Inequality with absolute value of a complex integral

Click For Summary

Discussion Overview

The discussion revolves around proving an inequality involving the absolute value of a complex integral. Participants are examining a lemma from Serre that involves the integral of an exponential function over a specified range, with parameters defined as 0

Discussion Character

  • Technical explanation, Debate/contested

Main Points Raised

  • One participant expresses difficulty in proving the inequality |\int_{a}^{b}e^{-tx}e^{-tiy}dt|\leq\int_{a}^{b}e^{-tx}dt, suggesting that a simpler approach than Cauchy-Schwartz is needed.
  • Another participant proposes using the inequality \left|\int_a^bf(x)dx\right|\le\int_a^b|f(x)|dx as a potential solution.
  • A third participant challenges this suggestion, stating that the inequality only holds for real functions.
  • A fourth participant counters that the inequality is valid for complex-valued functions, referencing the triangle inequality and approximation arguments for Riemann sums.

Areas of Agreement / Disagreement

The discussion contains disagreement regarding the applicability of the proposed inequality for complex functions, with some participants asserting it is valid while others contest this point.

Contextual Notes

Participants have not reached a consensus on the validity of the proposed inequality for complex functions, and there are unresolved assumptions regarding the nature of the functions involved.

bernardbb
Messages
4
Reaction score
0
I'm stuck trying to prove a step inside a lemma from Serre; given is

0<a<b
0<x

To prove:

|\int_{a}^{b}e^{-tx}e^{-tiy}dt|\leq\int_{a}^{b}e^{-tx}dt

I've tried using Cauchy-Schwartz for integrals, but this step is too big (using Mathematica, it lead to a contradiction); something simpler must do the trick.
Thanks in advance.
 
Physics news on Phys.org
Use

\left|\int_a^bf(x)dx\right|\le\int_a^b|f(x)|dx
 
As far as I know, that only holds if f(x) is real, which it is not.
 
It also holds for complex-valued functions. For Riemann sums this is just the triangle inequality, and in the general case one can use an approximation argument.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 19 ·
Replies
19
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K