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


22h
comment Solution to a system of linear equations containing some inequalities
There is, of course, a lot of software, but this is simple enough to be done by hand. Assuming the first two equations being independent, solve them parametrically and plug $x_j=b_j t+c_j$ into the last two inequalities, obtaining $t$ in a solution interval (possibly empty).
1d
comment Norm of swapped power series in the unit disk
If you have an interesting motivation, this is maybe worth another question.
1d
comment Norm of swapped power series in the unit disk
E.g. $f$ with all $a_k$ real and positive are OK, as they reach the norm at $z=1$.
1d
comment Norm of swapped power series in the unit disk
A question: what can be said about a function $f\in H^\infty(D)$ whose norm is invariant by exchange $a_1\leftrightarrow a_k$ for any $k\ge 1$?
1d
comment Norm of swapped power series in the unit disk
That's what I was also thinking; though passing to the limit from permutations with compact support, to all permutations, seems delicate to me, as I only see local uniform convergence.
1d
comment Continuity in banach space for non-linear maps
I think it is safer to have the family of balls locally finite too, and not only disjoint. Or also, to have the support of $f_n$ into the ball of radius $\epsilon/2$. Otherwise the resulting glueing function may fail to be continuous. For instance, in $\ell_\infty$ the unit balls centered at $(1+1/n)e_n$ are disjoint and well separated from each other, but any nbd of $0$ meets almost all of them, which allows the glued $f$ to be possibly discontinuous.
2d
reviewed Leave Open independent subset problems
2d
reviewed Leave Open Von Dyck Theorem
2d
reviewed Close About diagonal entries of the graph Laplacian
2d
comment Constructivity of zeros demanded by topological degree
In fact I 'm not sure I understand how this algorithm works. Case (A) may occur for a degree $1$ map, right?
Nov
20
comment A question about the duality principle
In other words, an $n\times m$ matrix $K$ and its transpose have the same operator norms w.r.to the Euclidean norms on $\mathbb{R}^n$ and $\mathbb{R}^m$, that is $\|K \|_{2,2}=\|K^T\|_{2,2}$. A unit vector $f$ maximizes $\|Kf\|_2$ iff it is an eigenvector of the positive symmetric matrix $K^TK$ w.r.to its maximum eigenvalue , aka maximum singular value of $K$, squared. ( I'm not quite sure why you are then dealing with $K^T$ and not $K$, and I guess you mean $g$ to be normalized).
Nov
15
comment Can any bounded area defined by polynomial inequality in $\mathbb{R}^n$ be partitioned into simply connected finite area such that
Actually there is a lot of material online. Chapter 3 of these notes: perso.univ-rennes1.fr/michel.coste/polyens/SAG.pdf
Nov
15
answered Can any bounded area defined by polynomial inequality in $\mathbb{R}^n$ be partitioned into simply connected finite area such that
Nov
15
comment Orthogonal projection
OK, by invertible operator I mean a linear homeomorphism $ A:H\to H$ , thus more than just $0\in\rho(A)$ . So yes, the answer was somehow misleading.
Nov
15
revised Orthogonal projection
added 363 characters in body
Nov
14
comment Orthogonal projection
But I understand $0\in\rho(G_0)$ as: $G_0$ being an invertible operator; therefore: $0\notin\sigma(G_0)$ and $\sigma(G_0)$ bounded... Why we don't need the latter?
Nov
14
comment Orthogonal projection
Yes, but I preferred to write it independently from the assumption "compact resolvent", to make a precise statement. Yes, "compact resolvent" implies 0 is automatically isolated, and $\sigma(G)$ unbounded.
Nov
14
revised Orthogonal projection
added 138 characters in body
Nov
14
answered Orthogonal projection
Nov
14
comment Does this condition imply a polynomial is a product of linear factors
Also, what if $H(\lambda):=\prod_{j=1}^\lambda (j^2+1)$, for $\lambda\in\mathbb{Z}_+$ ?