text:
\begin{document}

\begin{abstract}
We give an improvement of the Carath\'eodory theorem for strong convexity in $\mathbb R^n$, reducing the Helly number to $n$ in several cases; and show that the Carath\'eodory number cannot be smaller than $n$ for arbitrary gauge set $K$. We also give a simpler example with infinite Carath\'eodory number and an improved topological criterion for one convex body to be a Minkowski summand of another.
\end{abstract}

\maketitle

\section{Introduction}

In the works \cite{pol1996,balashov2000} (a recent English reference is \cite{holmsenkar2017}) a strengthening of the notion of convexity was studied. The first natural example is to call a convex body $A\subset\mathbb R^n$ \emph{$R$-strongly convex} for a positive real $R$, if $A$ is an intersection of balls of radius $R$. This notion seems to have been rediscovered many times (see the references in the cited works), and in \cite{pol1996,balashov2000} it was generalized to the following: For a fixed convex body $K\subset\mathbb R^n$, another convex body $A$ is \emph{$K$-strongly convex} if it is an intersection of translates of $K$.

It turned out that $K$-strong convexity inherits good properties from ordinary convexity and $R$-strong convexity (corresponding to the case $K$ is a Euclidean ball) under certain ``generating set'' assumption on $K$. We make some precise definitions and remind some notation from the cited works. We restrict ourselves to the case of a finite dimensional Euclidean space, this is the situation when one may expect Carath\'eodory-type results that we are going to discuss here.

\begin{definition}
\label{definition:genset}
A convex body $K\subset\mathbb R^n$ is a \emph{generating set} if any nonempty intersection of its translates
$$
K\gdiv T := \bigcap_{t\in T} (K - t)
$$
is a Minkowski summand of $K$, that is, $(K\gdiv T) + T' = K$ for some convex compactum $T'$.
\end{definition}

In \cite{kar2001gen} it was shown that it is sufficient to test the generating set property for the sets $T$ of two vectors; in \cite{ivanov2007} a non-trivial reformulation of this criterion was established and used to study other properties of generating sets. We will not use these observations here, but an interested reader may consult the provided references. We only use the relatively simple facts (see \cite{balashov2000}) that in dimension at most $2$ all convex bodies are generating sets and any Cartesian product of generating sets is a generating set.

\begin{definition}
\label{definition:strongconv}
Let $K\subset \mathbb R^n$ be a convex body. A set $C\subset \mathbb R^n$ is called {\em $K$-strongly convex} if it is an intersection of translates of $K$, that is
$$
C = K\gdiv T
$$
in the above notation.
\end{definition}

\begin{definition}
The minimal $K$-strongly convex set containing a given set $X$ is called its \emph{strongly convex hull}, and can be expressed by the following formula
$$
\conv_K X = K\gdiv (K\gdiv X).
$$
In case $X$ is not contained in any translate of $K$ we assume the $K$-strongly convex hull of $X$ undefined.
\end{definition}

In \cite{kar2001car} (as explained in \cite{holmsenkar2017} in English) it was shown that of $K$ is a generating set then a version of Carath\'eodory theorem (\cite{car1911}, see also the review \cite{eckhoff1993}) for strong convexity holds for finite point sets $X\subset\mathbb R^n$:
\[
\conv_K (X) = \bigcup_{Y\subseteq X,\ |Y|\le n+1} \conv_K (Y).
\]

Here we show that the Carath\'eodory number can be reduced to $n$ in some particular cases. First, assuming the generating set property of $K$, this holds for points outside $\conv X$ (the ordinary convex hull), which may be practically useful because the ordinary convex hull is always contained in the strong convex hull and the problem of determining the strongly convex hull is only interesting outside the ordinary convex hull.

\begin{theorem}
\label{theorem:caratheory-outside}
For a generating set $K\subset\mathbb R^n$ and $X\subset\mathbb R^n$, whose translate fits into $K$, we have
\[
\conv_K (X)\setminus \conv X \subseteq \bigcup_{Y\subseteq X,\ |Y|\le n} \conv_K (Y).
\]
\end{theorem}

Another result shows that the Carath\'eodory number is $n$ for all points for specific gauge bodies $K$.

\begin{theorem}
\label{theorem:product-gen}
If $L\subset\mathbb R^\ell$ and $M\subset \mathbb R^m$ are generating convex bodies and $K=L\times M\subset\mathbb R^n$, for $n=\ell+m$ with $\ell,m\ge 1$, then the Carath\'eodory number of $K$-strong convexity is precisely $n$.
\end{theorem}

A good example for this product theorem is a nontrivial product $K = L_1\times \dots \times L_k$ of convex bodies $L_i$ of dimension at most $2$ each. Since all convex bodies in dimension at most $2$ are generating sets, as well as their products, this theorem is applicable and the Carath\'eodory number of $K$-strong convexity equals $\dim K$ in this case.

The proof of Theorem \ref{theorem:caratheory-outside} starts from choosing a point in $\conv_K (X)\setminus \conv X $. If we translate everything so that the point in question becomes $0\in \conv_K X\setminus \conv X$ then the theorem is restated in the following alternative version (which is a Helly-type theorem on its own right):

\begin{theorem}[Alternative statement of Theorem \ref{theorem:caratheory-outside}]
\label{theorem:helly-type}
If $K\subset\mathbb R^n$ is a generating set and $X$ is a finite point set, the whole $X$ can be covered by a translate of $K$, $0\not\in\conv X$, and any $\le n$ points of $X$ can be covered by a translate of $K$ not covering the origin $0$. Then $X$ can be covered by a translate of $K$ without covering the origin.
\end{theorem}

It also makes sense to ask how small the Carath\'eodory number for $K$-strong convexity can be for some particular $K$, can it be smaller than the dimension $n$. For example, when $K$ is a parallelotope (affine image of a cube) then the Helly number of the system of its translates is $2$ (see \cite{boltyanski1997} for a deep study of the Helly number of a system of translates). Since the formulation for the Helly number of the system of translates is very close to the statement in Theorem \ref{theorem:helly-type} (see also its proof in Section \ref{section:helly-type} and \eqref{equation:empty-intersection-outside} in particular), we might expect a low strong convexity Carath\'eodory number for parallelotopes. But it fact this number is never smaller than the dimension $n$, since we establish:

\begin{theorem}
\label{theorem:at-least-n}
If $K\subset\mathbb R^n$ is a convex body (not necessarily a generating set) then the Carath\'eodory number of $K$-strong convexity is at least $n$.
\end{theorem}

As for the upper bounds, the example in \cite[Section 5]{holmsenkar2017} shows that in dimensions $n\ge 3$ there is no upper bound for the Carath\'eodory number for strong convexity, while in dimensions $n\le 2$ all convex bodies are generating sets.

Once we have started to consider the case when $K$ is not necessarily a generating set, we continue with giving a fact about the Carath\'eodory number of products.

\begin{theorem}
\label{theorem:product-any}
Let $K$ be the Cartesian product $L \times M$ of two convex bodies $L$ and $M$. If $k, \ell, m$ are the strong Carath\'eodory numbers for $K, L, M$ respectively, then $\ell + m - 2 \le k \le \ell + m$.
\end{theorem}

Below we give proofs of the theorems stated here. In Section \ref{section:infinite} we give a simple example of $K$ with an infinite Carath\'eodory number for strong convexity, compared to the example in \cite[Section 5]{holmsenkar2017}. Then in Section \ref{section:minkowski} we consider a related question of topological criteria for a convex compactum $A$ to be a Minkowski summand of a convex body $B$, giving a version of the corresponding result of \cite[Section 6]{holmsenkar2017}.

