# Unbounded Operator of Finite Rank

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?

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 $v$?

• Notice that $\operatorname{im}(T)$ is a finite-dimensional Banach space and $\| v \|' := |\alpha_1| + \cdots + |\alpha_n|$ defines another norm on $\operatorname{im}(T)$. Now you can utilize the fact that any two norms on a finite-dimensional Banach space are equivalent. – Sangchul Lee Oct 22 '15 at 9:55
• On any infinite dimensional space, one can construct an unbounded linear functional. For instance, take a countably infinite linearly independent set $\{e_n\}$, map $e_n \to n$ and extend linearly. – Prahlad Vaidyanathan Oct 22 '15 at 10:13

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 $n\in\mathbb{N}$.

• Thank you. How about an example where $Y$ is infinite dimensional? – PPR Oct 22 '15 at 10:18
• @PPR: you can just use the same example if $Y$ has larger dimension: fix $y \in Y$ and then define $Te_n = ny$, etc. – bakula Oct 22 '15 at 10:21
• In fact, $Y$ has to have at least dimension $1$. – gerw Oct 22 '15 at 10:37