For a bounded linear operator T, if $||Tx_o||<\epsilon$, then what can we say the norm of $Tx$ for $x \in$ the epsilon ball around $x_o$

I am not completely sure if this is correct, but I have seen that for a bounded linear operator T, if $||Tx_o||<\epsilon$, then the norm of $Tx$ for $x \in$ the $\epsilon$ ball around $x_o$ should be less than $2\epsilon$. I tried using linearity and the triangle inequality but coulnd't reach the conclusion. If x is in the $\epsilon$ ball around $x_o$, we can express it as $x_o + x_e$ where $x_e$ has norm less than $\epsilon$. However, the triangle inequality $||Tx||=||Tx_o+Tx_e||\leq||Tx_o||+||Tx_e||$ does not give us $2\epsilon$ because we still need to know the bound of the norm of $||Tx_e||$

Solution 1:

It is just not true. For example if $T: X\to X$, $T(x) = 2x$. Then $\|Tx_0\|<\epsilon$ implies that $\| x_0\| <\epsilon/2$, now $y_0=x_0+ 1.5 x_0$ is in the $\epsilon$-ball of $x_0$, since $$ \| y_0 - x_0\| = 1.5 \|x_0\| < 0.75 \epsilon.$$

But $\| T(y_0)\|= 2\|y_0\| = 5\|x_0\| <2.5 \epsilon$. That is, if $\|x_0\|$ is close to $\epsilon/2$, $\|y_0\|> 2\epsilon$.