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Haar's Tauberian theorem

Haar's Tauberian theorem is a mathematics topic covered in the lgStudy science library. This page brings together a partial reference excerpt, illustrations, worked examples, real-world applications and a short study plan, so you can understand Haar's Tauberian theorem rather than just read about it. In short: In mathematical analysis, Haar's Tauberian theorem 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.

Key takeaways

  • Haar's Tauberian theorem belongs to mathematics; place it in that map before memorising details.
  • Learn the definition first, then one example that makes the definition concrete.
  • Connect Haar's Tauberian theorem to a quantity you can measure, compute or draw — that is where exam questions come from.
  • Reproduce the core statement of Haar's Tauberian theorem from memory before moving on to harder problems.

Reference excerpt

In mathematical analysis, Haar's Tauberian theorem 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: Suppose that f ( t ) {\displaystyle f(t)} is a non-negative and continuous function for t ≥ 0 {\displaystyle t\geq 0} , having finite Laplace transform

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

for s > 0 {\displaystyle s>0} . Then F ( s ) {\displaystyle F(s)} is well defined for any complex value of s = x + i y {\displaystyle s=x+iy} with x > 0 {\displaystyle x>0} . Suppose that F {\displaystyle F} verifies the following conditions: 1. For y ≠ 0 {\displaystyle y\neq 0} the function F ( x + i y ) {\displaystyle F(x+iy)} (which is regular on the right half-plane x > 0 {\displaystyle x>0} ) has continuous boundary values F ( i y ) {\displaystyle F(iy)} as x → + 0 {\displaystyle x\to +0} , for x ≥ 0 {\displaystyle x\geq 0} and y ≠ 0 {\displaystyle y\neq 0} , furthermore for s = i y {\displaystyle s=iy} it may be written as

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

where ψ ( i y ) {\displaystyle \psi (iy)} has finite derivatives ψ ′ ( i y ) , … , ψ ( r ) ( i y ) {\displaystyle \psi '(iy),\ldots ,\psi ^{(r)}(iy)} and ψ ( r ) ( i y ) {\displaystyle \psi ^{(r)}(iy)} is bounded in every finite interval; 2. The integral

∫ 0 ∞ e i t y F ( x + i y ) d y {\displaystyle \int _{0}^{\infty }e^{ity}F(x+iy)\,dy}

converges uniformly with respect to t ≥ T {\displaystyle t\geq T} for fixed x > 0 {\displaystyle x>0} and T > 0 {\displaystyle T>0} ; 3. F ( x + i y ) → 0 {\displaystyle F(x+iy)\to 0} as y → ± ∞ {\displaystyle y\to \pm \infty } , uniformly with respect to x ≥ 0 {\displaystyle x\geq 0} ; 4. F ′ ( i y ) , … , F ( r ) ( i y ) {\displaystyle F'(iy),\ldots ,F^{(r)}(iy)} tend to zero as y → ± ∞ {\displaystyle y\to \pm \infty } ; 5. The integrals

∫ − ∞ y 1 e i t y F ( r ) ( i y ) d y {\displaystyle \int _{-\infty }^{y_{1}}e^{ity}F^{(r)}(iy)\,dy} and ∫ y 2 ∞ e i t y F ( r ) ( i y ) d y {\displaystyle \int _{y_{2}}^{\infty }e^{ity}F^{(r)}(iy)\,dy}

… excerpt ends here. Continue reading the full article.

Worked examples

Example 1 — a first encounter with Haar's Tauberian theorem

Start with the simplest possible case. Write down what Haar's Tauberian theorem claims or describes in one sentence, then invent the smallest concrete situation in which that sentence is true. In mathematics, the smallest case is usually a single object, a single equation or a single measurement. Check that every symbol or term in your sentence has a meaning in that case.

Example 2 — changing one variable

Take the situation from Example 1 and change exactly one quantity: double it, halve it, or set it to zero. Predict what should happen to Haar's Tauberian theorem before you calculate. Comparing your prediction with the result is the fastest way to find out whether you understand the idea or only the words.

Example 3 — an exam-style question

Typical questions about Haar's Tauberian theorem ask you to (a) state it precisely, (b) apply it to given data, and (c) explain a limitation. Practise writing all three answers in under five minutes; the third part is what separates a full-mark answer from an average one.

Applications of Haar's Tauberian theorem

In research
Haar's Tauberian theorem appears in mathematics research whenever the underlying quantities have to be modelled precisely. Papers usually cite it as a starting assumption and then explore where it breaks down.
In technology and industry
Engineering practice reuses Haar's Tauberian theorem in design rules, simulations and safety margins. Knowing the idea lets you read a specification sheet and understand why the numbers look the way they do.
In the classroom
Haar's Tauberian theorem is common in secondary-school and first-year university syllabi. It links to neighbouring topics Tauberian theorems, so understanding it makes those chapters shorter.
In everyday life
Look for Haar's Tauberian theorem outside the textbook — in sport, cooking, traffic, electronics or the sky above you. An example you found yourself is remembered far longer than one you were given.
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How to study Haar's Tauberian theorem in 20 minutes

  1. Read the reference excerpt below once, without taking notes.
  2. Close the page and write down what Haar's Tauberian theorem means in your own words.
  3. Compare your version with the excerpt and mark what you missed.
  4. Work through the three examples above with pen and paper.
  5. Explain Haar's Tauberian theorem out loud to somebody else — or to Teacher Smith in the lgStudy chat.

Frequently asked questions

What is Haar's Tauberian theorem in simple terms?

In mathematical analysis, Haar's Tauberian theorem 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.

Why does Haar's Tauberian theorem matter?

Because it connects several mathematics ideas at once: it gives you a definition you can apply, a quantity you can calculate, and a way to check whether a result is plausible.

How should I study Haar's Tauberian theorem?

Read the excerpt, restate it from memory, then work through the examples and applications listed on this page. The five-step study plan above takes about twenty minutes.

What does this page cover?

It gives you a compact reference excerpt plus original lgStudy explanations, examples, applications and study material on Haar's Tauberian theorem.

Tags

  • Tauberian theorems

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