Does f(x)=x^(-1/3) Belong in L2(0,1)?

  • Thread starter Thread starter iamalexalright
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around determining whether the function f(x)=x^(-1/3) is in the space L²(0,1). Participants explore the concept of square integrability and its implications for the function's membership in L² spaces.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the method of checking square integrability by evaluating the integral of the function squared over the interval (0,1). Questions arise regarding the implications of the result for the function's status in L²(0,∞).

Discussion Status

The discussion includes attempts to evaluate the integral and assess the function's membership in L² spaces. Some participants express agreement with the reasoning presented, while others raise questions about the validity of the method used and its implications for different intervals.

Contextual Notes

There is an indication that the original poster missed relevant lectures, which may affect their understanding of the topic. The discussion also touches on the distinction between L²(0,1) and L²(0,∞) without reaching a definitive conclusion.

iamalexalright
Messages
157
Reaction score
0

Homework Statement


Check whether the function f(x)=x^{-1/3}, 0<x<1, belongs to the space
L^{2}(0,1)

Homework Equations


Well, I missed this lecture so not really sure how to go about this but from what I gathered:

A function is in L2 if the function is square integrable

If that is the case then:

The Attempt at a Solution



\int^{1}_{0}(x^{-1/3})^{2}dx = \int^{1}_{0}(x^{-2/3})dx = 3x^{1/3}|^{1}_{0} = 3

Since the solution exists then the function is in L2, correct?
 
Physics news on Phys.org
if that method is correct then it would not exist in L2(0,infinity), correct?
 
that all sounds reasonable to me, though I'm not an expert in these things
 
iamalexalright said:
if that method is correct then it would not exist in L2(0,infinity), correct?

Correct.
 
Great, thanks!
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
4
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
7
Views
2K