I want to show a difference of inner products is small

  • #1

Main Question or Discussion Point

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...
 

Answers and Replies

  • #2
mathwonk
Science Advisor
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maybe add and subtract (x1,y2)? are you trying to show the dot product is continuous?
 
  • #3
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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?
 
  • #4
Office_Shredder
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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?
 
  • #5
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.
 
  • #6
329
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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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