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

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

Answer

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).

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