasing for $y \geq \sqrt {3}/2$ , from which we can conclude that

$$ \begin{align*} \frac{\pi \alpha e^{-\frac{\pi \alpha \left(4 y^2+1\right)^2}{64 y^3}}}{4096 y^8} \qquad \mbox{assumes its minimum at}~y = \frac{\sqrt{3}}{2} + \frac{1}{3\sqrt{\alpha}}. \end{align*} $$

An explicit computation shows that the exponential term satisfies

$$ \begin{align*} \exp \left(-\frac{\pi \left(4 \left(\frac{1}{3 \sqrt{a}}+\frac{\sqrt{3}}{2}\right)^2+1\right)^2 \alpha}{64 \left(\frac{1}{3 \sqrt{a}}+\frac{\sqrt{3}}{2}\right)^3}\right) \geq e^{-\frac{2 \pi }{27 \sqrt{3}}} e^{\frac{2\pi \alpha}{3 \sqrt{3}}} \geq 0.87 e^{-\frac{2\pi \alpha}{3 \sqrt{3}}}. \end{align*} $$

Combining all these estimates shows that

$$ \begin{align*} \frac{\partial^2}{\partial y^2} e^{-\pi \alpha \phi_{0,0}(y)} \leq 0.87 \, \frac{\pi \alpha e^{-\frac{2\pi \alpha}{3 \sqrt{3}}}}{4096 y^8} \left(1450 \pi \alpha \left( y - \frac{\sqrt{3}}{2} \right)^2 - \!990 - \!4500 \left( y - \frac{\sqrt{3}}{2} \right) \right). \end{align*} $$

Some quick computations show that, in the region of interest, the quadratic polynomial assumes its maximum value at $y = \sqrt {3}/2$ (that maximum being $-990$ ) as long as $\alpha < 72900/(841 \pi ^2) \sim 8.78$ and assumes its maximum at $y = \sqrt {3}/2 + 1/(3\sqrt {\alpha })$ for larger values of $\alpha $ . A standard computation of the quantities arising for $6 \leq \alpha \leq 20$ shows the desired inequality for these cases. As $\alpha $ becomes large, this allows us to prove the asymptotic estimate

$$ \begin{align*} \frac{\partial^2}{\partial y^2} e^{-\pi \alpha \phi_{0,0}(y)} \leq 0.87 \, \frac{\pi \alpha e^{-\frac{2\pi \alpha}{3 \sqrt{3}}}}{1296} \cdot (-480) \leq - \alpha e^{-\frac{2\pi \alpha}{3 \sqrt{3}}}.\\[-30pt] \end{align*} $$

Proof. An explicit computation shows that

$$ \begin{align*} \sum_{-1 \leq k,l \leq 1} e^{ - \pi \alpha \cdot \phi_{k,l}(y)} = e^{-\frac{\pi a \left(144 y^4+232 y^2+1\right)}{64 y^3}} \cdot X, \end{align*} $$

where

$$ \begin{align*} X = \left(e^{\frac{3 \pi \alpha}{y}}+e^{\frac{4 \pi \alpha}{y}}+e^{\pi \alpha \left(2 y+\frac{3}{2 y}\right)}+2 e^{\pi \alpha \left(2 y+\frac{5}{2 y}\right)}+3 e^{\pi \alpha \left(2 y+\frac{7}{2 y}\right)}+1\right). \end{align*} $$

The result follows from lengthy (but standard) estimates as in Lemma 5.9.

Proof. Collecting powers of y, we have

$$ \begin{align*} \phi_{k,l}'(y) = \frac{\left(16 y^4 (2 l+1)^2 - 8 y^2 \left(8 k^2+8 k (l+1)+2 l (l+1)+1\right)-3\right)}{64 y^4}. \end{align*} $$

As $y \geq \sqrt {3}/2$ , we see that

$$ \begin{align*} |\phi_{k,l}'(y)| &\leq \frac{(2 l + 1)^2}{4} + \left|\frac{\left(8 k^2+8 k (l+1)+2 l (l+1)+1\right)}{8 y^2}\right| + \frac{1}{12} \\ &\leq \frac{(2 l + 1)^2}{4} + \frac{\left|8 k^2+8 k (l+1)+2 l (l+1)+1\right|}{6} + \frac{1}{12}. \end{align*} $$

This should be compared to

$$ \begin{align*} Q_1(k,l) = \phi_{k,l}(\sqrt{3}/2) = \frac{2}{3 \sqrt{3}} + \frac{2}{\sqrt{3}} (k+l) + \frac{2}{\sqrt{3}}(k^2 + kl + l^2) \end{align*} $$

and a short computation shows that

$$ \begin{align*} \max_{y \geq \sqrt{3}/2} |\phi_{k,l}'(y)| \leq 2 \, Q_1(k,l). \end{align*} $$

We do the same thing for the second derivative. Collecting powers of y, we obtain

$$ \begin{align*} \phi_{k,l}"(y) = \frac{4 y^2 \left(8 k^2+8 k (l+1)+2 l (l+1)+1\right)+3}{16 y^5}. \end{align*} $$

We note that the algebraic expression $ \left (8 k^2+8 k (l+1)+2 l (l+1)+1\right )$ is comprised of a positive semidefinite quadratic form with lower order perturbations: the quantity can be negative but only linearly in $k,l$ (while it grows quadratically in $k,l$ ). As a fairly standard computation shows, we have

$$ \begin{align*} | \phi_{k,l}"(y)| &= \left| \frac{4 y^2 \left(8 k^2+8 k (l+1)+2 l (l+1)+1\right)+3}{16 y^5} \right| \leq 3 \, Q_1(k,l). \end{align*} $$

Again, as in many parts of the proof, stronger bounds are conceivable: an asymptotic analysis shows that as $|k|, |l|$ become large, the best constant converges to $8/3$ (which will not be needed for the argument).

