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location Pisa, Italy
age 30
visits member for 1 year, 9 months
seen Aug 25 at 0:56

Jun
25
comment Almost complex structures on spheres
@MichaelAlbanese: no, sorry, it's just a thing I heard somewhere and, when need came, I just did the computation of the Nijenhuis tensor … it's not very hard (also the covariant derivative of J is not hard to compute).
Dec
4
awarded  Yearling
Sep
19
comment Points of convergence
Your line of solution cannot work: a Borel set isn't necessarily open, it can consist also of isolated points. Take e.g. the functions $f_n(x)=n|x|$.The sum converges pointwise only in $x=0$, so your claim of finding $\epsilon$ such that all $y<x+\epsilon$ belong to the set of convergence cannot be true.
Sep
19
comment NullSpace of AB if $[A,B]=0$ and $AB\neq 0$?
null stands for kernel?
Aug
31
answered Function that maps straight lines into straight lines
Aug
31
comment Determining a conic section from points and tangent lines
yes, i wrote the answer supposing that it was as you said in this previous comment
Aug
31
revised Determining a conic section from points and tangent lines
added 17 characters in body
Aug
31
answered Determining a conic section from points and tangent lines
Aug
28
answered harmonic conjugate question
Aug
20
comment What type of function satisfies $f(x,y)=f\left(-x,\frac{1}{y}\right)$?
what are you taking as domain and codomain? do you impose any regularity on the function?
Aug
20
answered Need help with Trigonometric Identity in solution to problem on Napoleon Triangles.
Aug
20
answered Problem with determining the constants of the general solution of a differential equation
Aug
18
revised Negative self intersection and section of the conormal sheaf for a singular complex curve
Added link to MathOverflow
Aug
18
comment holomorphic functions uniformly convergent on any compact subset of an open set implies uniformly convergent on the whole open set.
Oh, yes, there was a "on compact sets" missing. Tnx
Aug
18
revised holomorphic functions uniformly convergent on any compact subset of an open set implies uniformly convergent on the whole open set.
added 2 characters in body
Aug
18
comment holomorphic functions uniformly convergent on any compact subset of an open set implies uniformly convergent on the whole open set.
I assumed what Stein proves after saying "without loss of generality we can assume that $f_n$ converges to $f$ uniformly on $\Omega$". Stein proves the theorem in the case when the convergence is uniform on the open set and deduces from this the general theorem, as I wrote, or as user89499 wrote.
Aug
18
comment holomorphic functions uniformly convergent on any compact subset of an open set implies uniformly convergent on the whole open set.
Fortunately for me, someone else explained it... let me just note that if Stein had proved (meaning that, he had given proof) of the fact you say in your first comment to this answer, you would have asked a very silly question..
Aug
17
comment holomorphic functions uniformly convergent on any compact subset of an open set implies uniformly convergent on the whole open set.
Stein proved the theorem with the hypotesis The functions converge uniformly on the open set. Therefore he proves that, whenever you have a sequence converging uniformly on an open set, the derivatives converge to the derivative of the limit. Given this result, I show that the general theorem (where the convergence is on the compacts of the open set and not on the whole open set) follows (rather easily).
Aug
17
answered I've solved this problem, but why is this differentiable?
Aug
17
answered holomorphic functions uniformly convergent on any compact subset of an open set implies uniformly convergent on the whole open set.