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I was thinking about this the other day.

Does there exists any unbounded linear transformations from $\ell_\infty(\mathbb{R}) \to \mathbb{R}$ ? Here $\ell_\infty$ represent the infinity norm.

N3buchadnezzar
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  • Ignore my (now deleted) answer. x_x I ballsed up twice in a row. – Rudy the Reindeer Oct 17 '12 at 09:23
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    The standard vectors $e_n$ are linearly independent. Extend them to an algebraic basis $(e_i)_{i \in I}$ of of $\ell^{\infty}$ where $\mathbb{N} \subset I$. A linear map is given by what it does on the basis. The map $T\colon \ell^{\infty}(\mathbb R) \to \mathbb{R}$ given by $T(e_n) = n$ and $T(e_i) = 0$ for $i \notin \mathbb{N}$ is unbounded. – commenter Oct 17 '12 at 10:05
  • @commenter: You should post that as an answer. – Asaf Karagila Oct 17 '12 at 10:16
  • See also Discontinuous linear functional. (But I guess this is basically the same solution as already posted in answers bellow - using Hamel basis.) – Martin Sleziak Oct 17 '12 at 13:51
  • It is known that this cannot be done if ZF (i.e., some form of choice is needed), see t.b.'s answer here: Nonnegative linear functionals over $l^\infty$ (and the links and references given in his answer). – Martin Sleziak Oct 17 '12 at 13:58
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    What do you mean by the notation $\ell_\infty(\mathbb R)$. Do you mean bounded functions from $\mathbb R$ to $\mathbb R$ or bounded real sequences? (The notation $\ell_\infty(M)$ is sometimes used to denote the bounded functions on $M$. In this notation we have $\ell_\infty=\ell_\infty(\mathbb N)$.) – Martin Sleziak Oct 17 '12 at 14:15
  • @commenter You were right in your comment under my answer (which is now deleted). The proof using Hahn-Banach Theorem, which I suggested there, is incorrect. (Thanks for noticing it and letting me know.) – Martin Sleziak Oct 17 '12 at 16:17

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If by unbounded you mean simply that the image of the operator is unbounded in $\mathbb R$, then every non-zero functional from $\ell_\infty$ to $\mathbb R$ has this property.

Suppose that $Tx=r\neq 0$, then $T\frac nrx=n$, and this approaches infinity as $n$ grows larger.

On the other hand, if you mean unbounded as $\|T\|=\infty$, namely $\sup\{Tx\mid \|x\|_\infty=1\}=\infty$, such operator cannot be written explicitly. It is consistent without the axiom of choice that there are no such operators, so we cannot write a formula for this. The equivalence between a bounded operator and a continuous operator is true in ZF, and there are models (e.g. Solovay's model, Shelah's model in which every set of reals have Baire property) in which the axiom of choice fails and every linear operator from a Banach space to a normed space is continuous. In particular, linear functionals from $\ell_\infty$ to $\mathbb R$.

Using the axiom of choice fix a Hamel basis (algebraic basis), it is obviously infinite, we may assume that it is normalized (namely all the basis elements have norm $1$). Take any function from this basis to the reals which is unbounded, it extends uniquely to an unbounded functional.

Asaf Karagila
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  • I think you can define such an unbounded functional pretty easily if you start from a hammel base (i.e. a vector space base, not a hilbert space base) of $\ell_\infty$ (this requires AC, of course). Then just map every element of that base to some real, and make sure there's a (countable suffices) subset where that mapping is unbounded. That should give you an unbounded linear functional if I'm not mistaken. – fgp Oct 17 '12 at 10:22
  • @fgp: Yes, that sounds a reasonable plan. – Asaf Karagila Oct 17 '12 at 10:55
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The following construction shows how to define such an unbounded functional, assuming that there's a Hamel base of $\ell_\infty$. A Hamel base is simply a base in the usual vector space sense, i.e. a set of vectors from which every other vector can be derived by a finite linear combination. Note that without the axiom of choice, such a base must not necessarily exist, but with the axiom of choice it always does.

Let $(b_i)_{i\in I}$ be a Hamel base of $\ell_\infty$, i.e. for every $l \in \ell_\infty$ there is a unique and finite $J \subset I$ and unique $(\lambda_j)_{j \in J} \in \mathbb{R}$ with $$ l = \sum_{J} \lambda_j b_j $$ Then pick a countable set $K \subset J$, let $N : K \to \mathbb{N}$ be its bijective mapping to $\mathbb{N}$, and set $$ T(l) = \sum_{J} \lambda_jc_j \text{ where } c_j = \begin{cases} 2^{N(j)}||b_k||_\infty &\text{if } j \in K \\ 1 &\text{otherwise} \end{cases} $$ $T$ is then obviously linear. Now, look at $T(\{b_i||b_i||_\infty^{-1} : i \in I\})$. That set is obviously bounded (in fact, all its members have norm $1$). But since $T(b_i||b_i||_\infty^{-1}) = 2^{N(i)}$ if $i \in K$, its image is unbounded.

fgp
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