Solving Composite Functions: Show INTEGRAL[Tof] = T(INTEGRAL[f])

Click For Summary
SUMMARY

The discussion focuses on proving the continuity of the composite function "T o f" where f is a continuous vector-valued function and T is a linear transformation on R^3. It is established that all linear transformations are continuous, thus confirming that T is continuous. The proof for the integral relationship INTEGRAL[T o f] = T(INTEGRAL[f]) is derived from the linearity of T and the properties of integrals, specifically utilizing the fundamental theorem of calculus.

PREREQUISITES
  • Understanding of continuity in vector-valued functions
  • Knowledge of linear transformations in R^3
  • Familiarity with the properties of integrals and the fundamental theorem of calculus
  • Basic concepts of composite functions and epsilon-delta proofs
NEXT STEPS
  • Study the properties of linear transformations in R^n
  • Learn about the fundamental theorem of calculus and its applications
  • Explore epsilon-delta definitions of continuity in depth
  • Investigate the relationship between differentiation and integration in composite functions
USEFUL FOR

Mathematicians, students studying calculus and linear algebra, and anyone interested in understanding the properties of composite functions and integrals in vector spaces.

jessawells
Messages
18
Reaction score
0
Hi,

I have to solve this problem:

Suppose that f is a continuous, vector-valued function and that T is a linear transformation on R^3. Show that "T o f" (stands for T composite f) is continuous and that INTEGRAL[T o f] = T(INTEGRAL[f]).
----------------------------------------------------------------------

I think i know the general epsilon - delta proof for continuity of composite functions. The theorem states that "If f:R^n -> R^m is continuous at vector "a" and T:R^m -> R^p is continuous at f(a) then Tof:R^n ->R^p is continuous at vector a."

The proof is as follows:
For any E>0, we want to show that there exists a d>0 such that |x-a|<d => |T(f(x)) - T(f(a)| < E.

Since T is continuous at f(a), then for E>0, there exists a d1 such that |f(x)- f(a)| < d1 => |T(f(x))-T(f(a))| < E

For f, use d1 as its "E value". Then there exists d2 > 0 such that |x-a|< d2 => |f(x)-f(a)|< d1.

Let d=d2. then:

|x-a| < d => |f(x) - f(a)| < d1

but from the continuity of T, we know |f(x)-f(a)| < d1 => |T(f(x))-T(f(a))| < E

so |x-a| < d => |f(x) - f(a)| < d1 => |T(f(x))-T(f(a))| < E
which proves that Tof is continuous at a.


however, this proof only works if both f and T are continuous. going back to the original question, it states that f is continuous, but it doesn't say anything about the continuity of T. so how do i find whether T is continuous or not? I was just wondering, are all linear transformations continuous? also, what is the significance of the statement "T is a linear transformation on R^3" - is the fact that it's "R^3" something i need to use in my proof? i must admit I'm not very familiar with linear transformations.

also, the proof i wrote above is for continuity at a specific point a. the question just states that f is continuous - I'm assuming it's continuous everywhere. would i still use the same proof for this question?

I don't really know how to do the second part of the problem: show that INTEGRAL[T o f] = T(INTEGRAL[f]). i suspect that it has something to do with the chain rule but i only know how to use the chain rule for derivatives. how do you deal with composite functions when it comes to integrals?

any help would be great - thanks in advance!
 
Physics news on Phys.org
:)</code>Yes, all linear transformations are continuous. The fact that it is R^3 is not important for the proof; it only means that T is a linear transformation mapping from R^3 to some other vector space. The proof for continuity of composite functions still works for f being continuous everywhere. For the second part of the problem, you can use the linearity of T and the fact that integrals are linear operators. That is, for any two functions g,h, INTEGRAL[g + h] = INTEGRAL[g] + INTEGRAL[h]. Thus, INTEGRAL[T o f] = INTEGRAL[T(f)] = T(INTEGRAL[f]).
 


Hello,

Thank you for your question. Let me first clarify a few things for you. You are correct in your understanding that the proof you wrote only works if both f and T are continuous. In this case, since the question states that f is continuous and T is a linear transformation on R^3, it is safe to assume that T is also continuous. This is because all linear transformations on R^3 are continuous.

The statement "T is a linear transformation on R^3" is significant because it tells us what type of transformation T is and on what space it is acting on. In this case, T is a function that takes vectors from R^3 and maps them to other vectors in R^3 in a linear manner. This is important in the proof because we are dealing with vector-valued functions and linear transformations.

For the second part of the problem, you are correct in thinking that it has something to do with the chain rule. In fact, the proof for this part relies on the fundamental theorem of calculus, which states that the integral of a function is equal to the difference of its antiderivative evaluated at the upper and lower limits of integration. In this case, we have the composite function Tof and we want to show that its integral is equal to T evaluated at the integral of f.

To do this, let's break down the integral of Tof into two parts: the integral of T and the integral of f. Since T is a linear transformation, we can bring it outside of the integral and use the linearity of integration to write it as T times the integral of f. This gives us T times the integral of f evaluated at the upper and lower limits of integration, which is exactly what we wanted to show: INTEGRAL[Tof] = T(INTEGRAL[f]).

I hope this helps. If you have any further questions, please don't hesitate to ask. Good luck!
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
12
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
2
Views
2K
  • · Replies 0 ·
Replies
0
Views
5K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K