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Let $X,Y$ be normed spaces and $T:X\to Y$ be a linear map.

I need to show that $T$ is continuous $\iff$ for all $x_n\to 0$ in $X$ , $sup_n||Tx_n|| \lt \infty$.

If $T$ is continuous then if $x_n\to 0$ then $T(x_n) \to T(0)$ and in particular $sup ||T(x_n)|| $ is finite.

Im not sure how to do the other direction.

Any ideas?

Thanks for helping.

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If $T$ is not continuous, then the set$$\{T(x)\,|\,\lVert x\rVert=1\}$$is unbounded. So, there is a sequence $(x_n)_{n\in\mathbb N}$ of unitary vectors such that $(\forall n\in\mathbb{N}):\bigl\lVert T(x_n)\bigr\rVert\geqslant n$. So, consider the sequence$$\left(\frac{x_n}{\sqrt n}\right)_{n\in\mathbb N}.$$

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  • $\begingroup$ Nice! and we have that sequence because we can take $x_n$ s.t $||x_n|| =1$ and $||T(x_n)|| \to \infty$ (by assumption ) and then take a sub sequence that satisfy $||y_n|| \ge n$ , right? $\endgroup$ – user335501 Feb 9 at 20:13
  • $\begingroup$ Since the set $\{T(x)\,|\,\lVert x\rVert=1\}$ is unbounded, then, for each natural $n$, there is a vector $x_n$ with norm $1$ such that $\bigl\lVert T(x_n)\bigr\rVert\geqslant n$. $\endgroup$ – José Carlos Santos Feb 9 at 20:50
  • $\begingroup$ Right ! thanks! $\endgroup$ – user335501 Feb 9 at 20:52

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