Are Norms in Vector Spaces Continuously Linked to Limits?

[Niles]

Hi guys

The norm is a continuous function on its vector space, but I a little unsure of how to interpret this. Does it mean that if we are in e.g. a Hilbert space with an orthonormal basis e_i (i is a positive integer), then we have

<br /> \left\| {\mathop {\lim }\limits_{N \to \infty } \sum\limits_{i = 1}^N {x_i e_i } } \right\| =\mathop {\lim }\limits_{N \to \infty } \left\| {\sum\limits_{i = 1}^N {x_i e_i } } \right\|<br />

for some vector x = Σ x_ie_i in the Hilbert space? Or does the above operation come from the continuity of taking limits?

 

[Landau]

Does it mean that if we are in e.g. a Hilbert space with an orthonormal basis e_i (i is a positive integer), then we have

<br /> \left\| {\mathop {\lim }\limits_{N \to \infty } \sum\limits_{i = 1}^N {x_i e_i } } \right\| =\mathop {\lim }\limits_{N \to \infty } \left\| {\sum\limits_{i = 1}^N {x_i e_i } } \right\|<br />

for some vector x = Σ x_ie_i in the Hilbert space?

Yes, it does.

Or does the above operation come from the continuity of taking limits?

I'm not sure what you mean by this. Maybe this will clear things up:

You know that for a continuous function f and a converging sequence x_n, we have
\lim_{n\to\infty}f(x_n)=f(\lim_{n\to\infty}x_n),
i.e. you can "pull the limit out of the argument of a continuous function".

Now, let f be the norm-function: f=\|...\|. Then the above becomes
\lim_{n\to\infty}\|x_n\|=\|\lim_{n\to\infty}x_n\|
i.e. you can "pull the limit out of the (argument of) the norm", which is what you did.