Is the Fourier Transform of an L1 Function Always Continuous and Bounded?

  • MHB
  • Thread starter mathmari
  • Start date
  • Tags
    Function
Then use the fact that $L^1(\Bbb R)$ is the dual space of $L^\infty (\Bbb R)$. In summary, we have a function $f$ in $L^1$ and we define $\hat{f}(t)$ as the integral of $f(x)$ multiplied by $e^{ixt}$ in $\mathbb{R}$. We need to prove that $\hat{f}$ is continuous, has a limit of 0 as $t$ approaches positive or negative infinity, and that the norm of $\hat{f}$ is less than or equal to the norm of $f$. We can use the dominated convergence theorem and the fact that continuous functions with compact support are dense
  • #1
mathmari
Gold Member
MHB
5,049
7
Hey! :eek:

In $\mathbb{R}$ with Lebesgue measure, we take $f\in L^1$ and we set $\hat{f}(t)=\int f(x) e^{ixt} dx$, for each $x$ $\ \ \ (i^2=-1)$

Show that:

1. $\hat{f}$ is continuous
2. $\lim_{t\rightarrow \pm \infty}\hat{f}(t)=0$
3. $||\hat{f}||_{\infty}\leq ||\hat{f}||_{1}$

Could you give me some hints how I could do that?? (Wondering)
 
Physics news on Phys.org
  • #2
1. Let $ \in \Bbb R$ and let $t_n$ be a sequence of real numbers converging to $t$. Show that $\hat{f}(t_n) \to \hat{f}(t)$ by the dominated convergence theorem applied to the sequence $f_n(x) = f(x)e^{ixt_n}$.

2. First prove the result when $f$ is a continuous function with compact support. For the general case, use the fact that space of continuous functions on $\Bbb R$ with compact support is dense in $L^1(\Bbb R)$.

3. Show that $|\hat{f}(t)| \le \|f\|_1$ for every $t$.
 

FAQ: Is the Fourier Transform of an L1 Function Always Continuous and Bounded?

What is a Lebesgue integrable function?

A Lebesgue integrable function is a type of function in mathematics that can be integrated using the Lebesgue integral. This is a more general form of integration that allows for a wider range of functions to be integrated compared to the Riemann integral.

How is a Lebesgue integrable function different from a Riemann integrable function?

The main difference between a Lebesgue integrable function and a Riemann integrable function is the type of integral used. The Lebesgue integral allows for a wider range of functions to be integrated, including functions that are not continuous or have discontinuities. The Riemann integral, on the other hand, can only be applied to continuous functions.

What is the significance of Lebesgue integrable functions in mathematics?

Lebesgue integrable functions are important in mathematics because they provide a more general and powerful tool for integration. They allow for the integration of a wider range of functions, making it easier to solve problems in calculus, analysis, and other areas of mathematics.

How is the Lebesgue integral defined?

The Lebesgue integral is defined as the limit of a sequence of integrals of simple functions. A simple function is a function that takes on a finite number of values on a given interval. The Lebesgue integral is defined in this way to allow for the integration of functions that are not continuous or have discontinuities.

What are some applications of Lebesgue integrable functions?

Lebesgue integrable functions have many applications in mathematics, physics, and other fields. They are used to solve problems in calculus, probability, statistics, and signal processing. They are also used in the study of dynamical systems, differential equations, and measure theory.

Back
Top