Proof. This follows quickly from the previous inequalities. Note that we have

$$ \begin{align*} \left| \frac{d}{dy} e^{- \pi \alpha \, \phi_{k,l}(y)} \right| &= \pi \alpha \left| \phi_{k,l}'(y) \right| e^{- \pi \alpha \, \phi_{k,l}(y)}\\ &\leq 2 \pi \alpha \, Q_1(k,l) \, e^{-\frac{\pi \alpha }{2} \sqrt{Q_1(k,l)}}. \end{align*} $$

Taking second derivatives, we have

$$ \begin{align*} \frac{d^2}{dy^2} e^{- \pi \alpha \, \phi_{k,l}(y)} = \left( \pi^2 \alpha^2 \phi_{k,l}'(y)^2 - \pi \alpha \, \phi_{k,l}"(y) \right) e^{- \pi \alpha \, \phi_{k,l}(y)} \end{align*} $$

and applying the bounds from (5.1) and (5.2) leads to the desired result.

Proof. Lemma 5.11 shows that, under the above assumptions,

$$ \begin{align*} \frac{d^2}{d y^2} \sum_{-1 \leq k,l \leq 1} e^{- \pi \alpha \, \phi_{k,l}(y)} \leq - 2.5 \, \alpha \, e^{- \frac{2 \pi \alpha}{3\sqrt{3}}}. \end{align*} $$

It therefore suffices to ensure that

$$ \begin{align*} X = \frac{d^2}{d y^2} \sum_{\substack{(k,l) \in \mathbb{Z}^2 \\ \max\left\{|k|,|l|\right\}> 1}} \hspace{-9pt} e^{- \pi \alpha \, \phi_{k,l}(y)} \leq 2.5\, \alpha \, e^{- \frac{2 \pi \alpha}{3\sqrt{3}}}. \end{align*} $$

Using the triangle inequality and Lemma 5.14, we have

$$ \begin{align*} X \leq \hspace{-9pt} \sum_{\substack{(k,l) \in \mathbb{Z}^2 \\ \max\left\{|k|,|l|\right\}> 1}} \hspace{-6pt} \max_{y \geq \sqrt{3}/2}\left| \frac{d^2}{d y^2} e^{- \pi \alpha \, \phi_{k,l}(y)} \right| \leq 5 \pi^2 \alpha^2 \hspace{-9pt} \sum_{\substack{(k,l)\in\mathbb{Z}^2 \\ \max\left\{|k|,|l|\right\} > 1}} \hspace{-6pt} Q_1(k,l)^2 e^{-\frac{\pi \alpha}{2} \sqrt{Q_1(k,l)}}. \end{align*} $$

This bound, as it turns out, is actually sufficient: we have

$$ \begin{align*} \min_{k,l \in \mathbb{Z}} Q_1(k,l) = \frac{2}{3\sqrt{3}} \sim 0.38\dots, \end{align*} $$

while

$$ \begin{align*} \min_{\substack{(k,l)\in\mathbb{Z}^2 \\ \max\left\{|k|,|l|\right\}> 1}} \hspace{-9pt} Q_1(k,l) = \frac{14}{3\sqrt{3}} \sim 2.69\dots \end{align*} $$

is larger by a factor of 7 (compare Figures 5 and 6). So, it is clear that for all $\alpha \geq \alpha _0$ sufficiently large,

$$ \begin{align*} 5 \pi^2 \alpha^2 \hspace{-9pt} \sum_{\substack{(k,l)\in\mathbb{Z}^2 \\ \max\left\{|k|,|l|\right\}> 1}} Q_1(k,l)^2 e^{-\frac{ \pi \alpha}{2} \sqrt{Q_1(k,l)}} \leq 2.5 \, \alpha \, e^{- \frac{2 \pi \alpha}{3\sqrt{3}}} \end{align*} $$

has to hold. We may pass to truncated geometric series to obtain closed expressions that can be compared. Numerically, we see that the desired inequality holds for $\alpha _0 \sim 5.96$ .

Proof. The combined value of the three dominant terms at $y = \sqrt {3}/2$ is $3 \exp \left (-2 \pi \alpha /(3 \sqrt {3})\right )$ . A simple algebraic computation shows that for one of the three dominant terms (meaning that $(k,l)$ is either $(0,0)$ or $(-1,0)$ or $(0,-1)$ ), $f_{k,l}$ is monotonically decreasing and

$$ \begin{align*} f_{k,l}\left(\tfrac{\sqrt{3}}{2} + \tfrac{1}{c \sqrt{\alpha}}\right) \leq e^{-\frac{2 \pi \left(9 c^2+10 \sqrt{3} c+9\right)}{9 c \left(\sqrt{3} c+2\right)^3}} f_{k,l}\left(\tfrac{\sqrt{3}}{2}\right). \end{align*} $$

Outside the $[\sqrt {3}/2, \sqrt {3}/2 + 1/(3\sqrt {\alpha })]$ interval, we can use this to argue that the contribution of the three dominant terms is at most, using $\alpha \geq 6$ ,

$$ \begin{align*} 3 \, e^{-\frac{2 \pi \sqrt{\alpha } \left(27 \alpha +10 \sqrt{3} \sqrt{\alpha }+3\right)}{9 \left(3 \sqrt{3} \sqrt{\alpha }+2\right)^3}} e^{-\frac{2 \pi \alpha}{3 \sqrt{3}}} \leq 2.7 \, e^{-\frac{2 \pi \alpha}{3 \sqrt{3}}}. \end{align*} $$

So, there is a substantial loss in the leading terms. The question is thus whether the remaining terms can contribute anything of size at least $0.3$ times the leading terms’ contribution at $y=\sqrt {3}/2$ . As it turns out, this does not happen: the inequality

$$ \begin{align*} \sum_{\substack{k,l \in \mathbb{Z} \\ Q_1(k,l) \geq \frac{8}{3\sqrt{3}}}} e^{- \frac{\pi \alpha}{2} \sqrt{Q_1(k,l)}} < 0.3 \, e^{- \frac{2 \pi \alpha}{3\sqrt{3}}} \end{align*} $$

is true for $\alpha \geq \alpha _0$ , where $\alpha _0 \sim 0.7$ .

