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How to prove there exists an unbounded linear operator $T:\ell^1\rightarrow \mathbb R^2$?

PtF
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3 Answers3

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This must require the use of the axiom of choice. It is consistent that the axiom of choice fails and every linear operator between Banach spaces is continuous, i.e. bounded. So you can't just write something like this down in a formula and you need to appeal to some intangible objects, in this answer we appeal to the existence of a Hamel basis.

Fix a Hamel basis $\cal B$ and without loss of generality assume that every $b\in\cal B$ is such that $\|b\|_1=1$. Let $\{b_i\mid i\in\Bbb N\}$ a sequence from $\cal B$, now we define the following function:

$$F(b)=\begin{cases} (i,i)& \exists i\in\Bbb N\text{ such that }b=b_i\\0 &\text{otherwise}\end{cases}$$

Extend this $F$ to a linear operator and it is easy to see that it is unbounded.

Asaf Karagila
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  • That's already my second stupid comment today. I read Hamel and I thought Schauder... Sorry. – Julien Feb 28 '13 at 15:11
  • Wolfram seems to think Hamel basis refers specifically to the basis for $\mathbb{R}$ over $\mathbb{Q}$, so I was confused at first. Anyways helpful answer. +1. – Caleb Stanford Apr 29 '14 at 00:15
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$$(a_1,a_2,a_3,...,a_n,...) \mapsto \left(\sum_{n\ge 1} n\,a_n\,,\ 0\right)$$ is an example, if you mean unbounded operator as this wikipedia article.

Berci
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    You really have to appeal to the axiom of choice here. It is consistent that there are none. So you probably want to check this answer for possible mistakes, or write a paper about the inconsistency of ZFC :-) – Asaf Karagila Feb 28 '13 at 14:23
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    So what value does this take when $a_n = 1/n^2$? – Nate Eldredge Feb 28 '13 at 14:26
  • What we called 'unbounded' linear operator, was usually just a partial function, defined only on a (dense) subspace. http://en.wikipedia.org/wiki/Unbounded_operator – Berci Feb 28 '13 at 14:28
  • @AsafKaragila: Would you expound your comment? – Berci Feb 28 '13 at 14:30
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    This operator is not defined for the sequence $(1/n^2)$. – Julien Feb 28 '13 at 14:31
  • Berci, clearly your operator is well defined in ZF. Since it is consistent that every linear operator between Banach spaces is bounded, either you made a mistake (e.g. what Nate said) or you found the long sought contradiction in set theory. :-) – Asaf Karagila Feb 28 '13 at 14:32
  • @Julien: Yes, exactly that's how I constructed it! – Berci Feb 28 '13 at 14:32
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    But, please, all, do you agree this part of the wikipedia article (linked above): 'the domain of the operator is a linear subspace, not necessarily the whole space (in contrast to "bounded operator");'? Also, the differentiation operator on formal power series (with $\ell^1$ coefficients) should be a good example. – Berci Feb 28 '13 at 14:36
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    @Berci An unbounded linear operator is a linear map $f: V \rightarrow W$ between normed vector spaces, such that $||f||$ is unbounded on the unit ball of $V$. – JSchlather Feb 28 '13 at 14:36
  • Probably everybody is right (in one or another interpretation).. – Berci Feb 28 '13 at 14:39
  • @JSchlather See the "contrary to the usual convention" here: http://en.wikipedia.org/wiki/Unbounded_operator – Julien Feb 28 '13 at 14:39
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    Why the downvotes? This is totally correct...+1. – Julien Feb 28 '13 at 14:41
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    This interpretation is quite common in PDEs, when considering the "unbounded operator" $\frac{\partial}{\partial x_i}$ on some subspace of a larger space. A common example is that the operator is defined on $H^1$ but treated as an operator on $L^2$. –  Feb 28 '13 at 17:19
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This is not a direct aswer, but may be helpful.

Berci's answer shows that one can explicitly construct an unbounded (everywhere defined) operator on some infinite-dimensional normed space $\,X$ (so without using AC). On the other side, Asaf told us that this is impossible if $\,X$ is Banach (i.e. complete with respect to the distance induced by the norm). Why is that so?

The reason is that every infinite-dimensional Banach space has no countable Hamel basis (this is a useful exercise) and so one cannot exhibit one without AC.

Ivo
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