Let $X,Y$ be normed vector spaces, X infinite dimensional. Can I find a linear map $T:X\rightarrow Y$ and a subset D of X such that D is dense in X, T is bounded in D (i.e. $\sup _{x\in D, x \neq 0} \frac{\Tx\}{\x\}<\infty $ but T is not bounded in X?
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.5.23 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 MüntzSzász 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 infinitedimension (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 $X$. 


X is infinitedimensional, so we can find $(e_n)$ a linearly independent sequence in X; let X' be the span. By rescaling, we can assume that $\e_n\=1$ for each n. Define $T:X'\rightarrow \mathbb R$ (or $\mathbb C$, or embed into Y if you wish) by $T(e_n) = n$ for each n. Clearly T is unbounded on X'. For each finite sum $x=\sum_{n=1}^N x_n e_n$ and $\epsilon>0$, we can choose $a\in\mathbb R$ and $m$ very large with $a<\epsilon$ and $T(x) = am$. Set $y=x+ae_m$, so $T(y)=T(x)+am=0$ and also $\xy\ = a<\epsilon$. Let D' be the collection of all such $y$; as such $x$ exhaust X', we certainly have that D' is dense in X'. Use Zorn to extend $E=\{e_n\}$ to $E'$ a basis of X. Extend T to X by setting $T(x)=0$ for $x\in E' \setminus E$. Let $D = D' + \text{span}(E'\setminus E)$, so D is dense in X, and T is bounded on D; actually T vanishes on D. Now, D is certainly not a subspace: if you want that as well, I don't know! 


This is just to make Nate Eldridge's answer selfcontained. For any normed vector space $V$ and any $r > 0$, write $V_r$ for the open ball of radius $r$ and center $0$ in $V$. Let $X$ be a normed vector space, $Y$ a closed space, $Z$ the quotient, $\pi$ the canonical projection, $\tau$ the quotient topology, $\nu$ the topology on $Z$ induced by the quotient norm $$\pi(x):=\inf_{y\in Y}x+y.$$ (It's easy to see that this is a norm.) We claim $\tau=\nu$. Both topologies are translation invariant. The set $\{\pi(X_r)\ \ r > 0\}$ is a basis for the $\tau$neighborhoods of $0$ in $Z$. The set $\{(Z_r)\ \ r > 0\}$ is a basis for the $\nu$neighborhoods of $0$ in $Z$. As $\pi(X_r)=Z_r$, we're done. 


No. Bounded means continuous, for linear operators. If a linear operator is bounded (continuous) on a dense subset, then you can extend it continuously to the whole space, which means it is bounded. 

