I am trying to prove or disprove

$$\sum_{k=1}^{\infty}e^{-\lambda_{k}t}c_{k} \xrightarrow{t\to 0} \sum_{k=1}^{\infty}c_{k} ,$$

where $\sum c_{k}<\infty, \sum c_{k}^{2}<\infty\text{ and }\frac{\lambda_{k}}{k}\to c$(Weyl's law). The $c_{k},\lambda_{k},t\in \mathbb{R}$ and

Update: the $\lambda_{k}$ are eigenvalues of any domain D in $\mathbb{R}^{2}$:

  1. $\lambda_{k}<\lambda_{k+1}<....$
  2. $\lambda_{k}=\frac{4\pi}{|D|}k+c_{2}\sqrt{k}+o(\sqrt{k})$
  3. $\sum^{k}_{j\geq 1}\lambda_{j}\geq \frac{2\pi}{|D|}k^{2}\Rightarrow \lambda_{k}\geq \frac{2\pi}{|D|}k$ (Li-Yau)

The counterexamples below were before the update, where we only assumed $\frac{\lambda_{k}}{k}\to c$ (but they seem to include monotonicity too).

Q:Since the line of convergence is $Re(z)=0$, is there any way to analytic continue this series in a neighbourhood of zero, in order to apply some of the known tauberian theorems.

Please, I prefer just hints to help me practice.


0)The $\lambda_{k}$ correspond to the Dirichlet-Laplacian eigenvalues of a domain in $\mathbb{R}^{2}$, so I was hoping to have them fixed and only modify the $c_{k}$ for a counterexample. Or come up with the $\lambda_{k}$ and the corresponding domain, that gives a counterexample.

1)The proof of Abel's theorem doesn't work because it requires linear error term $|\lambda_{k}/k-c|<\frac{c'}{k}$ (from Weyl's law we have $\frac{1}{\sqrt{k}}$ error): for $|s_{N}|=|\sum_{N}c_{k}|\leq \varepsilon$ we have $|\sum_{N} c_{k} e^{-\lambda_{k}t}|\leq \varepsilon \sum_{N} ( e^{-\lambda_{k}t}- e^{-\lambda_{k+1}t})\approx \varepsilon e^{-N c_{1} t}\frac{(1-e^{-\sqrt{N+1}t c_{2}})}{(1-e^{t c_{2}})}\to \varepsilon c \sqrt{N+1}$ assuming the unproved but plausible $\lambda_{k}\geq \frac{4\pi}{|D|}k$ for large k.

2)Littlewood tauberian theorems requiring $c_{k}k=o(1)$ don't apply because we might have $c_{k}=\frac{(-1)^{k}}{k}$.

3)By having $c_{k}=\frac{(-1)^{k}}{k}$ the line of convergence of abstract Dirichlet series

$$\sum e^{-\lambda z}c_{k}$$. So analytic continuation is not clear.

is $\{z:Re(z)=0\}$. The Ostrowski–Hadamard gap theorem doesn't apply because $\lambda_{k+1}/\lambda_{k}\to 1$.

4)Borel summation method might apply: Because $\sum c_{k}^{2}<\infty$ and so $$\sqrt{k}c_{k}=o(1).$$

Moreover, $e^{-x}\sum s_{k}\frac{1}{k!}z^{k}$ is weakly Borel summable. So indeed

$$\sum_{k=1}^{\infty}e^{-kt}c_{k} \xrightarrow{t\to 0} \sum_{k=1}^{\infty}c_{k}.$$

So maybe the proof can be modified for $\lambda_{k}$ instead.

5)Possible counterexample via letting $c_{k}=\frac{(-1)^{k}}{k}$ to disallow any dominated convergence.

6)Another Tauberian theorem requires showing for $f(z):=\sum_{k=1}^{\infty}e^{-\lambda_{k}z}c_{k}$ and some function F

$$\frac{f(z)-f(0)}{z}\to F(iIm(z)),$$

uniformly or in L1; as $Re(z)\to 0$ and $Im(z)\in [-\lambda,\lambda]$, where in our case $\lambda=\infty$. So assuming we can show convergence, the limit is


and since $\sum a_{k}<\infty$ we ask that $\sum_{k=1}^{\infty}c_{k}e^{-\lambda_{k}iy}$ is well-defined. However, I think we can concoct $c_k$ s.t. $\sum_{k=1}^{\infty}c_{k}e^{-\lambda_{k}iy_{0}}=\infty$ for some fixed $y_{0}$.

