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1. inner product spaces ('weight function'?)
I have always thoughts that :

<f,g> is defined as in the interval [a,b], but now in my course notes it says:

<f,g>= on the interval [a,b] defines an inner product on X[a,b] where w(x) is a given function called the weight function.

I am kinda stumped by this definition, surely that would be <w,f,g> not <f,g>?
And it goes on to say:
Let S = {f1 , f2 , ... , fm} be a set of functions (or polynomials) in an inner product space X[a,b]. Then the set is said to be orthogonal if:
<fi,fj> = w(x)fi(x)fj(x) dx = 0

I always thought that f and g were orthogonal when <f,g> = 0, without the weight function. I do not understand the concept of the weight function so any advice or info on where to find somewhere on the internet that uses simple enough explanations on what it is would be great because I can't seem to find anything about it on the internet or any of the books I have.

P.S. working on converting it to latex
Last edited by Lewk; 17-05-2012 at 18:44.
2. Re: inner product spaces ('weight function'?)
(Original post by Lewk)
I have always thoughts that :

<f,g> is defined as in the interval [a,b], but now in my course notes it says:

<f,g>= on the interval [a,b] defines an inner product on X[a,b] where w(x) is a given function called the weight function.

I am kinda stumped by this definition, surely that would be <w,f,g> not <f,g>?
And it goes on to say:
Let S = {f1 , f2 , ... , fm} be a set of functions (or polynomials) in an inner product space X[a,b]. Then the set is said to be orthogonal if:
<fi,fj> = w(x)fi(x)fj(x) dx = 0

I always thought that f and g were orthogonal when <f,g> = 0, without the weight function. I do not understand the concept of the weight function so any advice or info on where to find somewhere on the internet that uses simple enough explanations on what it is would be great because I can't seem to find anything about it on the internet or any of the books I have.

P.S. working on converting it to latex
A general inner product is defined with the weight function. I suppose you so far have only been exposed for the case where .

It's just a definition, so there's no magic to it, really.
3. Re: inner product spaces ('weight function'?)
Consider the real numbers as a one dimensional real vector space.

The standard inner product (i.e. the dot product) is then given by multiplication

i.e. .

Now, let be any positive scalar. Define

by setting .

It is easy to check that is an inner product on .

Now, if we say that is an inner product space - it is important to remember that there are two pieces of data: a vector space together with an inner product. In other words, in the example I gave above together with is a different inner product space to together with . In that example, the difference isn't so important since the two structures are isometrically isomorphic (i.e. isomorphic as vector spaces via a map that preserves the inner product) via the linear map .

Now, to get back to your example - the point is that if you start off with your inner product space with inner product given by ; given a function you can produce a new inner product space with the same underlying vector space but with the new inner product .

For example, if you take to be the constant function then you essentially get the same type of inner product as I discussed in my example - it is just a plain scaling of the original inner product. In general, the weight function will usually be more complicated then that. You can think of it as a combination of things like scaling and biasing certain vectors in your space.

Now, orthogonality is defined in any inner product space. The point is that you have to fix which inner product you are talking about first before you define what it means to be orthogonal.

In other words say we are talking about . You need to remember that with and with are different inner product spaces (in general not isometrically isomorphic) so to say what it means to be orthogonal you need to decide which inner product space you are talking about - in the former and are orthogonal if and in the latter and are orthogonal if
Last edited by Jake22; 17-05-2012 at 20:16.
4. Re: inner product spaces ('weight function'?)
(Original post by Jake22)
Consider the real numbers as a one dimensional real vector space.

The standard inner product (i.e. the dot product) is then given by multiplication

i.e. .

Now, let be any positive scalar. Define

by setting .

It is easy to check that is an inner product on .

Now, if we say that is an inner product space - it is important to remember that there are two pieces of data: a vector space together with an inner product. In other words, in the example I gave above together with is a different inner product space to together with . In that example, the difference isn't so important since the two structures are isometrically isomorphic (i.e. isomorphic as vector spaces via a map that preserves the inner product) via the linear map .

Now, to get back to your example - the point is that if you start off with your inner product space with inner product given by ; given a function you can produce a new inner product space with the same underlying vector space but with the new inner product .

For example, if you take to be the constant function then you essentially get the same type of inner product as I discussed in my example - it is just a plain scaling of the original inner product. In general, the weight function will usually be more complicated then that. You can think of it as a combination of things like scaling and biasing certain vectors in your space.

Now, orthogonality is defined in any inner product space. The point is that you have to fix which inner product you are talking about first before you define what it means to be orthogonal.

In other words say we are talking about . You need to remember that with and with are different inner product spaces (in general not isometrically isomorphic) so to say what it means to be orthogonal you need to decide which inner product space you are talking about - in the former and are orthogonal if and in the latter and are orthogonal if
thanks very much for the lengthy explanation, all this inner product stuff is beginning to make sense
5. Re: inner product spaces ('weight function'?)
Ah, I found a good definition in some book, if the integral of f(x)g(x)w(x) = 0 then f and g are orthogonal with respect to w(x)
6. Re: inner product spaces ('weight function'?)
(Original post by Lewk)
Ah, I found a good definition in some book, if the integral of f(x)g(x)w(x) = 0 then f and g are orthogonal with respect to w(x)
That is just exactly as I said before - if you want to say two vectors are orthogonal - you have to specify a structure of an inner product space. Fixing just specifies the inner product .