Can the Fourier Transform of an L^1 Function be Bounded by its L^1 Norm?

Click For Summary

Discussion Overview

The discussion revolves around the properties of the Fourier transform of functions in the space \( L^1(ℝ^n) \). Participants explore whether the Fourier transform \( \hat{f} \) can be bounded by the \( L^1 \) norm of the function \( f \), examining the implications of this relationship and the necessary conditions for it to hold.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that if \( f \in L^1(ℝ^n) \), then \( ||\hat{f}||_{C^0(ℝ^n)} \le ||f||_{L^1(ℝ^n)} \) can be shown using the dominated convergence theorem, but expresses uncertainty about the next steps.
  • Another participant suggests that evaluating \( \hat{f}(0) = \int f \) could lead to the desired result, noting that if \( f \ge 0 \), then \( \hat{f} \) is maximized at 0, but they are unsure how to conclude from this.
  • A subsequent reply questions the assertion that \( \hat{f} \) is maximized at 0 when \( f \ge 0 \) and expresses uncertainty about proving this statement.
  • One participant presents a solution involving the inequality \( |\hat{f}(ξ)| \le ||f||_{L^1(ℝ)} \) and claims to have shown that \( \max_{ξ \in ℝ} |\hat{f}(ξ)| \le ||f||_{L^1(ℝ)} \), seeking confirmation of correctness.

Areas of Agreement / Disagreement

Participants express differing views on the implications of certain properties of the Fourier transform, particularly regarding the maximization at 0 and the necessity of additional proofs. The discussion does not reach a consensus on the correctness of the claims or the completeness of the arguments presented.

Contextual Notes

Some participants reference a book that claims the result can be achieved without prior knowledge of the behavior of \( \hat{f} \) at infinity, indicating potential limitations in the assumptions made during the discussion.

dRic2
Gold Member
Messages
887
Reaction score
225
Hi, I have to show that if ##f \in L^1(ℝ^n)## then:
$$ ||\hat f||_{C^0(ℝ^n)} \le ||f||_{L^1(ℝ^n)}$$
Since ##|f(y)e^{-2 \pi i ξ ⋅y}| \le |f(y)|##, using the dominated convergence theorem, it is possible to show that ##\hat f \in C^0(ℝ^n)## but now I don't know how to go on.

Thanks is advance.
 
Physics news on Phys.org
##\hat{f}(0)=\int f## should give the result. If ##f\ge 0##, then ##\hat{f}## is max at 0. I'm not sure how to finish.
 
I'm very sorry for the late reply. I don't see why
mathman said:
If f≥0f≥0f\ge 0, then ^ff^\hat{f} is max at 0. I'm not sure how to finish.
.
If that is true I think I solved it, but I don't know how to prove your statement.

PS: In the book I'm reading the author says the result can be achieved without knowing that ##\hat f## goes to ##0## at ##\infty##, which he later proves using the result of this proof.
 
I think I got it! Here is my solution, hope you can tell me if it is correct. (Here I worked in ℝ for the sake of simplicity)

$$|\hat f(ξ)| = \left| \int_ℝ f(y)e^{-2\pi i ξ y}dy \right| \le \int_ℝ |f(y)e^{-2\pi i ξ y}|dy \le \int_ℝ|f(y)|dy = ||f||_{L^1(ℝ)}$$

So basically here I've proved that ##|\hat f(ξ)|## is less than a constant (##||f||_{L^1(ℝ)}##) then, even more so, ##\max_{ξ \in ℝ} |\hat f(ξ)| \le ||f||_{L^1(ℝ)}##. This proves the original statement and from here we can also conclude that ##\hat f## is limited.

Am I correct?
 
  • Like
Likes   Reactions: Delta2

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
2K