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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?

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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.

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