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seen May 18 '13 at 10:40

On extended leave from the site. I visit occasionally, but irregularly, mostly without logging in. For the time being it is impossible to predict if and when I will find the time to contribute again.


Dec
11
reviewed Leave Open How do mathematicians think about the existence of numbers?
Dec
10
comment Does the open mapping theorem imply the Baire category theorem?
@MattN. Thank you very much, but I agree with Asaf's assessment. A bounty would not be of use in this case. I doubt the answer is known at all, that's why I didn't take the question to MO either.
Dec
10
comment Does the open mapping theorem imply the Baire category theorem?
Thank you for your answer and I apologize for the confusion. As Asaf pointed out, this does indeed not address my question. I asked: "can one deduce the Baire category theory theorem for complete metric spaces (or the axiom of dependent choice) from the open mapping theorem (in whatever form)"? Indeed I linked to Schechter's book giving seven equivalent conditions on topological vector spaces (among which the open mapping theorem, the uniform boundedness, and so on). I know Jarchow's book, but my question is not "for which topological vector spaces does the OMT hold"?
Dec
2
awarded  Enlightened
Dec
2
awarded  Nice Answer
Nov
29
reviewed Close Convergence in measure and L^1 convergence
Nov
29
comment construction of a linear functional in $\mathcal{C}([0,1])$
Try again: $$f_\varepsilon(x) = \begin{cases} 1 & 0 \leq x \leq 1/2 - \varepsilon \\ \text{linear} & \text{on }1/2 - \varepsilon \lt x \lt 1/2+\varepsilon \\ -1 & 1/2 + \varepsilon \leq x \leq 1.\end{cases}$$ You should get $\varphi(f_\varepsilon) = 1 - \varepsilon$ while $\lVert f_\varepsilon \rVert = 1$, now let $\varepsilon \to 0$. Yes, no reflexivity needed: see here and here.
Nov
29
revised Is a convex function defined on a convex open subset of $\mathbb R^n$ continuous?
replaced broken ring by a working one
Nov
29
comment Is a convex function defined on a convex open subset of $\mathbb R^n$ continuous?
@Jeff: See e.g. Theorem 3.3.1 in these notes
Nov
29
comment A strange ring category
Sorry, I'm very late, but anyway: product of comm. rings = cartesian product, coproduct of comm. rings = tensor product. You were probably thinking of modules which is an additive category and finite coproducts and products are indeed equal.
Nov
29
comment construction of a linear functional in $\mathcal{C}([0,1])$
triangle inequality + $\lvert \int f \rvert \leq \int \rvert f\lvert$.
Nov
29
comment A question about Banach reflexive space
@Parakee: 1. Let $\iota_X\colon X \to X^{\ast\ast}$ and $\iota_{X^\ast}\colon X^\ast \to X^{\ast\ast\ast}$ be the canonical inclusions. Then we have that $(\iota_X)^\ast \circ (\iota_{X^\ast}) = \operatorname{id}_{X^\ast}$ by a direct verification. 2. If an operator is invertible then so is its adjoint $(\iota_X)^\ast$. 3. If an operator $S$ is right inverse to an invertible operator $T$ then $S$ is itself invertible and equal to the inverse: $S = T^{-1}$. 4. Apply this to $T = (\iota_X)^\ast$ and $S = \iota_{X^\ast}$.
Nov
29
comment Direct aproach to the Closed Graph Theorem
@AndréCaldas: Maybe you're interested in having a look at the proof of the closed graph theorem (Theorem 5.20 on page 166) in Kato's Perturbation theory. It doesn't quite follow your idea but gives a direct proof of it from Baire.
Nov
29
comment Direct aproach to the Closed Graph Theorem
@mlbaker: I mean that the other standard consequences the open mapping theorem, the closed graph theorem and the uniform boundedness principle follow quite quickly from Zabreiko's lemma, as I tried to indicate in the exercises. The question asked for something "but without the Closed Graph Theorem or any of its equivalent theorems" that's why I mentioned it as being something of a cheat.
Nov
29
comment Young's inequality for discrete convolution
@robjohn: yeah, right :) Thanks, hi and bye, Theo.
Nov
29
comment Dual of a dual cone
@Willie: thanks, yes it's what I meant, although I'm not sure whether one should really call it weak*-topology rather than weak topology when the duality is explicitly indicated. But I'm rather agnostic on this.
Nov
29
comment construction of a linear functional in $\mathcal{C}([0,1])$
$\leq$ is trivial, for $\geq$ take $f$ piecewise linear, constant equal to $1$ on $[0,1/2 - \varepsilon]$ and constant equal to $-1$ on $[1/2 +\varepsilon, 1]$.
Nov
29
revised linear subspace of dual space
incorporated the minor typo and undid the ghastly other things that were done to this answer.
Nov
29
revised linear subspace of dual space
rolled back to a previous revision
Nov
24
awarded  Enlightened