text:
\begin{document}

\begin{abstract}
Given $p$ polynomials of $n$ variables over a field $\k$
of characteristic $0$ and a point $a \in \k^n$, we
propose an algorithm computing the local
Bernstein-Sato ideal at $a$. Moreover with the same
algorithm we compute a constructible
stratification of $\k^n$ such that the local Bernstein-Sato
ideal is constant along each stratum. Finally, we present
non-trivial examples computed with our algorithm.
\end{abstract}

\maketitle

\section*{Introduction}

Let $n$ be a positive integer and $\k$ be a field of
characteristic zero and $a=(a_1, \ldots, a_n)$ be a fixed
point in $\k^n$. Let $x=(x_1, \ldots, x_n)$ be a set
of indeterminates.
In this introduction, $A$ shall be
one of the following rings: the polynomial ring $\k[x]$;
the localization $\k[x]_{a}$ of $\k[x]$ at $a$;
the formal power series ring
$\Ohat_{\k^n, a}=\k[[x-a]]:=\k[[x_1-a_1, \ldots, x_n-a_n]]$;
and when $\k=\C$, the ring $\O_{\C^n,a}$ of germs of
complex analytic functions at $a$.
Denote by $\dxi$ the partial differential operator
$\dxsur{}{i}$ and $D_A=A\langle \dx{1}, \ldots, \dx{n}
\rangle$ the ring of differential operators with
coefficients in $A$.

Let $p\ge 1$ be an integer and let us
consider $f=(f_1,\ldots, f_p) \in A^p$.
Denote by $F$ the product $f_1 \cdots f_p$ and let us
introduce a set of indeterminates $s=(s_1, \ldots, s_p)$
then consider the $A[1/F, s]$-free module
\[\mathcal{L}_A= A[1/F, s] \cdot f^s\]
where $f^s$ is seen as $f_1^{s_1} \cdots f_p^{s_p}$.
The set $\mathcal{L}_A$ is naturally endowed with a
$D_A[s]$-module structure.
Indeed, given $g \in A[1/F,s]$, we have:
\[\dxi \cdot g f^s=(\dxsur{g}{i} + g \sum_{j=1}^p
s_j \dxsur{f_j}{i} f_j^{-1}) f^s.\]
A Bernstein-Sato polynomial (or $b$-function) of $f$
with respect to $A$ is a polynomial $b(s) \in \k[s]$ such that
\[b(s) f^s \in D_A[s] \cdot F f^s.\]
Such $b(s)$ form the ideal $\B_A(f)$, the Bernstein-Sato
ideal of $f$ (with respect to A). When $p=1$, the monic generator (if
it is not zero) is called the Bernstein-Sato polynomial of $f$
(with respect to A).

When $f \in \k[x]^p$, $\B_{\k[x]}(f)$ is called the
global Bernstein-Sato ideal and $\B_{\k[x]_a}(f)$ is called
the local Bernstein-Sato ideal at $a \in \k^n$. It is
well-known that $\B_{\k[x]_a}(f)$ is equal to
$\B_{\Ohat_{\k^n, a}}(f)$ and in the case where $\k=\C$,
it is equal to $\B_{\O_{\C^n, a}}(f)$
(see Brian\c{c}on, Maisonobe \cite{bm90} and the general
arguments in Mebkhout, Narv\`aez-Macarro \cite[4.2]{meb-nar}
for these equalities).
When $f \in \O_{\C^n,a}$, $\B_{\O_{\C^n,a}}(f)$ is called
the analytic Bernstein-Sato ideal of $f$, and it is equal
to $\B_{\Ohat_{\C^n,a}}(f)$.
Finally, when $f \in \Ohat_{\k^n, a}$, we call
$\B_{\Ohat_{\k^n,a}}(f)$ the formal Bernstein-Sato ideal
of $f$ (at $a$).

Let us recall some historical facts.
The global or analytic or local Bernstein-Sato ideal is not
zero (see Bernstein \cite{bernstein} and Sabbah \cite{sabbah}
and also \cite{b-compos}). When $p=1$, the formal
$b$-function is not zero (see Bj\"ork \cite{bjork}).
However for $p\ge 2$, it is still an open question to know
whether $\B_{\Ohat_{\k^n, a }}(f)$ is zero or not.

When $f$ is a polynomial mapping,
a general algorithm for the global Bernstein-Sato ideal
was given by Oaku and Takayama \cite{ot}
(see also Bahloul \cite{b-JSC} and Brian\c{c}on,
Maisonobe \cite{bm02}). When $p=1$, Oaku \cite{oaku} gave
an algorithm for the local $b$-function (see also the recent
work by H. Nakayama \cite{nakayama}).
The first goal of the present paper is, given $f\in \k[x]^p$,
to present an algorithm for computing $\B_{\k[x]_a}(f)$
for $p \ge 1$.
Secondly, by works of Brian\c{c}on and Maisonobe \cite{bm02}
we know that there exists a constructible stratification of
$\k^n$ such that for $a$ running over a given stratum the local
Bernstein-Sato ideal at $a$ is constant. With our
algorithm, we obtain such a stratification.

In section 1 we present all the main results. For the sake of
clarity, all the proofs are postponed to section 2. Section 3
contains non-trivial examples treated with our algorithm.
For example we treat $f=(x^3+y^2,x^2+y^3)$ which was intractable
until today. We also treat $f=(z, x^4+y^4+2zx^2y^2)$ taken from
Brian\c{c}on and Maynadier \cite{bmay}. They proved
that $\B_{\C[x,y,z]_0}(f)$ is not principal by general geometric
arguments without computing it.

\section{Statement of the main results and description
of the algorithm}

In this section we shall recall the main lines of the
(known) algorithm for the global Bernstein-Sato ideal
(\cite{ot} and \cite{bm02}). At the same time we shall
describe our local algorithm emphasizing the common points
and the differences with the global case.

Let us fix a polynomial mapping $f \in \k[x]^p$.
We are interested in $\B_{\k[x]_a}(f)$.
As we recalled, the formal, analytic and local
Bernstein-Sato ideals of $f$ at $a$ are all the same, so we
shall use the notation: $\B_{loc,a}(f)= \B_{\k[x]_a}(f)$. This
will be compared with the global Bernstein-Sato ideal
$\B_{glob}(f)=\B_{\k[x]}(f)$.

Moreover, we shall use the following notations:
$D=\k[x]\langle \ddx \rangle$, $D_a=\k[x]_a\langle \ddx \rangle$,
$\Dhat_a=\Ohat_{\k^n, a}\langle \ddx \rangle$,
and when $\k=\C$, $\D_a=\O_{\C^n, a}\langle \ddx \rangle$.

Following Malgrange \cite{malgrange}, let us introduce
new variables $t_1,\ldots, t_p$ together with the associated
partial derivation operators $\dt{1}, \ldots, \dt{p}$; then
let us consider the ring
$\Dhat_a\langle t, \ddt \rangle=\Dhat_a \otimes_\k
\k[t]\langle \ddt \rangle $
where $t=(t_1,\ldots, t_p)$ and $\ddt=(\dt{1}, \ldots,
\dt{p})$. We then have the subrings
$D\langle t, \ddt \rangle$, $D_a\langle t, \ddt \rangle$
and (when $\k=\C$) $\D_a \langle t, \ddt \rangle$.

The free module $\mathcal{L}_{\Ohat_{\k^n, a}}=
\Ohat_{\k^n,a}[1/F,s] f^s$ has a
$\Dhat_a \langle t, \ddt \rangle$-module structure:
\[t_j \cdot g(s) f^s= g(s_1, \ldots, s_j+1, \ldots, s_p)
f_j f^s\]
and
\[\dt{j} \cdot g(s) f^s= -s_j g(s_1, \ldots, s_j-1,
\ldots, s_p) f_j^{-1} f^s\]
where $g(s) \in \Ohat_{\k^n,a}[1/F, s]$. It also has
a module structure over the subrings of $\Dhat_a \langle t,\ddt
\rangle$. Moreover, we have that $-\dtj t_j$ acts on
$\mathcal{L}_{\Ohat_{\k^n, a}}$ as $s_j$. Thus we shall
identify $s_j$ with $-\dtj t_j$ and the rings
$D[s]$, $D_a[s]$, $\D_a[s]$ and $\Dhat_a[s]$ shall be
seen as subrings of $\Dhat_a \langle t, \ddt \rangle$.

Let us consider the following elements:

\begin{equation}\label{eq:I}
t_j - f_j, \ j=1,\ldots, p\ ; \  \ \dxi+
\sum_{j=1}^p \dxsur{f_j}{i} \dtj , \ i=1, \ldots, n.
\end{equation}
One can easily check that these elements annihilate $f^s$,
in fact we have:

\begin{lem}\label{lem}
The annihilating ideals\footnote{In a non commutative ring,
ideal shall always mean left ideal.}
$\mathrm{ann}_{\Dhat_a\langle t, \ddt \rangle}(f^s)$,
$\mathrm{ann}_{\D_a\langle t, \ddt \rangle}(f^s)$,
$\mathrm{ann}_{D_a\langle t, \ddt \rangle}(f^s)$,
$\mathrm{ann}_{D\langle t, \ddt \rangle}(f^s)$
are all generated by the elements in (\ref{eq:I}).
\end{lem}

\noindent
Let us introduce the following ideals:
\begin{itemize}
\item
$I=\mathrm{ann}_{D\langle t, \ddt \rangle}(f^s)$,\\
$J=\mathrm{ann}_{D_a\langle t, \ddt \rangle}(f^s)$.
\item
$I_1= I \cap D[s] \subset D[s]$,\\
$J_1=J \cap D_a[s] \subset D_a[s]$.
\item
$I_2 = (I_1 + D[s] \cdot F) \cap \k[x,s] \subset
\k[x_1, \ldots, x_n, s_1, \ldots, s_p]$,\\
$J_2 = (J_1 + D_a[s] \cdot F) \cap \k[x]_a[s] \subset
\k[x_1, \ldots, x_n]_a[ s_1, \ldots, s_p]$.
\item
$I_3=I_2 \cap \k[s] \subset \k[s_1, \ldots, s_p]$,\\
$J_3=J_2 \cap \k[s] \subset \k[s_1, \ldots, s_p]$.
\end{itemize}

Obviously, $I_1$ (resp. $J_1$) is the annihilator of
$f^s$ in $D[s]$ (resp. in $D_a[s]$).

\begin{prop}\label{p1}
\[I_3=\B_{glob}(f) \qquad \text{and}\qquad J_3=\B_{loc,a}(f).\]
\end{prop}

In both cases (global and local) we start with the ``same''
ideals $I$ and $J$ (``same'' means that $J$ and $I$ admit
a common set of generators); then we construct in parallel
the ideals $I_k$ and $J_k$, with $k=1,2,3$, and finally get
the global Bernstein-Sato ideal $I_3=\B_{glob}(f)$
and the local Bernstein-Sato ideal $J_3=\B_{loc,a}(f)$.
It is natural to ask whether $I_k$ and $J_k$ are the
``same'' (in the above sense). Here is the beginning
of the answer.

\begin{prop}\label{p2}
We have:
\begin{itemize}
\item
$J_1= D_a[s] \cdot I_1$,
\item
$J_2=\k[x]_a[s] \cdot I_2$.
\end{itemize}
\end{prop}

This proposition says that the global and the local constructions
coincide (in the above sense) at least up to step $2$.
Going from $I_2$ to $I_3$ consists in a usual elimination.
However, going from $J_2$ to $J_3$ is different. Here it is:

\begin{prop}\label{p3}
Let $\Up$ be an ideal in $\k[x,s]$.
Let $\Up=\Up_1 \cap \cdots \cap \Up_r$
be a primary decomposition of $\Up$.
Let $\sigma_a \subset \{1,\ldots, r\}$ be such that

$i \in \sigma_a \iff a \in V(\Up_i \cap \k[x])$ (here $V( \cdot )$
means ``the zero set of''). Then,
\begin{itemize}
\item
if $\sigma_a =\emptyset$ then
$(\k[x]_a[s] \cdot \Up)\cap \k[s]= \k[s]$,
\item
otherwise: $(\k[x]_a[s] \cdot \Up) \cap \k[s]=
(\bigcap_{i \in \sigma_a} \Up_i) \cap \k[s]$.
\end{itemize}
\end{prop}

Thus applying this to $\Up=I_2$ tells us that combining
a primary decomposition, a finite intersection of ideals and
an elimination of variables, we obtain $J_3$.

The results above say that the computation
of the local Bernstein-Sato ideal coincides with
that of the global Bernstein-Sato ideal
up to a point where everything is commutative.
Computing $I_1$ and $I_2$ is well-known. For $I_1$
one can follow Oaku and Takayama \cite{ot}
or Brian\c{c}on and Maisonobe \cite{bm02} and for
$I_2$ it is a usual elimination of global
variables.

Until now, $a$ was a fixed point in $\k^n$.
In the following result we are concerned with the
behaviour of $\B_{loc,a}(f)$ when $a$ runs over $\k^n$.
Let us apply the previous proposition to a primary
decomposition $\Up_1 \cap \cdots \cap \Up_r$ of $I_2$,
then the following holds.

\begin{cor}\label{cor:strat}
For each subset $\sigma \subset \{1,\ldots,r\}$, let
\begin{itemize}
\item
$W_\sigma=\k^n \setminus
(\bigcup_{i=1}^n V(\Up_i \cap \k[x]))$
if $\sigma= \emptyset$,
\item
$W_\sigma=\bigcap_{i=1}^n V(\Up_i \cap \k[x])$
if $\sigma=\{1,\ldots, n\}$,
\item
$W_\sigma=(\bigcap_{i \in \sigma} V(\Up_i \cap \k[x]))
\setminus (\bigcup_{i \notin \sigma} V(\Up_i \cap \k[x]))$
otherwise.
\end{itemize}
Then $\bigcup_{\sigma} W_\sigma$ is a constructible
stratification of $\k^n$ such that the map $\k^n \ni a
\mapsto \B_{loc,a}(f)$ is constant on each $W_\sigma$.
\end{cor}

As another consequence of our algorithm, we recover this
well-known result:
\begin{cor}\label{cor:glob-loc}
Assume that $\k$ is algebraically closed. Then
\[\B_{glob}(f) = \bigcap_{a \in \k^n} \B_{loc,a}(f).\]
\end{cor}

\begin{rem*}
With a slight generalisation of the construction and
similar proofs one can obtain the following result
when $\k$ is not supposed to be algebraically
closed (see \cite[Prop. 1.4]{bm02}):
\[\B_{glob}(f)
= \bigcap_{m \in \mathrm{SpecMax(\k[x])}} \B_{loc,m}(f),\]
where $\B_{loc,m}(f)$ is the set of $b(s) \in \k[s]$ such that
there exists $a(x) \in \k[x] \smallsetminus m$ such that
$a(x) b(s) f^s \in D[s] F  f^s$.
\end{rem*}

Let us end this section with some additionnal results.

Again, $a$ is a fixed point in $\k^n$.
The first statement in Proposition \ref{p2} says
that the annihilators of $f^s$ in $D[s]$ and
in $D_a[s]$ have a common set of generators.
Similarly we have:

\begin{prop}\label{p4}
\[ \mathrm{ann}_{\Dhat_a[s]}(f^s)= \Dhat_a[s] \cdot \mathrm{ann}_{D[s]}(f^s)
\qquad \text{and} \qquad
\mathrm{ann}_{\Dhat_a}(f^s)= \Dhat_a \cdot \mathrm{ann}_{D}(f^s).\]
\end{prop}

\begin{rem} \
\begin{itemize}
\item
With the same proof we get: if $\k=\C$ then
\[\mathrm{ann}_{\D_a[s]}(f^s)= \D_a[s] \cdot \mathrm{ann}_{D[s]}(f^s)
\qquad \text{and} \qquad
\mathrm{ann}_{\D_a}(f^s)= \D_a \cdot \mathrm{ann}_{D}(f^s).\]
\item
Set $\hat{I}_1=\mathrm{ann}_{\Dhat_a[s]}(f^s)$;
$\hat{I}_2=(\hat{I}_1+ \Dhat_a[s] F) \cap \Ohat_{\k^n,a}[s]$;
$\hat{I}_3=\hat{I}_2 \cap \k[s]$.

With standard bases methods, one can prove that:
$\hat{I}_2=\Ohat_{\k^n, a}[s] \cdot I_2$.

