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?
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.