Integral of vector valued function

In summary, the conversation discusses the definition of the integral \int_A f(x) d\mu(x) on a measure space (X,d\mu) with a vector space Y and a function f:X\to Y. It is suggested that if Y has a basis e_1,e_2,e_3,..., the integral can be defined as a vector in this basis by integrating the function component-wisely. However, the question arises as to whether the integral will be independent of the basis. It is determined that if \textrm{dim}(Y)<\infty, the integral will be independent of the basis due to the transformation of representations and the commutativity with finite sums. However, this may not
  • #1
jostpuur
2,116
19
Let [itex](X,d\mu)[/itex] be measure space, [itex]Y[/itex] be a vector space, and [itex]f:X\to Y[/itex] some function. If [itex]A\subset X[/itex] is some measurable set, how could we define (or try to define) the integral

[tex]
\int\limits_A f(x) d\mu(x)
[/tex]

?

If the [itex]Y[/itex] has a basis [itex]e_1,e_2,e_3,\ldots[/itex], I can define the integral as a vector in this basis by integrating the function component-wisely:

[tex]
\int\limits_A f(x)d\mu(x) = \Big( \int\limits_A f_1(x)d\mu(x),\; \int\limits_A f_2(x)d\mu(x),\; \ldots \Big) \;\in\; Y
[/tex]

But is it certain that the integral will be independent of the basis?

I see that the integral will be independent of the basis if [itex]\textrm{dim}(Y)<\infty[/itex], because the transformations of the representations of the function and integral are going to be represented by the same matrices, and because the integral commutes with finite sums.

But integrals don't always commute with infinite sums.

So if we have no preferred basis for [itex]Y[/itex], and if [itex]\textrm{dim}(Y)=\infty[/itex], what can we do?
 
Last edited:
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  • #2
I believe you would need some restriction on f(x).
 
  • #3
Might your problem might be covered in "Hilbert space" theory?
 

1. What is the definition of the integral of a vector valued function?

The integral of a vector valued function is a mathematical operation that represents the area under the curve of a vector function in a given interval. It is denoted by ∫f(x)dx and can be thought of as the sum of infinitely many small rectangles under the curve.

2. What are the properties of the integral of a vector valued function?

Some of the properties of the integral of a vector valued function include linearity, additivity, and the fundamental theorem of calculus. These properties make it easier to evaluate integrals and perform calculations involving them.

3. How is the integral of a vector valued function calculated?

The integral of a vector valued function is calculated using various techniques such as substitution, integration by parts, and partial fractions. These techniques can be used to simplify the integrand and make it easier to evaluate the integral.

4. What is the significance of the integral of a vector valued function in physics?

In physics, the integral of a vector valued function is used to calculate quantities such as displacement, velocity, and acceleration. It is also used to calculate work, energy, and other physical quantities in various applications.

5. Can the integral of a vector valued function be negative?

Yes, the integral of a vector valued function can be negative if the curve is below the x-axis in certain intervals. This indicates that the area under the curve is negative, which can have physical significance in certain scenarios.

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