By using the faithfull flatness of $\Ohat_{\k^n, a}$ over $\k[x]_a$
one can prove that: $\hat{I_2} \cap \k[x]_a[s] = J_2$
and thus $\hat{I_3}=J_3$. This is another way to recover
the well-known equality between the local and the formal
Bernstein-Sato ideals.
\end{itemize}
\end{rem}

We shall leave the proof of the second statement of this remark
since it is not the goal of our paper.

\section{Proofs}

In this section we give all the proofs. All the proofs, except
for Cor. \ref{cor:strat} and Cor. \ref{cor:glob-loc}, concern a
fixed point $a$; so for these proofs, we shall assume that $a=0$.

First let us prove lemma \ref{lem}.

\begin{proof}
Let us do the proof for $\mathrm{ann}_{\Dhat_0\langle t,
\ddt \rangle}(f^s)$. The other cases are similar.
Recall that $\mathrm{ann}_{\Dhat_0 \langle t, \ddt
\rangle}(f^s)$ is the left ideal
$\{P \in \Dhat_0\langle t, \ddt \rangle \ | \ P\cdot f^s=0\}$.
Let $P$ be in this ideal. Modulo the elements in (\ref{eq:I})
we may assume that $P\in \k[[x]][\ddt]$. Let
us write $P=\sum_{\nu} c_{\nu}
\ddt^\nu$ with $\nu \in \N^p$ and
$\ddt^\nu=\prod_{j=1}^p \dtj^{\nu_j}$
and $c_\nu \in \k[[x]]$. Then
\[
0=P f^s=\sum_\nu (-1)^{|\nu |} c_\nu
\prod_{j=1}^p (s_j \cdots (s_j-\nu_j+1) f_j^{-\nu_j}) f^s.
\]
This equality takes place in the free module
$\k[[x]][1/F, s] \cdot f^s$, thus all the terms
in the sum are zero, which implies that all the $c_\nu$
are zero. This concludes the proof.
\end{proof}

Now let us prove the propositions.
Let us begin with the
\begin{proof}[Proof of Proposition \ref{p1}]
Let us prove the second equality (the proof is the same for
the first one).
Let $b(s)$ be in $\k[s]$. If $b(s) \in \B_{loc,0}(f)$ then
$b(s) f^s = P f^{s+1}$ for some $P \in D_0[s]$. Thus
$b(s) -P F$ annihilates $f^s$, i.e. $b(s) \in (D_0[s] F
+J_1) \cap \k[s]=J_3$. The converse uses the same
arguments. We leave the details to the reader.
\end{proof}

Let us go on with the
\begin{proof}[Proof of Proposition \ref{p2}]
We have $I \subset J$ so $I_1 \subset J_1$ and then
$D_0[s] I_1 \subset J_1$. Let us show the converse inclusion.
Take $P$ in $J_1=(D_0\langle t, \ddt \rangle \cdot I) \cap
D_0[s]$. Writing $P$ as an element in $D_0 \langle t, \ddt
\rangle I$ and as an element of $D_0[s]$ we may clear the
denominators and obtain the existence of $c(x) \in \k[x]$
with $c(0) \ne 0$ such that $c(x) P \in I \cap D[s]$.
Thus $P$ is in $D_0[s] (I \cap D[s])=D_0[s] I_1$.
This ends the proof for the first equality.
For the second one the arguments are exactly the same.
\end{proof}

Proposition \ref{p3} is an obvious consequence
of the following lemma.

\begin{lem}\label{lem:p3}\
\begin{itemize}
\item[(i)]
If $\Up \subset \k[x,s]$ is an ideal with
$0 \notin V(\Up \cap \k[x])$ then

$ (\k[x]_{0}[s] \cdot \Up) \cap \k[s]= \k[s]$.
\item[(ii)]
If $\Up \subset \k[x,s]$ is a primary ideal with
$0 \in V(\Up \cap \k[x])$ then

$ (\k[x]_{0}[s] \cdot \Up) \cap \k[s]=
\Up \cap \k[s]$.
\item[(iii)]
Given ideals $\Up_1, \ldots, \Up_r$ in $\k[x,s]$, we have:

$ \k[x]_0[s] \cdot (\bigcap_{i=1}^r \Up_i)=
\bigcap_{i=1}^r (\k[x]_0[s] \cdot \Up_i)$.
\end{itemize}
\end{lem}

\begin{proof}
\begin{itemize}
\item[(i)]
If $0 \notin V(\Up \cap \k[x])$ then there exists
$g \in \Up \cap \k[x]$ such that $g(0) \ne 0$ which
implies that $1 \in \Up$.
\item[(ii)]
Let $f \in (\k[x]_0[s] \cdot \Up) \cap \k[s]$,
then there exists $c \in \k[x]$ with $c(0) \ne 0$
such that $c f \in \Up$. Assume, by contradiction,
that $f \notin \Up$. Then since $\Up$ is primary,
$c^l \in \Up$ for some $l \in \N$. This implies
$c(0)=0$: contradiction. Thus $f \in \Up \cap \k[s]$.
We proved the left-right inclusion. The reverse one
is trivial.
\item[(iii)]
Since the left-right inclusion is trivial, let us prove
the other one. Let $f$ be in $\bigcap_{i=1}^r
(\k[x]_0[s] \cdot \Up_i)$. Then for each $i$,
$c_i f \in \Up_i$ for some $c_i \in \k[x]$ satisfying
$c_i(0) \ne 0$. As a consequence, $(\prod_1^r c_i)
f \in \bigcap_1^r \Up_i$ and then $f \in \k[x]_0[s]
\cdot (\bigcap_{i=1}^r \Up_i)$.
\end{itemize}
\end{proof}

Now, let us work with arbitrary points $a \in \k^n$ and
prove the two corollaries.

\begin{proof}[Proof of Corollary \ref{cor:strat}]
First, it is clear that each $W_\sigma$ is locally closed.
Moreover, it is clear that any $a \in \k^n$ belongs to some
$W_\sigma$ (indeed, $a \in W_{\sigma_a}$ with the notations
of Prop. \ref{p3}).
Thus we have a constructible stratification of $\k^n$.
The constancy of the map $(a \mapsto \B_{loc,a}(f))$ on each
$W_\sigma$ follows from the obvious observation that if
$a$ and $a'$ are two points in a $W_\sigma$ then
$\sigma_a=\sigma_{a'}$, which implies, by using the whole
algorithm and in particular Prop. \ref{p3}, that $\B_{loc,a}(f)=
\B_{loc,a'}(f)$.
\end{proof}

\begin{proof}[Proof of Corollary \ref{cor:glob-loc}]
First, it is obvious from the definitions that
$\B_{glob}(f)$ is included in any $\B_{loc,a}(f)$ so we
have the inclusion: $\B_{glob}(f) \subset
\cap_a \B_{loc,a}(f)$. Let us prove the converse one.
We follow the notations of Prop. \ref{p3} and Cor.~\ref{cor:strat}.
Let us fix $i \in \{1, \ldots, r\}$.
Notice that since $\Up_i \subset \k[x,s]$ is primary,
$\Up_i \cap \k[x]$ is also primary in $\k[x]$ so
$V(\Up_i \cap \k[x])$ is irreducible.
Let
\[\tau_i=\{ k \in \{1, \ldots, r\} |
V(\Up_i \cap \k[x]) \subset V(\Up_k \cap \k[x]) \}.\]

Assume, by contradiction, that $W_{\tau_i}=\emptyset$. Then
$V(\Up_i \cap \k[x]) \subset \cup_{k \notin \tau_i} V(\Up_k \cap \k[x])$
and by irreducibility of $V(\Up_i \cap \k[x])$ it would be contained
in some $V(\Up_k \cap \k[x])$ for some $k \notin \tau_i$, which
is impossible. So let $a_i \in W_{\tau_i}$. Then we have:
$\B_{loc,a_i}(f) \subset \Up_i \cap \k[s]$.
As a consequence, we get:
\begin{eqnarray*}
\bigcap_{a \in \k^n} \B_{loc,a}(f) & \subset &
\bigcap_{i=1,\ldots, r} \B_{loc, a_i}(f) \\
& \subset & \bigcap_{i=1,\ldots,r} (\Up_i \cap \k[s])\\
& = & I_2 \cap \k[s]\\
& = & \B_{glob}(f).
\end{eqnarray*}
\end{proof}

Let us end this section with the proof of Proposition~\ref{p4}
(again, we assume that $a=0$).

\begin{proof}
We shall prove only the first statement. The arguments
are the same for the second equality. Only one inclusion
is non trivial, namely
$\mathrm{ann}_{\Dhat_0[s]}(f^s) \subset \Dhat_0[s] I_1$.
First we have a natural isomorphism
\begin{equation}\label{seq1}
\k[[x]] \otimes_{\k[x]} D[s] f^s
\simeq
\Dhat_0[s] \otimes_{D[s]} D[s] f^s.
\end{equation}
This gives a natural left $\Dhat_0[s]$-module structure on
the tensor product of the left-hand side.

Now let us start with the following exact sequence
of $D[s]$-modules:
\[
0 \rightarrow I_1 \rightarrow D[s] \rightarrow D[s] f^s
\rightarrow 0.
\]
By flatness of $\Dhat_0[s]$ over $D[s]$, we get an exact
sequence of $\Dhat_0[s]$-modules:

\[
0 \rightarrow \Dhat_0[s] I_1 \rightarrow \Dhat_0[s] \rightarrow
\Dhat_0[s] \otimes D[s] f^s \rightarrow 0.
\]

\noindent
Thanks to the isomorphism (\ref{seq1}),
it remains to prove that $\k[[x]] \otimes D[s] f^s$
is naturally isomorphic to $\Dhat_0[s] f^s$.

We have an injective $D[s]$-morphism:
\[0 \to D[s] f^s \to \k[x][1/F,s] f^s.\]
Flatness of $\k[[x]]$ over $\k[x]$ implies the exactness of

\[
0 \to \k[[x]] \otimes_{\k[x]} D[s] f^s \stackrel{\varphi}{\to}
\k[[x]] \otimes_{\k[x]} \k[x][1/F,s]f^s.
\]

On the other hand, there is a natural homomorphism
\[
\k[[x]] \otimes_{\k[x]} \k[x][1/F,s]f^s \stackrel{\psi}{\to}
\k[[x]][1/F,s]f^s.
\]
An arbitrary element of $\k[[x]] \otimes_{\k[x]} \k[x][1/F,s]f^s$ is
written in the form $\sum_\mu \hat{c}_\mu(x)\otimes s^\mu F^{-m}f^s$
with $\hat{c}_\mu(x) \in \k[[x]]$ and $m \in \N$, and it is sent to
$\sum_\mu \hat{c}_\mu(x)s^\mu F^{-m}f^s$ by $\psi$.
The latter is zero if and only if $\hat{c}_\mu(x) = 0$ for any $\mu$.
This shows that $\psi$ is an isomorphism.

An arbitrary element
$\sum_{\mu,\beta} \hat{c}_{\mu,\beta}(x)\otimes s^\mu \partial_x^\beta f^s$ of
$\k[[x]] \otimes_{\k[x]} D[s] f^s$ with $\hat{c}_{\mu,\beta}(x) \in \k[[x]]$
is sent to
$\sum_{\mu,\beta} \hat{c}_{\mu,\beta}(x) s^\mu \partial_x^\beta f^s$
by $\psi\circ\varphi$.
This implies that the image of $\psi\circ\varphi$ coincides with
$\Dhat_0[s]f^s$.

Thus $\psi\circ\varphi$ gives a natural isomorphism $\k[[x]] \otimes
D[s] f^s \simeq \Dhat_0[s] f^s$ and it is naturally
$\Dhat_0[s]$-linear.
Hence we get an exact sequence of $\Dhat_0[s]$-modules
\[0 \rightarrow \Dhat_0[s] I_1 \rightarrow \Dhat_0[s]
\rightarrow \Dhat_0[s] f^s \rightarrow 0\]
with natural maps. This completes the proof.
\end{proof}

\section{Examples}

Here let us start with a result ``well-known to specialists''.
Let us assume $f$ to be in $\k[[x]]^p$.

\begin{lem} \label{lem:classic}
\begin{itemize}
\item[(i)]
Let $u_1, \ldots, u_p$ be units in $\k[[x]]$. Then
$\B_{\k[[x]]}(f)$ is equal to
$\B_{\k[[x]]}(u_1f_1, \ldots, u_p f_p)$.
\item[(ii)]
If $f_{k+1}, \ldots, f_p$ are units in $\k[[x]]$ then
$\B_{\k[[x]]}(f)$ is equal to $\k[s_1, \ldots, s_p] \cdot
\B_{\k[[x]]}(f_1, \ldots, f_k)$.
\item[(iii)]
Let $\K \supset \k$ be a field extension of $\k$. Then
$\B_{\K[x]}(f)=\K[s] \cdot \B_{\k[x]}(f)$ and for any $a \in \k^n$,
$\B_{\K[[x-a]]}(f)=\K[s] \cdot \B_{\k[[x-a]]}(f)$.
\item[(iv)]
Suppose $f \in \k[x]^p$. Let $a \in \k^n$ be such that $f(a)=0$
and $f$ is smooth at $a$.
Then $\B_{loc, a}(f)$ is generated by $\prod_{j=1}^p (s_j+1)$.
\end{itemize}
\end{lem}

\begin{proof}
\begin{itemize}
\item[(i)]
Let us sketch the proof. One can see that it is enough to
make the proof with $f'=(u_1f_1, f_2, \ldots, f_p)$.
By considering the change of variables
$(x,t) \mapsto (x, u^{-1} t_1, t_2, \ldots, t_p)$ one can
prove that the annihilators of $f^s$ and ${f'}^s$ coincide
in $\Dhat_0 \langle t, \ddt \rangle$ from which one
can conclude.
\item[(ii)]
Thanks to (i), we may assume that $f_{k+1}=\cdots=f_p=1$.
Moreover it is enough to make the proof with $k=p-1$.
Set $f'=(f_1, \ldots, f_{p-1})$. The inclusion
$\k[s]\B(f') \subset \B(f)$ is trivial.
Let us prove the converse one.
Set $s'=(s_1,\ldots,s_{p-1})$ and let
$b(s',s_p) \in \B(f)$. We have:
\[b(s',s_p) {f'}^{s'} 1^{s_p} \in
\Dhat_0[s', s_p] f_1 \cdots f_{p-1} {f'}^{s'} 1^{s_p}.\]
Thus we see that for any $\lambda \in \k$, $b(s', \lambda)$
belongs to $\B(f')$. Let us write: $b(s', s_p)=
\sum_{l=0}^d c_k(s') s_p^l$. Let $\lambda_0, \ldots, \lambda_d$
be pairwise distinct elements in $\k$. We thus obtain the
existence of $b_0(s'), \ldots, b_d(s') \in \B(f')$ such that
\[
\left(
\begin{array}{cccc}
1 & \lambda_0 & \cdots & \lambda_0^d \\
\vdots &   &   &  \vdots \\
1 & \lambda_d & \cdots & \lambda_d^d
\end{array}
\right)
\left(
\begin{array}{c}
c_0(s') \\
\vdots \\
c_d(s')
\end{array}
\right)
=
\left(
\begin{array}{c}
b_0(s') \\
\vdots \\
b_d(s')
\end{array}
\right).
\]
This is an invertible Vandermonde matrix from which we deduce
that each $c_l$ is in $\B(f')$. This implies $b(s', s_p) \in
\k[s] \B(f')$.
\item[(iii)]
See e.g. Brian\c{c}on, Maisonobe \cite[Prop. 1.5]{bm02}.
\item[(iv)]
Let $C \subset \k$ be the (finite) set of the coefficients
of the $f_j$'s and let $\K=\Q(C \cup \{a\})$. Then
$\K$ is identified with a subfield of $\C$. So, thanks
to (iii), $\B_{loc,a}(f)=\B_{\C[x]_a}(f)$ and it is well-known
that it is equal to $\B_{\C\{x-a\}}(f)$. Thus we can conclude
with Brian\c{c}on-Maynadier \cite[Prop. 1.2]{bmay}.
\end{itemize}
\end{proof}

\noindent
In the following, $\langle G \rangle$ denotes the
ideal generated by the set $G$.

The computations were made using Kan/sm1 \cite{kan}
and Risa/Asir \cite{asir}. Moreover the
computations were made over the field $\Q$.
In the last paragraph of this section, we check that
the results are valid over $\C$.

