Why if \$T\$ is not continuous, then for each \$nin N\$, there exists \$x_nin X\$ such that \$||Tx_n||ge n||x_n||\$ Code Answer

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 \$nin N\$, there exists \$x_nin X\$ such that \$\$||Tx_n||ge n||x_n||\$\$.

Why is this true? Could someone kindly explain it? Thanks!

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 \$Cin mathbb{R}\$ such that \$Vert TxVert leq CVert xVert\$ for all \$xin X\$. Here it is:

Assume we have \$C in mathbb{R}\$ so that
\$\$
forall{xin Xcolon } Vert TxVert leq CVert xVert.
\$\$
Our goal is to show continuity using the \$varepsilon\$-\$delta\$-criterion. Thus, we want to show
\$\$
forall{xin X } forall{varepsiloninmathbb{R}_{>0} } exists{deltainmathbb{R} } forall{yin Xcolon } Vert x-yVert < delta Longrightarrow Vert Tx – Ty Vert < varepsilontext{.}
\$\$
Firstly, we have for all \$x,yin X\$ by using our assumption from the beginning and the linearity of \$T\$
\$\$
CVert x-yVert geq Vert T(x-y)Vert = Vert Tx – TyVerttext{.}
\$\$
In particular, for all \$varepsiloninmathbb{R}_{>0}\$ we get by choosing \$delta := frac{varepsilon}{C}\$ that we have for all \$x,y in X\$ with \$Vert x-yVert < delta\$ that
\$\$
Vert Tx – TyVert leq CVert x-yVert < Cdelta = varepsilontext{.}
\$\$
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 \$ninmathbb{N}\$ an \$xin X\$ such that \$Vert TxVert geq nVert xVert\$ (because the constant can be chosen arbitrarily large).