# Is the space of real sequences normable

Intuition says the vector space of real sequences $$R^N$$ ($$N$$ the natural numbers, pointwise addition of real coordinates) is not normable. I have found this surprisingly hard to prove.

I am aware that $$R^N$$ in the product (Tychonoff) topology is not normable (but metrizable i.e. is a Frechet space). This is proven for instance in Aliprantis–Border: Infinite Dimensional Analysis (2006, p. 207), a reference which I found on Is a metrizable topological vector space normable? in the answer by @triple_sec which I understand and appreciate.

However, this leaves open whether a TVS topology on $$R^N$$ stronger than the product topology could be normable. Here are my attempts to prove there is no norm defined everywhere on $$R^N$$ making it a TVS with topology at least as strong as the product topology i.e. with continuous coordinate functions.

I aim for an indirect proof: assume there is a norm on $$R^N$$ and try to get a contradiction.

a) Banach space arguments: complete the normed space and apply Banach space technology like Uniform Boundedness, Open Mapping etc.. I could not find any proof along this line. (In particular, I found it hard to control whether the completion of $$R^N$$ in the assumed norm would not make the original normed space "bigger" i.e. adding ideal points outside of $$R^N$$). But since experts in the field might well know a good argument, I mention the attempt.

b) Geometric argument - construct a vector with an infinite norm: begin with the "unit" vectors $$e_n = (0,..., 1, 0, ...) \in R^N, n\in N$$ with a $$1$$ in the n-th coordinate and $$0$$ elsewhere. Rescale every $$e_n$$ to have the norm $$||e_n||=1$$ which is just a coordinate change. Now try $$v=(1, 2, 3, ...) \in R^N$$ to obtain a vector $$v$$ with norm $$||v||\geq n$$ $$\forall n$$ - are we done? I think no, because the norm $$||v||$$ of a vector $$v=(p_1, p_2, p_3, ...)$$ with non-negative real coordinates $$p_i, i\in N$$ is in general (unlike in the standard sequence spaces $$l^p$$) not a non-decreasing function of its real coordinates. So I wonder how to argue strictly that $$||v||\geq n$$ $$\forall n$$. I tried flipping the signs of each coordinate i.e. consider $$v=(\pm 1, \pm 2, \pm 3, ...)$$ and use the fact that for arbitrary $$v, x$$ in a normed space one of $$||v+x||$$ and $$||v-x||$$ must be not smaller than $$||v||$$ and $$||x||$$ (triangle inequality). But to generate an optimal $$v$$ with maximal (infinite) norm $$||v||$$ I face (countably) infinitely many flips depending all on each other. I have no idea how to control (countably) infinitely many sign flips to make such an argument rigorous, but perhaps somebody else has.

The idea seems to be included in the answer of @David C. Ullrich to Can the real vector space of all real sequences be normed so that it is complete ? . There is also an answer and a comment by @paul garrett which indicates what I am trying to prove here is an obvious matter of fact to the experts. However LF-spaces are currently too advanced for me, so I tried the same geometric argument one more time, but now "weakly":

c) Geometric argument - construct a "weakly unbounded" vector: we start again with unit vectors $$e_n, n\in N$$ like in b), and take the dual sequence of linear functionals $$p_n\in (R^N)^*, \forall n \in N$$ projecting an arbitrary vector $$v$$ to its $$n$$th coordinate. These coordinate functionals are continuous by assumption, and since $$p_n(e_n)=1$$, the functional norm $$||p_n||$$ is $$\geq 1$$. Now we consider the vector $$v=(1||p_1||, 2||p_2||, 3||p_3||, ..., n||p_n||, ...)$$ and estimate $$||v||$$ from below: $$||v|| \geq |p_n(v)|/||p_n|| =n$$ $$\forall n \in N$$.

My question: is the geometric argument in (c) good enough for a reasonable proof?

Remark upon my original motivation: reading that a product of barreled spaces is barrelled - "Un produit d'espaces tonnelés est tonnelé." on the french https://fr.wikipedia.org/wiki/Espace_tonnel%C3%A9 (the fact is not yet on the english page https://en.wikipedia.org/wiki/Barrelled_space) I tried to prove this. I looked at $$R^N$$ as one of the simplest non-trivial examples of a product of barrelled spaces: are in there any other barrels except closures of finite products of convex open balls=intervals (all remaining factors equal $$R$$)? Answer: there are not.

For the ingeneous Bourbaki this was so obvious to write in http://www.numdam.org/article/AIF_1950__2__5_0.pdf on page 6 mere one sentence:

"On montre sans peine que ... tout produit d'espaces tonnelés est tonnelé."

• doesnt the answer from @David C. Ullrich precisely answers your question? He shows that there is no norm on $R^N$ such that all coordinate projections are continuous. Commented Mar 28, 2019 at 10:11
• Dear @supinf, I might be ignorant, but it does not answer my question. In $$\sup_N\left|\sum_{n=1}^N2^{-n}c_n^{-1}x_n\right|=\infty$$ he is assuming implicitly the $l^1(R)$ norm $$\sum_{n=1}^\infty |x_n|$$ on $R^N$. Therefore I wrote that a norm on the sequence space is in general not a monotone non-decreasing function of (the absolute values of) its real coordinates. Example: $||(x_1,x_2)||=|x_1|+|x_1-x_2|$ at $(1,0)$ on $R^2$. I am asking for any possible norm depending continuously on the coordinates, not only monotone ones. Commented Mar 28, 2019 at 10:48
• He is not assuming implicitly the $l^1$-norm on $R^N$. If you want to know why the $\sup_N=\infty$ is true, just use the sequence $x_n:=c_n2^n$. Then you can omit the absolute value, and the claim becomes $\sup_N N=\infty$. Commented Mar 28, 2019 at 10:56
• In the $l^1$ norm yes, in the norm $$|x_1|+\sum_{n=1}^\infty |2^{-n}c_n^{-1}x_n-2^{-n-1}c_{n+1}^{-1}x_{n+1}|$$ not. Commented Mar 28, 2019 at 11:15

## 1 Answer

Yes, your argument in (c) is good enough for a reasonable proof.

Some comments regarding your attempt in (b): The strategy to consider an element of the form $$v=(\pm 1,\pm 2, \pm 3,\dots)$$ and show that $$\|v\|$$ can be arbitrarily large will probably not work so easily. Consider the function $$f(x)= \sum_{n\in N} 2^{-n} |x_n|.$$ On the subspace where $$f$$ is finite this is a norm. Moreover, $$f(v)$$ would be finite for all $$v$$ that you considered. The reason that $$f$$ is not a norm on the full space is that there are sequences that grow arbitrarily fast, not your sequence $$v$$ whose elements grow at a fixed rate.

• I fully agree with you. For finitely many signs to choose of, say, the first n coordinates of $v$, it is clear that there is an optimal choice to get a maximum $||v||$, keeping the (infinitely many) following ones fixed. But for all the (countably infinitely many) signs at the same time it is not obvious. Commented Mar 28, 2019 at 11:05