\subsection{Example 1}

This first example is trivial in the sense that all the
local Bernstein-Sato ideals can be computed using the previous
lemma. Let us define $f \in \Q[x,y]^3$ by:
\[(f_1,f_2,f_3)(x,y)=(x , y , 1-x-y).\]
Only with the lemma above one can say that given
$a \in \Q^2$, $\B_{loc,a}(f)$ is equal to:
\begin{itemize}
\item
$\Q[s]=\Q[s_1,s_2,s_3]$ for $a \notin \{x=0\} \cup \{y=0\}
\cup \{x+y=1\}$,
\item
$ \langle (s_1+1) \rangle$ for $a \in \{x=0\} \smallsetminus
\{(0,0), (0,1)\}$,
\item
$\langle (s_2+1)\rangle$ for $a \in \{y=0\} \smallsetminus
\{(0,0), (1,0)\}$,
\item
$\langle (s_3+1)\rangle$ for $a \in \{x+y=1\}
\smallsetminus \{(0,1), (1,0)\}$,
\item
$\langle (s_1+1)(s_2+1)\rangle$ if $a=(0,0)$,
$\langle (s_1+1)(s_3+1)\rangle$ if $a=(0,1)$,
$\langle (s_2+1)(s_3+1)\rangle$ if $a=(1,0)$.
\end{itemize}
By using Cor. \ref{cor:glob-loc} one has:

$\B_{glob}(f)=\langle(s_1+1)(s_2+1)(s_3+1) \rangle$.\\
We notice that the global Bernstein-Sato ideal is different
from all the local ones.
By using our algorithm, we found the following primary
decomposition for $I_2 \subset \Q[x,y,s_1,s_2,s_3]$:
\[I_2=\Up_1 \cap \Up_2 \cap \Up_3 \text{ where }
\Up_j=\langle s_j+1, f_j \rangle\]
which obviously enables to recover all the results above.

\subsection{Example 2}

Here, $f \in \Q[x,y]^3$ is given by:
\[(f_1,f_2,f_3)(x,y)= (y , y-2x+1 , y-x^2).\]
A computation by hand of all the local Bernstein-Sato
ideals is far from being easy. The computed primary decomposition
of the ideal $I_2$ has seven primary components $\Up_i$.
For each of them, we present:
$\sqrt{\Up_i}$, $\sqrt{\Up_i \cap \Q[x,y]}$,
$\Up_i \cap \Q[s_1,s_2,s_3]$.
\[
\begin{array}{llll}
(1) &
\langle  s_1+1, y\rangle &
\langle y \rangle &
\langle s_1+1 \rangle \\

(2) &
\langle  s_2+1, y-2x+1 \rangle &
\langle y-2x+1 \rangle &
\langle s_2+1 \rangle \\

(3) &
\langle s_3+1, y-x^2 \rangle &
\langle y -x^2 \rangle &
\langle s_3+1 \rangle \\

(4) &
\langle 2s_1+2s_3+3, x,y \rangle &
\langle x,y \rangle &
\langle 2s_1+2s_3+3 \rangle \\

(5) &
\langle 2s_2+2s_3+3, x-1, y-1 \rangle &
\langle x-1, y-1 \rangle &
\langle 2s_2+2s_3+3 \rangle \\

(6) &
\langle 2s_1+2s_3+5, x, y \rangle &
\langle x,y \rangle &
\langle 2s_1+2s_3+5 \rangle \\

(7) &
\langle 2s_2+2s_3+5, x-1, y-1 \rangle &
\langle x-1, y-1 \rangle &
\langle 2s_2+2s_3+5 \rangle \\
\end{array}
\]
We get:
$\B_{glob}(f)=\langle  (s_1+1) (s_2+1) (s_3+1) (2s_1+2s_3+3)
(2s_1+2s_3+5) (2s_2+2s_3+3) (2s_2+2s_3+5) \rangle$. This ideal
is different from all the local Bernstein-Sato ideals.

\subsection{Example 3}

Here, $f\in \Q[x,y]^2$ is given as follows:
\[(f_1,f_2)=(x^3+y^2, x^2+y^3).\]
This example has been stimulating several authors for several
years (in the global case and in the local case see
e.g. Bahloul \cite{b-JSC, b-kyushu},
Castro-Jim\'enez and Ucha-Enr\'{\i}quez \cite[\S 3.4]{castro-ucha}
and Gago-Vargas et al. \cite[Rem. 2]{gago}).

The obtained primary decomposition of the ideal $I_2$ is made of
twelve primary components $\Up_i$. In the following we present,
for each $i$, the ideals $\sqrt{\Up_i}$,
$\sqrt{\Up_i \cap \Q[x,y]}$ and $\Up_i \cap \Q[s_1, s_2]$.
\[
\begin{array}{llll}
(1) &
\langle x^3+y^2 , s_1+1 \rangle &
\langle x^3+y^2 \rangle &
\langle s_1+1 \rangle \\

(2) &
\langle x^2+y^3 , s_2+1 \rangle &
\langle x^2+y^3 \rangle &
\langle s_2+1 \rangle \\

(3) &
\langle x,y , 4s_1+6s_2+5 \rangle &
\langle x,y \rangle &
\langle 4s_1+6s_2+5 \rangle\\

(4) &
\langle x,y , 4s_1+6s_2+7 \rangle &
\langle x,y \rangle &
\langle 4s_1+6s_2+7 \rangle \\

(5) &
\langle x,y , 4s_1+6s_2+9 \rangle &
\langle x,y \rangle &
\langle 4s_1+6s_2+9 \rangle \\

(6) &
\langle x,y , 4s_1+6s_2+11 \rangle &
\langle x,y \rangle &
\langle 4s_1+6s_2+11 \rangle \\

(7) &
\langle x,y , 4s_1+6s_2+13 \rangle &
\langle x,y \rangle &
\langle 4s_1+6s_2+13 \rangle \\

(8) &
\langle x,y , 6s_1+4s_2+5 \rangle &
\langle x,y \rangle &
\langle 6s_1+4s_2+5 \rangle \\

(9) &
\langle x,y , 6s_1+4s_2+7 \rangle &
\langle x,y \rangle &
\langle 6s_1+4s_2+7 \rangle \\

(10) &
\langle x,y , 6s_1+4s_2+9 \rangle &
\langle x,y \rangle &
\langle 6s_1+4s_2+9 \rangle \\

(11) &
\langle x,y, 6s_1+4s_2+11 \rangle &
\langle x,y \rangle &
\langle 6s_1+4s_2+11 \rangle \\

(12) &
\langle x,y , 6s_1+4s_2+13 \rangle &
\langle x,y \rangle &
\langle 6s_1+4s_2+13 \rangle \\
\end{array}
\]

We get that $\B_{glob}(f)$ coincides with $\B_{loc,0}(f)$ and
is generated by the following polynomial:

\noindent
$(s_1+1) (s_2+1)
(4s_1+6s_2+5)(4s_1+6s_2+7)(4s_1+6s_2+9)(4s_1+6s_2+11)(4s_1+6s_2+13)
(6s_1+4s_2+5)(6s_1+4s_2+7)(6s_1+4s_2+9)(6s_1+4s_2+11)(6s_1+4s_2+13)$.

\subsection{Example 4}

Here $f \in \Q[x,y,z]^2$ is given by:
\[(f_1,f_2)=(z, x^4 + y^4 + 2 z x^2 y^2).\]
This important example is taken from
Brian\c{c}on, Maynadier \cite{bmay} where the authors proved
for the first time that local Bernstein-Sato ideals are not
principal in general.
They did not calculate $\B_{loc,0}(f)$; they only proved that it is
not principal.

The computation of a primary decomposition of $I_2$ outputs
nine primary components $\Up_i$ and
as before, we give, for each $i$:
$\sqrt{\Up_i}$,
$\sqrt{\Up_i \cap \Q[x,y,z]}$ and $\Up_i \cap \Q[s_1, s_2]$.
\[
\begin{array}{llll}
(1) &
\langle z , s_1+1 \rangle &
\langle z \rangle &
\langle s_1+1 \rangle \\

(2) &
\langle f_2 , s_2+1 \rangle &
\langle f_2 \rangle &
\langle s_2+1 \rangle \\

(3) &
\langle x,y , s_2+1 \rangle &
\langle x,y \rangle &
\langle (s_2+1)^2 \rangle \\

(4) &
\langle x,y , 2s_2+1 \rangle &
\langle x,y \rangle &
\langle 2s_2+1 \rangle \\

(5) &
\langle x,y , 4s_2+3 \rangle &
\langle x,y \rangle &
\langle 4s_2+3 \rangle \\

(6) &
\langle x,y , 4s_2+5 \rangle &
\langle x,y \rangle &
\langle 4s_2+5 \rangle \\

(7) &
\langle x,y,z , s_1+2, 2s_2+3 \rangle &
\langle x,y,z \rangle &
\langle s_1+2, 2s_2+3 \rangle \\

(8) &
\langle x,y, z-1 , 2s_2+3 \rangle &
\langle x,y, z-1 \rangle &
\langle 2s_2+3 \rangle \\

(9) &
\langle x,y,z+1 , 2s_2+3 \rangle &
\langle x,y,z+1 \rangle &
\langle 2s_2+3 \rangle
\end{array}
\]

As a consequence, $\B_{loc,0}(f)$ is generated by two elements:
\begin{eqnarray*}
\B_{loc,0}(f) &= &
\langle (s_1+1) (s_2+1)^2 (2s_2+1) (4s_2+3) (4s_2+5) (s_1+2),\\
 & & (s_1+1) (s_2+1)^2 (2s_2+1) (4s_2+3) (4s_2+5) (2s_2+3) \rangle
\end{eqnarray*}
and $\B_{glob}(f)$ is principal:
\[\B_{glob}(f)=\langle
(s_1+1) (s_2+1)^2 (2s_2+1) (2s_2+3) (4s_2+3) (4s_2+5)
\rangle\]

\subsection{Example 5}

Here $f \in \Q[x,y,z]^2$ is given by:
\[(f_1,f_2)=(z, x^5 + y^5 + z x^2 y^3).\]
The computed primary decomposition of $I_2$ is made of
twelve terms $\Up_i$. Again, for each $i=1$, we give
$\sqrt{\Up_i}$,
$\sqrt{\Up_i \cap \Q[x,y,z]}$ and $\Up_i \cap \Q[s_1, s_2]$.
\[
\begin{array}{llll}
(1) &
\langle z , s_1+1 \rangle &
\langle z \rangle &
\langle s_1+1 \rangle \\

(2) &
\langle f_2 , s_2+1 \rangle &
\langle f_2 \rangle &
\langle s_2+1 \rangle \\

(3) &
\langle x,y , s_2+1 \rangle &
\langle x,y \rangle &
\langle (s_2+1)^2 \rangle \\

(4) &
\langle x,y , 5s_2+2 \rangle &
\langle x,y \rangle &
\langle 5s_2+2 \rangle \\

(5) &
\langle x,y , 5s_2+3 \rangle &
\langle x,y \rangle &
\langle 5s_2+3 \rangle \\

(6) &
\langle x,y , 5s_2+4 \rangle &
\langle x,y \rangle &
\langle 5s_2+4 \rangle \\

(7) &
\langle x,y , 5s_2+6 \rangle &
\langle x,y \rangle &
\langle 5s_2+6 \rangle \\

(8) &
\langle x,y,z , s_1+2, 5s_2+7 \rangle &
\langle x,y,z \rangle &
\langle s_1+2, 5s_2+7 \rangle \\

(9) &
\langle x,y,z , s_1+3, 5s_2+7 \rangle &
\langle x,y,z \rangle &
\langle s_1+3, 5s_2+7 \rangle \\

(10) &
\langle x,y,z , s_1+4, 5s_2+7 \rangle &
\langle x,y,z \rangle &
\langle s_1+4, 5s_2+7 \rangle \\

(11) &
\langle x,y,z , s_1+5, 5s_2+7 \rangle &
\langle x,y,z \rangle &
\langle s_1+5, 5s_2+7 \rangle \\

(12) &
\langle x,y,z , s_1+2, 5s_2+8 \rangle &
\langle x,y,z \rangle &
\langle s_1+2, 5s_2+8 \rangle
\end{array}
\]

We get that $\B_{glob}(f)$ and $\B_{loc,0}(f)$ are equal and generated
by these three elements:

$(s_1+1)(s_2+1)^2 (5s_2+2) (5s_2+3) (5s_2+4) (5s_2+6)
(s_1+2) (s_1+3) (s_1+4) (s_1+5)$,

$(s_1+1)(s_2+1)^2 (5s_2+2) (5s_2+3) (5s_2+4) (5s_2+6)
(5s_2+7) (s_1+2)$,

$(s_1+1)(s_2+1)^2 (5s_2+2) (5s_2+3) (5s_2+4) (5s_2+6)
(5s_2+7) (5s_2+8)$.

\subsection{Validity of the computations over $\C$}

First, let us state some general results useful in the sequel.

Let $\k \subseteq \K$ be two fields of characteristic zero.

\begin{lem}\label{lem:ex_elimin}
Let $J$ be an ideal in $\k[y,z]=\k[y_1,
\ldots, y_q, z_1, \ldots, z_r]$ then
\[
(\K[y,z] \cdot J) \cap \K[y] = \K[y] \cdot (J \cap \k[y]).
\]
\end{lem}

\begin{proof}
Let us consider $g \in (\K[y,z] \cdot J) \cap \K[y]$.
Let $f_1, \ldots, f_s$ be a system of generators of $J$
and let us write $g=\sum_j u_j f_j$ with $u_j \in \K[y,z]$.
Let $e_l$, $l\in L$, be a basis of the $\k$-linear space
generated by all the coefficients of the $u_j$'s.
So one can write $g \in \oplus_l J e_l \subset \oplus_l
\k[x,y] e_l$. Since $g \in \K[y]$, we obtain
$g\in \oplus (J \cap \k[y]) e_l$.
The left-right inclusion is proven. The other one being
trivial, the proof is complete.
\end{proof}

\begin{lem}\label{lem:ex_primarity-K-k}
Let $J$ be an ideal in $\k[x]=\k[x_1,\ldots,x_n]$.
\begin{enumerate}
\item
$(\K[x] J) \cap \k[x]=J$.
\item
If $\K[x] J$ is primary in $\K[x]$ then $J$
is primary in $\k[x]$.
\end{enumerate}
\end{lem}

\begin{proof}
The right-left inclusion of (1) is trivial while the
left-right one can be proven by arguments similar to those
used for proving lemma \ref{lem:ex_elimin}.
Claim (2) is a direct consequence of the definitions and (1).
\end{proof}

We shall denote by $\Gal$ the Galois group of the extension
$\K / \k$. If $\tau \in \Gal$ then we shall also denote
by $\tau$ the induced ring automorphism of
$\K[x]=\K[x_1,\ldots,x_n]$.

\begin{lem}\label{lem:ex_Galois_fixed}
Assume that $\K / \k$ is a Galois extension.
Let $J \subset \K[x]=\K[x_1,\ldots,x_n]$ be an ideal.
Suppose that for any $\tau \in \Gal$, $\tau(J) \subset J$.
Then there exists an ideal $J_0 \subset \k[x]$
such that $J=\K[x] J_0$.
\end{lem}

\begin{proof}
Given $\tau \in \Gal$, we have $\tau^{-1}(J) \subset J$ from
which we deduce that $\tau(J)=J$.

Let $G$ be the reduced Gr\"obner basis of $J$ with
respect to a fixed well-order. In view of Buchberger's criterion,
we get that $\tau(G)$ is also the reduced Gr\"obner
basis of $J$ for any $\tau\in \Gal$.
Therefore, for any $g \in G$, $\tau(g)=g$.
Thus, $\tau$ fixes each coefficient of $g$. Since $\tau$ is
arbitrary and the extension $\K/\k$ is Galois, we get
$g \in \k[x]$.
\end{proof}

\begin{prop}\label{prop:ex_main}
Let $\Upsilon$ be a primary ideal of $\k[x]= \k[x_1,\dots,x_n]$
and $\K$ be a field containing the algebraic closure of $\k$.
If the radical of $\K[x]\Upsilon$ is a prime ideal of $\K[x]$,
then $\K[x]\Upsilon$ is a primary ideal of $\K[x]$.
\end{prop}

\begin{proof}
Take an irredundant primary decomposition
\begin{equation}\label{eq:minimal}
 \K[x]\Up = \bigcap_{j=0}^m \Up^{\K}_j
\end{equation}
in $\K[x]$. We assume by contradiction that $m\ge 1$.

