Why is a set of functions v(t) dense in L^2

In summary, the conversation discusses the density of functions of the form v(t) in L^2 and the request for a proof or explanation of this statement. The person requesting help is advised to provide a definition of v(t) in their post in order to receive assistance from both experts and non-experts in the topic.
  • #1
4real4sure
26
0
Hello,

I was going through the following paper: http://www.emis.de/journals/HOA/AAA/Volume2011/142128.pdf

In page 6, immediately after equation (3.15), its written that "functions of the form v(t) are dense in L^2". I have been looking for proofs online which verifies the above statement but unable to find one. I would appreciate if someone can direct me to a link or explain with proof of how the functions of the form v(t) are dense in L^2.

Thank you,
 
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  • #2
Since you haven't received any replies yet, I suggest that you at least sketch a definition of the functions v(t) in a post. An expert in the topic might breeze through the paper in your link, but you may need to appeal to non-experts in the topic to get an answer.
 
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Likes pasmith and Fredrik
  • #3
But it seems like natural courtesy that if you want others' help that you should be the one to lay out the
definitions for those whose help you seek.
 

What is the definition of a set of functions v(t) being "dense" in L^2?

A set of functions v(t) is considered dense in L^2 if any function in L^2 can be approximated by a sequence of functions from the set. In other words, the set of functions v(t) is dense in L^2 if it contains "almost all" possible functions in L^2.

Why is it important for a set of functions v(t) to be dense in L^2?

Having a set of functions v(t) that is dense in L^2 allows for a wider range of functions to be approximated, making it a more flexible and useful tool in many areas of mathematics and science.

How can we prove that a set of functions v(t) is dense in L^2?

A common method of proving density is by using the Stone-Weierstrass theorem, which states that a subset of a function space is dense if it satisfies certain conditions, such as being closed under multiplication and containing a non-vanishing function. Other methods may also be used, depending on the specific set of functions v(t) being considered.

Can a set of functions v(t) be dense in L^2 but not in other function spaces?

Yes, it is possible for a set of functions v(t) to be dense in L^2 but not in other function spaces. This is because the conditions for density may vary for different function spaces, and a set of functions may satisfy the conditions for one space but not for another.

How does the density of a set of functions v(t) in L^2 relate to the convergence of Fourier series?

The density of a set of functions v(t) in L^2 is closely related to the convergence of Fourier series. In fact, the set of trigonometric functions is dense in L^2, which is why Fourier series can be used to approximate any function in L^2. This is also why Fourier series are commonly used in signal processing and other applications involving L^2 functions.

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