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seen Mar 20 at 20:56

Mar
20
accepted Are constants the only continuous functions with “symmetric derivative” zero?
Mar
20
comment Are constants the only continuous functions with “symmetric derivative” zero?
@ Umberto P.: I already found it! The result I need is contained in Theorem 1.4 on page 6 and Corollary 1.5 on page 7. Thanks!
Mar
20
comment Are constants the only continuous functions with “symmetric derivative” zero?
@ Umberto P.: From books.google.fi/…, I can see the table of contents. Could you please be a bit more specific? There is section 1.7 "Borel symmetric derivative". Is it possibly there?
Mar
20
comment Are constants the only continuous functions with “symmetric derivative” zero?
In fact, in a way, I extracted it thereof!
Mar
20
comment Are constants the only continuous functions with “symmetric derivative” zero?
@ geodude: Your deduction does not prove differentiability of $f$ . It is unfounded to write $2\lim_{z\to 0} \frac{f(r+z)-f(r)}{z} = 2\lim_{z\to 0} f'(r)$ since differentiability of $f$ is not known. Note that your $r$ depends on $s$ , hence on $z$ .
Mar
20
asked Are constants the only continuous functions with “symmetric derivative” zero?
Dec
8
comment Riemann integrability in not sequentially complete LCS?
The question has been answered in mathoverflow.net/questions/151148/… .
Nov
17
asked Riemann integrability in not sequentially complete LCS?
Oct
28
awarded  Tumbleweed
Oct
21
awarded  Commentator
Oct
21
comment Series converging almost everywhere.
See e.g. Richard M. Dudley's book Real Analysis and Probability (or any other similar) the pages where he proves completeness of the $L^p$ spaces (if I recall correctly)
Oct
21
asked Why multivalued maps instead of relations?
Oct
21
answered Why do some people use $+\infty$ instead of $\infty$?
Sep
23
comment Does $\mathsf{Man}$ possess countable products?
At the present, I cannot see the details, but I think this should somehow lead to a contradiction by noting that $f_i$ goes "very fast" around $\mathbb S^{\,1}$ when $i$ becomes large.
Sep
23
comment Does $\mathsf{Man}$ possess countable products?
My intuition is that there is no locally convex modelled manifold which would be the product of the family ${\mathbb N}_0\times\{\,{\mathbb S}^{\,1}\}\,$. If there were one such $M$ with the projections $p_i:M\to\mathbb S^{\,1}$, then considering the maps $f_i:\mathbb R\to\mathbb S^{\,1}$ given by $t\mapsto{\rm e}\,^{i\,2\,\pi\,{\rm i}\,t}$ one would get existence of a (at least continuous) map $f:\mathbb R\to M$ with $p_i\circ f=f_i$. So given any neigbourhood $V$ in $M$ of $f(0)$, there would exist a neighbourhood $U$ in $\mathbb R$ of $0$ with $f[U]\subseteq V$. (cont.)
Sep
23
revised A sequence with no almost everywhere converging subsequence
added 633 characters in body
Sep
22
comment A sequence with no almost everywhere converging subsequence
Now I am wondering why in the earth I did so much work for nothing! One of the first things I noticed was that the set $C=\{\,t:\text{ there is $i$ with }\varphi_i(t)=0\,\}$ is countable, and hence has measure zero. Also only for $t\in C$ can $\langle\kern.6mm\varphi_{i_j}(t):j\in\mathbb N_0\kern.6mm\rangle\to 0$ hold true.
Sep
22
accepted A sequence with no almost everywhere converging subsequence
Sep
22
comment A sequence with no almost everywhere converging subsequence
@ Nate Eldredge: This is really simple. Thanks.
Sep
22
comment Help with a convergence almost everywhere (Rademacher)
@ Jonas Teuwen: For $f_n(t)=|n^{-1}\sum_{i=1}^n r_i(t)|^4$, Fatou's lemma only gives $\int_0^1\liminf_{\,n\to\infty}f_n(t){\kern.4mm\rm d\kern.4mm}t = 0$ from which you can conclude that $\liminf_{\,n\to\infty}f_n = 0$ almost everywhere. Wherefrom do you conclude that $\liminf_{\,n\to\infty}f_n(t)< \limsup_{\,n\to\infty}f_n(t)$ holds only for $t$ in a set of measure zero?