\section{Proof of Theorem \ref{theorem:helly-type}}
\label{section:helly-type}

Following \cite{holmsenkar2017}, we assume that $0$ does not belong to any $\conv_K Y$ for $Y\subseteq X$ with $|Y|\le n$, that is
\[
0\in K\gdiv (K\gdiv Y) \Leftrightarrow K\gdiv Y\subseteq K \Leftrightarrow \bigcap_{x\in Y} (K - x)\setminus K = \emptyset.
\]
In order to have a contradiction we need to show that
\begin{equation}
\label{equation:empty-intersection-outside}
\bigcap_{x\in X} (K - x)\setminus K = \emptyset.
\end{equation}
We introduce the system of sets
\[
F_x = (K-x)\setminus K,\quad x\in X.
\]

Note that if we were not subtracting $K$ from the translates then the question we study would be just the question of the Helly number of the system of translates of $K$ (well studied in \cite{boltyanski1997}, for example), this observation will be used in Section \ref{section:by-segment} below.

The proof in \cite{kar2001car,holmsenkar2017} was exploiting the topological Helly property for the family of sets $\{F_x\}$ with Helly number $n+1$, which we need to reduce now to $n$. In order to reduce the Helly number we assume without loss of generality $0\in\inte K$, take a slightly inflated $K_\epsilon = (1+\epsilon) K$ and consider $G_x = (K - x)\cap \partial K_\epsilon$ instead of $F_x$ trying to show the following for sufficiently small $\epsilon>0$:

\begin{enumerate}
\item
Any $\le n$ of $\{G_x\}$ have a common point;
\item
They do not cover the whole topological $(n-1)$-sphere $\partial K_\epsilon$;
\item
They satisfy the intersection property needed to apply Helly's theorem with number $n$.
\end{enumerate}

We first decode what it means $u\in G_x$, that is $u \in \partial K_\epsilon$ and $u\in K-x$. We rewrite $0\in \partial K_\epsilon - u$ and $x\in K - u$ and note that $K_\epsilon$ translated by $-u$ has origin on its boundary and $K$ translated by $-u$ has $x$ inside.

The first claim can be established as follows. $F_{x_1}\cap \dots \cap F_{x_n}\neq \emptyset$ by the assumption of the theorem. This means that $Y = \{x_1,\ldots, x_n\}$ can be covered by a translate of $K-u$ not touching the origin. For sufficiently small $\epsilon$ we will have that $K_\epsilon-u$ does not touch the origin either. The theorem assumes that the whole $X$ can be covered by some $K-v$ and therefore by $K_\epsilon-v$. If $0\not\in K-v$ then this is what we need to prove; we proceed assuming the contrary, $0\in K-v$. Interpolating between $K-u$ and $K-v$ we find $K-w$ that contains $Y$ such that $0$ is on the boundary of $K_\epsilon-w$. Therefore $w\in \partial K_\epsilon$ and $w\in K - x_i$ for every $x_i\in Y$. In this argument we needed sufficiently small $\epsilon$ for every choice of $Y\subseteq X$ and eventually we are able to choose $\epsilon>0$ so small that it is suitable for all choices.

The second claim follow from the assumption $0\not\in\conv X$. Taking linear $\lambda$ such that $\lambda(X)>0$ we see that the point of $K_\epsilon$ with largest $\lambda$ cannot be covered by any $K-x$, $x\in X$.

The third claim need the generating set property. The proof of \cite[Lemma 2.1]{holmsenkar2017} actually shows that for any $Y\subseteq X$ the set
\[
(K\gdiv Y) \setminus \inte K_\epsilon
\]
is acyclic. Indeed, we can trace the argument in the proof of the Lemma, or just apply the Lemma to closed convex bodies
\[
(K\gdiv Y) \setminus (1 + \delta_n) K
\]
with $\delta_n$ strictly increasing to $\epsilon$ and use the continuity property of the \v{C}ech cohomology.

It remains to note that
\[
G_Y = (K\gdiv Y)\cap \partial K_\epsilon = \bigcap_{x\in Y} G_x
\]
is obtained as a deformation retraction of $(K\gdiv Y)\setminus \inte K_\epsilon$, if we choose a center in $(K\gdiv Y) \cap \inte K_\epsilon$ and centrally project $(K\gdiv Y)\setminus \inte K_\epsilon$ onto $\partial K_\epsilon$.

The three claims ensure that we may apply the topological Helly's theorem for $G_x$ with number $n$ (see for example \cite{debrunner1970}). We strongly recommend the book \cite[\S 5]{godement1958} for further references on \v{C}ech cohomology (in which we mean the acyclicity), the covering resolution for the sheaf cohomology and other classical ideas that imply the topological Helly's theorem.

We conclude from the topological Helly theorem that all the $G_x$ intersect, that means that a translate $K-u$ contains the whole $X$ and $K_\epsilon-u$ has $0$ on its boundary at the same time. Hence $K-u\subseteq \inte (K_\epsilon - u)$ does not cover the origin and the proof is complete.

\begin{remark}
The case of $R$-strong convexity, that is $K$ equal to the Euclidean ball, is done with the application of the Helly theorem for convex spherical caps not requiring any topology: A family of convex spherical caps not covering the whole sphere $\mathbb S^{n-1}$ has Helly number $n$.
\end{remark}

\section{Proof of Theorem \ref{theorem:product-gen}}
\label{section:by-segment}

We start from a lemma that may be interesting on its own right.

\begin{lemma}
Let $K\subset\mathbb R^n$ be a generating set and let $X$ be a set of $n+2$ points. Then any point $p\in \conv_K X$ is contained in $\conv_K X'$ and $\conv_K X''$ for at least two distinct $X',X''\subset X$ of $n+1$ points each.
\end{lemma}

\begin{proof}
First note that the same thing is true for ordinary convexity: Any point $p\in \conv X$ is in $\conv X'$ and $\conv X''$ for at least two distinct $X',X''\subset X$ of at most $n+1$ points each. To show this we consider the linear map $f : \partial\Delta^{n+1}\to \mathbb R^n$, mapping the vertices of the simplex $\Delta^{n+1}$ to the points of $X$. This map has degree zero on the boundary of the simplex and assuming the contrary, $p$ lying in the image of only one facet of the simplex, we see that the degree cannot be zero, a contradiction.

Now we proceed to the case of $p\in \conv_K X$. If $p$ is in the ordinary convex hull then it is in the ordinary convex hulls of at least two distinct proper subsets of $X$, which in turn lie in the $K$-strongly convex hulls, and we are done. Otherwise by Theorem \ref{theorem:caratheory-outside} $p$ is in a convex hull of some $X'''\subset X$ of at most $n$ points, and we can choose two distinct $X',X''\supset X'''$ of $n+1$ points each.
\end{proof}

Now we are going to prove the theorem, take $K=L\times M$, take $X\subset\mathbb R^n$ and $p\in \conv_K X$. Denote by $P$ and $Q$ the projections of $\mathbb R^n$ onto the complementary $\mathbb R^\ell$ and $\mathbb R^m$ respectively, corresponding to the decomposition $L\times M$. Evidently,
\[
\conv_K X = \conv_L P(X)\times \conv_M Q(X).
\]
By the Carath\'eodory theorem for $L$-strong convexity we find a subset $Y\subseteq X$ of at most $\ell+1$ points so that $P(p)\in \conv_L P(Y)$; by the Carath\'eodory theorem for $M$-strong convexity we find a subset $Z\subseteq X$ of at most $m+1$ points so that $Q(p)\in \conv_M P(Z)$.

Totally we have at most $\ell + m + 2 = n+ 2$ points in $Y\cup Z$ and $p\in \conv_K(Y\cup Z)$.

