1
$\begingroup$

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 $v$?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Sangchul Lee Oct 22 '15 at 9:55
  • $\begingroup$ 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. $\endgroup$ – Prahlad Vaidyanathan Oct 22 '15 at 10:13
3
$\begingroup$

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 $n\in\mathbb{N}$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.