1
$\begingroup$

Hi does anyone know how to show the result that if we have a relexive Banach space $X$ and a mapping $A: X \rightarrow X^{*}$ (not necessarily linear), which is strongly continuous, which means $$u_{n} \rightharpoonup u~~~\text{in }X\implies A(u_{n}) \rightarrow A(u)~~\text{ in }X^{*}$$ Then does it follow that $A$ is also bounded, in the sense that $A$ takes bounded sets to bounded sets?

Thanks for any help.

$\endgroup$
  • $\begingroup$ @DavidMitra No it's necessarily linear. I changed it to 'mapping'. $\endgroup$ – user103184 Dec 30 '14 at 12:14
  • $\begingroup$ @DavidMitra But does it follow trivially that if you have a bounded set $B$ in $X$ then $A(B)$ is bounded in $X^{*}$? $\endgroup$ – user103184 Dec 30 '14 at 12:25
  • $\begingroup$ @DavidMitra Sorry! I meant it is NOT necessarily linear. $\endgroup$ – user103184 Dec 30 '14 at 12:34
  • $\begingroup$ Ok, deleting comments... $\endgroup$ – David Mitra Dec 30 '14 at 12:35
  • 2
    $\begingroup$ Suppose $(u_n)$ is bounded with $(Au_n)$ unbounded. Passing to a subsequence, if necessary, we may suppose $\Vert Au_n\Vert>n$. Use the fact that $(u_n)$ has a weakly convergent subsequence to obtain a contradiction. $\endgroup$ – David Mitra Dec 30 '14 at 12:45
2
$\begingroup$

Thanks for responses in the comments. Is this then okay?

Proof: Assume there is some bounded set $B \subset X$, where $A(B)$ is unbounded in $X^{*}$. Then choose a sequence $\{ A(u_{n}) \}_{n \in \mathbb{N}} \subset A(B)$ such that $\| A(u_{n}) \| > n$. Note then that $\{ u_{n}\}_{n} \subset B$ is bounded in $X$, so by Kakutani's Theorem and Emerlein-Smulian Theorem it follows that there exists a subsequence $\{ u_{n_{k}} \}_{k}$ such that $u_{n_{k}} \rightharpoonup u$ in $X$. It follows then from the strong continuity assumption that $A(u_{n_{k}}) \rightarrow A(u)$, the sequence is therefore bounded. But we also have $\| A(u_{n_{k}}) \| \geq k$, which contradicts $\{ A(u_{n_{k}}) \}_{k}$ being bounded.

$\square$

$\endgroup$
  • 1
    $\begingroup$ $\| A(u_{n_k})\|\geq n_k$ with $n_k$ a strictly increasing sequence, but otherwise it looks correct. $\endgroup$ – Jonas Dahlbæk Dec 30 '14 at 14:04
  • 1
    $\begingroup$ Looks good!${}$ $\endgroup$ – David Mitra Dec 30 '14 at 14:33

Your Answer

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