Consider the case $|Y\cup Z| = n+2$ first, this means $Y$ and $Z$ are disjoint. What we want to do is to drop one point $y$ from $Y$ and one point $z$ from $Z$ so that the remaining $n$ points of
\[
X' = (Y\setminus \{y\})\cup (Z\setminus\{z\})
\]
still possess the property that $P(p)\in \conv_L P(X')$ and $Q(p)\in \conv_M Q(X')$, implying $p\in \conv_K(X')$

If we take any point $y \in Y$ then by the lemma applied to $M$, $Q(p)$, $Q(Z\cup\{y\})$ there is a corresponding point $z \in Z$ such that
\[
Q(p) \in \conv_M Q\left( Z \cup \{y\} \setminus \{z\}\right).
\]
Let us choose a map $f : Y \to Z$ taking any $y$ to such $z$. In the other direction, consider the corresponding map $g$ (arising from the projection $P$ and the $L$-convexity) from $Z$ to $Y$.

It is sufficient to find $y^* \in Y$ and $z^* \in Z$ such that $f(y^*) \neq z^*$ and $g(z^*) \neq y^*$, dropping $f(y^*)$ and $g(z^*)$ we will keep $P(p)$ in $\conv_L P(X')$ and $Q(p)$ in $\conv_M Q(X')$ from the choice of $f$ and $g$. When
\begin{equation}
\label{equation:big}
(\ell + 1)(m + 1) > \ell + m + 2
\end{equation}
we just note that the $f$ and $g$ make edges of a bipartite graph on $Y\cup Z$ and we can take $y^*\in Y$ and $z^*\in Z$ not connected by an edge in any direction.

Inequality \eqref{equation:big} only fails for $\ell = m = 1$. In this case $K$ is a square in the plane and we can argue differently. Put $p$ to the origin and then it remains to find two points in $X$ that lie in the opposite quadrants relative to $p$ to have $p$ in their $K$-strong convex hull. But when it is impossible to do this it just means that a vertical or a horizontal line separates $p$ from $X$, which evidently contradicts the inclusion $p\in \conv_K(X)$.

We are done in case $|Y\cup Z|=n+2$, the case $|Y\cup Z|\le n$ is trivial since we do not need to drop points in this case. The remaining case $|Y\cup Z|=n+1$ is done similarly, we just drop a point $y^*$ from $Y$ and a point $z^*$ from $Z$, but note that since $Y$ and $Z$ may not be disjoint in this case, it may happen that we only drop one point in fact. But this is sufficient to reduce $|Y\cup Z|$ to $n$.

We have proved that the Carath\'eodory number is at most $n$, it is actually precisely $n$ because of Theorem~ \ref{theorem:at-least-n}.

\section{Proof of Theorem \ref{theorem:product-any}}

If $X$ is a set of points in the same space as $K$, denote by $P(X)$ and $Q(X)$ its corresponding projections to the spaces of $L$ and $M$.

Since directly from the definition
\[
\conv_K X = (\conv_L P(X)) \times (\conv_M Q(X)),
\]
if $p \in \conv_K X$, then there are subsets $Y$ and $Z$ of $X$ such that
\[
|Y| \le \ell,\quad P(p) \in \conv_L P(Y)
\]
and
\[|Z| \le  m,\quad  Q(p) \in \conv_M Q(Z).
\]
Therefore, $p \in \conv_K (Y \cup Z)$ thus proving $k \le \ell + m$.

Since $\ell$ is the strong Carath\'eodory number of $L$, there exists a set $Y$ of $\ell$ points and a point $p_Y$ such that $p_Y$ is in $\conv_L Y$ but not in $\conv_L Y'$ for any proper subset $Y'\subset Y$. Similarly, $Z$ and $p_Z$ are defined for $M$-strong convexity with $|Z| = m$.

Let $q_Y$ be any point in $Y$ and $q_Z$ be any point in $Z$. Consider the set of $\ell + m - 1$ points
\[
X = (Y \times q_Z) \cup (q_Y \times Z)
\]
and the point $p = p_Y \times p_Z$, against which we will test the $K$-strong Carath\'eodory theorem. Since $p_Y \in \conv_L Y$ and $p_Z \in \conv_M Z$, then,
\[
p = p_Y \times p_Z \in \conv_{L \times M} X.
\]

In order to show that $k\ge \ell + m - 2$ we need to check that $p$ does not belong to a convex hull of at most $\ell + m - 3$ points of $X$. Consider any subset $X' \subset X$ of at most $\ell + m - 3$ elements. It is easy to see that either $P(X')\neq Y$ or $Q(X')\neq Z$, otherwise $X'$ would have to be the whole $X$ possibly except the point $q_Y\times q_Z$ and $|X|$ would be at least $\ell + m - 2$.

Without loss of generality, assume $P(X')\neq Y$. Then by our choice of $p_Y$
\[
p_Y \notin \conv_L P(X') = P\left( \conv_K X' \right)
\]
and therefore $p = p_Y \times p_Z \notin \conv_K X'$.

\section{Proof of Theorem \ref{theorem:at-least-n}}

We assume $n\ge 2$, because for $n=1$ all convex bodies $K\subset \mathbb R$ are segments and the Carath\'eodory number is always $2$.

Consider the John ellipsoid of $K$, that is the ellipsoid of maximal volume contained in $K$. If we make an affine transformation to make the ellipsoid a unit ball $B$ centered at the origin then (see \cite{ball1997}) the set of tangency points $B\cap \partial K$ linearly spans the whole $\mathbb R^n$. Hence it contains certain points $v_1,\ldots,v_n$ that produce a base of $\mathbb R^n$, then normals to both $B$ and $K$ at those points are the same vectors $v_1,\ldots, v_n$.

Now make a linear transformation so that the points of $\partial K$ become some $p_1,\ldots, p_n$, while the outer normals to $K$ at those points become an orthonormal base $e_1,\ldots, e_n$. Since there is an inner tangent to $K$ ellipsoid at every $p_i$, we conclude that there is an inner tangent to $K$ ball of some radius at every point $p_i$. Scale $K$ so that all those balls have radius at least $R$, so that
\begin{equation}
\label{equation:big-radius}
R^2 \ge 1 + \left(R - \frac{1}{2n}\right)^2\Leftrightarrow \frac{R}{n} \ge 1  + \frac{1}{4n^2} \Leftrightarrow R \ge n + \frac{1}{4n}.
\end{equation}

Now let $X$ be just the set $\{e_1,\ldots,e_n\}$ and
\[
p = \frac{e_1 + \dots + e_n}{n}.
\]
Evidently $p\in \conv X \subseteq \conv_K X$ and in order to prove the theorem it remains to show that $p$ is not contained in the $K$-strongly convex hull of a proper subset $Y\subset X$. Without loss of generality assume $Y=\{e_2,\ldots,e_n\}$.

The hyperplane $H = \{x : x\cdot e_1 = 1/(2n)\}$ separates $p$ from $Y$. Translate $K$ by the vector
\[
\frac{1}{2n} e_1 - p_1,
\]
after this translation $H$ is a tangent plane to $K$ at $e_1/(2n)$ and therefore the translated $K$ does not contain $p$. Also, after the translation $K$ contains the ball $B_1$ of radius $R$ centered at
\[
\left(\frac{1}{2n} - R\right) e_1.
\]
Assumption \eqref{equation:big-radius} implies that $B_1$ also contains $e_2,\ldots, e_n$. Hence the translate of $K$ we consider contains $Y$ and does not contain $p$, proving that $p\not\in \conv_K Y$.

\section{Example with an infinite Carath\'eodory number}
\label{section:infinite}

Let $K$ be a cone in $\mathbb R^3$ with a unit disk base in the plane $A$. Let $X$ be the vertices of a regular polygon inscribed into a circle of radius $1/2$ in the plane $A$. Assume $|X|\ge 4$, since otherwise some of the pictures get too degenerate.

We will show that there exists a point $p\in\conv_K X$, which is not in any $\conv_K X'$ for any proper subset $X'\subset X$.

Put $H = \left( \conv_K X\right) \cap A.$ This planar set $H$ has boundary consisting of convex arcs between consecutive points of $X$ of radius equal one. This is so because the intersection of a translate $K+t$ with the plane $A$ is a disk of radius at most $1$ and it must contain thus described curvilinear regular polygon $H$.

