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Let $T\in L(X,Y)$, where $X, Y$ are Banach spaces. How can I conclude from $\ker(T) = \{0\}$ and $\mathrm{im}(T)$ closed that $T$ is bounded below, e.g. there exists a $c>0$ such that $||Tx||\geq c ||x||$?

rschwieb
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samsons
  • 91

2 Answers2

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$im(T)$ is a closed subspace of a banach space so itself is $banach$. $T:X\to im(T)$ is a surjective bounded linear operator between banach spaces, so open mapping theorem implies the result.

Hui Yu
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N(T) =0 => T is injective. R(T) is closed in Y and Y is banach space => R(T) is Banach. Now ,apply bounded inverse theorem on T :X ----> R(T) where T is bijective. Then, inverse of T will be also bounded. Thus T will be bounded below because T is injective and inverse of T is bounded.