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

Question

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.

Attempts

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

$$\frac{\sum_{k=1}^{\infty}c_{k}(e^{-\lambda_{k}iy}-1)}{iy}$$

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 $y_{0}$.

Answer 1

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.

Answer 2

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 $$\liminf_{t\to0_+} \sum_{k\ge1} e^{-\lambda_k t}c_k\ge1+\sum_{k=1}^{\infty}c_{k}\, . $$

Answer 3

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) $p>1$.