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

  • Thread starter iamalexalright
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In summary, the function f(x)=x^{-1/3}, 0<x<1, belongs to the space L^{2}(0,1) because it is square integrable. This can be shown by evaluating the integral of (x^{-2/3}) from 0 to 1, which results in a finite value of 3. Therefore, the function does not exist in L2(0,infinity).
  • #1
iamalexalright
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Homework Statement


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

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



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

Since the solution exists then the function is in L2, correct?
 
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  • #2
if that method is correct then it would not exist in L2(0,infinity), correct?
 
  • #3
that all sounds reasonable to me, though I'm not an expert in these things
 
  • #4
iamalexalright said:
if that method is correct then it would not exist in L2(0,infinity), correct?

Correct.
 
  • #5
Great, thanks!
 

1. Is f(x)=x^(-1/3) in L2(0,1)?

No, f(x)=x^(-1/3) is not in L2(0,1) because it does not satisfy the L2 norm, which requires the integral of the square of the function to be finite.

2. What is L2(0,1)?

L2(0,1) is a function space that contains all square integrable functions on the interval (0,1). It is a subset of the larger function space Lp(0,1), where p is any real number greater than or equal to 1.

3. How do you determine if a function is in L2(0,1)?

To determine if a function is in L2(0,1), you must calculate the L2 norm of the function. This involves taking the integral of the square of the function over the interval (0,1) and checking if the result is finite. If it is finite, then the function is in L2(0,1).

4. Can a function be in L2(0,1) but not in Lp(0,1) for p other than 2?

Yes, a function can be in L2(0,1) but not in Lp(0,1) for p other than 2. This is because the Lp norm is a more general concept that includes the L2 norm as a special case. Therefore, a function may satisfy the conditions for L2(0,1) but not for other values of p.

5. Why is L2(0,1) important in mathematics and science?

L2(0,1) is important in mathematics and science because it is a function space that contains many important and useful functions. It is often used in mathematical analysis, signal processing, and probability theory. Additionally, many physical phenomena can be described by functions in L2(0,1), making it a valuable tool in scientific research and applications.

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