Why if $T$ is not continuous, then for each $n\in N$, there exists $x_n\in X$ such that $||Tx_n||\ge n||x_n||$ If $X$ $Y$ are Banach space, $T$ is a linear operator between them.
I don't understand the following statement:
If $T$ is not continuous, then for each $n\in N$, there exists $x_n\in X$ such that $$||Tx_n||\ge n||x_n||$$.
Why is this true? Could someone kindly explain it? Thanks!
 A: By definition, to be continuous means there is a fixed $C>0$ s.t.
$$\Vert Tx\Vert\leq C\Vert x\Vert,\forall x\in X$$
So if $T$ is not continuous, then $\forall C>0$, $\exists x_C\in X$ (here $x_C$ depends on the choice of $C$) s.t.
$$\Vert Tx_C\Vert > C\Vert x_C\Vert$$
In particular, you can choose $C=n$ for any $n\in\mathbb{N}$, and in which case $x_n=x_C$. 
A: In addition to the other answers and comments here, I think it might be useful to quickly drop the proof here that the linear operator $T$ is continuous if there exists $C\in \mathbb{R}$ such that $\Vert Tx\Vert \leq C\Vert x\Vert$ for all $x\in X$. Here it is:
Assume we have $C \in \mathbb{R}$ so that
$$
\forall{x\in X\colon\ } \Vert Tx\Vert \leq C\Vert x\Vert.
$$
Our goal is to show continuity using the $\varepsilon$-$\delta$-criterion. Thus, we want to show
$$
\forall{x\in X\ } \forall{\varepsilon\in\mathbb{R}_{>0}\ } \exists{\delta\in\mathbb{R}\ } \forall{y\in X\colon \ } \Vert x-y\Vert < \delta \Longrightarrow \Vert Tx - Ty \Vert < \varepsilon\text{.}
$$
Firstly, we have for all $x,y\in X$ by using our assumption from the beginning and the linearity of $T$
$$
C\Vert x-y\Vert \geq \Vert T(x-y)\Vert = \Vert Tx - Ty\Vert\text{.}
$$
In particular, for all $\varepsilon\in\mathbb{R}_{>0}$ we get by choosing $\delta := \frac{\varepsilon}{C}$ that we have for all $x,y \in X$ with $\Vert x-y\Vert < \delta$ that
$$
\Vert Tx - Ty\Vert \leq C\Vert x-y\Vert < C\delta = \varepsilon\text{.}
$$
Consequently, $T$ is continuous. (Even uniformly continuous because $\delta$ didn't depend on $x$.)

So, as the others already said, if $T$ is not continuous, we get that no such constant $C$ can exist. Hence we always find for all $n\in\mathbb{N}$ an $x\in X$ such that $\Vert Tx\Vert \geq n\Vert x\Vert$ (because the constant can be chosen arbitrarily large).
A: It might be useful to point out that there is a standard technique at work here. Often one wants to prove or disprove an inequality with a multiplicative constant $C>0$: 
$$\tag{1}
A\le C B.
$$
The contrapositive of the statement 
I) "there exists $C>0$ such that for all values of $A$ and $B$ one has $A\le CB$" 
is
II) "for any constant $C>0$, however big, there exist at least one value of $A$ and one corresponding value of $B$ such that $A(C)>C\cdot B(C)$."
This last statement is true if and only if (1) does not hold. Usually one takes a sequential version of II) by making $C$ into a sequence $C_n\to \infty$: one gets that II) is equivalent to 
II-seq) "There exists a sequence $A_n$ and a sequence $B_n$ such that $A_n>C_n B_n$". 

In the case at hand the inequality one wants to disprove is $\|Tx\|\le C\|x\|$, which is a characterization of the continuity property of the operator $T$. Choosing $C_n=n$, the statement II-seq) becomes 
"There exists a sequence $x_n$ such that $\|Tx_n\|>n\|x_n\|$"
Thus, this statement is exactly the contrapositive of "$T$ is continuous".
