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Consider the sequence $(x^k)$ in $c_{00}$ defined by

$$\begin{align} x^1 &= (1,0,0,0,\ldots) \\ x^2 &= (1,1,0,0,0,\ldots) \\ x^3 &= (1,1,1,0,0,0,\ldots), \end{align}$$ and so on.

With respect to the sup norm, the sequence does NOT converge to the "obvious guess" $x = (1,1,1,1,1,1,\ldots)$ in the space $c_{00}$, because $(1,1,1,1,1,1,\ldots)$ is not in $c_{00}$. Neither does it converge to $x$ in a larger space with the sup norm. Neither does is converge to a larger space with a different norm that I can think of (e.g. the quadratic mean norm).

This is seemingly horrible. What is going on here!?! It SHOULD converge in SOME SPACE! Come on!

Is there a norm $\|\cdot\|$ so that $x^k$ converges to $x$ in a larger space?

My only guess is the "inf norm", but I'm not even sure that is actually a norm. (Actually, I'm quite certain it isn't because we would then have $\|(1,0,0,0,\ldots)\|_\inf = \|(2,0,0,0,\ldots)\|_\inf = 0$, contradicting one of the norm axioms). Actually, this is a stupid guess, but I'm out of ideas...

Also, intuitively, why do we not have $x^k \to x$ in $c_{00}$ w.r.t. all the standard norms mentioned above? Remember, humans introduced these norms, and why have we introduced all these norms if it doesn't have some of the most "obvious" convergence properties (like the one I am after). Is it something to do with the fact that $c_{00}$ is incomplete in all the standard norms? Or is it just because of the nature of the space $c_{00}$? Is the space $c_{00}$ not a "nice" one? By standard norms I mean the sup norm $\|\cdot\|_{\infty}$ and the norms $\|\cdot\|_p$ where p is an integer.


I think I have found an IDEA (one that I am not used to). Maybe we could think of limit inferior or limit superior in some way to come up with a suitable norm.

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This is mildly nitpicky, but some people regard typing in all-caps online as equivalent to yelling and would prefer that it not happen on places like math.SE. If you want to emphasize something, use italics or bold. – Qiaochu Yuan May 27 '12 at 15:45
lol, point taken – Adam Rubinson May 27 '12 at 17:59
Instead of writing in all-caps, put *asterisks* around the text you want to emphasize, and nobody will object. – MJD May 27 '12 at 22:31
up vote 5 down vote accepted

Also, intuitively, why do we not have $x^k \to x$ in $c_{00}$ w.r.t. all the standard norms mentioned above?

Because $||x^k||$ isn't even bounded in several of the norms you mentioned. In the sup norm we have $||x^k - x^m|| = 1$ for all $k \neq m$, so as far as the sup norm is concerned these points are equally spaced and asking for convergence is silly.

The topology you're intuitively looking for doesn't (I think) come from a norm: it's the topology of pointwise convergence on the space of functions $\mathbb{N} \to \mathbb{R}$, otherwise known as the product topology on $\mathbb{R}^{\mathbb{N}}$. Using the sup norm instead corresponds to talking about uniform convergence. Uniform convergence is a stronger notion in that fewer sequences converge, but on the other hand it is desirable because it preserves more properties (e.g. a uniform limit of continuous functions remains continuous).

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Yeah, in fact if we take any member, say y, of c_00, then ||x-y||_infinity = 1. So it is clear that no sequence in c_00 will ever converge to x in a larger space (containing x) with the sup norm. I guess the key here is "boundedness of x with respect to the norm under consideration". – Adam Rubinson May 27 '12 at 17:48

The norm $$||x||=\sum_{i=1}^\infty \frac{|x_i|}{i^2}$$ should work. The space is then the space of all sequences such that the limit of the series exists, which obviously contains all finite sequences.

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Yes, I didn't think of that. – Adam Rubinson May 27 '12 at 13:59
So is c_00 not a "not nice" space like I first thought. It is just the wrong space for the properties I am after. Is this correct? – Adam Rubinson May 27 '12 at 14:01

Similar to Simon's answer. For $\omega:\mathbb{Z}^+\to\mathbb{R}^+$ and $p\geq 1$, let $\ell^p_\omega(\mathbb{Z}^+)$ denote all complex sequences $(x_n)$ such that $$\|(x_n)\|_{\ell^p_\omega}=\left(\sum |x_n|^p\omega(n)\right)^{1/p}<\infty$$ Then $\ell^p_\omega$ is a Banach space (complete metric space for $0<p<1$ if we replace $\|(x_n)\|$ above by $\|(x_n)\|^p$).

It is easy to see that a sufficient condition for convergence of the sequence $x^k$ is that $\omega\in\ell^1$, that is if $\sum \omega(n)<\infty$.

Edit In fact it is necessary too: Suppose $x^k$ do converge in $\ell^p_\omega$, then it is a Cauchy sequence, i.e. given $\varepsilon>0$ there is $N$ such that
$$\|x^k-x^j\|^p<\varepsilon$$ provided $k,j>N$. But then for $N<k<j$ we must have $$\sum_{n=k}^j \omega(n)<\varepsilon$$ so $\omega\in\ell^1$.

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Another natural space is a Fréchet Space, which is not a normed space, but a locally convex one. The topology arises from this collection of seminorms:

$$\lVert x\rVert_{n} = \sum_{i=1}^n \lvert x_{i}\rvert, \qquad(x_{i}) \in \mathbb R^{\infty}.$$

A sequence converges in this topology if and only if it converges in all seminorms. This is a complete metric space since the collection of seminorms is countable.

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