Let us take a point $p$ to be in space above the center of $X$ (that is on the line through $X$ perpendicular to $A$) so that the maximum of the angle $\angle (py, A)$ over $y\in \partial H$ equals the angle $\alpha$ between the base of $K$ and any generatrix of $K$. Our choice of $p$ is to make sure that any translate of $K$ containing $X$ (whose intersection with $A$ is a disk of radius at most $1$) must contain the point $p$ as well. This is so, since otherwise a plane $B$ with $\angle (A,B)=\alpha$ would separate $H$ from $p$ and touch $H$, which we exclude by the choice of $p$.

If we drop a point to have $X'=X\setminus \{x\}$ instead of $X$, then we have $H'$ instead of $H$, with one arc of $\partial H'$ longer than others and having the removed point $x$ strictly outside, see Figure~\ref{figure:cone}. Considering a translate of $K$ with base in $A$ and with boundary of $K\cap A$ passing through this new arc (its base is the big circle in the picture), we have that $p$ is not in this translate of $K$, since the new arc is closer to the center of $X$ compared to $\partial H$. Hence the Carath\'eodory number for $K$-strong convexity is at least $|X|$, which can be arbitrarily high in our construction with the fixed cone $K$.

\begin{figure}
\includegraphics[scale=1.0]{cone}
\caption{The point $p$ is above the blue center of the circle.}
\label{figure:cone}
\end{figure}

\section{Better criterion for the Minkowski summand}
\label{section:minkowski}

Developing the property of acyclicity of $A\setminus (A + B)$ used in \cite[Lemma 2.1]{holmsenkar2017} we referenced above, in \cite{holmsenkar2017} it was shown that given a convex body $A\subset \mathbb R^n$ and a convex open bounded set in $B\subset \mathbb R^n$ such that for any vector $t\in\mathbb R^n$, the set $(A+t)\setminus B$ is either empty or acyclic, the set $A$ is a Minkowski summand of $B$. The latter means there exists an open convex $C$ such that $B = A + C$.

Here we give a version of that theorem where it is sufficient to test less sets like $(A+t)\setminus B$ for acyclicity and even exclude such unnatural things like open convex bodies from consideration. Again, the acyclicity is assumed in terms of the \v{C}ech cohomology.

\begin{theorem}
Let $A$ be a convex compactum in $\mathbb R^n$ and $B$ be a convex body in $\mathbb R^n$. Assume that a translate $A+t_0$ is contained in the interior of $B$, and for any vector $t\in\mathbb R^n$, if $(A+t)$ is contained in $B$ then the set $(A+t)\cap \partial B$ is either empty or acyclic. Then $A$ is a Minkowski summand of $B$.
\end{theorem}

Like it was in \cite{holmsenkar2017}, our proof will again use the following classical result \cite{begle1950,begle1955}:

\begin{lemma}[The Vietoris--Begle theorem]
\label{lemma:leray-map}
Let $f : X\to Y$ be a proper continuous map between metric spaces such that for every $y\in Y$, the fiber $f^{-1}(y)$ is acyclic. Then $f$ induces an isomorphism of the \v{C}ech cohomology of $X$ and $Y$.
\end{lemma}

\begin{proof}[Proof of the Theorem]
Put $C = B\gdiv A$, $C$ is a convex body since it contains $t_0$ in its interior, we will show that $A+C = B$. For any $t\in \partial C$ we have $A + t \subseteq B$, hence $(A+t)\cap \partial B$ is an acyclic compactum, it cannot be empty since otherwise $A+t\subseteq \inte B$ and from the positive distance between $A+t$ and $\mathbb R^n\setminus \inte B$ there would exist a neighborhood $U(t)$ such that $A+U(t)\subseteq B$, which would imply $t\in \inte C$, not in $\partial C$.

Now consider the set
\[
Z = \left\{ (x,t) : x\in \partial B,\ t\in \partial C,\ x \in A + t \right\},
\]
it is compact and its projection to $\partial C$ has compact acyclic fibers $(A+t)\cap \partial B$ as already shown. Hence by the Vietoris--Begle theorem $Z$ has the same \v{C}ech cohomology as $\partial C$, which is the $(n-1)$-sphere.

Consider now the projection of $Z$ to the first summand, $\partial B$. A fiber of this projection over $x\in \partial B$ is
\[
\left\{ t \in\partial C: A + t \ni x \right\} = \left\{ t \in C: A + t \ni x \right\} = C \cap (t - A),
\]
which is a possibly empty convex compactum. Let $S\subset \partial B$ be the image of this projection, the Vietoris--Begle theorem implies that the \v{C}ech cohomology of $S$ is the same as that of $Z$ and therefore its cohomology is the cohomology of the $(n-1)$-sphere. It $S$ is not the whole $\partial B$ then its $(n-1)$-dimensional cohomology would vanish, since it can be calculated as the limit of the cohomologies of its neighborhoods, open $(n-1)$-manifolds.

Hence $S$ must be the whole $\partial B$ and then $A+ C \supseteq \partial B$, hence $A+C = B$.
\end{proof}

\bibliography{../Bib/karasev}
\bibliographystyle{abbrv}
\end{document}


TITLE: Continuous bounded functions in $L^1$
QUESTION [2 upvotes]: Is a continuous function in $L^1$bounded?
I know that continouos functions are always bounded on a compact intervall.
But how do I prove it?

REPLY [5 votes]: If we study $L^1(0,\infty)$ or $L^1(\Bbb R)$, then you can have an unbounded continuous integrable function.
We can build it in the following way: let's start with $f= 0$. Then we add positive continuous "bumps" at $n=1,2$,etc  with each "bump" being higher, say, of magnitude $n$, but adding only $1/n^2$ to the integral. Like this our function remains integrable, continuous, yet unbounded.
One can explicitly build such a function, I'll outline the important details
First, we take $$g(x)=\begin{cases} e^{-\frac{1}{1-x^2}},&|x|<1,\\0,&|x|\ge 1.\end{cases}$$
It's possible to show that this function is $\mathcal C^{\infty}(\Bbb R)$, its support is $[-1,1]$, it's positive, and its integral is finite (let's call it $I$). Its supremum is $e^{-1}$.
Now let's study $$g_n(x):=ng\left( (x-n)n^3 \right).$$ It's still continuous, positive, its support is $[n-n^{-3}, n+n^{-3}]$. Its integral is $\frac{I}{n^2}$, and its supremum is $ne^{-1}$.
We take the sum $$G(x):=\sum_{k\ge 3}g_k(x).$$ It's possible to show that $G$ is continuous, unbounded, positive, integrable.


TITLE: Higher Order Multivariable Taylor Expansions
QUESTION [2 upvotes]: The quadratic multivariable Taylor approximation of a function $f(x, y)$ around a point $(a, b)$ is given by $f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) + \frac{1}{2}f_{xx}(a, b)(x - a)^2 + f_{xy}(a, b)(x - a)(y - b) + \frac{1}{2}f_{yy}(a, b)(y - b)^2$  There is a vectorized version of this expression which works for functions and points with an arbitrary number of variables, $f(\vec{a})\ + ∇f(\vec{a})\cdot(\vec{x} - \vec{a}) + \frac{1}{2}(\vec{x} - \vec{a})^TH(\vec{a})(\vec{x} - \vec{a})$, where the elements of $\vec{a}$ are the coordinates of the point around which we're approximating, the elements of $\vec{x}$ are the inputs of $f$, and $H$ is the Hessian matrix of $f$.
Following the pattern of single variable Taylor approximations, the cubic multivariable Taylor approximation should be $f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) + \frac{1}{2}f_{xx}(a, b)(x - a)^2 + f_{xy}(a, b)(x - a)(y - b) + \frac{1}{2}f_{yy}(a, b)(y - b)^2 + \frac{1}{3!}f_{xxx}(a, b)(x - a)^3 + \frac{3}{3!}f_{xxy}(a, b)(x - a)^2(y - b) + \frac{3}{3!}f_{xyy}(a, b)(x - a)(y - b)^2 + \frac{1}{3!}f_{yyy}(a, b)(y - b)^3$
The amount of simplification in my coefficients is inconsistent, and I'm intentionally not generalizing the approximation order by introducing additional variables or sigma notation, as this expression already requires more than enough mental unpacking in my opinion, and it can only get worse in vector notation.  Instead I'll use an image and words to describe the pattern I have in mind.