The algorithms for computing a primary decomposition imply
that there exists a finite Galois extension $\K'$ of $\k$
such that $\Upsilon^{\K}_j$ are defined over $\K'$
(see e.g. \cite[Chap. 4]{singular}).
Then the field extension from $\K'$ to $\K$ is trivial
in the sense that the primarity of each component is preserved.
Thus we may now assume $\K' = \K$.

Since the radicals $\sqrt{\Upsilon^{\K}_j}$ are distinct
and $\sqrt{\K[x]\Upsilon}$ is prime, we may assume,
without loss of generality that
$\sqrt{\Upsilon^{\K}_0} = \sqrt{\K[x]\Upsilon}$ and
the dimension of $\sqrt{\Upsilon^{\K}_j}$ is less than that of
$\sqrt{\K[x]\Upsilon}$ for $j = 1,\dots,m$.

Let $\tau$ be an element of the Galois group
$\mathrm{Gal}(\K/\k)$. Then
\begin{equation}\label{eq:tau}
 \K[x]\Upsilon = \bigcap_{j=0}^m \tau(\Upsilon^{\K}_j)
\end{equation}
is also an irredundant primary decomposition.
Since the non-embedded primary components are unique,
we have $\tau(\Upsilon^{\K}_0) = \Upsilon^{\K}_0$.
Since $\tau \in \Gal$ is arbitrary,
this implies, by lemma \ref{lem:ex_Galois_fixed},
that $\Upsilon^{\K}_0$ is defined over $\k$,
i.e., there is an ideal $\Upsilon_0$ of $\k[x]$
such that $\Upsilon^{\K}_0 = \K[x]\Upsilon_0$.
By lemma \ref{lem:ex_primarity-K-k}(2), $\Up_0$ is
primary in $\k[x]$.

For each $j = 1,\dots,m$, by lemma \ref{lem:ex_Galois_fixed},
there exists an ideal $\Upsilon_j$ of
$\k[x]$ such that
\begin{equation}\label{eq:Gal}
\K[x]\Upsilon_j = \bigcap_{\tau \in \mathrm{Gal}(\K/\k)}
\tau(\Upsilon^{\K}_j).
\end{equation}
Moreover, $\Upsilon_j$ is primary in $\k[x]$.
Indeed, assume $f,g \in \k[x]$ satisfy $fg \in \Upsilon_j$
and $f \not\in \Upsilon_j$.  Then there exists $\tau \in \Gal$
such that $f \not\in \tau(\Upsilon^{\K}_j)$.
Since $f$ is fixed by every element of $\Gal$, it follows that
$f \notin \tau(\Upsilon^{\K}_j)$ for any
$\tau \in \mathrm{Gal}(\K/\k)$.
In particular, $f \notin \Up_j^\K$. Hence there exists
an integer $\nu$ such that $g^\nu \in \Up_j^\K$. Since
$g^\nu=\tau(g^\nu) \in \tau(\Up_j^\K)$ for any $\tau \in \Gal$,
we have $g^\nu \in \K[x]\Up_j$, which implies, by
lemma~\ref{lem:ex_primarity-K-k}(1), $g^\nu \in \Upsilon_j$.

Combining equalities (\ref{eq:Gal}) and (\ref{eq:tau}), we get
\[
\K[x] \Up= \bigcap_{j=0}^m (\K[x] \Up_j).
\]
Using lemma \ref{lem:ex_primarity-K-k}(1), we obtain a
(not necessarily irredundant) primary decomposition in $\k[x]$:
\[
 \Upsilon = \bigcap_{j=0}^m \Upsilon_j.
\]

Since $\Up$ is primary and $\dim(\Up)=\dim(\Up_0)>
\dim(\Up_j)$ for $j\ge 1$, the uniqueness of the number of
components in irredundant primary decomposition implies that
$\Up_0 \subset \Up_j$ for $j=1, \ldots, m$. Hence we get
\[
\Up^\K_0 = \K[x] \Up_0 \subset \K[x] \Up_j=
\bigcap_{\tau \in \Gal} \tau(\Up_j^\K) \subset \Up_j^\K
\]
for $j=1, \ldots, m$. This contradicts the irredundancy
of (\ref{eq:minimal}).
\end{proof}

Let us return to our problem. We consider
the general situation: $f\in \Q[x]^p$ with $x=(x_1,\ldots,x_n)$.
The ideals $I$, $I_1$, $I_2$ introduced after lemma \ref{lem}
are defined over $\Q$.
Considering $f$ in $\C[x]^p$ one can define the ideals
$I^\C$, $I_1^\C$ and $I_2^\C$ over $\C$.

\begin{lem} \label{lem:ex_QvsC}\
\begin{enumerate}
\item
$I^\C= D_{\C[x]} \langle t, \ddt \rangle \cdot I$,
\item
$I_1^\C = D_{\C[x]} [s] \cdot I_1$,
\item
$I_2^\C = \C[x,s] \cdot I_2$.
\end{enumerate}
\end{lem}
Thus the construction is the same over $\Q$ and over $\C$
until step 2.

\begin{proof}
Statement (1) is a consequence of lemma \ref{lem}.
Statements (2) and (3) are proven exactly as
lemma~\ref{lem:ex_elimin}.
\end{proof}

Let $I_2=\Up_1 \cap \cdots \cap \Up_r$ be a
primary decomposition of $I_2$ (over $\Q$).

\begin{clai*}
In examples 1 to 5, the obtained primary decomposition
of $I_2$ is primary over $\C$.
\end{clai*}

Assume for the moment that this claim is proven, i.e.
the decomposition
$\C[x,s]  I_2= \bigcap_{i=1}^r (\C[x,s]  \Up_i)$ is
primary. Then, applying Proposition \ref{p3}, we get
the following conclusion: for any $\alpha \in \C^n$,
\begin{eqnarray*}
\B_{loc,\alpha}(f) & = & \bigcap_{i \in \sigma_\alpha}
  ((\C[x,s] \Up_i) \cap \C[s])\\
& = & \C[s] \cdot (\bigcap_{i \in \sigma_\alpha} (\Up_i \cap \Q[s]))
  \text{ by lemma } \ref{lem:ex_elimin},
\end{eqnarray*}
where $\sigma_\alpha=\{i \ | \ 1 \le i \le r, \alpha \in
V((\C[x,s] \Up_i) \cap \C[x])\}$. Notice that by
lemma~\ref{lem:ex_elimin}, $(\C[x,s] \Up_i) \cap \C[x]=
\C[x]( \Up_i \cap \Q[x] )$.

Consequently the computations made over $\Q$ shall be valid
over $\C$. Thus to end this section, it remains to prove the
claim.

\begin{proof}[Proof of the claim]
Let $\Up=\Up_i$ be a primary component (over $\Q$) of the
computed decomposition of $I_2$.
By Proposition \ref{prop:ex_main}, it is enough to prove that
$\C[x,s] \sqrt{\Up}$ is prime.

In all the examples, $\sqrt{\Up}$
is either generated by polynomials of degree $1$
(i.e. defines an intersection of affine hyperplanes)
or of the form $\langle s_j+1, f_j \rangle$.
In the first case, $\C[x,s] \sqrt{\Up}$ is obviously prime.
It remains to analyse the second case.

Assume for simplicity that $j=1$, then we have a ring isomorphism
$\C[x,s] / \langle s_1+1, f_1 \rangle \simeq
\C[x,s_2, \ldots,s_p]/ \langle f_1 \rangle$.
This relation shows that it is enough to prove that each
$f_j$ is irreducible over $\C$.

It is obviously the case in Ex. 1, 2 and 3 and for
$f_1$ in Ex. 4 and Ex. 5. Let us check
that $f_2$ in Ex. 4 is irreducible over $\C$. Suppose
that $f_2$ is a product of two polynomials. Regarding the
$z$-degree, we necessarily have a product of the form:
\[x^4 + y^4 +2 z x^2 y^2= u \cdot (v z + w),\]
where $u,v,w \in \C[x,y]$ but this is possible only if $u$ is
a constant.
Thus $f_2$ is irreducible. A similar proof shows
that $f_2$ in Ex. 5 is also irreducible over $\C$.
The proof is complete.
\end{proof}

config: arxiv


TITLE: Range of identity plus compact operator is closed
QUESTION [2 upvotes]: Suppose $K:H\to H$ is a compact linear operator on a Hilbert space $H$.
How do I show that the range of $I+K$ is closed in $H$? I believe this is equivalent to showing that $\{x_n\}\subset H$ and $(I+K)x_n\to y\in H \implies \exists x\in H$ such that $(I+K)x=y$.

REPLY [4 votes]: Hint 1: Reduce problem to the case when all $x_n$ are orthogonal to $\mathrm{Ker}(I+K)$.
Hint 2: Prove ad absurdum that ${x_n}$ is bounded. To start this subproof consider vectors $\Vert x_n\Vert^{-1} x_n$.
Hint 3: Find convergent subsequence in $(I+K)x_{n_k}$. Why it does exist?
Hint 4: Show that $\{x_{n_k}\}$ also have a limit (say $x$) and then show $y=(I+K) x\in \mathrm{Im}(I+K)$


TITLE: Exchanging probabilities
QUESTION [2 upvotes]: I have that $P(a < X_n | F_n) >p$, if I want to find $P(a > X_n | F_n)$, can I just reverse the (direction) probability and does $P(a>X_n | F_n) < 1-p$ or $P(a>X_n | F_n) > 1-p$? Note, $F_n$ is a filtration.
Also, if $P(a < X_n | F_n) >p$, can I say that $P(a < X_{cn} | F_{cn}) >p$ for a constant $c$?

REPLY [1 votes]: First part:
$$
\begin{align}
\mathbb{P}(a>X_n|\mathcal{F}_n)&= \mathbb{E}(\mathbb{1}(a>X_n)|\mathcal{F}_n)\\
&=1-\mathbb{E}(1-\mathbb{1}(a>X_n)|\mathcal{F}_n)\\
&=1-\mathbb{E}(\mathbb{1}(a\leq X_n)|\mathcal{F}_n)\\
&=1-\mathbb{P}(a\leq X_n|\mathcal{F}_n)
\end{align}
$$


TITLE: Intersection of decreasing sequence of compact connected set is connected.
QUESTION [2 upvotes]: Suppose that $F_1, F_2, F_3, . . .$ is a sequence of connected, compact subsets in $\mathbb{R}^n$ and $F_1 ⊃ F_2 ⊃ . . . .$ Show that $K= \bigcap F_n$ is connected.

My approach:If possible, suppose not.
Let, $f$ be a two valued continuous defined on $K$.
Now $K$ is a closed subset of $F_1$,so by Tietze extension theorem we can extend $f$ to a continuous function $f_1:F_1\to\mathbb{R}$.
Since, each of $F_i$ is connected so $f_1(F_i) ⊃ [0,1]$.
We claim $f_1(K)⊃ [0,1]$.
Let, $y\in [0,1]$ then $f_1(x_n)=y$ where $x_n\in F_n$.This $\{x_n\}$ has a convergent subsequence converging to some $l\in F_1$.Now as $F_n$ being a decreasing sequence of compact sets and $l\in F_n$ and hence $l\in K$.Since, $f_1$ is continuous, $f_1 (l)=y\in f_1 (K)$.Thus, our claim is proved which contracdicts our hypothesis as $f_1(K)=f(K)$ is a two valued function.Hence, done!
Is is correct?Any suggestion?Thanks.

REPLY [1 votes]: Your proof is correct. In fact, it generalizes to the case that $F_1$ is a metrizable subset of an arbitrary space $X$.


text:
\begin{document}

\title{Stability of a convex
feasibility problem}

\author{Carlo Alberto De Bernardi
}
\address{Dipartimento di Discipline Matematiche, Finanza Matematica ed Econometria, Universit\`{a} Cattolica del Sacro Cuore, Via Necchi 9, 20123 Milano, Italy}

\email{carloalberto.debernardi@unicatt.it}

\author{Enrico Miglierina
}
\address{Dipartimento di Discipline Matematiche, Finanza Matematica ed Econometria, Universit\`{a} Cattolica del Sacro Cuore, Via Necchi 9, 20123 Milano, Italy}

\email{enrico.miglierina@unicatt.it}

\author{Elena Molho
}
\address{Dipartimento di Scienze economiche e Aziendali, Universit\`{a} degli Studi di Pavia, Via San Felice 5, 27100 Pavia, Italy}

\email{elena.molho@unipv.it}

 \subjclass[2000]{Primary: 90C25; secondary: 90C31, 49J53}

 \keywords{convex feasibility problem, stability, set-convergence}

 \thanks{
}

\maketitle

\begin{abstract}
The 2-sets convex feasibility problem aims at finding a point in
the intersection of two closed convex sets $A$ and $B$ in a normed
space $X$.

More generally, we can consider the problem of finding (if
possible) two points in $A$ and $B$, respectively, which minimize
the distance between the sets.

In the present paper, we study some stability properties for the
convex feasibility problem: we consider two sequences of sets,
each of them converging, with respect to a suitable notion of set
convergence, respectively, to $A$ and $B$. Under appropriate
assumptions on the original problem, we ensure that the solutions
of the perturbed problems converge to a solution of the original
problem. We consider both the finite-dimensional and the
infinite-dimensional case. Moreover, we provide several examples
that point out the role of our assumptions in the obtained
results.
\end{abstract}

\section{Introduction}
The convex feasibility problem is the classical problem of finding
a point in the intersection of a finite collection of closed and
convex sets (see \cite[Section~4.5]{BorweinZhu} for the main
results on this subject). Many concrete problems in applications
can be formulated as a convex feasibility problem. As typical
examples, we mention solution of convex inequalities, partial
differential equations, minimization of convex nonsmooth
functions, medical imaging, computerized tomography and image
reconstruction. For some details and other applications see, e.g.,
\cite{BauschkeBorwein} and the references therein. Moreover, it is
worth to mention the recent annotated bibliography \cite{Censor},

about projection methods, containing several references to the
convex feasibility problem and its applications.

Many efforts have been devoted to the study of algorithmic
procedures to solve convex feasibility problems, both from a
theoretical and from a computational point of view  (see, e.g.,
\cite{BauschkeBorwein,BorweinSimsTam,Hundal,BCOMB} and the
references therein).

Often in concrete applications data are affected by some
uncertainties. Hence stability  of  solutions with respect to data
perturbations is a desirable property, also in view of the
development  of a computational approach to solve the convex
feasibility problem. Our paper is devoted to investigate some
stability properties of the  $2$-sets convex feasibility problem
by using set convergence notions. We will also consider the case
of a pair of closed and convex sets with empty intersection: in
this case a solution of the problem is a pair of minimal distance
elements  of the two sets.

In this paper, we investigate a sequence of perturbed convex
feasibility problems whose data are obtained by considering two
sequences of closed and convex sets $\left\lbrace A_n
\right\rbrace$ and $\left\lbrace B_n \right\rbrace$ converging
respectively to the sets $A$ and $B$. If the intersection of $A_n$
and $B_n$ is empty, we consider, as a solution of the $n$-th
perturbed problem, the pair of elements $a_n\in A_n$  and $b_n\in
B_n$ such that the distance between $A_n$ and $B_n$ is $\left\Vert
a_{n}-b_{n}\right\Vert$.

Our aim is to find some conditions that guarantee the convergence
of the solutions of the perturbed convex feasibility problems to a
solution of the original convex feasibility problem.

We obtain some stability results both in the finite-dimensional
and in the infinite-dimensional framework, using the
Kuratowski-Painlev\'{e} convergence notion in the
finite-dimensional case and the Attouch-Wets convergence in the
infinite-dimensional setting. Moreover, we give some examples
showing that the assumptions that we use to guarantee the
stability features of a given convex feasibility problem cannot be
avoided, both in the finite and in the infinite-dimensional case.

