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Assuming $E$ is a Banach space, consider linear operator $T:E\rightarrow E^*$, then for each Cauchy sequence in $E$, is it guaranteed that sequence $Tx_n$ is also Cauchy? i.e., exits $f\in E^*$ such that $Tx_n\rightarrow f$ ?

If we add new condition, say, $\langle Tx,x\rangle\geq 0, \forall x \in E$ ?

I know linear continuous operator can map cauchy to cauchy, but in my situation, it seems such $f$ doesn't exist.

original problem:

Let $E$ be a Banach space and let $T:E\rightarrow E^*$ be a linear operator satisfying $$\langle Tx,x\rangle\geq 0, \forall x \in E$$ Prove that $T$ is bounded operator. (From: functional analysis,...)

A possible counterexample: $Tx=\infty$ and let $E=\mathbb{R}$, then $T$ is not linear?

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    $\begingroup$ Is map $T$ continuous? What does the new condition mean in a Banach space? The dual space is always complete. $\endgroup$ – copper.hat Oct 18 '16 at 19:21
  • $\begingroup$ @copper.hat actually in the book, we want to prove $T$ is bounded(which is equivalent to continuous); by the way I also confusing about the inner product in the new condition. I will put the original problem here. $\endgroup$ – DuFong Oct 18 '16 at 19:25
  • $\begingroup$ If it is not continuous, there is some $x_n$ such that $x_n \to 0$ and $|T x_n| = 1$. $\endgroup$ – copper.hat Oct 18 '16 at 19:28
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There are discontinuous linear operators from $E$ to $E^\ast$ if and only if $E$ is infinite-dimensional (since $E$ and $E^\ast$ are complete, a map is continuous if and only if it maps Cauchy sequences to Cauchy sequences).

Regarding your original problem, it suffices to show that $T$ is closed. Then the closed graph theorem implies that $T$ is continuous.

Let $(x_n)$ be a sequence in $E$ such that $x_n\to 0$ and $Tx_n\to y$. For all $\alpha>0$, $z\in E$ we have \begin{align*} 0&\leq \langle T(x_n+\alpha z),x_n+\alpha z\rangle\\ &=\langle Tx_n,x_n\rangle+\alpha\langle Tz,x_n\rangle+\alpha\langle T x_n,z\rangle+\alpha^2\langle Tz,z\rangle\\ &\to \alpha\langle y,z\rangle+\alpha^2\langle Tz,z\rangle,\;n\to\infty. \end{align*} Dividing by $\alpha$ and letting $\alpha\to 0$ we obtain $$ 0\leq \langle y,z\rangle. $$ Since $z\in E$ was arbitrary, it follows that $y=0$. Thus, $T$ is closed.

This proof is taken from Kato's Perturbation Theory for Linear Operators. In this book the result is stated for operators on Hilbert spaces, but it carries over almost verbatim to your setting.

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  • $\begingroup$ I read this proof, and that's why I ask my question. I am not sure why we can assume $Tx_n\rightarrow y$, where is the $y$? $\endgroup$ – DuFong Oct 19 '16 at 12:49
  • $\begingroup$ $y$ is an element of $E^\ast$. As I said, we only have to prove that $T$ is closed due to the closed graph theorem. $\endgroup$ – MaoWao Oct 19 '16 at 13:06
  • $\begingroup$ I know. but do we need to prove the existence of $y$,cause no theorem say that $Tx_n$ convergence $\endgroup$ – DuFong Oct 19 '16 at 13:17
  • $\begingroup$ Yes, but I ASSUME that $(Tx_n)_n$ converges. $\endgroup$ – MaoWao Oct 19 '16 at 13:57
  • $\begingroup$ according to the condition of the original problem, I don't think we can assume the $Tx_n$ converges, do we? $\endgroup$ – DuFong Oct 19 '16 at 15:19

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