# When is the image of a linear operator closed?

Let $X$, $Y$ be Banach spaces. Let $T:X\rightarrow Y$ be a bounded linear operator.

Under what circumstances is the image of $T$ closed in $Y$ (except finite-dimensional image).

In particular, I wonder, under which assumptions $T:X\rightarrow T(X)$ is a bounded linear bijection between Banach spaces, so it is at least an isomorphism onto its image by bounded inverse theorem.

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Probably the most useful criterion is en.wikipedia.org/wiki/Closed_range_theorem but that may not be exactly what you're looking for. It also is a nice exercise to show that an operator whose image has finite codimension has closed range (this has some use in Fredholm theory). –  t.b. Mar 9 '11 at 23:10
An answer to your last question is that a bounded linear map $T$ between Banach spaces is injective with closed range if and only if it is bounded below, meaning that there is a constant $c>0$ such that for all $x$ in the domain, $\|Tx\|\geq c\|x\|$. You can read more about this in Chapter 2 of An invitation to operator theory by Abramovich and Aliprantis.