Uniform convergence of a parameter-dependent integral

In summary, a student of electrical engineering is seeking help in understanding a question about when a parameter-dependent integral uniformly converges. They have found a small note in their lecture scripts that mentions the necessity of uniform convergence for the existence of the integral, but they are unsure about how this relates to the continuity of the function. They have also shared their thought process and reasoning for trying to solve the question.
  • #1
Peter Alexander
26
3
Hello everyone!

I'm a student of electrical engineering, preparing for the theoretical exam in math which will cover stuff like differential geometry, multiple integrals, vector analysis, complex analysis and so on. So the other day I was browsing through the required knowledge sheet our professor sent us, and one question, in particular, is still perplexing to me.
The reason why I'm seeking help on this site is that I wasn't able to find the concrete answer to my question either in the textbook or online. This thread is more or less "theoretical" in nature. Any help would be very appreciated

1. The problem statement, all variables, and given/known data
The question goes like this:
When does the parameter-dependent integral $$F(y)=\int_a^b f(x,y)dx$$uniformly converge?

Homework Equations


In order to answer this question, one must be familiar with the definition of the parameter-dependent integral. Directly from my notes:
Suppose f( x,##y )## is a continuous function ##f:[ a,b ] \times [ c,d ] → R## , then $$F( y ) = \int_a^b f( x,y ) dx$$ is known as parameter dependent integral.

The Attempt at a Solution


I've found a small "note to self" when analyzing my lecture scripts that says
In order for $$F(y) = \int_a^\infty f(x,y)dx $$ to exist, uniform convergence is a mandatory condition
Unfortunately I never explained that claim, but from what I understand, the uniform convergence is a stronger type of convergence than pointwise convergence.

In terms of the parameter-dependent integral, if we have a difference$$F(y+h)-F(h)=\int_a^\infty f(x,y + h)dx - \int_a^\infty f(x,y)dx$$ and segregate the integration scope into two integrals, one with finite and other with infinite integration boundaries, then we get$$\int_a^m [f(x,y + h) - f(x,y)] dx + \int_m^\infty [f(x,y + h) - f(x,y)]dx$$

And here's the catch. The first integral will be arbitrarily small for a small, arbitrarily chosen constant ##h##. However, the second integral should somehow prove that uniform convergence is necessary. I know that the function ##f(x,y)## must be continuous.

From what I've read here, the parameter dependent integral is uniformly convergent for some ##y\in[c,d]## if for each ##\epsilon > 0## exists such ##\delta(\epsilon)## so that$$|F(y) = \int_a^\infty f(x,y)dx| < \epsilon$$ for all ##m \geq \delta(\epsilon)##

My question, therefore, is: from what I've provided, it is possible that uniform convergence is a necessary condition when evaluating the parameter-dependent improper integral simply because, without it, the continuity of function ##F(y)## cannot be ensured?

I really hope this post makes any sense to you, but regardless I wish you a happy Sunday!
 
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  • #2
Peter Alexander said:
I really hope this post makes any sense to you, but regardless I wish you a happy Sunday!

Uniform convergence relates to a sequence of functions:

https://en.wikipedia.org/wiki/Uniform_convergence

And I don't know what else it could mean.

Your post, I'm sorry to say, makes little sense to me.
 
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  • #3
Yeah, I was wondering why this question would even exist or how does it pertain to the parameter-dependent integrals. I was contemplating this task for awhile and I don't think I'm even going in the right direction.

I'm sorry for posting nonsense, but this is the line of reasoning I took trying to solve the task.
 
  • #4
Peter Alexander said:
Yeah, I was wondering why this question would even exist or how does it pertain to the parameter-dependent integrals. I was contemplating this task for awhile and I don't think I'm even going in the right direction.

I'm sorry for posting nonsense, but this is the line of reasoning I took trying to solve the task.

You learn something every day:

https://www.encyclopediaofmath.org/index.php/Parameter-dependent_integral

Although I'm still not sure what the question is asking.
 
  • #5
Peter Alexander said:
Hello everyone!

