Is <f,g> an Inner Product on C[a,b]?

  • Thread starter Shaunzio
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In summary: It would make it a lot easier for us to follow.Hi sorry. I didn't really do that. I just wanted to see if I could do it.I guess I'll have to try that then.
  • #1
Shaunzio
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Homework Statement


show that if we define the following operation:
let f=f(x) and g=g(x) be two functions in C[a,b] and define <f,g>=int(a to b) f(x)g(x)dx
show that the conditions of therom are satisified with this operation. Use h=h(x) to help with part b
this shows that this operation is an inner product
therom


Homework Equations


a. u dot v = v dot u
b. (u+v) dot w = u dot w + v dot w
c. (cu) dot v = c(u dot v) = u dot (cv)
d. u dot u >= 0, and u dot u = 0 if and only if u = 0


The Attempt at a Solution


I really have no idea where to begin here...
 
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  • #2
I assume you are trying to prove that your structure is an inner product. For reference, <f,g> is just a generalized way of writing a dot product (as a dot product is just one specific inner product). So what you need to prove is:

a. <f,g> = <g,f>
b. <f+g,h> = <f,h> + <g,h>
c. <cf,g> = c<f,g> = <f,cg> where c is a scalar, not a function

and so on. Plug and chug, almost.
 
Last edited:
  • #3
Hi sorry. I'm still pretty confused. Is there anyway you could elaborate?
 
  • #4
Shaunzio said:

Homework Statement


show that if we define the following operation:
let f=f(x) and g=g(x) be two functions in C[a,b] and define <f,g>=int(a to b) f(x)g(x)dx
show that the conditions of therom are satisified with this operation. Use h=h(x) to help with part b
this shows that this operation is an inner product
therom
It's rather confusing for us to use "<f, g>" above and "u dot v" below!


Homework Equations


a. u dot v = v dot u
So you want to prove that
[tex]<f, g>= \int_a^b f(x)g(x)dx= \int_a^b g(x)f(x)dx= <g, f>[/tex]

b. (u+v) dot w = u dot w + v dot w
So you want to prove that
[tex]<f+ g, h>= \int_a^b (f(x)+ g(x))h(x) dx= \int_a^b f(x)h(x) dx+ \int_a^b g(x)h(x) dx= <f, h>+ <g, h>[/tex]

c. (cu) dot v = c(u dot v) = u dot (cv)
So you want to prove that
[tex]<cf, g>= \int_a^b (cf(x))g(x)dx= c\int_a^b f(x)g(x)dx= \int_a^b f(x)(cg(x))dx[/tex]

d. u dot u >= 0, and u dot u = 0 if and only if u = 0
So you want to prove that
[tex]<f, f>= \int_a^b f^2(x)dx\ge 0[/tex]
and
[tex]<f, f>= \int_a^b f^2(x)dx= 0[/tex]
if and only if f(x)= 0 for all x between a and b.


The Attempt at a Solution


I really have no idea where to begin here...
Did you not consider writing out what you want to prove in terms of the given definition?
 

Related to Is <f,g> an Inner Product on C[a,b]?

1. What are orthogonal sets?

Orthogonal sets are a collection of vectors in a vector space that are all perpendicular to each other. This means that their dot products are equal to 0.

2. How are orthogonal sets useful in mathematics and science?

Orthogonal sets are useful in mathematics and science because they allow us to simplify complex systems by breaking them down into smaller, independent parts. They also help us to find solutions to systems of equations and perform transformations in geometry.

3. How can you determine if a set of vectors is orthogonal?

To determine if a set of vectors is orthogonal, you can calculate the dot product of each pair of vectors in the set. If all dot products are equal to 0, then the set is orthogonal.

4. What is the difference between orthogonal and orthonormal sets?

An orthonormal set is a subset of an orthogonal set where all vectors have a magnitude of 1. In other words, an orthonormal set is an orthogonal set where all vectors are also unit vectors.

5. Can orthogonal sets exist in spaces with more than 3 dimensions?

Yes, orthogonal sets can exist in spaces with any number of dimensions. In fact, they are commonly used in higher dimensions in fields such as quantum mechanics and computer graphics.

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