bio  website  

location  
age  
visits  member for  4 years, 11 months 
seen  5 hours ago  
stats  profile views  15,075 
I am a professor of Mathematics at the University of Pisa, Italy.
19h

comment 
Two questions on hyperspace of a metric space
As to the first question, you may also observe that if $(X,d)$ is a compact metric space, its hyperspace $H(X)$ (the set of closed sets of $X$ endowed with the Hausdorff distance) is a compact metric space too; moreover the construction extends to a functor from the category of compact metric spaces & continuous functions into itself, precisely : if $f:X\to Y$ is continuous the map $K\to f(K)$ is continuous from $H(X)$ to $H(Y)$ (just because $f$ is uniformly continuous). In particular, if $(X,d)$ and $(X',d')$ are homeomorphic compact metric spaces, so are their hyperspaces. 
2d

awarded  Good Answer 
Apr 22 
revised 
Functional minimization problem
added 16 characters in body 
Apr 21 
answered  Functional minimization problem 
Apr 20 
comment 
Boundary Value System.
Besides, the set of solutions of that ODE is a 2n linear space; the given BVC can't identify any. 
Apr 20 
comment 
Boundary Value System.
And if $A=B$, $y(x)$ does not depend on $x$? 
Apr 18 
revised 
Biggest parallelogram inside the union of two translated parallelograms
added 81 characters in body 
Apr 18 
revised 
Biggest parallelogram inside the union of two translated parallelograms
added 52 characters in body 
Apr 18 
comment 
Biggest parallelogram inside the union of two translated parallelograms
Even larger than the area of the parallelogram \ \ with the lower horizontal edge moved to the right (therefore, not the green one // ) ? 
Apr 18 
revised 
Biggest parallelogram inside the union of two translated parallelograms
added 167 characters in body 
Apr 18 
revised 
Biggest parallelogram inside the union of two translated parallelograms
deleted 15 characters in body 
Apr 18 
answered  Biggest parallelogram inside the union of two translated parallelograms 
Apr 17 
comment 
Biggest parallelogram inside the union of two translated parallelograms
Since linear maps preserves ratios of measures, it may be assumed wlog that the two parallelograms are squares. This simplifies the problem... 
Apr 16 
comment 
Old Peano theorem (demonstration is missing details)
It seems to me that the computation in the linked book ("Geometric applications of differential calculus") is clear and rigorous, as in Peano's style. Of course, a reader of a textbook is supposed to start from the beginning, so I suppose you need to refer to the material in the previous part of this textbook in order that notations, definitions and properties used are clear. 
Apr 16 
comment 
Fair cakecutting between groups
A remark on the existence result. Even with $k$ states and $N$ atomless probability measures, there exists a measurable $k$partition $\{A_j\}_{1\le j\le k}$ such that $\mu_i(A_j)=1/k$ for all $1\le i\le N$ and $1\le j\le k$. It's a consequence of the Lyapounov convexity theorem. 
Apr 11 
comment 
Invariant mesures for expanding maps of the circle
I don't understand... how can $T:\mathbb{S}^1\to \mathbb{S}^1$ have $T'(x)>1$ for all $x$ ? 
Apr 6 
comment 
Between Tietze's and Dugundji's Extension Theorems
@Włodzimierz Holsztyński. Sorry, I will not. 
Apr 5 
comment 
Between Tietze's and Dugundji's Extension Theorems
@Włodzimierz Holsztyński: thank you for your comment. However, the question is not about the original author of (what is known as) Tietze theorem. As you probably know, most theorems are known after a name which is different from the person who discovered it; often by the author's choice. en.wikipedia.org/wiki/Stigler%27s_law_of_eponymy 
Apr 3 
comment 
Between Tietze's and Dugundji's Extension Theorems
Thank you very much. Would you write this as an answer, possibly with some details, or a reference? 
Apr 3 
comment 
Between Tietze's and Dugundji's Extension Theorems
And I think at least in the case of a separable $E$, one can pass from a linear extensor for the scalar case to one for Evalued mappings, by the argument in Bill Johnson's answer. 