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Open mapping theorem says that bounded linear operator $T: X \to Y$ is an open mapping if $X$ and $Y$ are both Banach and $T$ is surjective. I am wondering what about unbounded linear operators? I guess it is not an open mapping but is there any counter example?

TH000
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1 Answers1

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From Taylor & Lay, Introduction to Functional Analysis

Theorem 5.5 (The Open Mapping Theorem): Let $X$ and $Y$ be normed linear spaces, and let $X$ be complete. Let $T$ be a closed linear operator with $\mathscr{D}(T)\subset X$ and $\mathscr{R}(T)$ a set of the second category in $Y$. Then $T$ is an open mapping; that is, if $W$ is an open set in $X$, then $T(\mathscr{D}(T)\cap W)$ is open in $Y$. Also, $\mathscr{R}(T)=Y$.

Disintegrating By Parts
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