  • $\begingroup$ To save us some time, maybe move point #5 closer to the top of the question, since that was my first reflex. $\endgroup$
    – Ian
    May 24, 2017 at 2:33
  • $\begingroup$ Are your numbers $t,c_k,\lambda_k$ real or complex? $\endgroup$ May 24, 2017 at 6:58
  • $\begingroup$ they are real, but we look t as complex to help apply tauberian theorems. $\endgroup$ May 24, 2017 at 12:50
  • $\begingroup$ Dear Kojar, I read your another problem about the Li-Yau inequality of the eigenvalue. But I do not understand why if we wish to get the result $\lim_{t\to 0^+}\sum_{k=1}^{\infty}e^{-\lambda_kt}c_k=\sum_{k=1}^{\infty}c_k $ the Li-Yau inequality is so crucial, could you give me some motivation? appreciate! $\endgroup$
    – Hu xiyu
    Nov 24, 2017 at 14:25
  • $\begingroup$ By the way if it is the case, my intuition is this could explain why some thing only make sense at dimensional 2, but not the higher dimension case, so there is a hope to get a proof of the inverse spectrum conjecture for dim 2? $\endgroup$
    – Hu xiyu
    Nov 24, 2017 at 14:28

3 Answers 3


The $\lambda_k$ correspond to the Laplacian eigenvalues of a domain in $\mathbb{R}^2$ implies that \begin{equation} 0<\lambda_1\le \lambda_2\le\cdot\cdot\cdot\le \lambda_n\le\cdot\cdot\cdot \end{equation} and $\lambda_n\rightarrow\infty $ as $n\rightarrow\infty$. Hence for any $t>0$ \begin{align} \sum_{n=1}^{\infty}e^{-\lambda_nt}c_n&=\sum_{n=1}^{\infty}[C(n)-C(n-1)]e^{-\lambda_nt}\\ &=\sum_{n=1}^{\infty}C(n)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right]\\ &=C(\infty)+\sum_{n=1}^{\infty}o_n(1)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right], \end{align} by Abel transform, where $$C(n)=\sum_{k=1}^{n}c_k ~\mbox{for}~ n\ge 1 $$ and $$o_n(1)=\sum_{k=1}^{\infty}c_k-C(n)=C(\infty)-C(n)\rightarrow 0~as~ n\rightarrow \infty.$$ Then it is obviously that $$\sum_{n=1}^{\infty}o_n(1)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right]\rightarrow 0~ as~t\rightarrow 0+.$$

In fact, for any $\varepsilon>0$, there exist an $N_{\varepsilon}\in\mathbb{N}$ such that for all $n>N_{\varepsilon}$, $|o_n(1)|< \varepsilon$ holds. Furthermore, $$\left|\sum_{n=1}^{\infty}o_n(1)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right]\right|\le \left|\sum_{n=1}^{N_{\varepsilon}}o_n(1)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right]\right|+\epsilon\sum_{n>N_{\varepsilon}}\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right].$$ Namely, $$\left|\sum_{n=1}^{\infty}o_n(1)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right]\right|\le \max_{1\le k\le N_{\varepsilon}}|o_k(1)|e^{\lambda_1t}+\epsilon.$$ The next let $t\rightarrow 0$ and then let $\varepsilon\rightarrow 0$, then the results is obvious.

  • $\begingroup$ why is the last step true? $\endgroup$ May 25, 2017 at 18:47
  • $\begingroup$ @ ChristianRemling, you let's also take $c_k$ with alternating signs and the increase property may not be satisfied. $\endgroup$
    – Zhou
    May 26, 2017 at 0:53
  • $\begingroup$ How did you show that $\varepsilon \sum_{n>N_{\varepsilon}}[e^{-\lambda_{n} t}-e^{-\lambda_{n+1} t}]\leq \varepsilon$? $\endgroup$ May 26, 2017 at 13:42
  • $\begingroup$ @Thomas Kojar, Since the $\lambda_n$ increase and $t>0$, so $e^{-\lambda_nt}>e^{-\lambda_{n+1}t}$ and $\sum_{n>N_{\epsilon}}|e^{-\lambda_nt}-e^{-\lambda_{n+1}t}|=e^{-\lambda_{N_{\epsilon}+1}t}\le 1$. $\endgroup$
    – Zhou
    May 26, 2017 at 14:15

