Consider $X$ and $Y$ normed spaces, $T: X \rightarrow Y$. Suppose that if a succession $\{x_n\}$ in $X$ such as $\{x_n\}\rightarrow 0$, then the following succession $\{T(x_n)\}$ is bounded. Prove that this operator is continuous.
To prove this, I am trying to see that the operator is bounded in the unit ball. To do that, I use that the unit ball is closed in $X$, so for all $x\in X$ I can find a succession $\{x_n\}\rightarrow x$. Now I can take $\{x_n-x\}\rightarrow 0$ and apply the operator, so $\{T(x_n-x)\}$ is bounded: $||T(x_n-x)||\leq M\in\mathbb{R}^+$.
I am not sure if that is enough. Any idea how to proceed?
If $T$ is not bounded on the unit ball then we can find $y_n$ such that $\|y_n\| \leq 1$ but $\|Ty_n\| >n^{2}$. Let $x_n =\frac 1 n y_n$. Then $x_n \to 0$ but $\|Tx_n\| >n$ for all $n$.
Rest of the proof:
Once you know that $T$ is bounded on the unit ball, you get a constant $C$ such that $\|Tx\| \leq C$ whenever $\|x\|\leq 1$. This implies that $\|Tx\| \leq C \|x\|$ for all $x$ . (Apply previous inequality to $\frac x {\|x\|}$). Hence $\|Tx-Ty\| =\|T(x-y)\|\leq C \|x-y\|$ for all $x,y$ and continuity of $T$ is immediate from this.