If for every $x_n$ such that $x_n \rightarrow x$, there exists a $x_{n_k}$ such that $Tx_{n_k} = Tx$, is $T$ continuous?

Let $X$ and $Y$ be Banach spaces and $T$ be the (possibly nonlinear) map $T\colon X \rightarrow Y$. $T$ is continuous if for every $x_n \in X$ such that $x_n \rightarrow x$, then $Tx_n \rightarrow Tx$. Is $T$ also continuous if instead, for every $x_n \in X$ such that $x_n \rightarrow x$, then there exists a subsequence $x_{n_k}$ such that $Tx_{n_k} = Tx$?

• Yes. This follows from the fact that if $(x_n)$ converges to $x$ and $(Tx_n)$ does not converge to $Tx$, then there is an $\epsilon>0$ and a subsequence $(x_{n_k})$ of $(x_n)$ such that $\Vert Tx_{n_k}-Tx\Vert>\epsilon$ for all $k$. That this fact holds follows from the definition of convergence. – David Mitra May 23 '13 at 20:53
• @DavidMitra, that one subsequence is bounded away from $Tx$ does not exclude that another subsequence could approach $Tx$. – Christopher A. Wong May 23 '13 at 21:01
• @ChristopherA.Wong Yes. But no subsequence of $(Tx_{n_k})$ could converge to $Tx$. – David Mitra May 23 '13 at 21:02
• Ah, right, of course. – Christopher A. Wong May 23 '13 at 21:07

I misunderstood the question first. Indeed the answer is yes. If $Tx_n$ does not go to $Tx$ one can choose subsequence $x_{n_k}$ that is bounded away from $Tx,$ e.g. $\|Tx_{n_k}-Tx\|\ge\varepsilon.$ This leads to a contradiction.