I want to show a difference of inner products is small

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Discussion Overview

The discussion revolves around the behavior of inner products in a separable Hilbert space as certain vectors approach each other under a parameter \( p \to 0 \). Participants explore whether the difference of inner products can be shown to converge to zero under these conditions, considering both finite and infinite dimensional cases.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes that since \( \| x_1 - y_1 \| \to 0 \) and \( \| x_2 - y_2 \| \to 0 \) as \( p \to 0 \), it might be possible to show that \( |(x_1,x_2) - (y_1,y_2)| \) also approaches zero.
  • Another participant suggests adding and subtracting \( (x_1,y_2) \) to explore the continuity of the dot product.
  • A different participant points out that the expression \( |(x_1-y_1,x_2-y_2)| \) approaches zero since both components converge to zero, but raises the concern that for the inner product to go to zero, the vectors must be moving towards orthogonality.
  • There is a clarification regarding the notation used for the inner product, with one participant confirming that they are using \( (x,y) \) to denote the inner product, despite some confusion about its interpretation.
  • One participant provides a specific example in \( \mathbb{R}^2 \) to illustrate their reasoning, suggesting that the convergence of the components implies the inner product difference approaches zero, while noting that the argument may be more complex in infinite dimensions.

Areas of Agreement / Disagreement

Participants express differing views on the conditions necessary for the inner product to converge to zero, particularly regarding the relationship between the vectors involved. There is no consensus on whether the original claim can be definitively shown to be true.

Contextual Notes

Some participants highlight potential confusion regarding notation and the implications of convergence in different dimensional spaces. The discussion reflects a range of assumptions about the properties of inner products in separable Hilbert spaces.

AxiomOfChoice
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Suppose I have a separable Hilbert space [itex]\mathcal H[/itex] and vectors [itex]x_1(p),x_2(p),y_1(p),y_2(p) \in \mathcal H[/itex] that depend on a parameter [itex]p>0[/itex] such that

[tex] \| x_1 - y_1 \| \to 0 \qquad \text{as $p \to 0$}[/tex]

and

[tex] \| x_2 - y_2 \| \to 0 \qquad \text{as $p \to 0$}.[/tex]

Can anything be said about [itex]|(x_1,x_2) - (y_1,y_2)|[/itex]? I'd like to be able to say it goes to zero as [itex]p\to 0[/itex], but I haven't been able to show that yet...
 
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maybe add and subtract (x1,y2)? are you trying to show the dot product is continuous?
 
As stated you are just asking for |(##x_1-y_1,x_2-y_2##)| which is the length of a vector and certainly goes to zero since both components are headed for 0.

Since you have tagged this as an inner product question consider this:
For the inner product to go to 0 the vectors have to be moving towards orthogonal. In the ordinary ##R^2## space it could be that ##x_1 - y_1## and ##x_2 - y_2## are parallel. They can go to zero all they want, but the inner product certainly won't.

Is this what you were asking?
 
I think there's some confusion about notation here. Is (x,y) the inner product of x and y, or is it a point in H2?
 
Office_Shredder said:
I think there's some confusion about notation here. Is (x,y) the inner product of x and y, or is it a point in H2?
Sorry; I should have clarified. I know some people use [itex]\langle \cdot,\cdot \rangle[/itex] exclusively to denote the inner product, but yes, I'm using [itex](\cdot,\cdot)[/itex] to denote the inner product.
 
Okay, I think I have it for ##R^2##; but it should be generalizable.

Let ##x_1 = (a_1,b_1), x_2 = (a_2,b_2), y_1 = (c_1,d_1), y_2= (c_2,d_2)##. We have

||##x_1 - y_1##|| ##\rightarrow ## 0 and the same for the 2's. ##\hspace{50px}## (1)

Looking at the inner products we have

##x_1 \cdot x_2 - y_1 \cdot y_2 ## = ##a_1a_2 + b_1b_2 - (c_1c_2 + d_1d_2)##.

By (1) we have ##c_1 \rightarrow a_1## and correspondingly for all the other components. So the term after the minus sign is going to the term before the minus sign and the whole thing goes to 0.

Life is a little harder if you are not in a finite dimensional space, but since your Hilbert space is separable, I think you can take the same idea and apply it.
 

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