Continuously smooth functions and Lp space

In summary, the two parts of the question can be answered as follows:1) If f ∈ C(Rn) and f has compact support, then f ∈ Lp(Rn) for every 1 ≤ p ≤ ∞.2) If f ∈ C(Rn), then f ∈ Lp_{loc}(Rn) for every 1 ≤ p < ∞.What is special about continuous functions on compact domains? (apply this to both parts)They are uniformly continuous.
  • #1
sdickey9480
10
0
How might I prove the following?

1) If f ∈ C(Rn) and f has compact support, then f ∈ Lp(Rn) for every 1 ≤ p ≤ ∞.

2) If f ∈ C(Rn), then f ∈ Lp_{loc}(Rn) for every 1 ≤ p < ∞.

(Where C(Rn) is the space of continuous functions on Rn)
 
Physics news on Phys.org
  • #2
What is special about continuous functions on compact domains? (apply this to both parts)
 
  • #3
They are uniformly continuous
 
  • #4
sdickey9480 said:
They are uniformly continuous

Yes, but what else?? Can functions on compact domains grow arbitrarly large??
 
  • #5
No, b/c they are bounded.
 
  • #6
Yes. So use that to find an estimate for the integral

[tex]\int_{\mathbb{R}^n} |f|^p[/tex]
 
  • #7
Not following. Could you provide a little more detail?
 
  • #8
So since we are dealing with bounded continuous functions, by finding an estimate to the aforementioned integral this will in turn justify that f must also belong to Lp?
 
  • #9
If we can show that

[tex]\int_{\mathbb{R}^n} |f|^p [/tex]

is not infinite, then the function is in [itex]L^p[/itex]. So we must find some real number C such that

[tex]\int_{\mathbb{R}^n} |f|^p\leq C[/tex]
 
  • #10
It might help to first answer the (hopefully easy) question

If f(x) is a continuous function on the reals, is [tex] \int_a^b |f(x)|^p dx[/tex] ever infinite?

If you can figure out the answer to this you can solve micromass's question
 
  • #11
Won't the integral always be finite? Hence do we even need to find a particular C, M, etc.? Can't we just assume there exists one, again b/c integral is finite? Am I using this same idea for 1) and 2)?
 
  • #12
Why do you think the integral will always be finite?? Where did you use compactness??
 
  • #13
Finite because it's bounded on a compact interval?
 
  • #14
Can I just prove the space of smooth continuous functions is dense in Lp, hence if a function belongs to C(R) it belongs to Lp(R). If so, what's the difference in the proof of 1) & 2)?
 

Related to Continuously smooth functions and Lp space

1. What is a continuously smooth function?

A continuously smooth function is a mathematical function that is defined on a continuous domain and has derivatives of all orders that exist and are also continuous. This means that the function is smooth and has no abrupt changes or breaks in its graph.

2. What is Lp space?

Lp space refers to a mathematical concept that represents a collection of functions with certain properties. These functions are defined on a specific domain and have a finite p-norm, which is a measure of their size and growth. The value of p determines the specific properties of the functions in the Lp space.

3. What is the significance of continuously smooth functions in Lp space?

Continuously smooth functions play a crucial role in Lp space because they are the most well-behaved and well-understood class of functions in this context. They form a dense subset of Lp space, meaning that any function in Lp space can be approximated by a sequence of continuously smooth functions. This makes them a useful tool for studying and analyzing more general functions in Lp space.

4. How is Lp space used in real-world applications?

Lp space has a wide range of applications in various fields, such as physics, engineering, and economics. For example, it is used in signal processing to analyze and manipulate signals, in control theory to design optimal control systems, and in econometrics to model economic data. Lp space also has applications in image and audio processing, data compression, and machine learning.

5. Are there any limitations to using continuously smooth functions in Lp space?

While continuously smooth functions are a useful tool in Lp space, they may not always be the most appropriate choice for certain problems. In some cases, functions with discontinuities or singularities may better represent the underlying phenomenon. Additionally, the use of continuously smooth functions in Lp space may lead to computational challenges due to the complex nature of these functions and the high dimensionality of Lp space.

Similar threads

  • Topology and Analysis
Replies
1
Views
826
Replies
4
Views
413
  • Topology and Analysis
Replies
8
Views
1K
Replies
1
Views
799
  • Topology and Analysis
Replies
4
Views
348
Replies
6
Views
1K
  • Topology and Analysis
Replies
14
Views
2K
  • Topology and Analysis
Replies
5
Views
1K
Replies
2
Views
444
Replies
4
Views
434
Back
Top