Every unique partial derivative up to the order of the approximation constitutes a term (third order approximation includes all zeroth, first, second, and third partial derivatives), and the numerator of the fraction coefficient for each term is the number of equivalent permutations of that partial derivative ($f_{xyy} = f_{yxy} = f_{yyx}$, so the numberator is $3$).
Is this the correct extension of the quadratic multivariable Taylor approximation to cubic approximations, and a correct pattern for writing quartic and higher approximations, and how can we put this into vector notation?  An optimal answer, in order to avoid layering on too much abstraction, will likely include the expression for a cubic approximation in vector notation (possibly also quartic, if helpful) and an explanation in words of how to continue the pattern into higher order approximations (i.e., do we require an increasingly higher order "tensor," a "Hyperhessian" of sorts, as we continue to add more terms?).

REPLY [1 votes]: Your expression is correct.
If you want to write it in "vector notation" we simply use the usual way of writing derivatives. Denote by $f^{(p)}$ the $p$th derivative of $f$ (which, by the way, is a $p$-linear continuous function) and write $h^{(p)}$ to mean the vector $(h, \ldots, h)$ ($h$ appearing $p$ times). Then, the Taylor polynomial of $f$ centred at $a$ of degree $n$ is
$$\mathbf{T}_nf(a) \cdot h = f(a) + f'(a) \cdot h + \dfrac{f''(a) \cdot h^{(2)}}{2!} + \ldots + \dfrac{f^{(n)}(a) \cdot h^{(n)}}{n!}.$$


TITLE: Does integral of function against all test function derivative imply function is equal to 0?
QUESTION [2 upvotes]: Howdy so I know that if
$\int fv = 0 $ for all test functions $v$ then $f=0$
If you have however
$\int fv_x = 0 $ for all test functions $v$ then does it mean that $f=0$?
I guess my thought is that the if you take the test function space and differentiate all elements, then its the same space again - right?
I just wanted someones opinion on this.

REPLY [0 votes]: If $f$ was a constant $f=c$, then $\int fv_x = c \int v_x = cv|_{-\infty}^{\infty} = 0$, which works for $c\neq 0$ also. Hence, there are non-zero solutions to the distributional equation  $f'=0$.
You can also prove that these are the only solutions.  As distributional solutions to $f_x = g$ for $f,g\in\mathcal D'$ are unique up to constants ("Distributions" by Duistermaat and Kolk, page 47, Theorem 4.3), the solutions to $f_x = 0$ are constants.
The uniqueness part of the proof in the book above proceeds by defining the map $p:\mathcal D\to\mathcal D$
$$ p (\phi)(x) := \int_{-\infty}^x\phi - \langle1,\phi\rangle \int_{-\infty}^x\chi$$
where $\chi$ is some fixed element of $\mathcal D$ with $\int\chi=1$. One can check that $p(\phi)$ is indeed always in $\mathcal D$ with $p(\phi)_x = \phi - \langle 1, \phi \rangle \chi$; the purpose of the second term is to make $p(\phi)$ vanish for large $x$.
Therefore, for all $\phi \in \mathcal D$, $$0 = \langle f_x , p(\phi)\rangle = \langle f, \phi - \langle 1, \phi \rangle \chi\rangle  = \langle f, \phi\rangle - \langle 1, \phi\rangle \langle f, \chi\rangle = \big \langle f -\langle f, \chi\rangle,\phi\big \rangle $$
So $f -\langle f, \chi\rangle$ is equal to $0$ in the sense of distributions, which is to say, $f = \langle f, \chi\rangle$ in $\mathcal D'$.


TITLE: Differential Geometry : Curve
QUESTION [0 upvotes]: Show that A regular curve between two points $p,q$ in $\mathbb{R}^n$ with minimal length is necessarily the line segment from $p$ to $q$

REPLY [0 votes]: It is demonstrated by the calculation of the variations that in Euclidee geometries, or of type $ \mathbb{R}^n\ $  , the minimum length curve that joins two points is "the segment". I emphasize that this result is not valid in non-Euclidean geometries as originally demonstrated by Gauss.


TITLE: Why this equation does not develop a boundary layer?
QUESTION [3 upvotes]: The equation I am talking about is
$$
\epsilon  y''(x)+y(x)+1=0,y(0)=0,y(1)=1
$$
The $+1$ is not essential as $y(x)$ can be decomposed into $1 + y_1$, but is kept here for a more direct comparison with the other example below. This equation takes a more innocent look if multiplied by $1/\epsilon $ that yields $y''(x) + 1/ \epsilon  y(x) + 1/\epsilon = 0$, sine/cosine function follows (the figure below shows solution with $\epsilon $ = 0.01, which corresponds to a period $2 \pi \sqrt{\epsilon} = 0.62$).

Some may say as $\epsilon $ decreases, the frequency gets higher and the curve get steeper, but there is no boundary layer for at least two reasons: first, this is a global behavior; second, typical boundary layers has an exponential rate of change that limit it to a narraw region (then merges smoothly with the outer solution). The example to be compared against is one with the second term $y(x)$ replaced by $y'(x)$, clearly boundary layer develops (figure also uses $\epsilon = 0.01$).

Buy why? The only distinction is there is no first order term in the first equation, but this argument appears to be very superficial.

REPLY [2 votes]: The easiest thing to do is to use your observation that the oscillation frequency will increase as $\epsilon$ gets smaller. Therefore, it seems useful to introduce the rescaled variable
\begin{equation}
\xi = \frac{x}{\sqrt{\epsilon}},
\end{equation}
such that the ODE becomes
\begin{equation}
 \frac{\text{d}^2 y}{\text{d}\xi^2} + y(\xi) +1 = 0,\; y(0) = 0, \;y(1/\sqrt{\epsilon}) = 1. \tag{1}
\end{equation}
In this 'desingularised' formulation, you see that the small parameter $\epsilon$ has vanished from the differential equation. It now only appears in the second boundary condition, which is now asymptotically 'far away', because we introduced the small scale variable $\xi$.
The solution to $(1)$ only including the boundary condition at $\xi=x=0$ is
\begin{equation}
 y(\xi) + 1 = \cos \xi + c_0 \sin \xi
\end{equation}
The second boundary condition now yields
\begin{equation}
 c_0  = \frac{2-\cos \frac{1}{\sqrt{\epsilon}}}{\sin \frac{1}{\sqrt{\epsilon}}}. \tag{2}
\end{equation}
As you can see, the problem $(1)$ has no solution if
\begin{equation}
 \epsilon = \epsilon_n := \frac{1}{4 \pi^2 n^2},
\end{equation}
because then every solution for which $y(0) = 0$ automatically satisfies $y(1) = 0$. Also, since $\sin$ and $\cos$ are periodic and bounded, $(2)$ does not have an asymptotic approximation for small values of $\epsilon$.
To answer your original question: not every singularly perturbed problem has a boundary layer. The appearance of the exact or approximate solution highly depends on the specific characteristics of the problem. A general approach to these questions is geometric singular perturbation theory, which combines methods from dynamical systems and multiple time scale analysis. For more information, see
C. Kuehn, Multiple Time Scale Dynamics, Springer, 2015.


text:
\begin{document}
\selectlanguage{english}
\author{D. Sytnyk, V. Makarov, V. Vasylyk}
\email{sytnik@imath.kiev.ua}
\urladdr{www.imath.kiev.ua/$\sim$sytnik}
\address{Institute of mathematics, National Academy of Sciences, Ukraine \\ 01601 Ukraine, Kiev-4,
3, Tereschenkivska st.}
\title{Existence of the solution to a nonlocal-in-time evolutional problem.}

\begin{abstract}

This work is devoted to the study of a nonlocal-in-time evolutional problem for the first order differential equation in Banach space.
Our primary approach, although stems from  the convenient technique based on the reduction of a nonlocal problem to its classical initial value analogue,  uses more advanced analysis.
That is a validation of the correctness in definition of the general solution representation via the Dunford-Cauchy formula.
Such approach allows us to reduce the given existence problem to the problem of locating zeros of a certain entire function.
It results in the necessary and sufficient conditions for the existence of a generalized (mild) solution to the given nonlocal problem.
Aside of that we also present new sufficient conditions which in the majority of cases generalize existing results.

