Inner Product (Infinite Dimensional)

In summary, the definition of the inner product of two functions on an interval is based on the form of a scalar product in finite dimensional space and in infinite dimensional space. In the finite dimensional space, the inner product is given by the dot product of two vectors, while in the infinite dimensional space, it is given by an integral of the two functions. The factor of dx in the integral represents the step in the Riemann sum approximation of the integral. However, this definition only works for certain representations of the vector space, and can be extended to infinite dimensions in a different form. The reason for using this definition is to apply calculus techniques to solve problems in linear algebra.
  • #1
Prologue
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I am trying to understand the definition of the inner product of two functions on an interval. I know that the form of a scalar product in finite dimensional space is given by

[tex]\vec{\phi} \bullet \vec{\psi}= \sum_{k} \phi_{k} \psi_{k}[/tex]

and in infinite dimensional space

[tex]\langle \phi , \psi \rangle = \int_{a}^{b} \phi(x) \psi(x) dx[/tex]

If we expand out the first definition we get

[tex]\vec{ \phi} \bullet \vec{ \psi} = \phi_{1} \psi_{1} + \phi_{2} \psi_{2} + \phi_{3} \psi_{3} ...[/tex]

and for the second we get

[tex]\langle \phi , \psi \rangle = \phi(a) \psi(a) dx \: + \: \phi(a+dx) \psi(a+dx) dx \: + \: \phi(a+2dx) \psi(a+2dx) dx ...[/tex]

What is up with the factor of dx multiplying everything is this last expansion? It has to be there because that is what an integral does, it takes the function evaluated at that specific x and multiplies it by dx, but what purpose does the dx factor serve in the definition of the inner product? It is pretty obvious that apart from the factor of dx it looks identical to the finite dimensional definition. So what gives?Edit: I edited this to make my question more clear. And I'd like to add that it seems to me that

[tex]\langle \phi , \psi \rangle = \phi(a) \psi(a) dx \: + \: \phi(a+dx) \psi(a+dx) dx \: + \: \phi(a+2dx) \psi(a+2dx) dx ...= \vec{\phi} \bullet \vec{\psi}dx[/tex]

Where [tex]\vec{\phi} \bullet \vec{\psi}[/tex] is supposed to just be a recognition of the form of the initial definition.
 
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  • #2
Prologue said:
What is up with the factor of dx multiplying everything is this last expansion?
They seem to be approximating the integral with a Riemann sum (the rectangle rule). IMO, dx is a poor name for the variable that expresses the step. Also, IMO, = is a poor symbol for "is approximately".
 
  • #3
Yeah, but I see the point of the argument, aside from the factor of dx.

That essentially

[tex]\langle \phi , \psi \rangle = dx(\vec{\phi} \bullet \vec{\psi)}[/tex]

is true, and then you just have this dot product on the right but there is still some crazy factor of dx.
 
  • #4
Oh, by the way, an inner product on a real vector space is merely any real-valued function of two vectors variables:
<x+y, z> = <x,z> + <y,z>
a<x,y> = <ax, y>
<x,y> = <y,x>
<x,x> is nonnegative
<x,x> = 0 if and only if x = 0

The forms you have described are valid only for very specific representations of your vector space!

It is true that for any finite dimensional real vector space, there is a way to present vectors so that your formula holds. But that only works in that specific presentation! The only kind of presentation that works is to use coordinates relative to an orthonormal basis

This can be directly extended to infinite dimensions1 -- but not as an integral!

That integral works only in a different kind of presentation where vectors are written as functions on the interval [a,b].


1: more precisely, countably infinite dimensional real Hilbert spaces
 
  • #5
Prologue said:
Yeah, but I see the point of the argument, aside from the factor of dx.

That essentially

[tex]\langle \phi , \psi \rangle = dx(\vec{\phi} \bullet \vec{\psi)}[/tex]

is true, and then you just have this dot product on the right but there is still some crazy factor of dx.
The dot product is crazier. It doesn't actually make any sense here.
 
  • #6
Well, let me tell you what my real desire is. I would like to have an intuitive reason for the definition given, rather than just accepting it as a definition. There has got to be some sort of, possibly 'loose', derivation of it.EDIT: I think I should edit the first post to maintain clarity in the discussion. I'll change the first definition to

[tex]\vec{\phi} \bullet \vec{\psi} = \sum_{k}\phi_{k}\psi_{k}[/tex]
 
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  • #7
I understand that I may be bastardizing the notation, but I still believe it is correct in the right context. If you think of the function as an infinite dimensional *vector*, that would then warrant the use of the vector symbol. And then why can't you just dot them like any other vector (thinking more along the lines of the *form* of the dot product, not necessarily what the meaning is in infinite dimensions) to get

[tex]\vec{\phi(x)} \bullet \vec{\psi(x)}=\phi(a)\psi(a)+\phi(a+dx)\psi(a+dx)+\phi(a+2dx)\psi(a+2dx)+\phi(a+3dx)\psi(a+3dx)[/tex]

EDIT: Sorry for the successive post.
 
