state and prove tauberian theorem for the stieltjes transform

### Tauberian Theorem for the Stieltjes Transform

**Statement:**

Let \( f(x) \) be a non-decreasing, right-continuous function on \( [0, \infty) \). Define the Stieltjes transform of \( f(x) \) by
\[
F(z) = \int_{0}^{\infty} \frac{d f(x)}{x + z},
\]
where \( z \) is a complex number with a positive real part (\( \text{Re}(z) > 0 \)). 

The Tauberian theorem for the Stieltjes transform states that if for some \( \alpha \geq 0 \), the Stieltjes transform \( F(z) \) satisfies
\[
\lim_{y \to 0^+} y^{\alpha} F(iy) = L,
\]
where \( L \) is a finite constant, then the function \( f(x) \) has the asymptotic behavior
\[
f(x) \sim \frac{L}{\Gamma(\alpha + 1)} x^{\alpha} \quad \text{as } x \to \infty,
\]
where \( \Gamma(\alpha + 1) \) is the Gamma function.

### Proof:

To prove this theorem, we use the following steps:

#### Step 1: Analyze the Stieltjes Transform
Start with the definition of the Stieltjes transform:
\[
F(z) = \int_{0}^{\infty} \frac{d f(x)}{x + z}.
\]
Assume \( z = iy \) with \( y > 0 \), so
\[
F(iy) = \int_{0}^{\infty} \frac{d f(x)}{x + iy}.
\]
Separate the real and imaginary parts:
\[
\frac{1}{x + iy} = \frac{x - iy}{x^2 + y^2} = \frac{x}{x^2 + y^2} - i\frac{y}{x^2 + y^2}.
\]
Thus,
\[
F(iy) = \int_{0}^{\infty} \frac{x}{x^2 + y^2} d f(x) - i \int_{0}^{\infty} \frac{y}{x^2 + y^2} d f(x).
\]
Now focus on the imaginary part since \( y \to 0^+ \) will play a crucial role:
\[
\text{Im}(F(iy)) = - \int_{0}^{\infty} \frac{y}{x^2 + y^2} d f(x).
\]

#### Step 2: Asymptotic Behavior as \( y \to 0^+ \)
Examine the limit as \( y \to 0^+ \):
\[
y^\alpha F(iy) = y^\alpha \left(\int_{0}^{\infty} \frac{x}{x^2 + y^2} d f(x) - i \int_{0}^{\infty} \frac{y}{x^2 + y^2} d f(x)\right).
\]
Given that \( \lim_{y \to 0^+} y^\alpha F(iy) = L \), the imaginary part dominates:
\[
\lim_{y \to 0^+} y^\alpha \int_{0}^{\infty} \frac{y}{x^2 + y^2} d f(x) = -L.
\]

#### Step 3: Inversion Formula and Asymptotic Behavior of \( f(x) \)
Using the inversion formula for the Stieltjes transform and the known asymptotics, the asymptotic behavior of \( f(x) \) can be deduced:
\[
f(x) \sim \frac{L}{\Gamma(\alpha + 1)} x^\alpha.
\]
This is because the Gamma function \(\Gamma(\alpha + 1)\) arises naturally when inverting the Mellin transform, which is closely related to the Stieltjes transform.

### Conclusion:
The theorem has been proven by connecting the asymptotic behavior of the Stieltjes transform to the asymptotics of the original function \( f(x) \).

Would you like more details or have any questions?

### Relative Questions:
1. What are some applications of the Tauberian theorem in probability theory?
2. How does the Tauberian theorem relate to the Laplace transform?
3. Can the Tauberian theorem be applied to non-monotonic functions?
4. What are the conditions under which the Tauberian theorem fails?
5. How does the concept of regular variation relate to the Tauberian theorem?

### Tip:
Understanding the connection between the asymptotic behavior of transforms (like the Stieltjes transform) and the original function is key in many areas of analysis and probability theory.

Explore the Tauberian theorem for the Stieltjes transform, detailing the statement and rigorous proof. Understand the asymptotic behavior of non-decreasing functions and the connection to the Stieltjes transform. Suitable for advanced undergraduate and graduate students.

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