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I am working on Problem 8, Chapter 6, in Luenberger's Optimization by Vector Space Methods. It states:

"Show that a linear transformation mapping one Banach space into another is bounded if and only if its nullspace is closed."

I am having a bit of trouble with the converse. In particular, if we let $f:X \rightarrow Y$ be a linear transformation, Luenberger doesn't assume that either $X$ or $Y$ be finite-dimensional. Do you have any idea on how to proceed? I have thought (without success) of considering the quotient space $\hat{X}/\ker f$

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This is not true in general:

Let $X$ be any infinite-dimensional banach space with (Hamel) basis $B$ and let $\{b_n\}_{n\in\ \mathbb N}$ be a countable subset of $B$. Then define $T:X\to X$ on the basis $B$ by $T(b_n)=nb_n$ and $T(b)=b$ for $b\in B\setminus \{b_n\}_{n\in\ \mathbb N}$ and extend linearly. Then $T$ is an unbounded operator, with $\ker T=\{0\}$ closed.

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  • $\begingroup$ I thought so as well, in the first moment -- but then I realized, is it indeed a map from a banach space into a banach space? Seems like sometime the image is not well-defined (as an element of the target space). $\endgroup$ Commented Jun 6, 2020 at 23:08
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    $\begingroup$ @PeterFranek: Sure it is well defined. Since $B$ is a Hamel basis, every vector in $X$ can be uniquely written as a finite linear combination of the elements of $B$. $\endgroup$ Commented Jun 6, 2020 at 23:11
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    $\begingroup$ Thanks a lot! I will accept this answer. I have one additional question however: is it possible to construct the $B$ in this counterexample without using the axiom of choice? $\endgroup$
    – charlus
    Commented Jun 6, 2020 at 23:14
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    $\begingroup$ It is not possible using just ZF, there are models of ZF + negation of AC in which every such linear map is continuous, see here $\endgroup$ Commented Jun 6, 2020 at 23:19
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    $\begingroup$ Very interesting answer, and very interesting link. Thank you all. $\endgroup$
    – charlus
    Commented Jun 6, 2020 at 23:22

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