Denseness of bounded funtions in L^2?

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In summary, C_b^\infty(\mathbb{R}^n) is not dense in L^2(\mathbb{R}^n), but the intersection of C_b^\infty(\mathbb{R}^n) and L^2(\mathbb{R}^n) is dense in L^2(\mathbb{R}^n). This is because there are functions in C_b^\infty(\mathbb{R}^n) that are not in L^2(\mathbb{R}^n), such as the function f(x)=1, but the space of infinitely differentiable functions with compact support, C_0^\infty(\mathbb{R}^
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Let [itex]C_b^\infty(\mathbb{R}^n)[/itex] be the space of infinitely differentiable functions f, such that f and all its partial derivatives are bounded.

Is [itex]C_b^\infty(\mathbb{R}^n)[/itex] dense in [itex]L^2(\mathbb{R}^n)[/itex]? I think the answer is yes, because [itex]C_b^\infty(\mathbb{R}^n)[/itex] contains [itex]C_0^\infty(\mathbb{R}^n)[/itex], the space of all infinitely differentiable functions with compact support, as a subset. And it's well known that [itex]C_0^\infty(\mathbb{R}^n)[/itex] is dense in [itex]L^p(\mathbb{R}^n)[/itex].

However, there appear to be functions in [itex]C_b^\infty(\mathbb{R}^n)[/itex] but are not in [itex]L^2(\mathbb{R}^n)[/itex], for example the function f(x)=1. So this means that instead, we have [itex]C_b^\infty(\mathbb{R}^n)\cap L^2(\mathbb{R}^n)[/itex] dense in [itex]L^2(\mathbb{R}^n)[/itex]?
 
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You seem to have answered your own question.
 

Related to Denseness of bounded funtions in L^2?

1. What is the definition of "denseness" in the context of bounded functions in L^2?

In mathematics, a set of points is considered dense if any point in the given space can be arbitrarily close to a point in the set. In the context of bounded functions in L^2, denseness refers to the property of a set of functions that allows any function in the space to be approximated by functions in the given set.

2. How is denseness related to the L^2 norm of a function?

The L^2 norm of a function is a measure of its "size" or "magnitude" in the L^2 space. In the context of denseness, it is used to determine how closely a function in the given set approximates a function in the L^2 space. A set of functions is considered dense in L^2 if the L^2 norm of any function in the space can be approximated by the L^2 norms of functions in the given set.

3. Can you give an example of a set of bounded functions that is dense in L^2?

One example is the set of continuous functions on a closed interval [a,b]. This set is dense in L^2 because any square-integrable function can be approximated by a continuous function on the same interval using techniques such as the Weierstrass approximation theorem.

4. How is denseness in L^2 important in practical applications?

In practical applications, denseness in L^2 is important because it allows for the approximation of functions in the L^2 space by functions in a given set. This is useful in fields such as signal processing, where it is necessary to approximate signals with simpler functions in order to analyze and manipulate them.

5. Are there any conditions that must be met for a set of functions to be dense in L^2?

Yes, there are several conditions that must be met for a set of functions to be dense in L^2. The functions in the set must be bounded, square-integrable, and must have a finite L^2 norm. Additionally, the set must be closed under addition, subtraction, and scalar multiplication in order for it to be considered a vector space.

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