The paper is organized as follows. Section~\ref{Notations and
preliminaries} is devoted to definitions and preliminary results,
mainly concerning the various notions of set-convergence.
Section~\ref{Finite-dimensional} presents  a stability result for
the convex feasibility problem when $A$ and $B$ are contained in a
finite-dimensional normed vector space and the sequences of closed
and convex sets $\left\lbrace A_n \right\rbrace$ and $\left\lbrace
B_n \right\rbrace$ converge in the Kuratowski-Painlev\'{e} sense
respectively to $A$ and $B$. Section~\ref{Infinite-dimensional} is
devoted to study the stability properties of a convex feasibility
problem in an infinite-dimensional setting. Here, we use the
Attouch-Wets convergence, that is stronger than the
Kuratowski-Painlev\'{e} convergence, even if they coincide in the
finite-dimensional setting. Moreover, it is worth to be noticed
that we obtain results concerning both weak and norm convergence
of the solutions of perturbed problems to a solution of the
original problem. In order to obtain the norm convergence of a
sequence of solutions of perturbed problems, we assume that $A$
has nonempty interior and it is locally uniformly rotund (LUR) at
a given solution $a$. Hence, we use a geometrical notion that
strengthens the convexity assumption used to prove the weak
convergence result. Finally, in Section~\ref{sectionesempi}, we
provide some rather involved examples in $\ell_2$ that point out
the role of our assumptions even in a Hilbert space framework.

\section{Notations and preliminaries}\label{Notations and preliminaries}

Throughout all this paper, $X$ denotes a real normed space with
the topological dual $X^*$. We
denote by $B_X$ and $S_X$ the closed unit ball and the unit sphere of $X$, respectively.

For $x,y\in X$, $[x,y]$ denotes the closed segment in $X$ with
endpoints $x$ and $y$, and $(x,y)=[x,y]\setminus\{x,y\}$ is the
corresponding ``open'' segment. For a subset $K$ of $X$,
$\alpha>0$, and a functional $x^*\in \SX$ bounded on $K$, let
$$S(x^*,\alpha,K)=\{x\in
K;\, x^* x\geq\sup x^*(K)-\alpha\}$$ be the closed slice of $K$
given by $\alpha$ and $x^*$.

For a subset $A$ of $X$, we denote by $\inte A$, $\conv(A)$ and
$\cconv(A)$ the interior, the convex hull and the closed convex
hull of $A$, respectively. Moreover,
$$\overline{\mathrm{cone}}(A)=\cconv{\bigl([0,\infty)\cdot
A\bigr)}$$ is the closed convex cone generated by the set $A$. We
denote by $$\textstyle \mathrm{diam}(A)=\sup_{x,y\in A}\|x-y\|,$$
the (possibly infinite) diameter of $A$. For $x\in X$, let
$$\dist(x,A) =\inf_{a\in A} \|a-x\|.$$ Moreover, given $A,B$
nonempty subset of $X$,  we denote by $\dist(A,B)$ the usual
``distance'' between $A$ and $B$, that is,
$$ \dist(A,B)=\inf_{a\in A} \dist(a,B).$$

\subsection*{Convergence of sets}

By $\C(X)$ we denote the family of all nonempty closed subsets of
$X$.

Let  $\{A_n\}$ be a sequence in $\C(X)$ and let us consider the
following sets:

$$\textstyle \mathrm{Li}\ A_n=\{x\in X;\, x=\lim_n x_n, x_n\in A_n\}$$
and

$$\textstyle \mathrm{Ls}\ A_n=\{x=\lim_k x_k\in X;\, x_k\in A_{n_k}, \{n_k\}\, \hbox{is a subsequence of the integers} \}.$$

\begin{definition}
\label{Kuratowski} Let $\{A_n\}$ be a sequence in $\C(X)$ and
$A\in\C(X)$.
\begin{enumerate}
\item $\{A_n\}$ converges to $A$ for the lower
Kuratowski-Painlev\'{e} convergence if{f} $A\subset\mathrm{Li}
A_n$. \item $\{A_n\}$ converges to $A$ for the upper
Kuratowski-Painlev\'{e} convergence if{f} $A\supset\mathrm{Ls}
A_n$. \end{enumerate} Moreover, we say that $\{A_n\}$ converges to
$A$ for the Kuratowski-Painlev\'{e} convergence
($A_{n}\stackrel{K}{\rightarrow}A$) if{f} $\{A_n\}$ converges to
$A$ for the upper and the lower Kuratowski-Painlev\'{e}
convergence.
\end{definition}

Now, let us introduce the (extended) Hausdorff metric $h$ on
$\C(X)$. For $A,B\in\C(X)$, we define the excess of $A$ over $B$
as

$$e(A,B) = \sup_{a\in A} d(a,B).$$

\noindent Moreover, if $A\neq\emptyset$ and $B=\emptyset$ we put
$e(A,B)=\infty$,  if $A=\emptyset$ we put $e(A,B)=0$. We define

$$h(A,B)=\max \bigl\{ e(A,B),e(B,A) \bigr\}.$$

\begin{definition} A sequence $\{A_j\}$ in $\C(X)$ is said to
Hausdorff converge to $A\in\C(X)$ if $$\textstyle \lim_j h(A_j,A)
= 0.$$
\end{definition}

Finally, we introduce the so called Attouch-Wets convergence (see,
e.g., \cite[Definition~8.2.13]{LUCC}), which can be seen as a
localization of the Hausdorff convergence.  If $N\in\N$ and
$A,B\in\C(X)$, define
\begin{eqnarray*}
e_N(A,C) &=& e(A\cap N B_X, C)\in[0,\infty),\\
h_N(A,C) &=& \max\{e_N(A,C), e_N(C,A)\}.
\end{eqnarray*}

\begin{definition} A sequence $\{A_j\}$ in $\C(X)$ is said to
Attouch-Wets converge to $A\in\C(X)$ if, for each $N\in\N$,
$$\textstyle \lim_j h_N(A_j,A)= 0.$$
\end{definition}

We recall that in the finite-dimensional case the Attouch-Wets
convergence and the Kuratowski-Painlev\'{e} convergence coincide
(see, e.g., \cite[Section 8.2]{LUCC}).

In the sequel, we use the following easy-to-prove fact. For the
convenience of the reader we provide a proof.

\begin{fact}\label{distanzanulla} Let $A$ and $B$
two closed and convex subsets of a normed space $X$ Let $\{A_n\}$
and $\{B_n\}$ be two sequences of closed convex sets such that
$A_n\rightarrow A$ and $B_n\rightarrow B$ for the lower
Kuratowski-Painlev\'{e} convergence. Then $$\textstyle \limsup_n
\dist(A_n,B_n)\leq \dist(A,B).$$ In particular, if $A\cap
B\neq\emptyset$ we have $\lim_n \dist(A_n,B_n)=0$.
\end{fact}

\begin{proof} Let $\epsilon>0$ and let $x\in A$ and $y\in B$ be
such that $\dist(A,B)\leq\|x-y\|+\epsilon$. Since $A_n\rightarrow
A$ and $B_n\rightarrow B$ for the lower Kuratowski-Painlev\'{e}
convergence, there exist two sequences $\{x_n\}$ and $\{y_n\}$
such that $x_n\to x$, $y_n\to y$ and, for each $n\in\N$, $x_n\in
A_n$, $y_n\in B_n$. In particular, it eventually holds
$\|x_n-x\|\leq \epsilon$ and $\|y_n-y\|\leq \epsilon$. Hence, the
following inequalities eventually hold:
$$\dist(A_n,B_n)\leq\|x_n-y_n\|\leq\dist(A,B)+3\epsilon.$$
By the arbitrariness of $\epsilon>0$, we have the thesis.
\end{proof}

\section{Convergence of minimal distance points of a pair of convex
sets: the finite-dimensional case}\label{Finite-dimensional}

In this section, we denote by $X$ a finite-dimensional normed
space.

\begin{definition} Let $A,B$ be nonempty
closed convex set in $X$. Let
$$\m(A,B)=\{a\in A;\, \dist(a,B)=\dist(A,B)\}.$$
\end{definition}

It is easy to see that $\m(A,B)$ is a closed convex set.

\begin{definition}
Let $C$ be a non empty closed convex set and $x\in C$. Let us
define
$$D(x)=\{d\in X;\, x+td\in C,\ \forall t>0\}.$$
\end{definition}

\begin{remark} By \cite[Proposition~2.1.5]{AT2003}, if $x,y\in C$ then
$D(x)=D(y)$. That is, the set $D(x)$ does not depend on $x\in C$.
We denote this set, called the {\em asymptotic cone} of $C$, by
$C_\infty$.
\end{remark}

We prove the following lemma that will be useful in the sequel (it
can be seen as a slight generalization of
\cite[Proposition~2.1.9]{AT2003}).

\begin{lemma}\label{lemma:conirecessione} Let $A$ and $B$ be nonempty
closed convex sets in $X$ such that $\m(A,B)$ is nonempty. Then
$$A_\infty\cap B_\infty=[\m(A,B)]_\infty.$$
\end{lemma}

\begin{proof} Let $a\in A$ and $b\in B$ be such that
$\|b-a\|=\dist(A,B)$.

Let us prove that $A_\infty\cap B_\infty\subset[\m(A,B)]_\infty$.
Let $d\in A_\infty\cap B_\infty=D(a)\cap D(b)$. Since, for each
$t>0$,
$$\|a-b\|=\|a+td-(b+td)\|=\dist(A,B),$$
we have that $a+td\in\m(A,B)$, whenever $t>0$. Hence $d\in
[\m(A,B)]_\infty$.

For the reverse inclusion, suppose that $a+td\in \m(A,B)$,
whenever $t>0$. Clearly, $d\in A_\infty$. Now, we prove that $d\in
B_\infty$. Let us fix $t>0$ and $n\in\N$, and let us observe that
$$\dist(b+ntd, B)\leq2\,\dist(A,B).$$
Hence, there exists $d_n\in B$ such that
$$\|b+ntd-d_n\|\leq2\,\dist(A,B).$$
Then,
$$\textstyle \|b+td-(b+\frac{d_n-b}n)\|\leq\frac2n\,\dist(A,B).$$
By the arbitrariness of $n\in\N$, since $b+\frac{d_n-b}n\in B$,
and since $B$ is closed, it holds that $b+td\in B$. By the
arbitrariness of $t>0$, the thesis is proved. \end{proof}

The following theorem is the main result of this section. It
proves that, under mild assumption, the 2-sets convex feasibility
problem has a considerable degree of stability.

\begin{theorem}
\label{thm:stability principale} Let $\{ A_{n}\} $ and $\{
B_{n}\}$ be two sequences of nonempty closed convex sets in $X$,
$A$ and $B$ two nonempty closed convex subsets of $X$ such that
$$A_{n}\to A\quad\text{and}\quad B_{n}\to B,$$
for the Kuratowski-Painlev\'{e} convergence. Suppose that
$\m(A,B)$ is a nonempty bounded set. Let $\left\{ a_{n}\right\} $
and $\left\{ b_{n}\right\}$ be  sequences such that $a_{n}\in
A_{n}$, $b_{n}\in B_{n}$ ($n\in\N$) and
\[
\text{dist}(A_{n},B_{n})=\left\Vert a_{n}-b_{n}\right\Vert.
\]
Then there exists a subsequence $\left\{ a_{n_{k}}\right\}$ such
that
\[
\lim_{k\rightarrow\infty}a_{n_{k}}=c\in \m(A\,,B).
\]
Moreover, if $\m(A,B)=\{a\}$ then $a_n\to a$.
\end{theorem}

\begin{proof} Let us prove the first part of the theorem. By
Fact~\ref{distanzanulla}, it holds
\begin{equation}\label{eq:distanza tende a zero}
\limsup_{n} \|a_n-b_n\|\leq\dist(A,B).
\end{equation}
We claim that $\{a_n\}$ and $\{b_n\}$ are bounded.

Suppose that this is not the case and let $a\in A$ and $b\in B$
such that $\|a-b\|=\dist(A,B)$. Without loss of generality, we can
suppose that $\|a_n\|,\|b_n\|\to\infty$. By the lower part of the
convergence of $\left\{ A_{n}\right\} $ there exists a sequence
$\left\{ a_{n}'\right\} $ such that $a_{n}'\in A_{n}$ and
$a_{n}'\rightarrow a$. Since $A_{n}$ is a convex set, for any
$\alpha\in\left[0,1\right]$ it holds:
\[
\alpha a_{n}+(1-\alpha)a_{n}'=a_{n}'+\alpha(a_{n}-a_{n}')\in
A_{n}.
\]

The sequence
\[
\left\{ \frac{a_{n}-a_{n}'}{\left\Vert a_{n}-a_{n}'\right\Vert
}\right\}
\]
has a subsequence converging to $d\neq0.$ There is no loss of
generality in assuming
\[
\lim_{n\rightarrow+\infty}\frac{a_{n}-a_{n}'}{\left\Vert
a_{n}-a_{n}'\right\Vert }=d.
\]
Therefore, it holds
\[
a+\beta
d=\lim_{n\rightarrow+\infty}\left(a_{n}'+\frac{\beta}{\left\Vert
a_{n}-a_{n}'\right\Vert }(a_{n}-a_{n}')\right).
\]

Since for every $\beta>0$ there exists $n_{2}(\beta)\in\N$ such
that $\frac{\beta}{\left\Vert a_{n}-a_{n}'\right\Vert
}\in\left[0,1\right],$ whenever $n>n_{2}(\beta)$, it holds
\[
a+\beta d\in A,
\]
for every $\beta>0.$ Hence, $d\in A_{\infty}.$

Analogously, we may prove that
\[
\lim_{n\rightarrow+\infty}\frac{b_{n}-b_{n}'}{\left\Vert
b_{n}-b_{n}'\right\Vert }=d'\in B_{\infty},
\]
where $\left\{ b_{n}'\right\} $ is a sequence such that $b_{n}'\in
B_{n}$ and $b_{n}'\rightarrow b$.

Let us observe that $\{a'_n\}$, $\{b'_n\}$ and $\{a_n-b_n\}$ are
bounded sequences in $X$. Since $\|a_n\|,\|b_n\|\to\infty$, we
have $\|a_n-a'_n\|\sim\|b_n-b'_n\|$ and hence
$$d=\lim_{n\rightarrow+\infty}\frac{a_{n}-a_{n}'}{\left\Vert
a_{n}-a_{n}'\right\Vert
}=\lim_{n\rightarrow+\infty}\frac{b_{n}-b_{n}'}{\left\Vert
b_{n}-b_{n}'\right\Vert }=d',
$$
Therefore we have
\[
0\neq d\in A_{\infty}\cap B_{\infty}.
\]
By Lemma~\ref{lemma:conirecessione}, we have
\[
A_{\infty}\cap B_{\infty}=[\m(A,B)]_{\infty}.
\]
Then $\m(A,B)$ is not a bounded set, a contradiction.

By the claim  and compactness, there exist two subsequences
$\left\{ a_{n_{k}}\right\} $ and $\left\{ b_{n_{k}}\right\} $,
respectively of $\left\{ a_{n}\right\} $ and of $\left\{
b_{n}\right\} $, such that
\[
\lim_{k\rightarrow+\infty}a_{n_{k}}=u,\quad\lim_{k\rightarrow+\infty}b_{n_{k}}=v,
\]
where $u\in A$ and $v\in B$. By Fact~\ref{distanzanulla},
$\|u-v\|=\dist(A,B)$ and the thesis is proved.

The second part of the theorem follows easily by the first part.
\end{proof}

\begin{remark}\label{remark:stabilityfunctions} The above theorem can be proved in an alternative way,
by using known results concerning stability theory for convex
optimization problem. However, we preferred to present a direct
and more geometrical proof. We give a sketch of the alternative
proof below. (See, e.g., \cite{LUCC} for definitions and main
results about convergence of functions and well-posed problems).

