That's the statement I'm trying to prove. It's the converse of the result mentioned in this post: "Every linear mapping on a finite dimensional space is continuous".

It seems that a good way to start would be to replace 'continuous' with 'bounded' because they're the same thing for linear maps, and then attempt to show that a linear functional on an infinite-dimensional space can't be bounded. But that's where I get stuck.

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    $\begingroup$ Proving the converse should be easier. Can you show that if a vector space is not finite dimensional then there is an unbounded linear functional? $\endgroup$ – Zestylemonzi Mar 3 '18 at 11:08
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    $\begingroup$ see: math.stackexchange.com/questions/288075/… $\endgroup$ – Emilio Novati Mar 3 '18 at 11:09
  • $\begingroup$ Going by these two comments, it looks like if we have a basis $\{e_1, e_2, ...\}$, we can insist that the basis is normalised so that all the $e_n$ have norm 1, and define the linear functional $f$ by $f(e_n) = n$. Then since there are infinitely many $n$ to choose from, there's no bound we can put on $f$. That sounds good to me, but please correct me if I'm wrong. $\endgroup$ – SPS Mar 3 '18 at 11:16
  • $\begingroup$ I am writing a fairly long answer. I request you to wait for it before accepting. $\endgroup$ – Teresa Lisbon Mar 3 '18 at 11:52
  • $\begingroup$ @SPS, your idea is right. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 3 '18 at 12:22

We must speak in the context of a topological vector space in this question. Such a topology is not necessarily normable, therefore you see that you cannot talk about boundedness, orthonormality etc. in the ordinary sense.

You have also assumed that the basis of $V$ is countable, which it may not be. But these are small issues : you grasped the heart of the matter.

A linear functional, then, is continuous if and only if it is bounded. Bounded? Where is the norm on the space to speak of boundedness?

So, we have a notion of boundedness, that does not even mention the word "norm". A set $E$ in a topological vector space is bounded if given any other open set $V$ around $0$ (remember : open sets of the topological vector space ensure that addition and scalar multiplication are continuous), there is a real number $t$ large enough, such that the set $tV = \{tv : v \in E\}$ contains $E$. You can see why this reflects boundedness in the ordinary sense.

Now, continuity is equivalent to this : for any bounded set $E$, $l(E)$ should also be bounded i.e. bounded in the ordinary sense, as a subset of the real numbers.

To be clean on notation, let's define $l$ as follows : take a countable linearly independent set $e_i$, and extend it to a basis $e_i \cup f_\alpha$. Now define $l(e_i) = i$ and $l(f_{\alpha}) = 0$.

Why is $l$ not continuous now? Of course, it is easy to see, though I won't explain it, that the set $E = \{\sum a_{i}e_i + \sum a_\alpha f_{\alpha} : |a_i|,|a_{\alpha}| \leq 1 \forall i,\alpha\}$ is a bounded set : it's sort of like the open box in finite dimensions.

But $l(E)$ is unbounded, because $l(e_i)$ itself is unbounded!

That, is the proof of your assertion, with a lot of background added.

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  • $\begingroup$ Thanks. I could have been clearer and explained that I'm only interested in countable-dimensional normed vector spaces, but as you noticed the answers are similar for that case and the more general case you outline here. $\endgroup$ – SPS Mar 3 '18 at 13:23
  • $\begingroup$ You are welcome! I am thinking of a more direct method to do this : in particular, a normed linear space is finite dimensional if and only if its unit ball is compact. So can I use the fact that every linear functional is continuous, to show that the unit ball is compact? I think I can, and I will edit the answer if I can. $\endgroup$ – Teresa Lisbon Mar 3 '18 at 13:26
  • $\begingroup$ Could you give me a quick hint to see why the set $E$ in the end is bounded? $\endgroup$ – Renan Maneli Mezabarba May 23 at 2:28
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    $\begingroup$ @RenanManeliMezabarba We must show that for every open neighbourhood $V$ of the origin, a scaling $tV$ of it covers $E$. Even open set around the origin contains an open set of the form $\{\sum a_ie_i + \sum a_{\alpha}f_{\alpha} : |a_i|,|a_{\alpha}| \leq \epsilon\}$ for some $\epsilon > 0$ because of the continuity of addition and the definition of the vector space topology. Finally, use scalar multiplication. $\endgroup$ – Teresa Lisbon May 23 at 8:34

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