I'm a student of electrical engineering, preparing for the theoretical exam in math which will cover stuff like differential geometry, multiple integrals, vector analysis, complex analysis and so on. So the other day I was browsing through the required knowledge sheet our professor sent us, and one question, in particular, is still perplexing to me.
The reason why I'm seeking help on this site is that I wasn't able to find the concrete answer to my question either in the textbook or online. This thread is more or less "theoretical" in nature. Any help would be very appreciated

1. The problem statement, all variables, and given/known data
The question goes like this:

Homework Equations


In order to answer this question, one must be familiar with the definition of the parameter-dependent integral. Directly from my notes:

The Attempt at a Solution


I've found a small "note to self" when analyzing my lecture scripts that says Unfortunately I never explained that claim, but from what I understand, the uniform convergence is a stronger type of convergence than pointwise convergence.

In terms of the parameter-dependent integral, if we have a difference$$F(y+h)-F(h)=\int_a^\infty f(x,y + h)dx - \int_a^\infty f(x,y)dx$$ and segregate the integration scope into two integrals, one with finite and other with infinite integration boundaries, then we get$$\int_a^m [f(x,y + h) - f(x,y)] dx + \int_m^\infty [f(x,y + h) - f(x,y)]dx$$

And here's the catch. The first integral will be arbitrarily small for a small, arbitrarily chosen constant ##h##. However, the second integral should somehow prove that uniform convergence is necessary. I know that the function ##f(x,y)## must be continuous.

From what I've read here, the parameter dependent integral is uniformly convergent for some ##y\in[c,d]## if for each ##\epsilon > 0## exists such ##\delta(\epsilon)## so that$$|F(y) = \int_a^\infty f(x,y)dx| < \epsilon$$ for all ##m \geq \delta(\epsilon)##

My question, therefore, is: from what I've provided, it is possible that uniform convergence is a necessary condition when evaluating the parameter-dependent improper integral simply because, without it, the continuity of function ##F(y)## cannot be ensured?

I really hope this post makes any sense to you, but regardless I wish you a happy Sunday!

Your note to self is wrong: in order for ##F(y) = \int_a^b f(x,y) \, dx## to exist, all you need is for ##f(x,y)## to be ##x##-integrable for each ##y##. If you want ##F(y)## to be continuous or have continuous derivatives, THEN you may need to assume more properties of ##f(x,y)##. Things may be a bit different when ##a## and ##b## are both finite than when one or both are infinite.
 
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  • #6
That helped a lot, thank you for your time and patience!
 
  • #7
Peter Alexander said:
That helped a lot, thank you for your time and patience!

Who are you responding to? In future, please try to respond to an individual reply, so that you can keep the attributions apparent.
 
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Related to Uniform convergence of a parameter-dependent integral

1. What is uniform convergence of a parameter-dependent integral?

Uniform convergence of a parameter-dependent integral refers to the property of a sequence of integrals, where the integrand depends on a parameter, converging uniformly to a limit. This means that the rate of convergence is independent of the parameter.

2. How is uniform convergence of a parameter-dependent integral different from pointwise convergence?

Uniform convergence is stronger than pointwise convergence because it guarantees that the entire sequence of integrals converges uniformly, while pointwise convergence only guarantees that each individual integral converges.

3. What are some applications of uniform convergence of a parameter-dependent integral?

Uniform convergence of a parameter-dependent integral is important in many areas of mathematics, including analysis, probability theory, and differential equations. It is also used in physics and engineering to model and solve various problems.

4. How is the concept of uniform convergence related to the concept of continuity?

Uniform convergence is closely related to the concept of continuity. A function is said to be continuous if it is continuous at every point in its domain. Similarly, a sequence of functions is said to converge uniformly if it is continuous at every point in its domain.

5. Are there any conditions that guarantee uniform convergence of a parameter-dependent integral?

Yes, there are several conditions that guarantee uniform convergence, including the Weierstrass M-test, the Dini's theorem, and the Cauchy criterion for uniform convergence. These conditions are useful for determining whether a given sequence of integrals converges uniformly or not.

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