1
$\begingroup$

I'm confused by the questions Discontinuous linear functional and Example of an unbounded operator which ask about unbounded linear functionals/operators on Banach spaces.

I don't understand how these can even exist.

Let $X$ be a Banach space. If $T$ is unbounded, then there exists a sequence $x_i \in X$ such that $\|x_i\|=1$ but $T(x_i)>i^3.$ Then we can let $x= \sum_i \frac{x_i}{i^2}.$ This is an element of $X$ by completeness but $T(x)$ is infinite. Hence $T$ isn't defined on $x.$

$\endgroup$
  • 4
    $\begingroup$ Unbounded operators are not continuous. $\endgroup$ – Daniel Fischer Jul 15 '15 at 20:43
  • 3
    $\begingroup$ In case you didn't catch the point to Daniel's comment: When you say $Tx$ is infinite you're assuming that $T$ of that sum is the sum of $T$ of the individual terms. There's no reason that should be if $T$ is not continuous. $\endgroup$ – David C. Ullrich Jul 15 '15 at 20:46
1
$\begingroup$

As the commenters said, your argument is flawed in the part where you conclude that (using notation $s_n=x_1+\dots+x_n$) $$s_n\to x \text{ and } Ts_n\to \infty \overset{?}{\implies} Tx=\infty $$

You did prove something, however: your argument shows that an unbounded operator must be discontinuous. (This isn't obvious; in fact, on an incomplete normed space one can have an unbounded continuous operator, such as $(x_n)\mapsto (nx_n)$ on the space of sequences that are eventually zero, equipped with any $\ell^p$ norm)

$\endgroup$
  • $\begingroup$ I think the last sentence is not true: A linear operator is bounded if and only if it is continuous. $\endgroup$ – gerw Feb 28 '18 at 7:55

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.