Proof. We proceed as in the case of the first sum. Using Lemma 5.17, we already know that there is a critical point at $y= \sqrt {3}/2$ . The argument is based on a case distinction.

Case 1. We first deal with the case $1 \leq \alpha \leq 5$ and y being close to the critical point in the sense of $\sqrt {3}/2 \leq y \leq 1.5$ . This case has the advantage of being a finite region in parameter space. We split the series (suppressing $\alpha $ in the notation) as follows:

$$ \begin{align*} g_{\alpha}(y) = \sum_{\substack{(k,l) \in \mathbb{Z}^2 \\ Q_2(k,l) \leq 1}} e^{- \pi \alpha \phi_{k,l}(y)} \psi_{k,l}(y) + \sum_{\substack{(k,l) \in \mathbb{Z}^2 \\ Q_2(k,l) \geq 3}} e^{- \pi \alpha \phi_{k,l}(y)} \psi_{k,l}(y) = g_{\text{main}}(y) + g_{\text{rem}}(y). \end{align*} $$

We set $u_{k,l}(y) = e^{-\pi \alpha \phi _{k,l}(y)} \psi _{k,l}(y)$ , and termwise differentiation yields the explicit second derivative:

(5.3) $$ \begin{align} u_{k,l}"(y) = e^{-\pi \alpha \phi_{k,l}(y)} \left( \psi_{k,l}" - 2\pi \alpha \phi_{k,l}'\psi_{k,l}' - \pi \alpha \phi_{k,l}"\psi_{k,l} + \bigl(\pi \alpha \phi_{k,l}'\bigr)^2 \psi_{k,l} \right). \end{align} $$

To establish that $g_\alpha "(y) < 0$ globally, the remainder must not disrupt the sign of $g_{\text {main}}"(y)$ . We further partition the $\alpha $ -interval, governed by the damping strength: $\alpha \in [1, 2]$ and $\alpha \in (2, 5]$ .

Case 1a: $\alpha \in [1,2]$ . A direct numerical evaluation of the finite main sum shows that actually $g_{\text {main}}"(y) \leq -1.5 \cdot 10^{-3}$ in this case. We split the remainder at a modest cutoff $R_1 = 6$ :

$$ \begin{align*} |g_{\text{rem}}"(y)| \leq \sum_{3 \leq Q_2(k,l) < 6} |u_{k,l}"(y)| + \sum_{Q_2(k,l) \geq 6} |u_{k,l}"(y)|. \end{align*} $$

Evaluating the finite sum ( $3 \leq Q_2(k,l) \leq 6$ ) yields a bound of $1.2 \cdot 10^{-4}$ . The infinite tail is bounded via a coarse integral comparison (analogous to the case below) by $4.5 \cdot 10^{-6}$ . Thus,

$$ \begin{align*} \sup_{(\alpha,y)\in[1,2]\times[\sqrt{3}/2,1.5]} |g_{\text{rem}}"(y)| \ll 1.5 \cdot 10^{-3}, \end{align*} $$

preserving the strict negativity of the second derivative.

Case 1b: $\alpha \in (2,5]$ . The main part is now bounded by

$$ \begin{align*} g"_{\text{main}}(y) < -1.1142 \cdot 10^{-5}. \end{align*} $$

We choose a cutoff threshold $R_2 = 10$ :

$$ \begin{align*} |g_{\text{rem}}"(y)| \leq \sum_{3 \leq Q_2(k,l) < 10} |u_{k,l}"(y)| + \sum_{Q_2(k,l) \geq 10} |u_{k,l}"(y)| = \sum_{\text{near}} + \sum_{\text{far}}. \end{align*} $$

Evaluating the finite sum using interval arithmetic yields $\sum \limits _{\text {near}} \leq 2.5 \cdot 10^{-7}$ .

We exploit the uniform ellipticity of the function $\phi _{k,l}(y) = \frac {(k + l/2)^2}{y} + l^2 y$ . For $y \in [\sqrt {3}/2, 1.5]$ , the coefficients are uniformly bounded from below, yielding the explicit relation to the quadratic form $Q_2(k,l) = k^2 + kl + l^2$ :

$$ \begin{align*} \phi_{k,l}(y) \geq c_0 \left(k^2 + kl + l^2\right) = c_0 Q_2(k,l), \end{align*} $$

where optimizing over the interval endpoints yields the sharp constant $c_0 = \frac {2}{3}$ .