Let $f,f_n:X\times X\to(\-\infty,\infty]$ ($n\in\N$) the convex
lower semicontinuous functions defined as follows. For each
$(x_1,x_2)\in X\times X$ and $n\in\N$, put
$$
f(x_1,x_2)=\begin{cases}
\|x_1-x_2\|\ \ \ \ \ \  &\text{if $x_1\in A$ and $x_2\in B$;}\\
\infty\ \ \ \ \ \  &\text{otherwise;}
\end{cases}
$$
and
$$
f_n(x_1,x_2)=\begin{cases}
\|x_1-x_2\|\ \ \ \ \ \  &\text{if $x_1\in A_n$ and $x_2\in B_n$;}\\
\infty\ \ \ \ \ \  &\text{otherwise.}
\end{cases}
$$
Since $A_{n}\to A$ and $B_{n}\to B$ for the
Kuratowski-Painlev\'{e} convergence (equivalently, for the
Attouch-Wets convergence), we have that $f_n\to f$ for the
Kuratowski-Painlev\'{e} convergence. Moreover, proceeding as in
the proof of Theorem~\ref{thm:stability principale}, we may prove
that $f$ is Tykhonov well-posed in the generalized sense. Hence,
we can apply \cite[Theorem~10.2.24]{LUCC} to obtain the thesis.
\end{remark}

Whenever the two limit sets are such that $A\cap B\neq\emptyset$,
we have the following corollary.

\begin{corollary}
\label{thm:stabilitycorollario} Let $\{ A_{n}\} $ and $\{ B_{n}\}$
be two sequences of nonempty closed convex sets in $X$, $A$ and
$B$ two nonempty closed convex subsets of $X$ such that
$$A_{n}\to A\quad\text{and}\quad B_{n}\to B,$$
for the Kuratowski-Painlev\'{e} convergence. Suppose that $A\cap
B$ is a nonempty bounded set. Let $\left\{ a_{n}\right\} $ and
$\left\{ b_{n}\right\}$ be  sequences such that $a_{n}\in A_{n}$,
$b_{n}\in B_{n}$ ($n\in\N$) and
\[
\text{dist}(A_{n},B_{n})=\left\Vert a_{n}-b_{n}\right\Vert.
\]
Then there exist two subsequences $\left\{ a_{n_{k}}\right\} $ and
$\left\{ b_{n_{k}}\right\} $ such that
\[
\lim_{k\rightarrow\infty}a_{n_{k}}=\lim_{k\rightarrow\infty}b_{n_{k}}=c\in
A\cap B.
\]
Moreover, if $A\cap B=\{c\}$ then $a_n,b_n\to c$.
\end{corollary}

The following examples show that both the assumptions in Theorem
\ref{thm:stability principale} play an independent role and each
of them cannot be deleted. The first one focuses on the role of
convexity assumptions.

\begin{example}
\label{exa:main theorem convex} Let us consider the sets
$(n\geq2)$:
\[
\begin{array}[t]{c}
\textstyle A_{n}=\left\{
\left(x_{1},x_{2}\right)\in\mathbb{R}^{2}:-1\leq
x_{1}\leq-\frac{1}{n},
\,{n}\frac{x_{1}+1}{n-1}\leq x_{2}\leq2+{n}\frac{x_{1}+1}{1-n}\right\} \cup\\
\textstyle \left\{
\left(x_{1},x_{2}\right)\in\mathbb{R}^{2}:-1\leq
x_{1}\leq-\frac{1}{2n},\,\frac{2}{x_{1}^{2}}\leq
x_{2}\leq8n^{2}\right\}
\end{array}
\]
and
\[
\begin{array}[t]{c}
B_{n}=\left\{ \left(x_{1},x_{2}\right)\in\mathbb{R}^{2}:\frac{1}{n}\leq x_{1}\leq1,\,{n}\frac{x_{1}-1}{n-1}\leq x_{2}\leq2+{n}\frac{x_{1}-1}{1-n}\right\} \cup\\
\left\{ \left(x_{1},x_{2}\right)\in\mathbb{R}^{2}:\frac{1}{2n}\leq
x_{1}\leq1,\,\frac{2}{x_{1}^{2}}\leq x_{2}\leq8n^{2}\right\} .
\end{array}
\]
The sequences $\left\{ A_{n}\right\} $ and $\left\{ B_{n}\right\}
$ converge respectively to
\[
\begin{array}[t]{c}
A=\left\{ \left(x_{1},x_{2}\right)\in\mathbb{R}^{2}:-1\leq x_{1}\leq0,\, x_{1}+1\leq x_{2}\leq1-x_{1}\right\} \cup\\
\left\{ \left(x_{1},x_{2}\right)\in\mathbb{R}^{2}:-1\leq
x_{1}<0,\, x_{2}\geq\frac{2}{x_{1}^{2}}\right\}
\end{array}
\]
and
\[
\begin{array}[t]{c}
B=\left\{ \left(x_{1},x_{2}\right)\in\mathbb{R}^{2}:0\leq x_{1}\leq1,\,1-x_{1}\leq x_{2}\leq1+x_{1}\right\} \cup\\
\left\{ \left(x_{1},x_{2}\right)\in\mathbb{R}^{2}:0\leq
x_{1}\leq1,\, x_{2}\geq\frac{2}{x_{1}^{2}}\right\} .
\end{array}
\]
It is easy to see that $A\cap B=\left\{ (0,1)\right\} $ and
\[
A_{n}\stackrel{K}{\rightarrow}A,\quad
B_{n}\stackrel{K}{\rightarrow B}.
\]

All the assumptions of Theorem \ref{thm:stability principale} are
satisfied except for the convexity of $A_{n}$ and $B_{n}$. The
minimal distance between the sets $A_{n}$ and $B_{n}$ is achieved
only at the pair of points
\[
a_{n}=\left(-\frac{1}{2n},8n^{2}\right)\in A_{n}\:\mathrm{and}\:
b_{n}=\left(\frac{1}{2n},8n^{2}\right)\in B_{n}.
\]
It is apparent that the sequences $\left\{ a_{n}\right\} $ and
$\left\{ b_{n}\right\} $ have no convergent subsequences. Hence
the thesis of Theorem \ref{thm:stability principale} does not
hold.
\end{example}

The second example proves that the boundedness assumption on the
set $\m(A,B)$ cannot be dropped.
\begin{example}
\label{exa:main theorem bounded assumption}Let $A_{n}$ and $B_{n}$
be defined as in Example~\ref{exa:main theorem convex}. Let us
consider the sets
\[
C_{n}=\mathrm{conv}(A_{n})\:\mathrm{and}\,
D_{n}=\mathrm{conv}(B_{n}).
\]
It is easy to see that
\[
C_{n}\stackrel{K}{\rightarrow}C,\quad
D_{n}\stackrel{K}{\rightarrow}D,
\]
where
\[
C=\left\{ \left(x_{1},x_{2}\right)\in\mathbb{R}^{2}:-1\leq
x_{1}\leq0,\, x_{1}+1\leq x_{2}\right\}
\]
and
\[
D=\left\{ \left(x_{1},x_{2}\right)\in\mathbb{R}^{2}:0\leq
x_{1}\leq1,\,1-x_{1}\leq x_{2}\right\} .
\]
Moreover, we have $C\cap D=\left\{
\left(x_{1},x_{2}\right)\in\mathbb{R}^{2}:x_{1}=0,\,
x_{2}\geq1\right\} $.

All the assumptions of Theorem~\ref{thm:stability principale} are
satisfied except for the boundedness of the set $C\cap D$. The
minimal distance between the sets $C_{n}$ and $D_{n}$ is achieved
only at the same pair of points $a_{n}\in C_{n}$ and $b_{n}\in
D_{n}$ as in Example~\ref{exa:main theorem convex}. Of course, as
in the previous example both the sequences $\left\{ a_{n}\right\}
$ and $\left\{ b_{n}\right\} $ have no convergent subsequences.
Therefore the thesis of Theorem~\ref{thm:stability principale}
does not hold.
\end{example}

\section{Convergence of minimal distance points of a pair of convex
sets: the infinite-dimensional case}\label{Infinite-dimensional}

In an infinite-dimensional setting, we need some strengthenings of
the assumptions to obtain stability results for our problems.
Indeed, Example~\ref{esempiocono-cono}, in
Section~\ref{sectionesempi}, shows that an analogue of
Theorem~\ref{thm:stability principale} does not hold, even if we
assume that the sequences of sets converge for the Hausdorff
convergence and that the space $X$ is a Hilbert space. In this
section, we prove that an additional geometric condition on the
limit sets ensures the stability result (see
Theorem~\ref{puntolur} below). Moreover, we use the Attouch-Wets
convergence of sets instead of the Kuratowski-Painlev\'{e}
convergence.

\medskip

We start with some definitions and preliminary results. Let us
recall that a {\em body} in $X$ is a closed convex set in $X$ with
nonempty interior.

\begin{definition}[{see, e.g., \cite[Definition~7.10]{FHHMZ}}] Let $A$ be a nonempty subset of a normed space $X$. A point $a\in A$
is called a strongly exposed point of $A$ if there exists  a
support functional $f\in X^*\setminus\{0\}$ for $A$ in $a$ (i.e.,
$f (a) = \sup f(A)$), such that  $x_n\to a$ for all sequences
$\{x_n\}$ in $A$ such that $\lim f(x_n) = \sup f(A)$. In this
case, we say that $f$ strongly exposes $A$ at $a$.
\end{definition}

Let us observe that $f\in S_{X^*}$ strongly exposes $A$ at $a$
if{f} $f(a)=\sup f(A)$ and
$$\mathrm{diam}\bigl(S(f,\alpha,A)\bigr)\to0 \text{ as } \alpha\to 0.$$

\begin{definition}
Let $A\subset X$ be a body. We say that $x\in\partial A$ is an
{\em LUR (locally uniformly rotund) point} of $A$ if for each
$\epsilon>0$ there exists $\delta>0$ such that if $y\in\partial A$
and $\dist(\partial A,(x+y)/2)<\delta$ then $\|x-y\|<\epsilon$. If
$A=B_X$, this definition coincides with the standard definition of
local uniform rotundity of the norm at $x$.

Moreover, we say that $A$ is an {\em LUR body} if each point in
$\partial A$ is an LUR point of $A$.
\end{definition}

\begin{lemma}\label{slicelimitatoselur} Let $A$ be a body in $X$
and suppose that $a\in\partial A$ is an LUR point of $A$. Then, if
$f\in S_{X^*}$ is a support functional for $A$ in $a$, $f$
strongly exposes $a$. Moreover, every slice $S$ of the form
$S=S(f,\alpha,A)$ is a bounded set.
\end{lemma}

The first part of the lemma is well-known in the case the body is
a ball (see e.g. \cite[Exercise~8.27]{FHHMZ}) and in the general
case the proof is similar. However, for the convenience of the
reader we include a proof.

\begin{proof} Without loss of generality, we can suppose that
$a=0$. Fix $w\in \inte A$ and observe that $f(w)<0$.

Let us prove the first part of the lemma. Let $\alpha>0$, $z\in
S=S(f,\alpha,A)$ and
$$\textstyle z'=z-\frac {f(z)}{f(w)}w.$$
Since $\frac z2\in A$ and $f(z')=0$, we have that $[\frac z2,\frac
{z'}2]\cap \partial A\neq\emptyset$. Hence
$$\textstyle \dist(\partial A,\frac z2)\leq \frac12\|z'-z\|\leq \frac12\frac{\|w\|}{|f(w)|}\alpha.$$
Since $a=0$ is an LUR point of $A$, if $\alpha\to0$ then
$\mathrm{diam}(S)\to0$ and the proof is concluded.

Now, the second part of the lemma follows easily. Suppose on the
contrary that there exists $\alpha>0$ such that $S=S(f,\alpha,A)$
is unbounded. Then there exists a sequence $\{y_n\}$ in
$S\setminus\{0\}$ such that $\|y_n\|\to\infty$. Put
$z_n=\frac{y_n}{\|y_n\|}$ and observe that $\|z_n\|=1$ and $z_n\in
S(f,\alpha/\|y_n\|,A)$, a contradiction by the first part of the
lemma.
\end{proof}

\begin{lemma}\label{convnorma} Let $X$ be a normed space. There exists a constant $\Gamma>0$ such
that if $R>1$, if $x,y,a,b\in X$ are such that $\|x\|,\|y\|< R$
and $\|a\|,\|b\|>2 R$, then, if $[x,a]\cap RS_X=\{a'\}$ and
$[y,b]\cap RS_X=\{b'\}$, it holds
$$\|b'-a'\|\leq \Gamma\max\{\|x\|,\|y\|,\|a-b\|\}.$$
\end{lemma}

\begin{proof} Let $\lambda,\mu\in(0,1)$ be such that $a'=\lambda
a+(1-\lambda)x$ and $b'=\mu b+(1-\mu)y$. By the triangle
inequality, it follows easily that
$$\textstyle \frac{R-\|y\|}{\|b\|-\|y\|}\leq\mu
.$$ Moreover, since

$$\textstyle R=\|\lambda
a+(1-\lambda)x\|\geq\lambda\|a\|-(1-\lambda)\|x\|,$$ we have
$$\textstyle \lambda\leq \frac{R+\|x\|}{\|a\|+\|x\|}.$$

Without loss of generality, we can assume that $\lambda>\mu$. If
we denote $$d=\max\{\|x\|,\|y\|,\|a-b\|\},$$ we have
\begin{eqnarray*}
\textstyle \|b'-a'\| &\leq& \textstyle \lambda
                   \|a-b\|+(1-\lambda)\|x-y\|+|\lambda-\mu|(\|y\|+\|b\|)\\
          &\leq& \textstyle
3d+\bigl(\frac{R+\|x\|}{\|a\|+\|x\|}-\frac{R-\|y\|}{\|b\|-\|y\|}\bigr)(\|y\|+\|b\|)\\
    &\leq& \textstyle
3d+\frac{R(\|b\|-\|a\|-\|y\|-\|x\|)+\|a\|\,\|y\|+\|b\|\,\|x\|}{(\|a\|+\|x\|)(\|b\|-\|y\|)}(\|y\|+\|b\|)\\
    &\leq& \textstyle
3d+\frac{R(\|b\|-\|a\|-\|y\|-\|x\|)+\|a\|\,\|y\|+\|b\|\,\|x\|}{(\|a\|+\|x\|)(\|b\|/2)}(2\|b\|)\\
    &\leq& \textstyle
3d+4\frac{R(\|b\|-\|a\|-\|y\|-\|x\|)}{2R}+4\frac{\|a\|\,\|y\|+\|b\|\,\|x\|}{\|a\|+\|x\|}\\
    &\leq& \textstyle
    3d+2\bigl(\bigl|\|b\|-\|a\|\bigr|\bigr)+\frac{4\|a\|\,\|y\|}{\|a\|+\|x\|}+\frac{4\|b\|\,\|x\|}{\|a\|+\|x\|}\\
&\leq& \textstyle
    5d+\frac{4\|a\|}{\|a\|+\|x\|}d+\frac{4\|b-a+a\|\,\|x\|}{\|a\|+\|x\|}\\
&\leq& \textstyle
    5d+4d+\frac{4\|x\|}{\|a\|+\|x\|}\|b-a\|+\frac{4\|a\|}{\|a\|+\|x\|}\|x\|\leq \textstyle 17d\\
    \end{eqnarray*}
 The proof is concluded if we set $\Gamma=17$

\end{proof}

The following theorem is the main result of this section.

\begin{theorem}\label{puntolur} Let $X$ be a normed space, $B$
a nonempty closed convex subsets of $X$, $A$ a body in $X$ and
$a\in\partial A$ an LUR point of $A$. Let $\{A_n\}$ and $\{B_n\}$
be two sequences of closed convex sets such that $A_n\rightarrow
A$ and $B_n\rightarrow B$ for the Attouch-Wets convergence.
Suppose that $\{a_n\}$ and $\{b_n\}$ are sequences in $X$ such
that $a_n\in A_n,\ b_n\in B_n$ ($n\in \N$) and
$$\dist(A_n,B_n)=\|a_n-b_n\|.$$
Suppose that at least one of the following conditions holds.
\begin{enumerate}
\item[(1)] $A\cap B=\{a\}$. \item[(2)] $A\cap B=\emptyset$ and
there exists $b\in B$ such that $\dist(A,B)=\|a-b\|$.
\end{enumerate}
Then $a_n\to a$ in the $\|\cdot\|$-topology.
\end{theorem}

\begin{proof} There is no loss of generality in assuming $a=0$.
Let us assume that (1) holds.

Since $\inte(A)\cap B=\emptyset$, by the Hahn-Banach theorem there
exists $f\in\SX$ such that $$\sup f(A)=0=\inf f(B).$$ In
particular, $f$ is a support functional for $A$ in $0$. Let
$\alpha>0$ and observe that, by Lemma~\ref{slicelimitatoselur},
there exists $r>1$ such that $S=S(f,3\alpha,A)\subset rB_X$. Put
$R=r+\alpha$.

We claim that $\{a_n\}$ and $\{b_n\}$ are eventually contained in
$2R B_X$. Suppose that this is not the case and let $\{a_{n_k}\}$
and $\{b_{n_k}\}$ be two subsequences such that $\|a_{n_k}\|>2R$
and $\|b_{n_k}\|>2R$ whenever $k\in\N$. Now, let $x_{n_k}\in
A_{n_k}$ and $y_{n_k}\in B_{n_k}$ be such that $\|x_{n_k}\|\to 0$
and $\|y_{n_k}\|\to 0$ as $k\to\infty$. Let $[x_{n_k},a_{n_k}]\cap
RS_X=\{a'_{n_k}\}$ and $[y_{n_k},b_{n_k}]\cap RS_X=\{b'_{n_k}\}$,
and observe that, by Lemma~\ref{convnorma}, it holds
$\|b'_{n_k}-a'_{n_k}\|\to0$ as $n\to\infty$.

