# Bounded below operators.

Suppose $X$ and $Y$ are Banach spaces and $T$ is a continuous operator between them. Suppose for any sequence $(x_n)_{n=1}^{\infty}$ with $\|x_n\|=1$, $\|Tx_n\| \nrightarrow 0$. Is it true $T$ is that $T$ must be bounded below?

Bounded below means that there is some positive constant $C$ such that $\|Tx\| \geq C \|x\|$.

• Do you want to say $lim_n\|T(x_n)\|\neq 0$ ? – Tsemo Aristide Dec 14 '17 at 1:11
• Yes, that is correct. – Keith Dec 14 '17 at 1:11

If it is not bounded below, for every $n$ there is $y_n$ such that $\|T y_n\| < \|y_n\|/n$. Take $x_n = y_n/\|y_n\|$.
Suppose that $T$ is not bounded below, for every $n$, there exists $y_n\neq 0$ such that $\|T(y_n)\|<{1\over n}\|y_n\|$, write $x_n={{y_n}\over{\|y_n\|}}$, $\|x_n\|=1$ and $\|T(x_n)\|=\|T({{y_n}\over{\|y_n\|}})\|={1\over{\|y_n\|}}\|T(y_n)\|\leq {1\over n}$. Contradiction.