We may take for a counterexample any convergent series $\sum_{k=1}^\infty c_k$ with $\sum_{k=1}^\infty |c_k|^2<\infty$ and $\delta_k:={1\over kc_k}\to0$. Take $\lambda_k:=k(1-\delta_k)$, so that ${\lambda_k\over k}\to1$. Then for any $t>0$ $$\sum_{k\ge1} e^{-\lambda_k t}c_k-\sum_{k\ge1} e^{-k t}c_k=\sum_{k\ge1} e^{-k t}\,{e^{\delta_k kt}-1\over {\delta_k kt }}\,t\ge\sum_{k\ge1} e^{-kt-|\delta_k|k t}\;t=$$$$= \int_0^{+\infty}e^{-x}dx+o(1)=1+o(1)\, ,$$ because ${e^u-1\over u}\ge e^{-|u|}$ for all $u\neq0$, and because $\delta_k=o(1)$ ). Taking into account the result you listed at point (4) we have $$\liminf_{t\to0_+} \sum_{k\ge1} e^{-\lambda_k t}c_k\ge1+\sum_{k=1}^{\infty}c_{k}\, . $$

  • $\begingroup$ I edited and wrote a simple bound from below $\endgroup$ May 25, 2017 at 11:08
  • $\begingroup$ For small enough $|D|<<1$ we will have by Li-Yau that $\lambda_{k}\geq \frac{c}{|D|}k>k$ and so we can just set $c_{k}:=\frac{(-1)^{k}}{(\lambda_{k}-k)^{1+2\varepsilon}}$. Since $\lambda_{k}-k\approx k^{1/2}$ we have $\sum c_{k}^{2}\approx \sum \frac{1}{k^{1+\varepsilon}}$. Maybe this can work. $\endgroup$ May 25, 2017 at 22:49

Let me try one more time, inspired by Pietro's idea.

Let's for now focus on $t=1/N$ and an interval $I_N$ of length $N$ located near $k\simeq (N/3)\log N$. Let's take $\lambda_k=(1\pm\delta)k$ on this interval, with $\delta=1/\log N$, and the sign is the same as that of $c_k$. Then the terms of $$ \sum_{k\in I_N} c_k(e^{-\lambda_k t}- e^{-kt}) = \sum e^{-k/N}c_k(e^{\pm\delta k/N}-1) \quad\quad\quad\quad (1) $$ are non-negative. In fact, let's also take $c_k$ with alternating signs. Then the sum is $\gtrsim N^{-1/3} \sum |c_k|$. We can easily make this large, for example by giving $|c|$ the constant value $|c_k|=N^{-1/2-\epsilon}$ on $I_N$ (note that this will keep $\sum_{I_N} |c_k|^2$ small, as required by the $\ell^2$ condition).

Now we define the whole sequence by first choosing $N_j$'s that increase very rapidly, and then defining $c_k$ as above on each of these intervals, and $c_k=0 $ otherwise. Notice that for $t=1/N_j$, if $k$ is taken from one of the other intervals $I_m$, $m\not= j$, then $e^{-\lambda_kt}-e^{-kt}$ is extremely small, because $tk$ is either very small or very large. So these intervals make negligible contributions to (1).

It follows that (1) does not go to zero as $t\to 0+$ along the sequence $t=1/N_j$, and as discussed earlier, this means that we have a counterexample.

Finally, an obvious modification also gives examples where $c\in\ell^p$ for any (or all) $p>1$.

  • $\begingroup$ This may also suggest how to strengthen conveniently the assumptions in order to get the quantity (1) infinitesimal $\endgroup$ May 25, 2017 at 7:02
  • $\begingroup$ In your example, is it possible to keep the $\lambda_{k}$ fixed but change the $c_{k}$ accordingly? $\endgroup$ May 25, 2017 at 14:58
  • $\begingroup$ Because your $\lambda_{k}$ don't match the weyl asymptotics $\lambda_{k}\approx c_{1} k+c_{2}\sqrt{k}+O(\sqrt{k})$. $\endgroup$ May 25, 2017 at 15:09
  • $\begingroup$ @ThomasKojar: Well, I focused on the question you actually asked in the OP. I'm not sure I can do $\delta=o(k^{-1/2})$ in this style; already $O(k^{-1/2})$ might be tricky, though $O(k^{-1/2+\epsilon})$ is easy. But from Pietro's example, you can easily get $\delta_k=k^{-3/4}$, say, which has the desired asymptotics with $c_2=0$. $\endgroup$ May 25, 2017 at 17:38

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