\textbf{Keywords:} nonlocal-in-time evolutional problem, unbounded operator coefficient, mild solution, zeros of polynomial.

\end{abstract}
\maketitle

\section{Introduction}
The Cauchy problem for differential equation with a nonlocal in time condition:
\begin{equation}\label{bp}
      \begin{array}{l}
    u'_t+Au=f(t), \quad t \in [0,T]
    \\[1em]
    u(0)+g(t_1,t_2,\ldots, t_k, u) =u_0, \quad 0\leq t_0< t_1<\ldots < t_k \leq T,
  \end{array}
\end{equation}
with $A$ being a densely defined strongly positive operator (the detailed definition will be given below) and $g:X\rightarrow D(A) \subseteq X$, is one of the important topics in the theory of differential equation theory and its application.
Interest in such problems stems mainly from the better effect of the nonlocal initial condition than the usual one in treating physical problems.
Actually, the nonlocal initial condition from (\ref{bp}) models many interesting nature phenomena \cite{Allegretto1999}, \cite{Christensen1982}, where the normal initial condition $u(0) = u_0$ may not fit in. In addition some models of the control theory \cite{nonlocal_time_numerVabishchevich1982} and economical management problems may be represented in form (\ref{bp}) as well.

Even in a much simpler form:
\begin{subequations}\label{bp1}
    \begin{equation}\label{bp1:eq}
    u'_t+Au=f(t), \quad t \in [0,T]
    \end{equation}
    \begin{equation}\label{bp1:nc}
    u(0)+\suml_{k=1}^{n}\alpha_k u(t_k) =u_0,\quad 0<t_1<t_2<\ldots < t_n\leq T,
    \end{equation}
\end{subequations}
the model reflects an important case when one is more likely to have  more information about the solution at the times $t = t_1,t_2, \ldots, t_n$ rather than as in the classical case just at $t=0$.
Such situation is common for physical systems where an observer was not able to witness (or measure ) characteristics of the system at the initial time.
It is also intrinsic for modelling certain physical measurements performed repeatedly by the devices having relaxation time comparable to the delay between the measurements.
A simple example of this kind is a recovery of movement captured by a multi-exposure camera (streak-camera).

The particular cases of (\ref{bp1}) covers many well-known physical phenomena such as: problems with periodic conditions
$u(0)-u(t_1)=0$,  problems with Bicadze-Samarskii conditions $u(0)+\alpha_1 u(t_1)=\alpha_2 u(t_2)$, regularized backward problems  etc.
One can not underestimate the role of nonlocal problems having the form \eqref{bp1} in the theory of ill--posed problems, where they appear as natural counterparts of the improperly posed problem in the course of quasi-regularization technique \cite{IllPosedTikhonov1995}.

Historically the first, to the authors best knowledge, work devoted to the problems with nonlocal-in-time conditions was the work of Dezin \cite{NonlocaAbstractDezin1965}.
In this work author studied the restrictions of abstract differential operators $u'(t)+A$ in a Hilbert space imposed by the non-local condition \eqref{bp1} ( see also \cite{bookDezin1980} and the references therein).
Similar theoretical technique in conjunction with a numerical scheme was used to solve nonlocal-in-space boundary value problem for elliptic partial-differential operators in the work of  Bicadze and Samarskii \cite{NonlocalBitsadze1969}.
Next addition to the theory was made by Gordeziani with co-authors \cite{nonloc_BS_Gordeziani1970}, \cite{nonloc_BS_Gordeziani1972},

They  rather generalized the previous results and proposed iterative problem-solving methods. Eigenvalue problems for elliptic operators with nonlocal conditions were considered in \cite{Bandyrskii2006} see also \cite{Sapagovas2012} for review of recent works in that direction.
In \cite{Byszewski1991}, \cite{Byszewski1992}, \cite{NonlocalAbsNonLinJackson1993}  authors stated the sufficient conditions for the existence and uniqueness of a solution to \eqref{bp} in a Banach space.
Initial analysis of \eqref{bp1} used in the present work was performed in \cite{Gavrilyuk2010}. Starting from the conditions similar to those proposed in \cite{Byszewski1992} authors of \cite{Gavrilyuk2010} develop  efficient parallel numerical methods for \eqref{bp1}. They use a rather general methodology developed for classical Cauchy problem for equation \eqref{bp1:eq}

Majority of later theoretical works are mainly devoted to the generalization of results, received in \cite{Byszewski1991}, to more wider classes of equations.

Aside of that, some efforts have been made in the direction of sharpening the sufficient existence and uniqueness conditions for various specific classes of a nonlocal condition from \eqref{bp} and an operator coefficient $A$ \cite{Liang2002}.

Most of previous research concerning the nonlocal problem \eqref{bp1}  were done under a strict constraint on quantities  $\alpha_i$, from the nonlocal condition \cite{Byszewski1991}, \cite{NonlocalAbsNonLinJackson1993}, \cite{Byszewski1992}:
$$
\suml_{i=1}^{n} |\alpha_i| \leq 1,
$$
allowing the transition from \eqref{bp1} to the classical Cauchy problem.
This inequality obviously gives only a sufficient condition.

Indeed, if one had a solution of ordinary Cauchy problem for equation from (\ref{bp}) he could simply choose such a nonlocal condition that mentioned constrain fails but the solution exist and unique.

In \cite{NonlocalAbsNonLinNtouyas1997} authors, using spectral characteristics of $A$, obtained  new slightly weaker constraint
\begin{equation}\label{estLiang2002}
\suml_{i=1}^{n} |\alpha_i|e^{-\rho t_i} \leq 1,
\end{equation}
here $\rho$ has the same meaning as in (\ref{spectrA}).

Here we will introduce more advanced and yet quite straightforward technique for the treatment of nonlocal problems.
To do that, in section \ref{sec1} we reduce the given problem \eqref{bp1} to the corresponding Cauchy problem  with ordinary initial value condition as in \cite{Byszewski1991}, \cite{NonlocalAbsNonLinNtouyas1997}.
The subsequent analysis, presented in section \ref{sec2}, study the correctness of the solution operator defined via the Dunford-Cauchy formula on a feasible integration contour.
It is equivalent to the calculation of zeros set for a certain entire function or the polynomial corresponding to it.
The resulting necessary and sufficient conditions for the existence and uniqueness of the solution take in to account both the nonlocal condition and the spectral information of $A$.

Section \ref{sec3} is devoted to the situation when zeros of the mentioned entire function can not be calculated directly.
That may be caused by a dependence of the nonlocal condition on some additional parameter like in applications of quasi-regularization technique, or because of a large number of time moments $t_k$ in \eqref{bp1}, etc.
Such situation as it will be shown,  can be circumvented using some available zeros estimates \cite{Milovanovic2000}, \cite{Milovanovic2000a} for the mentioned entire function in its reduced to the polynomial form. It results in the sufficient conditions for \eqref{bp1} which nonetheless outperform \eqref{estLiang2002} in many important applications.

