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I am confused with a part of the proof for the following theorem.

Theorem : Let $E$ and $F$ be normed spaces. Let $T: E\rightarrow F$ be linear. Then $T$ is continuous if it is bounded.

Let $T$ be continuous, if it is bounded we are done so suppose it is unbounded. I am confused with the following statement:

Since $T$ is unbounded for every $n$ we can find and $x_n$ such that $\|Tx_n\|>n$ and $\|x_n\|=1$.

Part of the proof that I am confused with:

Since $T$ is unbounded it is clearly through that we can find arbitrarily large $n$ with $\|Tx_n\|>n$, but I do not see how we can simultaneously require $\|x_n\|=1$.


Additional notes on my confusing that may help when explaining this to me: I believe I am missing a common idea or trick used in functional analysis as I am also confused when trying to show that the following are equal $$\sup_{x\in E,\|x\|\leq 1}|f(x)|=\sup_{x\in E,\|x\|= 1}|f(x)|$$

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You are dealing with linear operators between normed spaces $X$ and $Y$ here. And we say that such an operator $T$ is bounded if there is some $M\geqslant0$ such that$$(\forall x\in X):\bigl\lVert T(x)\bigr\rVert\leqslant M\lVert x\rVert.$$In particular, $\lVert x\rVert=1\implies\bigl\lVert T(x)\bigr\rVert\leqslant M$.

It is not the same thing as a bounded function in general.

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    $\begingroup$ This is a good reminder to always make sure your definitions are correct. $\endgroup$ – T. Stark Jul 27 at 20:43
  • $\begingroup$ Yes I just to wait until I could accept it. $\endgroup$ – T. Stark Jul 27 at 21:03

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