Since $A_n\rightarrow A$ for the Attouch-Wets convergence,
$a'_{n_k}\in A_{n_k}\cap RS_X$ and
$$A= S(f,3\alpha,A)\cup[A\cap\{x\in X;\,f(x)\leq -3\alpha\}]\subset rB_X\cup\{x\in X;\,f(x)\leq -3\alpha\},$$
it eventually holds $a'_{n_k}\in\{x\in X;\,f(x)\leq -2\alpha\}$.

Analogously, since $B_n\rightarrow B$ for the Attouch-Wets
convergence, $b'_{n_k}\in B_{n_k}\cap RS_X$ and
$$B\subset \{x\in X;\,f(x)\geq 0\},$$
it eventually holds $b'_{n_k}\in\{x\in X;\,f(x)\geq -\alpha\}$.

In particular, it eventually holds $\|b'_{n_k}-a'_{n_k}\|\geq
f(b'_{n_k}-a'_{n_k})\geq \alpha$, a contradiction. Therefore our
claim is proved.

Now, since $\{a_n\}$ and $\{b_n\}$ are bounded, there exist
sequences $\{w_n\}\subset A$ and $\{z_n\}\subset B$ such that
$\|w_n-a_n\|\to0$ and $\|z_n-b_n\|\to0$. Since clearly
$\lim_n\|z_n-w_n\|=0$, it holds
$$0\leq\liminf_n[f(z_n)-\|w_n-z_n\|]\leq\liminf_n[f(w_n)]\leq\limsup_n f(w_n)\leq0,$$
and hence that $f(w_n)\to0$ as $n\to\infty$. Since, by
Lemma~\ref{slicelimitatoselur}, $f$ strongly exposes $0$, we have
that $w_n\to0$ and hence that $a_n\to0$ in the
$\|\cdot\|$-topology. This concludes the proof in case (1).
\medskip

If assumption (2) holds, the proof is similar, but some additional
efforts are needed. Let $d=\dist(A,B)$ and observe that:
\begin{enumerate}
\item $\inte(A)\cap(B+dB_X)=\emptyset$; \item $0\in B+dB_X$; \item
$\limsup_n \|a_n-b_n\|\leq d$
\end{enumerate}
 Then there exists $f\in\SX$ such
that $$\sup f(A)=0=\inf f(B+dB_X).$$ In particular, $f$ is a
support functional for $A$ in $0$ and $\inf f(B)=d$. Let $\Gamma$
be the constant given by Lemma~\ref{convnorma} and let us consider
$S=S(f,(\Gamma+2)d,A)$ and observe that, by
Lemma~\ref{slicelimitatoselur}, there exists $r>1$ such that
$S\subset rB_X$. Let $R=r+d$.

We claim that $\{a_n\}$ and $\{b_n\}$ are eventually contained in
$2R B_X$. Suppose that this is not the case and let $\{a_{n_k}\}$
and $\{b_{n_k}\}$ be two subsequences such that $\|a_{n_k}\|>2R$
and $\|b_{n_k}\|>2R$ whenever $k\in\N$. Now, let $x_{n_k}\in
A_{n_k}$ $y_{n_k}\in B_{n_k}$ be such that $x_{n_k}\to a$ and
$y_{n_k}\to b$ as $k\to\infty$. Let $[x_{n_k},a_{n_k}]\cap
RS_X=\{a'_{n_k}\}$ and $[y_{n_k},b_{n_k}]\cap RS_X=\{b'_{n_k}\}$,
and observe that, by Lemma~\ref{convnorma}, it eventually holds
$\|b'_{n_k}-a'_{n_k}\|< (\Gamma+1)d$.

Since $A_n\rightarrow A$ for the Attouch-Wets convergence,
$a'_{n_k}\in A_n\cap RS_X$ and

\begin{eqnarray*}
A&=& S(f,(\Gamma+2)d,A)\cup[A\cap\{x\in X;\,f(x)\leq -(\Gamma+2)d\}]\\
&\subset& rB_X\cup\{x\in X;\,f(x)\leq -(\Gamma+2)d\},
\end{eqnarray*}

it eventually holds $a'_{n_k}\in\{x\in X;\,f(x)\leq
-(\Gamma+1)d\}$.

Analogously, since $B_n\rightarrow B$ for the Attouch-Wets
convergence, $b'_{n_k}\in B_n\cap RS_X$ and
$$B\subset \{x\in X;\,f(x)\geq d\},$$
it eventually holds $b'_{n_k}\in\{x\in X;\,f(x)\geq 0\}$.

In particular, it eventually holds $\|b'_{n_k}-a'_{n_k}\|\geq
f(b'_{n_k}-a'_{n_k})\geq (\Gamma+1)d$, a contradiction and our
claim is proved.

Now, since $\{a_n\}$ and $\{b_n\}$ are bounded, there exist
sequences $\{w_n\}\subset A$ and $\{z_n\}\subset B$ such that
$\|w_n-a_n\|\to0$ and $\|z_n-b_n\|\to0$. Let us observe that
$$d\leq\liminf\|z_n-w_n\|\leq\limsup_n\|z_n-w_n\|=\limsup_n\|a_n-b_n\|\leq d$$ and
$$0\leq\liminf_n[f(z_n)-\|w_n-z_n\|]\leq\liminf_n[f(w_n)]\leq\limsup_n f(w_n)\leq0.$$
Hence, we obtain $f(w_n)\to0$ as $n\to\infty$. Since, by
Lemma~\ref{slicelimitatoselur}, $f$ strongly exposes $0$, we have
that $w_n\to0$ and hence that $a_n\to0$ in the
$\|\cdot\|$-topology. \end{proof}

\begin{remark} As in the finite-dimensional case (see Remark~\ref{remark:stabilityfunctions}), the theorem above can be proved in an alternative way,
by using known results concerning stability theory for convex
optimization problem. However, the well-posedness of the involved
problems requires a proof with techniques similar to those used in
Theorem~\ref{puntolur}. As in the finite-dimensional case, we
preferred to present a direct and more geometrical proof.
\end{remark}

If the limit sets $A$ and $B$ satisfy a strong condition about
non-separation, we obtain a result similar to
Corollary~\ref{thm:stabilitycorollario}.

\begin{proposition}\label{internononvuoto} Let $A$ and $B$
two closed convex subsets of a reflexive Banach space $X$ such
that $A\cap B$ is bounded and such that $(\inte A)\cap
B\neq\emptyset$. Let $\{A_n\}$ and $\{B_n\}$ be two sequences of
closed convex sets such that $A_n\rightarrow A$ and
$B_n\rightarrow B$ for the Attouch-Wets convergence. Suppose that
$\{a_n\}$ and $\{b_n\}$ are sequences in $X$ such that $a_n\in
A_n,\ b_n\in B_n$ ($n\in \N$) and
$$\dist(A_n,B_n)=\|a_n-b_n\|.$$
Then there exist two subsequences $\{a_{n_k}\}$ and $\{b_{n_k}\}$
that weakly converge to a point of $A\cap B$.
\end{proposition}

\begin{proof} Let us observe that, since $(\inte A)\cap
B\neq\emptyset$, the sets $A_n\cap B_n$ ($n\in\N$) are eventually
nonempty and hence $a_n$ and $b_n$ eventually coincide. Since $X$
is reflexive, it suffices to prove that $\{a_n\}$ and $\{b_n\}$
are bounded. By \cite[Corollary~9.2.8]{LUCC}, the sequence
$\{A_n\cap B_n\}$ converges to $A\cap B$ for the Attouch-Wets
convergence. Since $A\cap B $ is bounded, the thesis holds.
\end{proof}

By combining the above proposition with Theorem~\ref{puntolur}, we
obtain the following corollary.

\begin{corollary}\label{LUR}
Let $X$ be a reflexive Banach space $X$. Let $A$ be an LUR body of
$X$ and $B$ a closed convex subset of $X$ such that $A\cap B$ is
nonempty and bounded. Let $\{A_n\}$ and $\{B_n\}$ be two sequences
of closed convex sets such that $A_n\rightarrow A$ and
$B_n\rightarrow B$ for the Attouch-Wets convergence. Suppose that
$\{a_n\}$ and $\{b_n\}$ are sequences in $X$ such that $a_n\in
A_n,\ b_n\in B_n$ ($n\in \N$) and
$$\dist(A_n,B_n)=\|a_n-b_n\|.$$
Then there exist subsequences $\{a_{n_k}\}$ and $\{b_{n_k}\}$ that
weakly converge to a point $c\in A\cap B$. Moreover, if $(\inte
A)\cap B=\emptyset$ then $a_n, b_n\to c$ with respect to the norm
convergence.

\end{corollary}

\section{Examples and final remarks}\label{sectionesempi}

In this section we provide two examples to illustrate the role of
the assumptions in the infinite-dimensional case. We point out
that both of them are in $\ell_2$. Therefore, the assumptions used
in Section~\ref{Infinite-dimensional} cannot be avoided even in
the ``simplest'' infinite-dimensional space.

The following example shows that an analogous of
Theorem~\ref{thm:stability principale} does not hold in the
infinite-dimensional setting.

\begin{example}\label{esempiocono-semispazio} Let $X=\ell_2$ and $\{e_n\}_n$ its standard basis. Let $A,B,A_n,B_n\subset X$ ($n\in\N$, $n\geq2$)
be defined as follows.
\begin{eqnarray*}
A   &=& \textstyle \ccone(\{ e_k+\frac{1}{k} e_1;\, k\in\N\});\\
B   &=& \{x\in X;\, e_1^*(x)=   0\};\\
A_n   &=& \textstyle\cconv\bigl(\{\ln n\, e_n+\frac1n e_1\}\cup(\frac1n e_1+A)\bigr);\\
B_n   &=& B.
\end{eqnarray*}
Let $a_n=\ln n\, e_n+\frac1n e_1\in A_n$ and $b_n=\ln n\, e_n\in
B_n$. Then:
 \begin{enumerate}
 \item $A\cap B=\{0\}$; \item $A_n\to A$ and $B_n\to B$ for
the Hausdorff convergence (and, hence, for the Attouch-Wets
convergence); \item $\dist(A_n,B_n)=\|a_n-b_n\|$; \item
$\|a_n\|,\|b_n\|\to\infty$.
\end{enumerate}
\end{example}

\begin{proof} We just have to prove (i) and (ii), since the
proofs of (iii) and (iv) are straightforward.

 (i) For $n\in\N\setminus\{1\}$, let
$f_n=n e_1^*-e_n^*$ and $g_n=e_n^*$ and observe that $$\textstyle
\{ e_k+\frac{1}{k} e_1;\, k\in\N\}\subset\{x\in X;\, f_n(x)\geq
0,\, g_n(x)\geq 0\}.$$ Then $A\subset \bigcap_{n=1}^\infty\{x\in
X;\, f_n(x)\geq 0,\, g_n(x)\geq 0\}$. Now, if $x\in A\cap B$, it
holds $e_1^*(x)=0$, $f_n(x)=-e^*_n(x)\geq0$ and
$g_n(x)=e^*_n(x)\geq0$. Then $x=0$.

 (ii) We just have to prove that $A_n\to A$ for the Hausdorff
convergence. Let us observe that $$\textstyle \dist(a_n,\frac1n
e_1+A)\leq\|\frac1n e_1+\ln n(e_n+\frac1n e_1)-a_n\|=\frac{\ln
n}n.$$ Hence, it holds
$$\textstyle d_H(A_n,A)\leq d_H(A_n,\frac1n e_1+A)+d_H(\frac1n e_1+A,A)\leq \frac{\ln n}{n}+\frac1n,$$
and the proof is concluded. \end{proof}

\medskip

Given two sets $A,B\subset X$, we say that $A$ and $B$ are {\em
separated} if{f} there exists $x^*\in X^*\setminus\{0\}$ such that
$$\textstyle \sup x^*(A)\leq\inf x^*(B).$$
The following example shows that, in
Proposition~\ref{internononvuoto}, the condition $$(\inte A)\cap
B\neq\emptyset$$ cannot be replaced by the weaker condition ``$A$
and $B$ are not separated''.

\begin{example}\label{esempiocono-cono} Let us consider $X=\ell_2$ and for $n\in\N$ let us consider the following subsets of $X$.
\begin{eqnarray*}
C_n   &=& \textstyle \ccone\bigl(\{e_{2n-1},e_{2n}+\frac1n e_{2n-1}\}\bigr)-\frac1n e_{2n-1};\\
D_n   &=& \textstyle \ccone\bigl(\{-e_{2n-1},e_{2n}-\frac1n e_{2n-1}\}\bigr)+\frac1n e_{2n-1};\\
C'_n   &=& \textstyle C_n-\frac1{\ln n}{e_{2n-1}};\\
D'_n   &=& \textstyle D_n-\frac1{\ln n}{e_{2n-1}};\\
A   &=& \textstyle \cconv\bigl(\bigcup_{n\in\N}C_n\bigr);\\
B   &=& \textstyle \cconv\bigl(\bigcup_{n\in\N}D_n\bigr);\\
A_n   &=& \textstyle \cconv\bigl(\bigcup_{k\in\N\setminus\{n\}}C_k\cup C'_n\bigr);\\
B_n   &=& \textstyle
\cconv\bigl(\bigcup_{k\in\N\setminus\{n\}}D_k\cup D'_n\bigr).
\end{eqnarray*}
 Then:
 \begin{enumerate}
\item $A$ and $B$ are not separated; \item $A\cap B$ is bounded;
 \item $A_n\to A$ and $B_n\to B$ for
the Hausdorff convergence (and, hence, for the Attouch-Wets
convergence); \item let $a_n=b_n=ne_{2n}\in A_n\cap B_n$ then
$\dist(A_n,B_n)=\|a_n-b_n\|=0$ and $\|a_n\|=\|b_n\|=n\to\infty$.
\end{enumerate}
\end{example}

Let us define $X_n=\span\{e_{2n-1},e_{2n}\}$, $B_{X_n}=B_X\cap
X_n$ and $Y_N=\span(\bigcup_{n=1}^N X_n)$. Observe that
$$Y_N=X_1\oplus_2\ldots\oplus_2 X_N,$$ where we denote by $X_1\oplus_2\ldots\oplus_2
X_N$ the direct sum $X_1\oplus\ldots\oplus X_N$ endowed with the
norm
$$\textstyle \|(x_1,\ldots,x_N)\|=(\|x_1\|^2+\ldots+\|x_N\|^2)^{\frac12}.$$

To exploit the features of Example~\ref{esempiocono-cono} we need
some preliminary lemmas. The easy proof of the following lemma is
left to the reader.

\begin{lemma}\label{intersezione} Let $C_n, D_n\subset X_n$ be
defined as above, then the following inclusion holds.
$$\textstyle (C_n+\frac1{\sqrt{n^2+1}} B_{X_n})\cap(D_n+\frac1{\sqrt{n^2+1}} B_{X_n})\subset 2B_{X_n}$$
\end{lemma}

\begin{lemma}\label{bolla-L2-L1} Let $W_n$ be convex subsets of $X_n$ containing the origin ($n=1,\ldots,N$) and
let $\epsilon>0$, then the following inclusion holds:
$$\textstyle \conv(\bigcup_{n=1}^N W_n+\epsilon B_{Y_N})\subset 2\,\conv(\bigcup_{n=1}^N [W_n+\sqrt N\epsilon B_{X_n}]).$$
\end{lemma}

\begin{proof} Since $Y_N=X_1\oplus_2\ldots\oplus_2 X_N$, it is
not difficult to prove that $$B_{Y_N}\subset \sqrt
N\conv(\bigcup_{n=1}^N B_{X_n}),$$ hence the following inclusions
hold:

\begin{eqnarray*}
\textstyle \conv(\bigcup_{n=1}^N W_n+\epsilon B_{Y_N}) &\subset&
\textstyle \conv(\bigcup_{n=1}^N W_n+\epsilon \sqrt
N\conv(\bigcup_{n=1}^N
B_{X_n}))\\
   &\subset&2\,\conv(\bigcup_{n=1}^N [W_n+\sqrt N\epsilon
   B_{X_n}]).
\end{eqnarray*}

\end{proof}

\begin{lemma}\label{scritturaquasiunica} For $n=1,\ldots,N$, let $W_n$ and $Z_n$ be convex subsets of $X_n$ containing the origin.
Then the following inclusion holds:
$$\textstyle \conv\bigl(\bigcup_{n=1}^N W_n\bigr)\cap\conv\bigl(\bigcup_{n=1}^N Z_n\bigr)\subset 2\conv\bigl(\bigcup_{n=1}^N
W_n\cap Z_n\bigr)$$
\end{lemma}

\begin{proof} Let $x\in \conv\bigl(\bigcup_{n=1}^N
W_n\bigr)\cap\conv\bigl(\bigcup_{n=1}^N Z_n\bigr)$, then there
exist $\alpha_n, \beta_n\in[0,1]$, $w_n\in W_n$ and $z_n\in Z_n$
($n=1,\ldots,N$) such that
$$\textstyle x=\sum_{i=1}^N \alpha_n w_n=\sum_{i=1}^N \beta_n z_n.$$
Since $Y_N=X_1\oplus\ldots\oplus X_N$, it holds $\alpha_n
w_n=\beta_n z_n$ ($n=1,\ldots,N$). Now suppose that
$\alpha_n\geq\beta_n>0$, then $w_n=\frac{\beta_n}{\alpha_n}z_n\in
W_n\cap Z_n$. Analogously, if $0<\alpha_n\leq\beta_n$, then
$z_n=\frac{\alpha_n}{\beta_n}w_n\in W_n\cap Z_n$. Hence
$$x\in(\alpha_1+\beta_1)(W_1\cap Z_1)+\ldots+(\alpha_N+\beta_N)(W_N\cap Z_N)\subset2\conv\bigl[\bigcup_{n=1}^N
(W_n\cap Z_n)\bigr].$$ \end{proof}

\begin{proof}[Proof of Example~\ref{esempiocono-cono}.]

