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I'm using the definition of a bounded from below linear operator as a linear operator $X\overset{T}{{\to}}Y$, where $X$ and $Y$ are Banach spaces and the operator satisfies: for all $x \in X$, $\lVert T(x) \rVert \geq C \lVert x \rVert$, for some constant $C \gt 0$.

In this question i will refer only to such an operator in a Banach space $B$, $ T \colon B{{\to}}B$.

Note that I don't assume it to be continuous. With only these conditions, can we guarantee that the image of $T$ is closed in $B$? (Which is equivalent to asking if these assumptions imply continuity of the operator, by the open mapping theorem applied to the inverse of the operator, which exists since $\lVert T(x) \rVert = 0 \Rightarrow \lVert x \rVert = 0$).

I have tried to prove that it is closed with only these conditions and checked out other posts that have similar questions, but they don't seem to answer this one, at least in an immediate way. At this point, I believe that it is not generally true, but I also haven't been able to write a counter-example.

I appreciate any help and thank you in advance!

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  • $\begingroup$ I suspect that, if false, this might be a very difficult question. The reason is that the only tool we have to exhibit discontinuous linear maps defined on Banach spaces is the axiom of choice, and I suspect it must be very difficult to control for boundedness below. $\endgroup$
    – Ruy
    Dec 16, 2020 at 13:16
  • $\begingroup$ Yeah, I also suspect that it is difficult to prove (if possible) that it is wrong or even that it is true, but I could be wrong. $\endgroup$
    – Daàvid
    Dec 16, 2020 at 13:53
  • $\begingroup$ @Ruy Is that true (needing full choice vs say countable which most people will accept)? Because given a separable Banach space, the map given by $Te_n = n e_n$ and extended linearly is discontinuous. Edit: oh lol I just saw your answer. Nevermind :D $\endgroup$ Dec 16, 2020 at 15:02
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    $\begingroup$ @Cameron, regardless of my answer I think you do have a point, but personally I am as scared of countable choice as I am of its full version (-: $\endgroup$
    – Ruy
    Dec 16, 2020 at 15:07
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    $\begingroup$ The ambiguity was intended! $\endgroup$
    – Ruy
    Dec 16, 2020 at 15:10

1 Answer 1

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Reardless of my comment above, the result is false, and it is in fact not so difficult to give a counter example.

Let $H$ be a separable, infinite dimensional Hilbert space with basis $$ E=\{e_n\}_{n\in \mathbb N}. $$

Consider a Hamel basis for $H$ containing the linearly independent set $E$, which we will write as the disjoint union $$ E\cup F. $$

Defining $$ N = \{(e_n, ne_n):n\in \mathbb N\}, $$ we leave for the reader to verify that the subset $C$ of $H\times H$ defined below is linearly independent: $$ C = (0\times E) \cup (0\times F) \cup N. $$

Next consider a Hamel basis $ D$ of $H\times H$ containing $ C$, which we write as the disjoint union $$ D =\underbrace{(0\times E) \cup (0\times F) \cup N}_{ C} \ \cup \ M, $$ and define $$ K= \text{span} (N\cup M). $$ where we insist that the span refers to the linear span, no closure. It is then easy to see that $$ H\times H = (0\times H) \oplus K, $$ where "$\oplus $" is to be interpreted as algebraic direct sum, as opposed to orthogonal direct sum.

For that reason, the canonical projection $\pi _1: H\times H\to H\times 0$, once restricted to $K$ becomes a bijection, and hence we may speak of its inverse $$ \pi _1^{-1}: H\times 0 \to K\subseteq H\times H. $$

Being the inverse of a bounded bijective map, it is clear that $\pi _1^{-1}$ is bounded below, but it is certainly discontinuous since $$ \|\pi _1^{-1}(e_n)\| = \|(e_n, ne_n)\| = \sqrt{1+n^2}. $$

As noted by the OP, the range of $\pi _1^{-1}$ cannot be closed.

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  • $\begingroup$ Why does $\pi_1$ become bijective if we restrict it to K? $\endgroup$
    – Daàvid
    Dec 16, 2020 at 18:02
  • $\begingroup$ Injective because the kernel of $\pi_1$, namely $0\times H$, has trivial intersection with $K$. Surjective because you may write any $(x,0)$ as $(0,y)+(k_1,k_2)$, with $(k_1,k_2)\in K$, and then necessarily $k_1=x$, so $\pi_1(k_1,k_2) = (x,0)$. $\endgroup$
    – Ruy
    Dec 17, 2020 at 0:17
  • $\begingroup$ I think the same proof works if $H$ is any infinite dimensional Banach space and $\{e_n\}_n$ is any linear independent set. $\endgroup$
    – Ruy
    Dec 17, 2020 at 0:41
  • $\begingroup$ Yeah, I think I got! $\endgroup$
    – Daàvid
    Dec 17, 2020 at 8:29
  • $\begingroup$ I appreciate your help. Thanks a lot! $\endgroup$
    – Daàvid
    Dec 17, 2020 at 8:29

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