
#1
Nov1709, 09:38 PM

P: 185

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. 



#2
Nov1709, 10:47 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101





#3
Nov1709, 11:02 PM

P: 185

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
Nov1709, 11:03 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

Inner Product (Infinite Dimensional)
Oh, by the way, an inner product on a real vector space is merely any realvalued function of two vectors variables:
<x+y, z> = <x,z> + <y,z>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 dimensions^{1}  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
Nov1709, 11:04 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101





#6
Nov1709, 11:06 PM

P: 185

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] 



#7
Nov1709, 11:08 PM

P: 185

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. 



#8
Nov1709, 11:23 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

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
Nov1709, 11:28 PM

P: 185

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
Nov1709, 11:38 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

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 welldefined. Such "continuous linear combinations" are generally made rigorous using integrals. In fact, your expansion simply inverted that process. If you are willing to learn nonstandard analysis, you could instead mutter something about standard parts and approximating the interval with a lattice of hyperfinitely many points. 



#11
Nov1709, 11:45 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

*: at least, every case we are interested in 



#12
Nov1709, 11:46 PM

P: 185

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/hebcs/Books?ac...chapterId=4825 Where in equation 2A6 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. 



#13
Nov1809, 04:59 AM

P: 685

An integral [tex]\int_{a}^{b} f(x) dx[/tex] can be approximated by [tex]\sum_{k=0}^{N1} 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
Nov1809, 05:16 AM

P: 986

[tex]\langle \phi , \psi \rangle = \lim_{dx \rightarrow 0} \sum_{n=0}^{(ba)/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
Nov1909, 02:00 PM

P: 185

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{ba}{\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(bdx)\psi(bdx)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{ba}{\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(bdx)\psi(bdx)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] 


Register to reply 
Related Discussions  
Infinite dimensional representation of su(2)  Linear & Abstract Algebra  6  
Infinite dimensional counterexample  Calculus & Beyond Homework  4  
Infinite dimensional PDE  General Math  3  
Infinite dimensional vectorial (dot) product)  Calculus  1  
Is 11 dimensional space infinite?  Beyond the Standard Model  1 