Furthermore, the algebraic prefactors in equation (5.3) are safely enveloped by a polynomial growth condition. Maximizing over $y \in [\sqrt {3}/2, 1.5]$ and setting $\alpha = 5$ for the polynomials yields the uniform bound (compare Lemma 5.21)

$$ \begin{align*} |u_{k,l}"(y)| \leq C_1 \, Q_2(k,l)^2 e^{-\pi \alpha \phi_{k,l}(y)} \leq C_1 \, Q_2(k,l)^2 e^{-\pi \alpha c_0 Q_2(k,l)}, \end{align*} $$

where a direct worst-case coefficient collection over the interval bounds yields the explicit constant $C_1 = 157.8$ . For the tail region $Q_2(k,l) \geq 10$ , and since $\alpha> 2$ , we dominate the series by an integral over the domain $\Omega = \{(x,y) \in \mathbb {R}^2 : x^2 + xy + y^2 \geq 10\}$ :

$$ \begin{align*} \sum_{\text{far}} \leq \iint_{\Omega} C_1 (x^2 + xy + y^2)^2 e^{-2\pi c_0 (x^2 + xy + y^2)} \,dx \,dy. \end{align*} $$

Applying a linear transformation to diagonalize the quadratic form, followed by a change to polar coordinates, simplifies the area integral directly to

$$ \begin{align*} \sum_{\text{far}} \leq \frac{4\pi C_1}{\sqrt{3}} \int_{\sqrt{10}}^{\infty} r^5 e^{-2\pi c_0 r^2} \, dr = \frac{2\pi C_1}{\sqrt{3}(2\pi c_0)^3} \left( (20\pi c_0)^2 + 2(20 \pi c_0) + 2 \right) e^{-20\pi c_0}. \end{align*} $$

Substituting $c_0 = \frac {2}{3}$ and $C_1 = 157.8$ provides the explicit analytical bound:

$$ \begin{align*} \sum_{\text{far}} \leq \frac{157.8 \cdot 2\pi}{\sqrt{3} (4\pi/3)^3} \left[ \left(\frac{40\pi}{3}\right)^2 + \frac{80\pi}{3} + 2 \right] e^{-40\pi/3} < 10^{-14}. \end{align*} $$

Combining the estimates, we obtain

$$ \begin{align*} |g_{\text{rem}}"(y)| \leq 2.5 \cdot 10^{-7} + 10^{-14} < 3 \cdot 10^{-7}. \end{align*} $$

Consequently, the second derivative is strictly negative in this case.

Case 2a. The case $y \geq 1.5$ and $1.5 \leq \alpha \leq 5$ will follow from Lemma 5.25 further below. Lemma 5.25 shows that the maximum is confined to a small neighborhood of $\sqrt {3}/2$ , so it cannot occur in $y\geq 1.5$ when $\alpha \geq 1.5$ .

Case 2b. It remains to analyze $1 \leq \alpha \leq 1.5$ , $y \geq 1.5$ . This domain is straightforward to handle since $y \geq 1.5$ enforces a strict separation from the critical point. In this regime, the first seven terms of the lattice sum, corresponding to the origin $(k,l) = (0,0)$ and the six pairs satisfying $k^2 + kl + l^2 = 1$ , provide a very good approximation of the series.

As y increases, the term $l^2 y$ in $\phi _{k,l}(y)$ grows linearly, which causes all terms with $l \neq 0$ to decay exponentially fast. A direct interval evaluation of these dominant terms confirms a substantially uniform gap. The computations for the first seven terms ( $Q_2(k,l)\leq 1$ ) yield

$$ \begin{align*} g_{\text{main}}\left( \frac{\sqrt{3}}{2}\right) - g_{\text{main}}(1.5) \geq \frac{1}{20} \quad \text{for all } 1 \leq \alpha \leq 1.5. \end{align*} $$

The remaining unbounded tail, where $y> 1.5$ , can then be handled using the integral estimates from the previous case or standard geometric series arguments.

Proof. Let $l \neq 0$ . The only interesting case is when k and l have opposite signs; if they have the same sign, we can flip one of the signs to obtain a function with a smaller minimum. Therefore, we now assume that k and l have opposite signs and that $l \neq 0$ . The function is positive in the region of interest since $k^2 + k l + l^2 \geq 1$ . First, observe that the function is convex:

$$ \begin{align*}\frac{d^2}{dy^2} ~ \frac{1}{y} \left(k^2 + k l + \left(\frac{1}{4} + y^2 \right) l^2\right) = \frac{(2k+l)^2}{2y^3} \geq 0. \end{align*} $$

This shows that the minimum is either at $y = \sqrt {3}/2$ or at a point where the derivative vanishes. The derivative vanishes at

$$ \begin{align*}y_0 = \left|\frac{1}{2} + \frac{k}{l}\right|.\end{align*} $$

We recall that k and l have opposite signs, and, therefore, the term $k/l$ is negative. There is now a simple case distinction: if $k/l \leq -\sqrt {3}/2 - 1/2$ , then $y_0 \geq \sqrt {3}/2$ and the minimum is assumed at the point where the derivative vanishes. Otherwise, the derivative vanishes earlier, and the minimum is assumed at $y = \sqrt {3}/2$ . This second case is trivial, since then

$$ \begin{align*}\frac{1}{y} \left(k^2 + k l + \left(\frac{1}{4} + y^2 \right) l^2\right) \geq \frac{2}{\sqrt{3}} Q_2(k,l).\end{align*} $$

If, on the other hand, $k/l \leq -\sqrt {3}/2 - 1/2$ , then a short computation shows that

$$ \begin{align*}\min_{y \geq \sqrt{3}/2} \quad \frac{1}{y} \left(k^2 + k l + \left(\frac{1}{4} + y^2 \right) l^2\right) = -l(2k+l).\end{align*} $$

