In a book I'm reading, it says:If beta is orthogonal to the set A,

[yifli]

In a book I'm reading, it says:

If beta is orthogonal to the set A, then beta is orthogonal to the closure of the linear span of A

It's easy to see beta is orthogonal to the linear span of A, but I don't understand why it has to mention closure here?

 

[micromass]

Hi yifli!

What is your question exactly?

Do you wonder why they wrote this here? Well, because it's true. Of course it's also true that x is orthogonal to span(A), so they could have written this too. However, in most applications, you need x orthogonal to the closure of span(A). This is why they've written this here.

Do you ask why it is true? Well, in general, if x is orthogonal to a set A, then x is orthogonal to the closure of A. Indeed, pick a in the closure of A, then there exists a sequence a_n that converges to a. Since a_n is in A, we know that <x,a_n>=0. And thus &lt;x,a&gt;=\lim_{n\rightarrow +\infty}{&lt;x,a_n&gt;}=0. Thus x is orthogonal to a.

 

[yifli]

Hi yifli!

What is your question exactly?

Do you wonder why they wrote this here? Well, because it's true. Of course it's also true that x is orthogonal to span(A), so they could have written this too. However, in most applications, you need x orthogonal to the closure of span(A). This is why they've written this here.

Do you ask why it is true? Well, in general, if x is orthogonal to a set A, then x is orthogonal to the closure of A. Indeed, pick a in the closure of A, then there exists a sequence a_n that converges to a. Since a_n is in A, we know that <x,a_n>=0. And thus &lt;x,a&gt;=\lim_{n\rightarrow +\infty}{&lt;x,a_n&gt;}=0. Thus x is orthogonal to a.

micromass, thank you so much for your answer!

 

[lavinia]

Hi yifli!

What is your question exactly?

Do you wonder why they wrote this here? Well, because it's true. Of course it's also true that x is orthogonal to span(A), so they could have written this too. However, in most applications, you need x orthogonal to the closure of span(A). This is why they've written this here.

Do you ask why it is true? Well, in general, if x is orthogonal to a set A, then x is orthogonal to the closure of A. Indeed, pick a in the closure of A, then there exists a sequence a_n that converges to a. Since a_n is in A, we know that <x,a_n>=0. And thus &lt;x,a&gt;=\lim_{n\rightarrow +\infty}{&lt;x,a_n&gt;}=0. Thus x is orthogonal to a.

Do you need to show that the inner product with x is a continuous function?

 

[micromass]

Do you need to show that the inner product with x is a continuous function?

Of course! But that's not that hard, I believe...