Unbounded operator bounded in a dense subset

Matthew's answer reminded me of a fact that makes this easy: if $X$ is a normed space (say, over $\mathbb{R}$) and $f : X \to \mathbb{R}$ is a linear functional, then its kernel $\ker f$ is either closed or dense in $X$, depending on whether or not $f$ is continuous (i.e. bounded). The proof is trivial: $\ker f$ is a subspace of $X$ of codimension 1. Its closure is a subspace that contains it, so must either be $\ker f$ or $X$. And of course, a linear functional is continuous iff its kernel is closed. This is Proposition III.5.2-3 in Conway's A Course in Functional Analysis.

So let $f$ be an unbounded linear functional on $X$ (which one can always construct as in Matthew's example), and take $D = \ker f$. $D$ is dense by the above fact, and $f$ is identically zero on $D$.

Let $X$ be the space of real polynomials, normed as functions in $C[a,b]$. Here we want $0 < a < b$ fixed. Now define $T \colon X \to X$ so that $T(x^n) = 0$ if $n$ is even and $T(x^n) = nx^n$ if $n$ is odd. Then $T$ is unbounded, but it vanishes on the set of even polynomials. That set is dense by the Müntz-Szász theorem. http://en.wikipedia.org/wiki/M%C3%BCntz%E2%80%93Sz%C3%A1sz_theorem

I think I have NOT used the Axiom of Choice.

Take a dense linear subspace $Z$ of $X$ such that $X/Z$ is infinite-dimension (algebraically). Let $x_1,x_2,\ldots$ be a sequence of elements of $X$ whose images in $X/Z$ are linearly indepdendent. We can define a linear map on the algebraic span on $Z$ and the $x_j$ which is zero on $Z$ and satisfies $\|Tx_n\|=n\|x_n\|$. Extend this to all $X$ by Zorn's lemma. Then $T$ is zero on the dense space $Z$ but unbounded on $X$.