# Conditions for bounded below Operator being continuous.

Let $$X$$ and $$Y$$ be Banach spaces, $$T: X\mapsto Y$$, be a bounded below operator, meaning $$\|Tx\|\geq m\|x\|$$. Under what conditions is $$T$$ continuous (bounded from above)? Here it was stated that it is implied directly, but how can one prove that $$\operatorname{im} T$$ is closed in $$Y$$ as it was used in that answer? Am I overlooking something? Or are there details missing?

## 1 Answer

Let $$X$$ be an infinite-dimensional Banach space and $$(a_i)_{i\in I}$$ an algebraic basis of $$X$$. Define $$T:X\to \ell^1(I)$$ via the linear extension of $$T(a_i)=\|a_i\| i$$ (here $$i$$ denotes the $$\ell^1$$-Basis element). Note that $$\left\|T\left(\sum_i v_i a_i\right)\right\|_{\ell^1(I)} = \left\|\sum_i v_i\|a_i\|_X\ i\right\|_{\ell^1(I)}=\sum_i |v_i|\, \|a_i\|_X ≥ \left\|\sum_i v_i a_i\right\|_X$$ and thus $$T$$ is bounded from below.

However this map has no chance of being continuous, and the image isn't closed, it appears to me the claim is wrong.