This can only be small when $2k+ l \sim 0$ , but this cannot happen because $k/l \leq -\sqrt {3}/2 - 1/2$ , which implies $ -l(2k+l) \geq \sqrt {3} \, l^2$ . This then proves the result unless

$$ \begin{align*}\left( \sqrt{3} \, l^2 \right)^2 = 3 \, l^4 \leq k^2 +kl + l^2 = Q_2(k,l).\end{align*} $$

If this is the case, then certainly also $2 \, l^4 \leq k^2 + kl $ , which implies

$$ \begin{align*}|k| \geq \frac{1}{2} \left(\sqrt{8 \, l^4+l^2}-l\right) \geq l^2.\end{align*} $$

In that regime, we have

$$ \begin{align*}\min_{y \geq \sqrt{3}/2} \quad \frac{1}{y} \left(k^2 + k l + \left(\frac{1}{4} + y^2 \right) l^2\right) = -l(2k+l) \sim 2 |k| |l|,\end{align*} $$

while our lower bound is clearly smaller:

$$ \begin{align*}\sqrt{Q_2(k,l)} = \sqrt{k^2 + kl + l^2} \sim |k|.\\[-36pt] \end{align*} $$

Proof. This follows immediately from Lemma 5.19 and $|\psi _{k,l}(y)| \leq 1$ .

Proof. The product rule yields

$$ \begin{align*} \frac{d}{dy} \left(e^{- \pi \alpha \, \phi_{k,l}(y)} \, \psi_{k,l}(y) \right) = \left(- \pi \alpha \, \phi_{k,l}'(y) \, \psi_{k,l}(y) + \psi_{k,l}'(y)\right) e^{- \pi \alpha \, \phi_{k,l}(y)}. \end{align*} $$

For the first term, we need an upper bound on

$$ \begin{align*}|\phi_{k,l}'(y)| = \left| l^2 - \frac{(2k+l)^2}{4y^2} \right| \leq l^2 + \frac{(2k + l)^2}{3} \leq \frac{4}{3} Q_2(k,l).\end{align*} $$

For the second term, we need an upper bound on

$$ \begin{align*}| \psi_{k,l}'(y)| \leq 2\pi \left(\frac{k}{4y^3} + \frac{l}{8 y^3} \right) \leq 2.5 |k| + 1.25 |l| \leq 4\sqrt{Q_2(k,l)}.\end{align*} $$

Combining both, we get

$$ \begin{align*}\left| \frac{d}{dy} \left( e^{- \pi \alpha \, \phi_{k,l}(y)} \, \psi_{k,l}(y) \right)\right| \leq \left( \frac{4}{3} \pi \alpha \, Q_2(k,l) + 4\sqrt{Q_2(k,l)} \right)e^{-\pi \alpha \sqrt{Q_2(k,l)}} .\end{align*} $$

Since $\alpha \geq 5$ , we have

$$ \begin{align*}4\sqrt{Q_2(k,l)} \leq 4 Q_2(k,l) \leq \left( \frac{4}{5 \pi} \right) \pi \alpha Q_2(k,l)\end{align*} $$

and the result follows. The product rule further yields

$$ \begin{align*} \frac{d^2}{dy^2} \left(e^{- \pi \alpha \, \phi_{k,l}(y)} \, \psi_{k,l}(y)\right) = \; & \Big( \psi_{k,l}"(y) - \pi \alpha \, \phi_{k,l}"(y) \, \psi_{k,l}(y)\\ & \quad + \pi^2 \alpha^2 \phi_{k,l}'(y)^2 \, \psi_{k,l}(y) - 2\pi \alpha \, \phi_{k,l}'(y) \, \psi'_{k,l}(y) \Big) e^{- \pi \alpha \, \phi_{k,l}(y)}. \end{align*} $$

We need two new ingredients: the first is

$$ \begin{align*}|\phi_{k,l}"(y)| = \frac{(2k+l)^2}{2y^3} \leq 4 \cdot Q_2(k,l).\end{align*} $$

The second is

$$ \begin{align*} | \psi_{k,l}"(y)| & \leq 4\pi^2 \left(\frac{k}{4y^3} + \frac{l}{8 y^3} \right)^2 + 2\pi \left( \frac{3 |k|}{4y^4} + \frac{3|l|}{8 y^4}\right) \\ & \leq 16 \, Q_2(k,l) + 2 \pi \left( \frac{4 |k|}{3} + \frac{2|l|}{3 } \right) \\ & \leq 16 \, Q_2(k,l) + 13 \, Q_2(k,l) = 29 \, Q_2(k,l). \end{align*} $$

It remains to combine all the estimates. We have

$$ \begin{align*} \left| \pi \alpha \, \phi_{k,l}"(y) e^{- \pi \alpha \, \phi_{k,l}(y)} \, \psi_{k,l}(y) \right| & \leq 4 \pi \alpha \, Q_2(k,l) e^{-\pi \alpha \sqrt{Q_2(k,l)}} \\ \left| \pi^2 \alpha^2 \phi_{k,l}'(y)^2 e^{- \pi \alpha \, \phi_{k,l}(y)} \,\psi_{k,l}(y) \right| & \leq \tfrac{16}{9} \pi^2 \alpha^2 Q_2(k,l)^2 e^{-\pi \alpha \sqrt{Q_2(k,l)}}\\ \left| -2 \pi \alpha \, \phi_{k,l}'(y)e^{- \pi \alpha \, \phi_{k,l}(y)} \, \psi'_{k,l}(y) \right| & \leq \tfrac{32}{3} \pi \alpha \,Q_2(k,l)^{3/2} e^{-\pi \alpha \sqrt{Q_2(k,l)}}\\ \left| e^{- \pi \alpha \, \phi_{k,l}(y)} \, \psi_{k,l}"(y) \right| & \leq 29 \, Q_2(k,l) e^{-\pi \alpha \sqrt{Q_2(k,l)}}\\ \left| - \pi \alpha \, \phi_{k,l}'(y) e^{- \pi \alpha \, \phi_{k,l}(y)} \, \psi_{k,l}(y) \right| & \leq \tfrac{4}{3} \pi \alpha \, Q_2(k,l) e^{-\pi \alpha \sqrt{Q_2(k,l)}}. \end{align*} $$

Using $Q_2(k,l) \geq 1$ and $\alpha \geq 5$ , by adding all these terms and bounding them from above, we end up getting the desired bound.

