Unbounded and linear transformation 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.  
 A: 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.
A: 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.
