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This is a homework question, so I'd just like to gather some hints on how to get started.

Here is the problem statement:

Show that there is an unbounded linear operator from $L^{\infty}(\mathbb{R})$ to itself.

In other words, we want to exhibit a $T: L^{\infty}(\mathbb{R}) \to L^{\infty}(\mathbb{R})$ such that for any $C > 0$, there exists some $f \in L^{\infty}(\mathbb{R})$ with \begin{align*} \|T(f)\|_{L^{\infty}} > C\|f\|_{L^{\infty}} \end{align*} I was thinking to use the derivative operator, so that when we consider the functions $\sin(nx)$, the $L^{\infty}$ norm of their derivative can be made arbitrarily large while $\|\sin(nx)\|_{L^{\infty}} = 1$. The problem with this is that the derivative operator isn't defined on every element of $L^{\infty}$, for instance $1_{\mathbb{Q}}$ doesn't have a derivative but is bounded above by 1.

Any hints would be helpful.

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    Any infinite-dimensional Banach space admits an unbounded operator, which can be shown very easily using Zorn's Lemma. I'm assuming that the point here is to construct one without using the Axiom of Choice. – 2017-02-09

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Typically unbounded operators include those which are not necessarily defined on the whole space, in which case differentiation is an excellent example of an unbounded operator. On the other hand, perhaps you want an unbounded operator defined on the entirety of $L^\infty(\mathbb R)$.

Hint 1: Look for an (algebraic) splitting $L^\infty(\mathbb R) = E \oplus F$ for which $E$ is not a closed subspace; equivalently, the associated projection operator $\pi$ onto $E$ parallel to $F$ is unbounded Proving the equivalence is a nice exercise.

Hint 2: To find such a splitting $L^\infty(\mathbb R) = E \oplus F$, it suffices merely to find a nonclosed subspace $E \subset L^\infty(\mathbb R)$ (note: any Hamel basis for $E$ can be extended to a Hamel basis for all of $L^\infty(\mathbb R)$).

Hint 3: Consider the subspace of finite linear combinations of indicator functions of Borel subsets of $\mathbb R$.