Proof. For small values of t, we have the Gaussian short-time asymptotic

$$ \begin{align*} u(t,x) \sim \tfrac{1}{\sqrt{4 \pi t}} \, e^{-\frac{x^2}{4t}}, \end{align*} $$

placing the inflection point asymptotically at $x_0 \sim \sqrt {2t} \ll 0.3$ . As time becomes large, we have dominance of the leading order term and expect

$$ \begin{align*} u(t,x) \sim 1 + 2e^{-4 \pi^2 t} \cos{(2 \pi x)}, \end{align*} $$

having the inflection point at $0.25 < 0.3$ . For intermediate times $t \sim 1$ , some basic estimates show that the critical point actually wanders continuously from 0 to $1/4$ . Since we do not require sharp bounds, the case can be dealt with by basic asymptotics.

Proof. We write the expression as $u(t(y), x(y))$ with

$$ \begin{align*} t(y) = \frac{\alpha}{4 \pi y} \qquad \mbox{and} \qquad x(y) = \frac{1}{2} - \frac{1}{8y^2}. \end{align*} $$

Note that, since $y \geq \sqrt {3}/2$ , we have $1/3 \leq x(y) \leq 1/2$ and, in particular, Lemma 5.22 is always applicable. We note that, using the chain rule,

$$ \begin{align*}\frac{d}{dy} u(t(y), x(y)) = \frac{d}{dt} u(t(y), x(y)) \frac{dt(y)}{dy} + \frac{d}{dx} u(t(y), x(y)) \frac{dx(y)}{dy}.\end{align*} $$

Since $u(t,x)$ is the heat kernel, it satisfies the heat equation

$$ \begin{align*}\frac{d}{dt} u(t,x) = \frac{d^2}{dx^2} u(t,x).\end{align*} $$

We have some information about the second derivatives of the heat kernel in the region of interest, thanks to Lemma 5.22, and can conclude that this term is positive. Moreover, we have

$$ \begin{align*}\frac{dt(y)}{dy} = - \frac{\alpha}{4 \pi y^2} < 0 \qquad \mbox{and} \qquad \frac{dx(y)}{dy} = \frac{1}{4y^3}> 0.\end{align*} $$

Therefore,

$$ \begin{align*}\frac{d}{dy} u(t(y), x(y)) = \underbrace{\frac{d^2}{dx^2} u(t(y), x(y))}_{>0} \underbrace{\frac{dt(y)}{dy}}_{<0} + \underbrace{\frac{d}{dx} u(t(y), x(y))}_{<0} \underbrace{\frac{dx(y)}{dy}}_{>0} < 0.\\[-30pt]\end{align*} $$

Proof. The first part of the statement,

$$ \begin{align*} \max_{y \geq \frac{\sqrt{3}}{2} + \frac{1}{4\sqrt{\alpha}}} G_{\alpha}(y) = G_{\alpha}\left(\tfrac{\sqrt{3}}{2} + \tfrac{1}{4\sqrt{\alpha}} \right), \end{align*} $$

follows immediately from Lemma 5.23. We use Lemma 5.23 once more: instead of evaluating the heat kernel at $(t(\sqrt {3}/2), x(\sqrt {3}/2))$ and comparing it to

$$ \begin{align*} \text{the heat kernel at} \quad \left(t\left(\tfrac{\sqrt{3}}{2} + \tfrac{1}{4\sqrt{\alpha}}\right), x\left(\tfrac{\sqrt{3}}{2} + \tfrac{1}{4\sqrt{\alpha}}\right)\right), \end{align*} $$

we may as well use the monotonicity in x once more to compare $(t(\sqrt {3}/2), x(\sqrt {3}/2))$ with $(t(\sqrt {3}/2 + 1/(4\sqrt {\alpha })), x(\sqrt {3}/2))$ . It suffices to estimate the difference between

$$ \begin{align*} \sum_{k \in \mathbb{Z}} e^{- \frac{\pi\alpha k^2 }{\sqrt{3}/2}} \, \cos \left( \tfrac{ 2\pi k}{3} \right) \quad \text{ and } \quad \sum_{k \in \mathbb{Z}} e^{- \frac{\pi\alpha k^2 }{\sqrt{3}/2 +1/(4\sqrt{\alpha})}} \, \cos \left( \tfrac{ 2\pi k}{3} \right). \end{align*} $$

Both sums are, due to the rapid decay of their terms, essentially given by their first term as soon as $\alpha $ is sufficiently large and

$$ \begin{align*} \sum_{k \in \mathbb{Z}} e^{- \pi\alpha k^2 t} \, \cos \left( \tfrac{ 2\pi k}{3} \right) \sim 1 - e^{- \pi\alpha t} + \mbox{l.o.t.} \end{align*} $$

This requires us to estimate $\exp (- \pi \alpha t)$ for two nearly adjacent values of t, these being

$$ \begin{align*}t_1 = \frac{1}{\sqrt{3}/2} \qquad \mbox{and} \qquad t_2 = \frac{1}{\sqrt{3}/2 + 1/(4\sqrt{\alpha})}.\end{align*} $$

