In mathematical analysis, Haar's Tauberian theorem^[1] named after Alfréd Haar, relates the asymptotic behaviour of a continuous function to properties of its Laplace transform. It is related to the integral formulation of the Hardy–Littlewood Tauberian theorem.

Simplified version by Feller

William Feller gives the following simplified form for this theorem:^[2]

Suppose that ${\displaystyle f(t)}$ is a non-negative and continuous function for ${\displaystyle t\geq 0}$, having finite Laplace transform

${\displaystyle F(s)=\int _{0}^{\infty }e^{-st}f(t)\,dt}$

for ${\displaystyle s>0}$. Then ${\displaystyle F(s)}$ is well defined for any complex value of ${\displaystyle s=x+iy}$ with ${\displaystyle x>0}$. Suppose that ${\displaystyle F}$ verifies the following conditions:

1. For ${\displaystyle y\neq 0}$ the function ${\displaystyle F(x+iy)}$ (which is regular on the right half-plane ${\displaystyle x>0}$) has continuous boundary values ${\displaystyle F(iy)}$ as ${\displaystyle x\to +0}$, for ${\displaystyle x\geq 0}$ and ${\displaystyle y\neq 0}$, furthermore for ${\displaystyle s=iy}$ it may be written as

${\displaystyle F(s)={\frac {C}{s}}+\psi (s),}$

where ${\displaystyle \psi (iy)}$ has finite derivatives ${\displaystyle \psi '(iy),\ldots ,\psi ^{(r)}(iy)}$ and ${\displaystyle \psi ^{(r)}(iy)}$ is bounded in every finite interval;

2. The integral

${\displaystyle \int _{0}^{\infty }e^{ity}F(x+iy)\,dy}$

converges uniformly with respect to ${\displaystyle t\geq T}$ for fixed ${\displaystyle x>0}$ and ${\displaystyle T>0}$;

3. ${\displaystyle F(x+iy)\to 0}$ as ${\displaystyle y\to \pm \infty }$, uniformly with respect to ${\displaystyle x\geq 0}$;

4. ${\displaystyle F'(iy),\ldots ,F^{(r)}(iy)}$ tend to zero as ${\displaystyle y\to \pm \infty }$;

5. The integrals

${\displaystyle \int _{-\infty }^{y_{1}}e^{ity}F^{(r)}(iy)\,dy}$ and ${\displaystyle \int _{y_{2}}^{\infty }e^{ity}F^{(r)}(iy)\,dy}$

converge uniformly with respect to ${\displaystyle t\geq T}$ for fixed ${\displaystyle y_{1}<0}$, ${\displaystyle y_{2}>0}$ and ${\displaystyle T>0}$.

Under these conditions

${\displaystyle \lim _{t\to \infty }t^{r}[f(t)-C]=0.}$

Complete version

A more detailed version is given in.^[3]

Suppose that ${\displaystyle f(t)}$ is a continuous function for ${\displaystyle t\geq 0}$, having Laplace transform

${\displaystyle F(s)=\int _{0}^{\infty }e^{-st}f(t)\,dt}$

with the following properties

1. For all values ${\displaystyle s=x+iy}$ with ${\displaystyle x>a}$ the function ${\displaystyle F(s)=F(x+iy)}$ is regular;

2. For all ${\displaystyle x>a}$, the function ${\displaystyle F(x+iy)}$, considered as a function of the variable ${\displaystyle y}$, has the Fourier property ("Fourierschen Charakter besitzt") defined by Haar as for any ${\displaystyle \delta >0}$ there is a value ${\displaystyle \omega }$ such that for all ${\displaystyle t\geq T}$

${\displaystyle {\Big |}\,\int _{\alpha }^{\beta }e^{iyt}F(x+iy)\,dy\;{\Big |}<\delta }$

whenever ${\displaystyle \alpha ,\beta \geq \omega }$ or ${\displaystyle \alpha ,\beta \leq -\omega }$.

3. The function ${\displaystyle F(s)}$ has a boundary value for ${\displaystyle \Re s=a}$ of the form

${\displaystyle F(s)=\sum _{j=1}^{N}{\frac {c_{j}}{(s-s_{j})^{\rho _{j}}}}+\psi (s)}$

where ${\displaystyle s_{j}=a+iy_{j}}$ and ${\displaystyle \psi (a+iy)}$ is an ${\displaystyle n}$ times differentiable function of ${\displaystyle y}$ and such that the derivative

${\displaystyle \left|{\frac {d^{n}\psi (a+iy)}{dy^{n}}}\right|}$

is bounded on any finite interval (for the variable ${\displaystyle y}$)

4. The derivatives

${\displaystyle {\frac {d^{k}F(a+iy)}{dy^{k}}}}$

for ${\displaystyle k=0,\ldots ,n-1}$ have zero limit for ${\displaystyle y\to \pm \infty }$ and for ${\displaystyle k=n}$ has the Fourier property as defined above.

5. For sufficiently large ${\displaystyle t}$ the following hold

${\displaystyle \lim _{y\to \pm \infty }\int _{a+iy}^{x+iy}e^{st}F(s)\,ds=0}$

Under the above hypotheses we have the asymptotic formula

${\displaystyle \lim _{t\to \infty }t^{n}e^{-at}{\Big [}f(t)-\sum _{j=1}^{N}{\frac {c_{j}}{\Gamma (\rho _{j})}}e^{s_{j}t}t^{\rho _{j}-1}{\Big ]}=0.}$

References