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  • #8
Prologue said:
Well, let me tell you what my real desire is. I would like to have an intuitive reason for the definition given, rather than just accepting it as a definition. There has got to be some sort of, possibly 'loose', derivation of it.
Oh! I didn't notice the expansion was your own doing -- I thought you were citing something.


I already gave the actual definition of inner product. Algebraically, I think the definition directly captures what is really important.

IMO, the reason for using a function space with an inner product given by that integral is not a matter of gaining intuition about inner products. The reason is to do calculations, and to let us apply our great knowledge of calculus to solve problems of linear algebra.

(Just like in finite dimensional inner product spaces, we use orthonormal coordinates to let us apply our great knowledge of Euclidean coordinate geometry to solve linear algebra problems)
 
  • #9
Yeah, the argument is all mine. I am trying to find the reason behind the definition. You don't just define things like

[tex]Work=\int_{\vec{a}}^{\vec{b}}\vec{F}\bullet d\vec{r}[/tex]

They have a reason, you know?Edit: And to comment on the topic of the usefulness of the abstract definiton, I am not calling that into question at all. It is very useful to think of 'lengths', 'magnitudes', and what not of functions, it is great! I already know all of that from finite dimensional stuff.
 
  • #10
Prologue said:
If you think of the function as an infinite dimensional *vector*
...
And then why can't you just dot them like any other vector
Hrm. Are you mixing up the following two terms?
  1. vector meaning an element of a vector space
  2. vector meaning a list of scalar values (I'll call these "coordinate vectors")

I've always been using vector in the former sense...

Dot products only make sense for coordinate vectors, and even then only when the components can be summed over...

(e.g. even if you rewrote your function as a "list" of values at each point, there are too many points for infinite sums to make sense)


What you wrote is a possibly useful heuristic -- but in the end, it needs to be translated back into something well-defined. Such "continuous linear combinations" are generally made rigorous using integrals. In fact, your expansion simply inverted that process.

If you are willing to learn non-standard analysis, you could instead mutter something about standard parts and approximating the interval with a lattice of hyperfinitely many points.
 
  • #11
Prologue said:
They have a reason, you know?
My claim is that there isn't any deep reason. It's a form we can compute with, and every* case of interest can be rewritten to turn it into this form. In fact, some interesting cases were already in this form. I believe one of the earliest examples and most important driving forces of this setting were the study of Fourier series.


*: at least, every case we are interested in
 
  • #12
Ok, I hear what you are saying. This problem of acceptance of mine could be traced back to reading the supplement to Chapter 2 in Gasiorowicz's Quantum Physics book.

http://bcs.wiley.com/he-bcs/Books?action=chapter&bcsId=1533&itemId=0471057002&chapterId=4825

Where in equation 2A-6 he 'introduces' the factor of delta n. I know it is not an identical case, but if I can figure out the justification of the factor of delta n, then I bet there is an argument for a delta k in the very first definition of the dot product. And if the delta k sneaks into that definition, then the dx is perfectly explained.
 
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  • #13
Prologue said:
I am trying to understand the definition of the inner product of two functions on an interval. I know that the form of a scalar product in finite dimensional space is given by

[tex]\vec{\phi} \bullet \vec{\psi}= \sum_{k} \phi_{k} \psi_{k}[/tex]

and in infinite dimensional space

[tex]\langle \phi , \psi \rangle = \int_{a}^{b} \phi(x) \psi(x) dx[/tex]

If we expand out the first definition we get

[tex]\vec{ \phi} \bullet \vec{ \psi} = \phi_{1} \psi_{1} + \phi_{2} \psi_{2} + \phi_{3} \psi_{3} ...[/tex]

and for the second we get

[tex]\langle \phi , \psi \rangle = \phi(a) \psi(a) dx \: + \: \phi(a+dx) \psi(a+dx) dx \: + \: \phi(a+2dx) \psi(a+2dx) dx ...[/tex]

What is up with the factor of dx multiplying everything is this last expansion? It has to be there because that is what an integral does, it takes the function evaluated at that specific x and multiplies it by dx, but what purpose does the dx factor serve in the definition of the inner product? It is pretty obvious that apart from the factor of dx it looks identical to the finite dimensional definition. So what gives?


Edit: I edited this to make my question more clear. And I'd like to add that it seems to me that

[tex]\langle \phi , \psi \rangle = \phi(a) \psi(a) dx \: + \: \phi(a+dx) \psi(a+dx) dx \: + \: \phi(a+2dx) \psi(a+2dx) dx ...= \vec{\phi} \bullet \vec{\psi}dx[/tex]

Where [tex]\vec{\phi} \bullet \vec{\psi}[/tex] is supposed to just be a recognition of the form of the initial definition.

You seem to be irritated by the dx in the second definition. As Hurkyl said, they approximate the integral by a Riemann sum. Let me elaborate:

An integral [tex]\int_{a}^{b} f(x) dx[/tex] can be approximated by [tex]\sum_{k=0}^{N-1} f(a+k\Delta x) \Delta x=f(a) \Delta x+ f(a+\Delta x) \Delta x + f(a+2\Delta x) \Delta x...[/tex]. Try to visualize this (see here).

