If the image of a linear transformation of normed spaces is finite dimensional, is the map bounded? Let $V, W$ be normed spaces.  If $T: V \rightarrow W$ is such that $T(V)$ is finite dimensional, does it follow that $T$ is bounded?
Edit: This isn't a homework question, I'm just asking because I'm curious.  Good lord.  
Edit: I have edited the question and title.  I originally called a $T$ such as one in the hypothesis "finite rank," whereas in the literature a transformation of finite rank is by definition a bounded linear transformation whose image is finite dimensional.  
 A: There can be discontinuous linear functionals on an infinite-dimensional normed space. Such functions have one-dimensional ranges. You can use a few of these functions $\{ f_k \}_{k=1}^{n}$ with vectors $\{ x_k \}_{k=1}^{n}$ to define all kinds of discontinuous linear maps with finite-dimensional ranges:
$$
                Tx = f_1(x)x_1+f_2(x)x_2+\cdots+f_n(x)x_n.
$$
Finding a discontinuous linear function requires the axiom of choice. An easy method involves using the Axiom of Choice to find a Hamel basis $\{ e_{\alpha} \}$ for $X$, letting $\{ c_{\alpha} \}$ be an unbounded set of scalars, and defining a functional $f$ on this basis by
$$
                 f(e_{\alpha})=\frac{c_{\alpha}}{\|x_{\alpha}\|}.
$$
It's easy to check that $f$ is unbounded.
A: Yes.  You can think of this as a generalization of the fact that linear operators on finite vector spaces are bounded.
edit: nevermind, this was a misconception
A: Not sure why I'm being downvoted, but as indicated by Najib Idrissi and Vincent Boelens, the claim is false.  If $V$ is any infinite dimensional normed space, let $v_1, v_2, ...$ be a countably infinite collection of linearly independent elements (normalized to have norm $1$).  Extend $\{v_i\}$ to a normalized basis $\{v_i\} \cup \{ w_j : j \in X\}$, and define $F: V \rightarrow \mathbb{C}$ by $F(v_n) = n$ and $F(w_j) = 1$.  
The quotients $\frac{|F(v_n)|}{||v_n||} = n$ are unbounded, so it follows that $F$ is not bounded.  We used the axiom of choice, since this is equivalent to the fact that every vector space has a basis.  
