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let $T: (X,||\cdot||_X) \to (Y, ||\cdot||_Y)$ be a linear transformation then

$T$ is continuous in $x_0 \in X$ iff $T$ is bounded in $\beta_1(0) = \{x \in X / ||x|| \leq 1 \}$

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  • $\begingroup$ What is your answer exactly? $\endgroup$ – GSF Feb 18 '17 at 1:14
  • $\begingroup$ I know to prove (->) when $x_0 = 0$ but when $x_0 $ is not zero, i dont know how to proceed, i need help... :/ $\endgroup$ – jfruizc273 Feb 18 '17 at 1:20
  • $\begingroup$ Hope my answer below worked for you. $\endgroup$ – GSF Feb 18 '17 at 1:39
  • $\begingroup$ Thank you very much is a very good answer. :) $\endgroup$ – jfruizc273 Feb 18 '17 at 1:41
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If $T$ is bounded in the unit ball, there exists an $M>0$ such that $\|Tx\|_Y\leq M$ for all $x$ in the unit ball.

Now, for each $0<\epsilon <1 $, $\|T(\frac{x}{\|x\|}(1-\epsilon))\|_Y\leq M$, because $\frac{x}{\|x\|}(1-\epsilon)$ lies in the unit ball, therefore, using the properties of $\|\cdot\|_Y$ and of linear transormations, $\|Tx\|_Y\leq \frac{M}{1-\epsilon}\|x\|_X$. Making $\epsilon\rightarrow 0$, $\|Tx\|_Y\leq M\|x\|_X$.

Finally, if we fix $x\in X$, for any $y\in X$ we have $$\|Tx-Ty\|_Y=\|T(x-y)\|_Y\leq M\|x-y\|_Y$$ hence $T$ is continous at any $y\in X$ and you have the implication you wanted.

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  • $\begingroup$ Great answer! Any sense on if, and how, this (or something similar) generalizes beyond linear transformations? $\endgroup$ – David Feb 18 '17 at 3:14

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