Continuity of linear transformation

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 $\beta_1(0) = \{x \in X / ||x|| \leq 1 \}$

• What is your answer exactly? – GSF Feb 18 '17 at 1:14
• I know to prove (->) when $x_0 = 0$ but when $x_0$ is not zero, i dont know how to proceed, i need help... :/ – jfruizc273 Feb 18 '17 at 1:20
• Hope my answer below worked for you. – GSF Feb 18 '17 at 1:39
• Thank you very much is a very good answer. :) – jfruizc273 Feb 18 '17 at 1:41

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.