0

I keep forgetting it, so...

Given Banach spaces $X$ and $Y$.

Then it is wrong: $$\dim\mathcal{R}F<\infty:\quad F\in\mathcal{L}(X,Y)\implies F\in\mathcal{C}(X,Y)$$

Can I construct such?

C-star-W-star
  • 16,275
  • 1
    Any unbounded linear functional is a counterexample. (If you want an example for arbitrary $Y$: Say $\lambda$ is an unbounded functional on $X$, let $y\in Y$, $y\ne0$ and let $Fx=y\lambda x$.) – David C. Ullrich Aug 03 '15 at 17:51
  • @DavidC.Ullrich: Yes, I should have said: Are there nice explicit counterexamples? As far as I know there exist some. But the proof I know shows there existence by a cardinality argument exploiting axiom of choice. – C-star-W-star Aug 03 '15 at 18:05
  • Can't be answered precisely without a precise definition of "explicit". But ignoring that: No. There are no explicit examples of an unbounded operator linear operator on a Banach space. (Here's something that seems to be a fact, although I can't say it's a theorem without that definition of "explicit": If one write down an "explicit" operator, and if it's actually defined on the whole Banach space, then it seems that the Closed Graph Theorem is always enough to show it's bounded...) – David C. Ullrich Aug 03 '15 at 18:10
  • @DavidC.Ullrich: Here the older question with construction: Discontinuity – C-star-W-star Aug 03 '15 at 18:11
  • @DavidC.Ullrich By explicit I mean up to Hamel basis for a special Banach space like $\ell^2(\mathbb{N})$ and a functional given like $\lambda(\delta_n)=n$ and extended via Hamel basis. Wait that's it, isn't it? :D – C-star-W-star Aug 03 '15 at 18:12
  • @DavidC.Ullrich: Are there Banach spaces where one knows an explicit Hamel basis? I mean similarly to: For nonseparable Hilbert spaces one cannot construct an ONB, in general. However, there are nonseparable Hilbert spaces where one can still find an explicite ONB like $\ell^2(\mathbb{R})$. – C-star-W-star Aug 03 '15 at 18:16
  • 1
    (Of course we're only talking about the infinite dimensional case.) No, there are no Banach spaces with a Hamel basis known explicitly. An orthonormal basis is not a Hamel basis. – David C. Ullrich Aug 03 '15 at 18:20
  • @DavidC.Ullrich: Sure, ONB and Hamel basis are different concepts. Really? Not a single one?? Ah yes right, restricting to infinite dimensional spaces. ^^ – C-star-W-star Aug 03 '15 at 18:22
  • For the third time (the first being implicit in my assertion that there is no explicit unbounded operator): No. You can ask again if you want. – David C. Ullrich Aug 03 '15 at 18:26
  • @DavidC.Ullrich: Taking aside the precise definition of explicit. (Not sure how to define.): How can we know that there is no explicit example? I mean just think of the analogue by ONB's on Hilbert spaces. There one has an analoguous proof for the existence of an ONB. As that exploits AC it is nonconstructive. Still that doesn't imply that one cannot find an explicit ONB. But I think you would agree with me that $\delta_x$ for all $x\in\mathbb{R}$ is an explicit example for an ONB of $\ell^2(\mathbb{R})$. – C-star-W-star Aug 03 '15 at 18:35
  • This is the last time I'm going to answer a question that I've already answered: How can we know that there is no explicit example? I didn't say we can know this. I said there is no explicit example. To know there is no explicit example we can't ignore the question of a precise definition of "explicit". – David C. Ullrich Aug 03 '15 at 18:39
  • @DavidC.Ullrich: True, I cannot pass around the precise definition of explicit if I want to decide the question. Alright! But just for future times: Please don't take questions personal here; it seems you got quite upset. And if you're honest, I was just adressing your statement: "No, there are no Banach spaces with a Hamel basis known explicitely." (I marked the comment.) – C-star-W-star Aug 03 '15 at 18:48

1 Answers1

0

Ok satisfactory enough:

Given the Hilbert space $\ell^2(\mathbb{N})$.

Extend to a Hamel basis: $$\mathcal{B}:=\{\delta_n:n\in\mathbb{N}\}\cup\{\beta,\beta',\ldots\}$$

Define an operator by: $$T(\delta_n):=n\quad(n\in\mathbb{N})$$ $$T(\beta):=0\quad(\beta,\beta',\ldots)$$

Then it is unbounded.

C-star-W-star
  • 16,275