# There is a bounded linear transformation $T$ such that $\|Tx\|\geq b\|x\|$, show that $T$ is $1-1$ mapping.

Let $T$ be a bounded linear transformation from a normed space $X$ onto a normed space $Y$. If there is a positive $b$ such that $\|Tx\|\geq b\|x\|$ for all $x$ in $X$, show that $T$ is a one-one mapping.

You don't need surjectivity. Being bounded below makes it automaticaly one-to-one, because the inequality $\|Tx\|\geq b\|x\|$ gives you that if $Tx=0$, then $x=0$.