Question on operator norm $\vert \vert \cdot \vert \vert$ I have just been introduced to the operator norm $\vert \vert \cdot \vert \vert$ on $L(X,Y)$ where $X, Y$ are respective normed spaces and any $T \in L(X,Y)$ is simply a linear map from $X$ to $Y$. 
The norm is defined as $\vert \vert T\vert\vert:=\sup_{\vert\vert x\vert \vert \leq 1}\vert\vert T x\vert\vert$
I have the following question (coming from an exemplarly proof I was reading): 
Say we have a Cauchy sequence $(T_{n})_{n}\subseteq L(X,Y)$, then (without any further explanation) I am told that $(T_{n}x)_{n}$ is also a cauchy sequence, for any $x \in X$. 
I am confused by this statement, as we are looking at a Cauchy sequence with respect to the operator norm $\sup_{\vert\vert x\vert \vert \leq 1}\vert\vert T x\vert\vert$, so $x$ is bounded in some kind of unit circle. But, the sequence $(T_{n}x)_{n}$ can have $x \in X$ where $\vert \vert x \vert \vert>1$. I do not see how we can immediately assume that $(T_{n}x)_{n}$ is indeed a Cauchy sequence.  
 A: Elaborating the hint of  Reveillark: For $x \in X$ with $\Vert x \Vert = 1$ you clearly have $$\Vert Tx \Vert \leq \sup_{\Vert x \Vert \leq 1} \Vert Tx \Vert = \Vert T \Vert = \Vert T \Vert \Vert x \Vert. $$
For $x = 0$ the inequality holds trivially. So let $x \neq 0$. Then by the above inequality you have for $y = x / \Vert x \Vert$ that
$$\frac{1}{\Vert x \Vert} \Vert T x \Vert = \Vert T y \Vert \leq \Vert T \Vert \Vert y \Vert = \Vert T \Vert \frac{\Vert x \Vert}{\Vert x \Vert} = \Vert T \Vert$$
since $\Vert y \Vert = 1$. Multiplying the inequality with $\Vert x \Vert$ yields that $\Vert T x \Vert \leq \Vert T \Vert \Vert x \Vert$ for all $x \in X$. 
Now suppose that $(T_n)_{n \in \mathbb N}$ is a Cauchy sequence in $\mathcal L(X, Y)$. Then for all $\varepsilon > 0$ there is $N \in \mathbb N$ such that $\Vert T_n - T_m \Vert < \varepsilon$ for all $n, m \geq N$. Let $x \in X$. Then the above inequality yields
$$\Vert T_n x - T_m x\Vert = \Vert (T_n - T_m)x \Vert \leq \Vert T_n - T_m \Vert \Vert x \Vert < \varepsilon \Vert x \Vert$$
for all $n, m \geq N$ which shows that $(T_n x)_{n \in \mathbb N}$ is indeed a Cauchy sequence in $X$ for any $x \in X$. I hope the arguments got clear :)