Now, to get to your definition we replace f(x) by [tex]\phi(x) \psi(x)[/tex], i.e. we let [tex]f(x)=\phi(x) \psi(x)[/tex].

Then, [tex]\int_{a}^{b} \phi(x) \psi(x) dx[/tex] can be approximated by [tex]\phi(a) \psi(a) \Delta x + \phi(a+\Delta x) \psi(a+\Delta x) \Delta x + \phi(a+2\Delta x) \psi(a+2\Delta x) \Delta x ...[/tex]
 
  • #14
Prologue said:
and for the second we get

[tex]\langle \phi , \psi \rangle = \phi(a) \psi(a) dx \: + \: \phi(a+dx) \psi(a+dx) dx \: + \: \phi(a+2dx) \psi(a+2dx) dx ...[/tex]

It really should be

[tex]\langle \phi , \psi \rangle = \lim_{dx \rightarrow 0} \sum_{n=0}^{(b-a)/dx} \phi(a+n*dx) \psi(a+n*dx) dx [/tex]and the limit only exists if you have dx on the right side. Otherwise it diverges, you get an infinity.
 
  • #15
So, it is safe to say that the dx is a scaling factor that brings the otherwise divergent expression

[tex]\vec{\phi(x)}\bullet\vec{\psi(x)}[/tex]

back to a finite value.
This leads to my bastardized notation where you define (for functions), *thanks to hamster for making it more precise*

[tex]\lim_{\Delta x \rightarrow 0}\vec{\phi(x)} \bullet \vec{\psi(x)}=\lim_{\Delta x \rightarrow 0} \sum_{n=0}^{\frac{b-a}{\Delta x}} \phi(a+n\Delta x) \psi(a+n\Delta x)[/tex]

Which, as the limit goes to zero, becomes

[tex]\vec{\phi(x)} \bullet \vec{\psi(x)}= \phi(a) \psi(a) + \phi(a+dx) \psi(a+dx)+\phi(a+2dx) \psi(a+2dx)+...+\phi(b-dx)\psi(b-dx)dx+\phi(b)\psi(b)dx[/tex]

This expression definitely diverges, though. If we multiply both sides by a delta x in the original equation, before the limit was applied, we get

[tex]\lim_{\Delta x \rightarrow 0}\vec{\phi(x)} \bullet \vec{\psi(x)} \Delta x=\lim_{\Delta x \rightarrow 0} \sum_{0}^{\frac{b-a}{\Delta x}} \phi(a+n\Delta x) \psi(a+n\Delta x) \Delta x[/tex]

And applying the limit

[tex]\vec{\phi(x)} \bullet \vec{\psi(x)}dx= \phi(a) \psi(a)dx + \phi(a+dx) \psi(a+dx)dx+\phi(a+2dx) \psi(a+2dx)dx+...+\phi(b-dx)\psi(b-dx)dx+\phi(b)\psi(b)dx[/tex]

Which can be recognized to be

[tex]\vec{\phi(x)} \bullet \vec{\psi(x)}dx= \int_{a}^{b}\phi(x) \psi(x)dx[/tex]

This integral does not diverge as long as the functions are well behaved in the interval. Then, we can define

[tex]\langle \phi(x) , \psi(x) \rangle =\vec{\phi(x)} \bullet \vec{\psi(x)}dx=\int_{a}^{b}\phi(x) \psi(x)dx[/tex]
 
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1. What is an inner product in an infinite dimensional space?

An inner product in an infinite dimensional space is a mathematical operation that takes two vectors as input and produces a scalar value as output. It is similar to the dot product in finite dimensional spaces, but it is defined for an infinite number of dimensions.

2. How is the inner product defined in an infinite dimensional space?

The inner product in an infinite dimensional space is defined as the sum of the products of the corresponding components of two vectors. This sum is taken over all possible dimensions, which can be infinite. It is denoted by <x, y> and is also known as the scalar product or the dot product.

3. What are the properties of an inner product in an infinite dimensional space?

An inner product in an infinite dimensional space has the following properties:

  • Linearity: the inner product is linear in its first argument.
  • Symmetry: the inner product is symmetric, meaning <x, y> = <y, x>.
  • Positive definiteness: the inner product is positive definite, meaning <x, x> > 0 for all nonzero vectors x.

4. What is the significance of the inner product in an infinite dimensional space?

The inner product in an infinite dimensional space is a fundamental concept in functional analysis and plays a crucial role in many areas of mathematics and physics. It allows for the definition of important concepts such as orthogonality, norm, and distance in infinite dimensional spaces.

5. How is the inner product used in practical applications?

The inner product has many practical applications, particularly in fields such as signal processing, quantum mechanics, and machine learning. It is used to measure the similarity between vectors, project them onto other vectors, and find the best approximation of a vector in a given space. It also allows for the development of efficient algorithms for solving problems involving infinite dimensional spaces.

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