Can one find an example of an unbounded operator on a Banach space whose rank is finite?
In particular, let $X, Y$ be two Banach spaces and and $T:X\to Y$ be a linear map between them such that $T(X)\subseteq Y$ is of finite dimension. Is it possible that $T$ is unbounded?
Attempt at answer:
No:
$||T||\equiv \sup(\{||Tv||\,|\,||v||\leq1\})$. Let $T(X)=span(v_1,\dots,v_n)$ for some $(v_i)_{i=1}^n$ fixed vectors in $Y$. Then $||Tv||=||\sum_{i=1}^n\alpha_i v_i||\leq\sum_{i=1}^n|\alpha_i|||v_i||\leq(\sum_{i=1}^n|\alpha_i|)\max(\{||v_i||\,|\,i\in\{1,\dots,n\}\})$ for some $(\alpha_i)_{i=1}^n$. However, I am not sure how to show this is finite. Somehow one would have to get an estimate on $(\sum_{i=1}^n|\alpha_i|)$ using the fact that $||v||\leq1$. But how to relate $(\alpha_i)_i$ to $v$?
The answer is yes (as long as $X$ has infinite dimension):
In fact, one can find such an operator for any $Y$.
For example, if $X$ is a complex Banach space, we can take $Y = \mathbb{C}$. The problem then reduces to finding an unbounded functional (which has finite rank, of course); such functionals exist iff $X$ is infinite-dimensional.
To construct an unbounded functional, pick an infinite independent set in $X$, say $\{e_1, e_2, e_3, \dots\}$ with $\lVert e_n \rVert = 1, \forall n\in \mathbb{N}$.
Define $Te_n = n, \forall n\in \mathbb{N}$. This defines $T$ on Span$\{e_1, e_2, e_3, \dots\}$.
Now complete the set $\{e_1, e_2, e_3, \dots\}$ to a basis and define $T$ to be $0$ on other basis elements.
We have thus defined a linear functional $T$ on $X$; it is easy to check that it is not bounded.
Namely, since $\lVert e_n \rVert = 1$ and $Te_n = n$, we have $\lVert T \rVert \geq n$, and this holds for every $n\in\mathbb{N}$.