\section{Reduction of nonlocal problem to classical Cauchy problem.}  \label{sec1}
From now on we assume that the coefficient  $A$ from \eqref{bp}  is a densely defined strongly positive operator acting in Banach space
$X \supseteq D(A) \rightarrow X$ \cite{bHilPhil1997}, \cite{bGavrilyuk2011}.
Its spectrum $\Sigma (A)$ lies in a sector
\begin{equation}\label{spectrA}
\Sigma = \left\{
z=\rho+r e^{i\varphi}:\quad r \in [0,\infty), \ \left|\varphi\right| \leq \theta < \frac{\pi}{2}, \rho > 0
\right\},
\end{equation}
while the resolvent $R\left(z,A\right)$ of $A$ satisfies the following estimate on the boundary of spectrum $\Gamma_\Sigma$  and outside of it

\begin{equation}\label{estrez}
\left\|(zI-A)^{-1}\right\|\leq \frac{M}{1+\left|z\right|},
\end{equation}
with $\left\| \cdot \right\|$ being an operator norm.

The class of strongly positive operators plays an important role in  applications of functional analysis to the theory of partial differential equations, dynamical systems, numerical analysis, etc.  Strongly--elliptic partial differential operator defined on a bounded Lipschitz domain is a strongly positive operator with spectral parameters that can be estimated from the coefficients of elliptic operator \cite{fujita}, similar is true for a general elliptic pseudo-differential operator.

The theory of sectorial operators in its present form was developed in the works of Hille, Dunford, Philipps  \cite{b_cp_Fattorini1984}.
According to this theory every closed strongly-positive operator generates a one parameter semi-group
$T(t)= e^{-t A}$
which acts as a propagator for a solution to \eqref{bp1:eq}.
That is any solution $u(t) \in D(A)$ of the differential equation \eqref{bp1:eq} has the following representation
\begin{equation}\label{bp1int}
  u(t)=e^{-At}u(0)+\intl_0^t{e^{-A(t-\tau)}f(\tau)}d\tau.
\end{equation}
Recall that the opposite is not true in general \cite{b_cp_Fattorini1984}, since even if $u(t)$ satisfies \eqref{bp1int} it does not need to belong to the domain of $A$. It is also well know that $D(A)$ is dense in X so there always exists a sequence of elements from $D(A)$ converging to $u(t)\in X$ defined by \eqref{bp1int}.
A function $u(t)$ satisfying \eqref{bp1int} is called a \textit{generalized }\cite[p. 30]{b_cp_Fattorini1984} or sometimes \textit{mild } \cite[p. 117]{arendt} \textit{solution} to problem \eqref{bp1}. Formula \eqref{bp1int} becomes more convenient than the original equation \eqref{bp1:eq} in cases where the classical Cauchy problem for \eqref{bp1:eq} with a given $u(0)$ is considered. In our case of problem \eqref{bp1} that distinction between \eqref{bp1int} and \eqref{bp1:eq} is not so obvious because $u(0)$ is not given directly.

Here we intend to derive a direct representation of the solution to \eqref{bp1}.

Let us start from representation \eqref{bp1int} and assume the existence of an initial value $u(0)$ such that  $u(t)$ defined by \eqref{bp1int} satisfies nonlocal condition \eqref{bp1:nc} (the precise conditions for the existence of such $u(0)$ will be stated below).

By substituting  the formula for $u(0)$ from \eqref{bp1:nc} into representation \eqref{bp1int} and evaluating the result at  $t_i, i=\overline{1,n}$ we obtain the system of equations
\begin{equation}
\label{bp1LS}
  \begin{array}{l}
    u(t_i)=e^{-At_i}\left[u_0-\suml_{k=1}^{n}\alpha_k u(t_k) \right]+\intl_0^{t_i}{e^{-A(t_i-\tau)}f(\tau)}d\tau,\\
    i=\overline{1,n}.
  \end{array}
\end{equation}

Next we multiply each part of $i$-th equation by $\alpha_i$ and then sum up the resulting equalities. It gives us the following:

\begin{equation}
  \label{sum_t_i}
  \begin{split}
  \suml_{i=1}^n \alpha_i u(t_i) =& \suml_{i=1}^n \alpha_i e^{-At_i} u_0-\suml_{i=1}^n \alpha_i e^{-At_i}
  \suml_{k=1}^n \alpha_k u(t_k)+
  \\
  &+\suml_{i=1}^n \alpha_i \intl_0^{t_i} e^{-A (t_i-\tau)}f(\tau) d\tau.
    \end{split}
\end{equation}
 Now let us put $\suml_{i=1}^n \alpha_i u(t_i)=w$, then  (\ref{sum_t_i}) can be represented as follows:
 $$
 w = {-\suml_{i=1}^n \alpha_i e^{-At_i} w} +\suml_{i=1}^n \alpha_i e^{-At_i} u_0+ \suml_{i=1}^n \alpha_i \intl_0^{t_i} e^{-A (t_i-\tau)}f(\tau) d\tau,
 $$
 the last equality can be regarded as an operator equation with respect to $w$. Using the notation
 $B(A)=I +\suml_{i=1}^n \alpha_i e^{-At_i}$, we rewrite it to get
\begin{equation}\label{eqnoneq1}
B w = B u_0 -u_0 +\suml_{i=1}^n \alpha_i \intl_0^{t_i} e^{-A (t_i-\tau)}f(\tau) d\tau.
\end{equation}
Here unknown $w$ appears only on the left-hand side of \eqref{eqnoneq1}, hence in order  to solve \eqref{eqnoneq1} for $w$ we need to assume the existence and boundedness of the operator valued function $B^{-1}(A)$ inverse to $B(A)$ which we will call reduction operator.

As it will turn out later (see Theorem \ref{thmNCExist}) such assumptions with respect to $B^{-1}(A)$ is the only thing we need to reduce the question about the existence of solution to \eqref{bp1} to the question about the existence of corresponding solution to a classical Cauchy problem associated with \eqref{bp1:eq}.
For the time being let us assume that $B^{-1}(A)$ is a properly defined bounded operator valued function, in that case
\[
 w = u_0- B^{-1} u_0 + B^{-1}\suml_{i=1}^n \alpha_i \intl_0^{t_i} e^{-A (t_i-\tau)}f(\tau) d\tau.
 \]
Our last step is to substitute $u(0) = u_0-w$ \eqref{bp1int}, and finally get a direct representation of solution to \eqref{bp1} free of the unknown values $u(t_k)$:
 \begin{equation}
   \label{bp1IntRed}
    \begin{split}
   u(t)=&e^{-At}\left[B^{-1} u_0 - B^{-1}\suml_{i=1}^n \alpha_i \intl_0^{t_i} e^{-A (t_i-\tau)}f(\tau) d\tau\right] +
   \\
   &+\intl_0^t{e^{-A(t-\tau)}f(\tau)}d\tau.
    \end{split}
 \end{equation}

For further analysis of relationship between the solution to \eqref{bp1} and the initial data $\alpha_i, t_i$ we will need a following definition from the operator function calculus \cite[p. 167]{bHilPhil1997},  \cite{Clem}.
\begin{definition}\label{thmDunford}
Let $f(z)$ be a complex valued function  analytic in the neighbourhood of the spectrum $\Sigma(A) \subset \C$ and at the infinity. Suppose that there exist an open set $V \supset \Sigma(A)$ with the boundary $\Gamma$ consisting of a finite number of rectifying Jourdan curves such that $f(z)$ is analytic in $V\cap \Gamma$,  then $f(A)$ can be defined as follows

\begin{equation}
		\label{reprDunford}
		f(A) x=f(\infty)I+\frac{1}{2\pi i} \intl_{\Gamma}f(z) R(z,A)x dz,
\end{equation}
here we assume that the integral is taken over the positively oriented contour $\Gamma$ which may pass trough $\infty$.
\end{definition}
The integral in \eqref{reprDunford} is called a Dunford-Cauchy integral.