We see that the leading order difference is of the order given by

$$ \begin{align*}e^{- \pi\alpha t_2} - e^{- \pi\alpha t_1}.\end{align*} $$

The derivative of $\exp \left (- \pi \alpha t\right )$ is $ -\pi \alpha \exp \left (- \pi \alpha t\right )$ and thus, using the mean value theorem,

$$ \begin{align*} e^{- \pi\alpha t_2} - e^{- \pi\alpha t_1} &\geq \pi \alpha \, (t_1-t_2) e^{- \pi \alpha t_1}\\ &\geq \frac{\pi \alpha}{4 \sqrt{\alpha}} \frac{1}{\frac{\sqrt{3}}{2} \left( \frac{\sqrt{3}}{2} + \frac{1}{4\sqrt{\alpha}}\right)} \, e^{-\frac{2 \pi \alpha}{\sqrt{3}}}\\ &= \frac{2 \pi \alpha}{6 \sqrt{\alpha}+\sqrt{3}} \, e^{-\frac{2 \pi \alpha}{\sqrt{3}}}\\ &\geq \frac{2 \sqrt{\alpha}}{3} \, e^{-\frac{2 \pi \alpha}{\sqrt{3}}}. \end{align*} $$

The inequality is easily verified to be effective for $1 \leq \alpha \leq 10$ , after which the asymptotic analysis is more than accurate.

Proof. We decompose the indices into three sets:

$$ \begin{align*} \mathbb{Z}^2 = & \left\{(k,0): k \in \mathbb{Z} \right\} \\ \cup & \left\{(k,l): Q_2(k,l) = 1 ~\mbox{and}~l \neq 0\right\} \\ \cup & \left\{(k,l): Q_2(k,l)> 1 ~\mbox{and}~l \neq 0\right\}. \end{align*} $$

The first set gives rise to the quantity resembling a heat kernel, which has been analyzed in Lemma 5.22 and is well behaved: it decays away from $y = \sqrt {3}/2$ at a controlled rate. The second set is small and completely explicit:

$$ \begin{align*}\left\{(k,l): Q_2(k,l) = 1 ~\mbox{and}~l \neq 0\right\} = \left\{(0,-1), (1,-1), (-1,1), (0,1) \right\}.\end{align*} $$

The third set is not small but will amount to only a small contribution. We first sum over the four terms in the second set:

$$ \begin{align*}\sum_{\substack{(k,l) \in \mathbb{Z}^2 \\ Q_2(k,l) = 1 , l\neq0}} \hspace{-9pt} e^{- \pi \alpha \, \phi_{k,l}(y)} \psi_{k,l}(y) = -2 \, e^{-\frac{\pi a}{y}} \cos \left(\tfrac{\pi }{4 y^2}\right).\end{align*} $$

This term is clearly decreasing in y. It remains to bound the rest. We note

$$ \begin{align*} e^{- \pi \alpha \, \phi_{k,l}(y)} \, \psi_{k,l}(y) &\leq e^{- \pi \alpha \, \phi_{k,l}(y)} \leq e^{- \pi \alpha \min_{y \geq \sqrt{3}/2} \phi_{k,l}(y)}. \end{align*} $$

Appealing to Lemma 5.19, we can bound the remaining sum from above by

$$ \begin{align*} \sum_{\substack{(k,l) \in \mathbb{Z}^2, \\ Q_2(k,l)> 1, l \neq 0}} e^{- \pi \alpha \, \phi_{k,l}(y)} \, \psi_{k,l}(y) \leq \sum_{\substack{(k,l) \in \mathbb{Z}^2, \\ Q_2(k,l) > 1, l \neq 0}} e^{- \pi \alpha \sqrt{Q_2(k,l)}}. \end{align*} $$

The difference in the asymptotic decay shows that

$$ \begin{align*} \sum_{\substack{(k,l) \in \mathbb{Z}^2, \\ Q_2(k,l)> 1, l \neq 0}} e^{- \pi \alpha \sqrt{Q_2(k,l)}} \leq \tfrac{2 \sqrt{\alpha}}{3} \, e^{-\frac{2 \pi \alpha}{\sqrt{3}}} \end{align*} $$

has to hold for $\alpha \geq \alpha _0$ sufficiently large. Basic numerics show that $\alpha _0 \sim 1.3$ .

Proof. The first inequality is simple since, using $y \leq 1$ ,

$$ \begin{align*} \left| e^{- \frac{ \pi \alpha k^2}{y}} \cos\left(2\pi k \left( \tfrac{1}{2} - \tfrac{1}{8y^2}\right) \right) \right| \leq e^{-\pi \alpha k^2}. \end{align*} $$

The second statement follows from an explicit computation and basic estimates.

