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I am a professor at the University of Pisa, Italy.


1d
awarded  Good Answer
Apr
13
revised For which maps $S^1\to S^1$ is the winding number defined?
edited body
Apr
13
comment Rising Sun Inequality (Dunford-Schwartz maximal inequality)
@Xiao: this question is likely to be closed soon. Instead you may have a look e.g. to Wheeden-Zygmund's Measure and Integral: An Introduction to Real Analysis for an elementary and clear proof (to be arranged for your one-side version)
Apr
13
comment solving an exponential inequality with parameters
Or put $y:=b^{x/2}$ and study $y^A<By+C$.
Apr
13
revised solving an exponential inequality with parameters
added 11 characters in body
Apr
13
awarded  Enlightened
Apr
12
awarded  Nice Answer
Apr
12
revised For which maps $S^1\to S^1$ is the winding number defined?
added 146 characters in body
Apr
12
answered For which maps $S^1\to S^1$ is the winding number defined?
Apr
12
comment Deterministic shifts
In the definition it's " $x_n=y_n$ for every $n<0$ " (not " for some $n<0$ " ), right?
Apr
11
comment Integer roots of a polynomial
Choose e.g. $a_1=a_2=\dots=a_{p-1}:=0$ and $-a_p:=1-2^{-p}$, so (2) is $2^p=1-a_p 2^p$.
Apr
11
comment Integer roots of a polynomial
For instance, if $\sum_{i=1}^p|a_i| < 1$ there are of course no solutions $X$ with $|X|\le 1$ to (1), $\sum_{i=1}^p a_i X^i = 1$. But $\sum_{i=1}^p|a_i| < 1$ is certainly not an obstruction in order that $X=2$ solves (2) for some $j$, e.g. for $j=p$, that is $2^p=1 - \sum_{i=1}^p a_i 2^i$.
Apr
11
comment Applications of really large numbers
"it might be that Archimedes used problems with practically incalculable numbers to discover who(…)" Sounds interesting, could you elaborate on this? Is there any source that suggests that?
Apr
10
comment Comparing numbers $a \uparrow^b c$ and $d \uparrow^e f$
The first two rules, as they are written, can't be generally true, because even if $b>e$ and $c>f$ , $d$ may be arbitrarily large.
Apr
9
comment Smooth convex extensibility of combination of two line segments
Yep. You can also take it if the form $\phi(x,y)=\psi(x)\psi(y)$ as above.
Apr
9
comment Smooth convex extensibility of combination of two line segments
(in other words: take, in your last picture, the straight segments not necessarily parallel, but let each of them join a couple of points with the same slope wrto $y$).
Apr
9
comment Smooth convex extensibility of combination of two line segments
A slight variant of your function: define, on the vertical edges, $f(0,y)$ and $f(1,y)$ to be strictly convex, with $\partial_y f(0,0)=\partial_y f(1,0)$ and $\partial_y f(0,1)=\partial_y f(1,1)$. Then join by segments $(0,u,f(0,u))$ and $(1,v,f(1,v))$ whenever $\partial_y f(0,u)=\partial_y f(1,v)$. This should give the graph of a smooth convex function.
Apr
8
comment Smooth convex extensibility of combination of two line segments
I like your suggested construction , that maybe could be formalized as a solution of a suitable PDE with boundary conditions? A doubt on the picture: any $C^1$ function $f$ on the square, such that $f(0,t)=f(t,0)=0$ for $0\le t\le 1$, also verifies $Df(0,0)=0$. If it is convex, this implies that $(0,0)$ is a global minimum, so $f \ge0$ .
Apr
8
revised Smooth convex extensibility of combination of two line segments
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Apr
8
comment Smooth convex extensibility of combination of two line segments
Yes, say $\phi(x)=c\exp(-1/(1-|4x|^2))$ if $|x|<1/4$, and $0$ otherwise.