Let us prove assertions (i), (ii) and (iii); the proof of (iv) is
obvious.

 (i) Let us observe that, for each $n\in\N$, the segments
$[-\frac1n e_{2n-1},\frac1n e_{2n-1}]$ and $[0, e_{2n}]$ are
contained in $A\cap B$. Now, suppose that there exists $f\in X^*$
such that $\sup f(A)\leq\inf f(B)$, then $f$ is constant on $A\cap
B$ and, by the above remark, it holds $f(e_n)=0$ whenever
$n\in\N$. Hence $f=0$.

(ii) Let us prove that $A\cap B$ is bounded. For $k\in\N$, let us
denote by $P_k$ the canonical projection on the first $k$
coordinates. Let $x\in A\cap B$ and let $N\in\N$ be such that
$\|x-P_{2N}x\|\leq1$. To conclude the proof it suffices to show
that $\|P_{2N}x\|\leq8$

We claim that $P_{2N} x\in \cconv\bigl(\bigcup_{n=1}^N C_n\bigr)$.
Indeed, since $x\in A$, there exists a sequence $\{y_k\}$,
converging in norm to $x$, such that
$y_k\in\conv\bigl(\bigcup_{n=1}^k C_n\bigr)$. Then the sequence
$\{P_{2N}y_k\}\subset \cconv\bigl(\bigcup_{n=1}^N C_n\bigr)$
converges in norm to $P_{2N}x$ and the claim is proved.

Analogously, it holds $P_{2N} x\in \cconv\bigl(\bigcup_{n=1}^N
D_n\bigr)$ and hence,
\begin{eqnarray*}
P_{2N} x   &\in& \textstyle [\conv\bigl(\bigcup_{n=1}^N
C_n\bigr)+\frac1{\sqrt{N^3+N}}B_{Y_N}]\cap[\conv\bigl(\bigcup_{n=1}^N
D_n\bigr)+\frac1{\sqrt{N^3+N)}}B_{Y_N}]\\
   &{\subset}& \textstyle 2\, \conv\bigl(\bigcup_{n=1}^N
[C_n+\frac1{\sqrt{N^2+1}}B_{X_n}]\bigr)\cap2\,
\conv\bigl(\bigcup_{n=1}^N
[D_n+\frac1{\sqrt{N^2+1}}B_{X_n}]\bigr)\\
   &{\subset}& \textstyle 4\, \conv\bigl(\bigcup_{n=1}^N
[C_n+\frac1{\sqrt{N^2+1}}B_{X_n}]\cap
[D_n+\frac1{\sqrt{N^2+1}}B_{X_n}]\bigr)\\
   &{\subset}& \textstyle 4\, \conv\bigl(\bigcup_{n=1}^N
2B_{X_n}\bigr)\subset 8B_X, \end{eqnarray*}

\noindent where the above inclusions hold by
Lemma~\ref{bolla-L2-L1}, Lemma~\ref{scritturaquasiunica} and
Lemma~\ref{intersezione}, respectively.

(iii) Let us prove that $A_n\to A$ for the Hausdorff convergence,
the proof that $B_n\to B$ for the Hausdorff convergence is
similar. Let us observe that $d_H(C_n,C'_n)=\frac1{\ln n}$, hence
we have:
$$\textstyle d_H(A,A_n)=d_H(\cconv\bigl(\bigcup_{k\in\N\setminus\{n\}}C_k\cup C'_n\bigr),
\cconv\bigl(\bigcup_{k\in\N\setminus\{n\}}C_k\cup
C_n\bigr))\leq\frac1{\ln n},$$ and the proof is concluded.
\end{proof}

\section*{Acknowledgments.}
The research of the first and second authors is partially
supported by GNAMPA-INdAM. The research of the second and third
authors is partially supported by Ministerio de Econom\'{i}a y
Competitividad (Spain), MTM2015-68103-P, Plan Nacional de
Matem\'{a}ticas, (2016-2018).


TITLE: Solid settling to the top of solution when centrifuged
QUESTION [0 upvotes]: I have a mixture of dense particles (about 4x that of water) and not-so-dense particles (maybe the same density of water or slightly less or more dense). When I centrifuge a tube filled with a solution of these particles in water, the tube is on an angle and I find that the denser particles settle to the outer wall of the tube. But the less dense particles also settle - they settle on the inner wall of the tube and form a thick layer.
I am not sure why this happens. I would assume that rather than rise to the top, the particles would remain suspended in the solution as the centrifugal force is not strong enough to allow them to settle. However, there still appears to be particles in the liquid afterwards. If interested, I use 7000 rpm for 30 minutes.

REPLY [1 votes]: If I understand your description ("the tube is on an angle") correctly, then the "outer" wall of the tube is where the centrifugal force is pointing - and the "inner" wall is the opposite side.
When the tube is spinning, there will be an apparent radial force on all the particles - the lighter particles will experience a "buoyancy" towards the inner wall, and the heavier particles a "gravity" towards the outer wall.
Once they get there, they may not want to slide since to do so, they would have to displace other particles that are already there - or perhaps the friction with the wall is such that they don't want to move. Without a diagram / photo, it's hard to tell you exactly what is going on.


TITLE: Is the set of points in $\mathbb R^n$ with $\sum_{j=1}^n x_j^k = 0$ a submanifold?
QUESTION [7 upvotes]: Consider the set
$$A:= \{x\in \mathbb R^n :\sum_{j=1}^n x_j^k = 0\}$$
for $k$ an odd integer. Is this a submanifold of $\mathbb R^n$ for every $n$? For $n=1$, it is just 0; for $n = 2$, it is the anti-diagonal $\{(x_1 , -x_1) : x_1 \in \mathbb R\}$, which is a submanifold. However, I cannot find a way to determine this in higher dimensions. Any suggestions?

REPLY [4 votes]: Your set is scale-invariant: if $x\in A$ and $t\in\mathbb R$ then $tx\in A$. If such a set is a submanifold, then
it is a linear subspace. If $n>2$, $A$ is not a linear subspace, hence not a manifold.


TITLE: Prove that $G$ has a cycle of length $\leq4$
QUESTION [1 upvotes]: Let $G$ be an undirected simple graph with $n\leq9$ vertices and each vertex has $m\geq3$ neighbours. Show that $G$ has a cycle of length $\leq4$.
My first idea was to just assume that $G$ has exactly $9$ vertices and each vertex has exactly $3$ neighbours. If the statement is true for $n=9$, then it must be true for $n<9$, too, same thing for $m>3$. How can I now show mathematically, that $G$ with $n=9$ and $m=3$ actually contains a cycle of length $\leq4$?

REPLY [2 votes]: The claim is false if we allow graphs with $n=0$ vertices.
So to proof the claim, we may assume $n\ge1$, each vertex has degree $\ge 3$, all cycles have length $\ge 5$, and from this we try to show that $n>9$.
Pick a vertex $v$, and from its at least three neighbours pick $w_1,w_2,w_3$. For each $w_i$, pick two neighbours $u_{i,1}, u_{i,2}$ distinct from $v$.
Note that $u_{i,j}\ne w_k$ as otherwise we'd have a cycle of length $3$.
Also, $u_{i,j}\ne u_{i',j'}$ for $(i',j')\ne(i,j)$ because otherwise, we'd have a cycle of length $4$.
Thus the ten vertices
$$ v,w_1,w_2,w_3,u_{1,1},u_{1,2},u_{2,1},u_{2,2},u_{3,1},u_{3,2}$$
are pairwise distinct, and so $n\ge10$.


TITLE: Selecting 3 non-consecutive days of the week
QUESTION [0 upvotes]: We need to find how many groups of $3$ days of the week there can be, given that no two days should be consecutive. The answer should be $7$, but I do not know how to get to the answer.
Thanks in advance.

REPLY [0 votes]: For each day of the week $X$, there is exactly one way to select an arrangement so that $X$ is not one of the three days nor one of the three days immediately following them. So there are exactly seven possible arrangements, one for each choice of $X$.


TITLE: Does almost surely convergence implies convergence in $L^p$?
QUESTION [0 upvotes]: I have the following excercise:
Does almost surely convergence imply convergence in $L^p$? Prove or give a counterexample.
For $0 < p < ∞$, we say $X_n$ converges in $L^p$ to $X$ if $X ∈ L^p(Ω)$ and $\Bbb E|Xn − X|^p → 0$ as $n → ∞$
I think it's not true, but I can't get a counterexample.
Could you provide any?
I know it is not true that convergence in probability implies almost surely convergence.

REPLY [0 votes]: This is certainly not true. For example, take a sequence of measurable sets $\Omega \supseteq A_1 \supseteq A_2 \supseteq A_3 \supseteq \cdots$ Such that $\mathbb{P}(A_n) = \frac{1}{n}$. Now let $X_n$ be the random variable given by $X_n := n^{1/p} \cdot \mathbf{1}_{A_n}$, where $\mathbf{1}$ denotes an indicator function.
It is easy to see that $X_n \rightarrow 0$ a.s., but for all $n$ we have that:
$$\begin{align} \mathbb{E}|X_n - 0|^p & = \mathbb{E}(n \cdot \mathbf{1}_{A_n})\\
& = n \cdot \mathbb{P}(A_n) \\
& = 1  \end{align}$$


TITLE: An integral which is related to Biharmonic extension
QUESTION [1 upvotes]: In my research, I need to evaluate an integral:
$$\int_{R^{3}}\frac{y^{3}}{(|x-\xi|^{2}+y^{2})^{3}}\log(|\xi^{2}|+\frac{1}{4})d\xi$$
where $x\in R^{3}$, $y\geq0$.  Moreover, I want to see whether it is a biharmonic function on
the up-half space or not.
I also want to know will
$$ \log(|x|^{2}+(y+\frac{1}{2})^{2})-c\int_{R^{3}}
\frac{y^{3}}{(|x-\xi|^{2}+y^{2})^{3}} \log(|\xi^{2}|+\frac{1}{4})d\xi$$ be bounded on the up half space? Here $c$ is a constant such that
$$c\int_{R^{3}}\frac{y^{3}}{(|x-\xi|^{2}+y^{2})^{3}}d\xi=1$$

REPLY [2 votes]: If you use spherical coordinates, the integral splits in a radial part and an integral over the unit sphere in $\mathbb{R}^3$. The spherical integral is
$$
\int_{S^2} \frac{1}{y^2+r^2+|x|^2 - 2 r \langle x , \xi' \rangle} d \sigma(\xi')
$$where $\xi= r \xi'$, $\xi' \in S^2$.
Now you can apply the Funk-Hecke theorem to compute this spherical integral, because the integrand only depends on the inner product of $\xi'$ with a fixed vector.
(you can find the Funk-Hecke theorem in books on spherical harmonics).
| haven't carried out this computation explicitly, but this is how I would compute this integral.


TITLE: Strong separating hyperplane - check proof
QUESTION [0 upvotes]: Let $S,T \subset \mathbb R^n$ be non empty convex sets.
We say that the hyperplane $H: a^Tx = b$  strongly separates $S,T$ if $\exists \epsilon$ such that $S + \epsilon B_0(1) \subset \{x \in \mathbb R^n: a^Tx > b\}$ and $T+\epsilon B_0(1) \subset \{x \in \mathbb R^n :a^Tx < b\}$
Where $B_0(1) = \{x \in \mathbb R^n: |x| \leq 1\}$ is the unit ball centered at zero.
We are asked to show that if $T,S$ can be strongly separated, then $0 \notin \text{cl}(S-T)$, or in other words, $\text{inf}_{t \in T, s \in S}\|t-s\| > 0$
Here's what I tried.
Suppose that $\text{inf}_{t \in T, s \in S}\|t-s\| = 0$ and that there's a strong separation. Then there are two sequences $\{t_k\} \subset T, \{s_k\} \subset S$ that converge to the same limit, $t_k \to v, s_k \to v$.
Let $\epsilon$ be the epsilon from the strong separation. Then there's a number $N$ such that $\|t_k - v\| < \frac{\epsilon}{4}, \| s_k - v\| < \frac{\epsilon}{4}$ for all $k > N$. Let $t_l, s_l$ be such vectors.
It follows that $\epsilon B_{t_l}(1) \bigcap \epsilon B_{s_l}(1) \neq \emptyset$, and so $T + \epsilon B_{0}(1) \bigcap S + \epsilon B_{0}(1) \neq \emptyset$
It's impossible for an element $x$ in the intersection to be both $a^Tx > b$ and $a^Tx < b$, hence we arrive at a contradiction.
Is this solution correct?

REPLY [0 votes]: There's a  flaw in your assumption that the sequences $\ \left\{t_k\right\}\ $ and $\ \left\{s_k\right\}\ $ converge. It's possible that $\ 0\in\text{cl}(S-T)\ $, without any such convergent sequences existing. Let $\ S=\left\{(x,y)\in\mathbb{R}^2\left|y\ge e^x\right.\right\}\ $, $\ T=\left\{(x,y)\in\mathbb{R}^2\left|y\le -e^x\right.\right\}\ $, for instance. Then $\ S-T= $$\left\{(x,y)\in\mathbb{R}^2\left|y>0\right.\right\}\ $, and $\ (0,0)\in\text{cl}(S-T)\ $, but the only sequences $\ \left\{t_k\right\}\subset T\ $, $\ \left\{s_k\right\}\subset S\ $ for which $\ \lim_\limits{k\rightarrow\infty}\left\|s_k-t_k\right\|=0\ $ have $x$-coordinates that diverge to $\ -\infty\ $. While such unbounded pairs of convex sets like this aren't strongly separated, your proof fails to account for them.
Nevertheless, I think you can easily modify your proof to avoid the assumption that  $\ \left\{t_k\right\}\ $ and $\ \left\{s_k\right\}\ $ converge. All you really need to use is that $\ s_k-t_k \rightarrow0\ $.


TITLE: Limiting Value Equation
QUESTION [0 upvotes]: Do I just plug in the numbers in the equation and solve like a normal differential equation? I have just started to learn this stuff, and am new to the whole process.

REPLY [1 votes]: As a physicist, I should prefer to solve the problem for the general case. The integration of the given differential equation is simple and leads to the general formula
$$I(t)=\frac{V \left(1-e^{-\frac{R t}{L}}\right)}{R}$$ This allows later to consider any possible case for which this model applies. For your specific case, this for sure will lead to the value of $5$; however, this $5$ is the ratio of $V$ to $R$ which is a more general result.
Furthermore, having the general formula would allow to analyse the impact of each involved quantity computing for example the derivatives of $I(t)$ with respect to $V$, $R$, $L$ or $t$.