Proof of continuity: Spherical mean function

In summary: M_{h}(x,r) is continuous in r for fixed x.In summary, we have shown that M_{h}(x,r) is continuous in both x and r, which proves the continuity of the spherical mean.
  • #1
Mugged
104
0
This is one of my homework problems.

If h(x) is continuous in x, show that the spherical mean:

M[itex]_{h}[/itex](x,r) = [itex]\frac{1}{w_{n}}[/itex][itex]\int[/itex][itex]_{|\xi|=1}[/itex] h(x+r[itex]\xi[/itex]) dS[itex]_{\xi}[/itex]

is continuous for all x and r [itex]\geq[/itex] 0.

A lot of PDE textbooks state this fact (in regards to the wave equation in 3 dimensions) - like Evans, McOwen, etc. But none I've looked at provide a proof of continuity.

This should be simply enough but my analysis skills lack. The strategy one would need to take is first take r to be constant, prove continuity in x. Then fix x, and prove continuity in r.

Please help me
 
Physics news on Phys.org
  • #2
prove this. Thanks!Proof: We will prove continuity of M_h(x,r) by showing that the function is both continuous in x and r. First, let r be constant and consider the limit as x approaches x_0. Note that h(x) is continuous in x so that h(x) -> h(x_0) as x -> x_0. Additionally, since |ξ| = 1 for all ξ, the surface integral of h(x+rξ) is also continuous with respect to x. Therefore, as x -> x_0, we haveM_{h}(x,r) -> M_{h}(x_0,r)where M_{h}(x_0,r) is a finite number since h(x_0) is a finite number. Thus, M_{h}(x,r) is continuous in x for fixed r.Next, let x be constant and consider the limit as r -> 0. As r -> 0, the spherical mean can be written asM_{h}(x,r) = \frac{1}{w_{n}}\int_{|\xi|=1} h(x) dS_{\xi} + \frac{1}{w_{n}}\int_{|\xi|=1} (h(x+r\xi)-h(x)) dS_{\xi}Since h(x) is a finite number, the first term on the right hand side goes to a finite number as r -> 0. The second term on the right hand side can be written as\frac{1}{w_{n}}\int_{|\xi|=1} \left(\frac{h(x+r\xi)-h(x)}{r}\right) r dS_{\xi}.Note that the integrand can be written as a product of a continuous function and r. Since the integral is taken over a bounded domain and both terms go to zero as r -> 0, we can apply the Dominated Convergence Theorem to conclude that the integral, and thus the second term, goes to zero as r -> 0. Hence,
 

Related to Proof of continuity: Spherical mean function

1. What is the spherical mean function?

The spherical mean function is a mathematical function that takes the average value of a given function over all directions from a fixed point in space.

2. How is the spherical mean function related to continuity?

The spherical mean function is a tool used to prove the continuity of a function at a particular point. If the spherical mean function is continuous at a point, then the original function is also continuous at that point.

3. How is the spherical mean function calculated?

The spherical mean function is calculated by integrating the original function over the surface of a sphere centered at the fixed point and dividing by the surface area of the sphere.

4. What is the significance of the spherical mean function in mathematics?

The spherical mean function is important in various fields of mathematics, such as harmonic analysis, partial differential equations, and geometric measure theory. It also has applications in physics, such as in the study of wave propagation.

5. Are there any limitations to using the spherical mean function to prove continuity?

While the spherical mean function is a useful tool, it may not be applicable in all cases. It works best for functions that exhibit symmetry, and may not be effective for functions that are highly irregular or discontinuous at the point in question.

Similar threads

  • Calculus and Beyond Homework Help
Replies
13
Views
1K
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
12
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
865
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top