Proof. An explicit computation shows that

$$ \begin{align*} X = \sum_{-1 \leq k,l \leq 1} \exp\left( - \pi \alpha \cdot \phi_{k,l}(y)\right)\psi_{k,l}(y) \end{align*} $$

simplifies to

$$ \begin{align*} X = e^{-\pi \alpha y-\frac{9 \pi \alpha }{4 y}} \left(2 \sin \left(\tfrac{3 \pi }{8 y^2}\right)-4 \, e^{\frac{2 \pi \alpha }{y}} \sin \left(\tfrac{\pi }{8 y^2}\right)\right)- 2 e^{-\frac{\pi \alpha }{y}} \cos \left(\tfrac{\pi }{4 y^2}\right)+1. \end{align*} $$

We now differentiate each term twice in y. We first observe that

$$ \begin{align*}- \frac{\pi \alpha}{y} \qquad \mbox{and} \qquad - \pi \alpha y - \frac{9 \pi \alpha}{4y} + \frac{2\pi \alpha}{y}\end{align*} $$

coincide at $y = \sqrt {3}/2$ ; they both result in $-2\pi \alpha /\sqrt {3}$ . In contrast, the remaining term is many orders of magnitude smaller. It thus suffices to understand

$$ \begin{align*} X_2 &= -4 \, e^{-\pi \alpha y-\frac{9 \pi \alpha }{4 y}} e^{\frac{2 \pi \alpha }{y}} \sin \left(\tfrac{\pi }{8 y^2}\right)-2 \, e^{-\frac{\pi \alpha }{y}} \cos \left(\tfrac{\pi }{4 y^2}\right)\\ &= -4 \, e^{-\pi \alpha y-\frac{\pi \alpha }{4 y}} \sin \left(\tfrac{\pi }{8 y^2}\right)-2 \, e^{-\frac{\pi \alpha }{y}} \cos \left(\tfrac{\pi }{4y^2}\right). \end{align*} $$

Differentiating twice and simplifying, we end up with

$$ \begin{align*}\frac{d^2}{dy^2} X_2 = \frac{\pi e^{-\frac{\pi \alpha \left(y^2+1\right)}{y}}}{4 y^6} \left( Y_1 + Y_2 +Y_3 + Y_4\right),\end{align*} $$

where, for $\alpha \geq 5$ and $\sqrt {3}/2 \leq y \leq 1$ , all terms are negative:

$$ \begin{align*} Y_1 &= -2 \, y e^{\frac{3 \pi \alpha }{4 y}} \left(\pi \alpha \left(4 y^2-1\right)+6 y\right) \cos \left(\tfrac{\pi }{8 y^2}\right) \leq 0 \\ Y_2 &=-2 \, e^{\pi \alpha y} \left(4 \pi \alpha ^2 y^2-8 \alpha y^3-\pi \right) \cos \left(\tfrac{\pi }{4 y^2}\right) \leq 0\\ Y_3 &= -e^{\frac{3 \pi \alpha }{4 y}} \left(\pi \alpha ^2 \left(4 y^3-y\right)^2-8 \alpha y^3-\pi \right) \sin \left(\tfrac{\pi }{8 y^2}\right)\leq 0 \\ Y_4 &= 4 y \, e^{\pi \alpha y} (3 y-2 \pi \alpha ) \sin \left(\tfrac{\pi }{4 y^2}\right) \leq 0. \end{align*} $$

Therefore, in that regime, we have

$$ \begin{align*} \frac{d^2}{dy^2} X_2 \leq \tfrac{\pi e^{-\frac{\pi \alpha \left(y^2+1\right)}{y}}}{4 y^6} Y_2 \quad \text{ and } \quad Y_2 \leq -2.4 \, \pi \alpha^2 e^{\pi \alpha y}. \end{align*} $$

Simultaneously, we have, in the same regime,

$$ \begin{align*}\tfrac{\pi e^{-\frac{\pi \alpha \left(y^2+1\right)}{y}}}{4 y^6} \geq \tfrac{\pi}{4} \, e^{-2 \pi \alpha}.\end{align*} $$

Combining these two inequalities yields the result.

Proof. It remains to derive an upper bound on the second derivatives of the remaining terms,

$$ \begin{align*} X = \frac{d^2}{d y^2} \sum_{\substack{(k,l) \in \mathbb{Z}^2 \\ \max\left\{|k|, |l| \right\}> 1}} \hspace{-9pt} e^{- \pi \alpha \, \phi_{k,l}(y)} \, \psi_{k,l}(y), \end{align*} $$

and to show that they cannot compensate for the negative term derived for the leading terms in Lemma 5.27. We use Lemma 5.21 and Lemma 5.26 to argue that

$$ \begin{align*} X &\leq \sum_{\substack{|k|> 1 \\ l =0}} \frac{d^2}{dy^2} e^{- \pi \alpha \, \phi_{k,l}(y)} \, \psi_{k,l}(y) + \hspace{-18pt} \sum_{\substack{(k,l) \in \mathbb{Z}^2 \\ \max\left\{|k|, |l| \right\} > 1, l \neq 0}} \hspace{-18pt} \frac{d^2}{dy^2} e^{- \pi \alpha \, \phi_{k,l}(y)} \, \psi_{k,l}(y) \\ &\leq 40 \sum_{k=2}^{\infty} \alpha^2 k^4 e^{-\pi \alpha k^2} + 3 \pi^2 \alpha^2 \hspace{-18pt} \sum_{\substack{(k,l) \in \mathbb{Z}^2 \\ \max\left\{|k|, |l| \right\} > 1, l \neq 0}} \hspace{-18pt} Q_2(k,l)^2 \, e^{-\pi \alpha \sqrt{Q_2(k,l)}}. \end{align*} $$

We note that, again, this upper bound decays asymptotically like

$$ \begin{align*}40 \sum_{k=2}^{\infty} \alpha^2 k^4 e^{-\pi \alpha k^2} \sim 40 \alpha^2 \cdot 16 \, e^{-4 \pi \alpha},\end{align*} $$

which decays faster than our upper bound. Therefore, as before, it is clear that for all $\alpha \geq \alpha _0$ , which by basic numerics is $\sim 1.1$ , we will be able, using this argument, to deduce that

$$ \begin{align*}X \leq 0.6 \, \pi^2 \alpha^2 \, e^{\pi \alpha (\frac{\sqrt{3}}{2}-2)}.\\[-30pt]\end{align*} $$