Representation \eqref{reprDunford} once applied to $B^{-1}(A)$ lead us to the formula
\begin{equation}
		\label{reprDelta}
		B^{-1}(A) u=I+\frac{1}{2\pi i}\intl_{\Gamma_A} \frac{1}{1+\sum_{k=1}^n{\alpha_k e^{(-t_k z)}}} R(z,A) u dz,
\end{equation}
from which it is clear that the only possible source of singularities of reduction operator $B^{-1}(z)$ would be a set of zeros of denominator from \eqref{reprDelta}.
Thus the function $B^{-1}(A)u$ is properly defined in the sense of Definition \ref{thmDunford} if and only if all the zeros of
\begin{equation}
	  \label{zerosExp}
	 B(z)=1+\sum_{k=1}^n{\alpha_k e^{(-t_k z)}},\quad z \in \mathbb{C}_+,
\end{equation}
belong to a set  $\mathbb{C} \backslash \Sigma $.
Now we can formalize our previous analysis as a theorem.
\begin{theorem}\label{thmNCExist}
Let  $A$ be a strongly positive linear operator with the spectral parameters $(\rho, \theta)$, and  $f(t) \in L^1((0;T),X)$ be a given function. Then the generalized solution \eqref{bp1IntRed} exists if and only if the set of zeros $ \Ker(B(z))\equiv\left\{z:B(z)=0, z \in C \right\}$, of $B(z)$ associated with nonlocal condition \eqref{bp1:nc}, satisfy the inclusion
\begin{equation}\label{zerosB}
\Ker(B(z))\subset \mathbb{C} \backslash \Sigma.
\end{equation}
\end{theorem}
\begin{proof}
\textit{1. Necessary conditions.}

We assume that the solution to \eqref{bp1} exists.
Then this solution should satisfy formula \eqref{bp1int} (see \cite[ Proposition 3.1.16]{arendt}) which in the homogeneous case $f(t) \equiv 0$ is reduced to
\begin{equation*}\label{exp_sol_repr}
u(t)=e^{-At}u(0).
\end{equation*}

Once the validity of \eqref{bp1int} (or \eqref{exp_sol_repr}) for $u(t)$ is established we can use it along with \eqref{bp1:nc} to get operator equation \eqref{eqnoneq1}, in a way described above.
This equation, as we already discovered,  has a nontrivial solution only if the inclusion \eqref{zerosB} is valid.

\newline
\textit{ 2. Sufficient conditions.}

To prove the sufficiency let us focus our attention on function $u(t)$ defined by  \eqref{bp1IntRed}.
Operator valued function $e^{-A (t-s)}$ from the integrands appearing in \eqref{bp1IntRed} is differentiable so the integrals  are convergent for any $f(t) \in L^1((0;T),X)$, and therefore $u(t)$ is properly defined and bounded once $B^{-1}(A)$ is properly defined. Next from the Dunford-Cauchy integral representation \eqref{reprDelta} we observe that $B^{-1}(A)x$  is properly defined and bounded $\forall x \in X$ as long as the propositions of the theorem regarding the spectrum of $A$ are fulfilled and \eqref{zerosB}  is valid.
It remains to show that  $u(t)$ defined by  \eqref{bp1IntRed} is indeed a generalized solution to \eqref{bp1}. This can be easily done by substitution of the representation of $u(t)$ into \eqref{bp1:nc}.
\end{proof}
This theorem answers the question regarding the existence of solution to problem \eqref{bp1}. Such result alone have a limited theoretical and practical utility.  Two supplementary questions regarding the uniqueness of $u(t)$  and its stability with respect to the initial data are need to be studied as well.

We discuss the stability first. In order to show that assumptions stated in Theorem \ref{thmNCExist} imply the continuous dependence of solution $u(t)$ defined by \eqref{bp1IntRed} on $u_0$ $f(t)$ and the parameters of nonlocal condition we recall similar results for classical Cauchy problem.
If the theorem's assumptions regarding $A$ are valid then the classical Cauchy problem associated with equation \eqref{bp1:eq} is \textit{well posed} in the sense of \cite[p. 29]{b_cp_Fattorini1984}. Its solution represented by \eqref{bp1int} exist for any initial state  $u(0) \in X$ and for $u(0) = u_0 - w$ in particular. The last expression continuously depends on the initial data.

The uniqueness of the solution to \eqref{bp1}  can be justified in similar manner as for the ordinary abstract Cauchy problem associated with equation \eqref{bp1:eq}.
Clearly any function $v(t)$ satisfying \eqref{bp1:eq} can be represented in the form
$$
v(t)=e^{-At}v(0).
$$
Assume that $u(t)$ is a solution to nonlocal problem \eqref{bp1}  $\forall t\geq 0$  $u(t) \in D(A)$, and consider a function
$$
w(t)=e^{-At}u(0).
$$
Both $w(t)$, $u(t)$  satisfy differential equation \eqref{bp1:eq}, and so does their difference $p(t)= w(t)-u(t)$.
The corresponding Cauchy problem is well-posed thus
it has a unique solution for any initial state $u_0 \in D(A)$ \cite[Theorem 23.8.1]{bHilPhil1997}.
We take $p(0)=0$ as such state. Then the difference $p(t)\equiv 0$ everywhere since it the only  solution to \eqref{bp1:eq} with the zero initial condition.

\begin{example}\label{ex1}
To demonstrate the application of Theorem \ref{thmNCExist} let us consider the following nonlocal problem
\begin{equation}\label{exNCP1p}
  \begin{array}{l}
    u'_t+Au=f(t), \quad t \in [0,T]\\[1em]
    u(0)+\alpha_1 u(t_1) =u_0,\quad 0<t_1\leq T.
  \end{array}
\end{equation}
For such nonlocal condition  $B(z)$  will have the form
$$
B(z) = 1 +\alpha_1e^{-zt_1}.
$$
Its representation permits us to write the set of zeros $\Ker{B(z)}$ in a closed form
\begin{equation}
\begin{split}\label{kerB1p}
\Ker(B(z)) = & -\frac{1}{t_1}\ln\left(-\frac{1}{\alpha_1}\right)=
\\
= &-\frac{1}{t_1}\left[\ln\left|\frac{1}{\alpha_1}\right|+i\left( \Arg\left(-\frac{1}{\alpha_1}\right)+2\pi m\right) \right], \quad m\in \Z
\end{split}
\end{equation}
here $\Arg\left(\cdot\right)$ stands for the principal value of argument.
Assuming that the operator $A$  has the spectral parameters  $(\rho, \theta)$
$$
z=x+iy \in \mathbb{C} \backslash \Sigma \Leftrightarrow |y|>(x-\rho)\tan\theta.
$$
One can observe from the last inequality that if the principal value of logarithm satisfies condition \eqref{kerB1p} the same is true for the entire set of the logarithm values, so one can safely put $m=0$ in \eqref{kerB1p}.

Condition \eqref{zerosB} for problem \eqref{exNCP1p}, therefore is equivalent to the following inequality
\begin{equation}\label{estNC1p}
\left|\Arg\left(-\frac{1}{\alpha_1}\right)\right|>\left(\ln\left|{\alpha_1}\right|-t_1\rho\right)\tan\theta.
\end{equation}
Note that unlike \eqref{estLiang2002} or earlier estimates by Byszewski this inequality takes into account both spectral parameters of $A$. Closed form representation for $\Ker{B(z)}$ and the proposition of Theorem \ref{thmNCExist} guarantee that \eqref{estNC1p} forms a necessary and sufficient conditions for the existence of generalized solution to \eqref{exNCP1p}.

Example \ref{ex1} is instructive in a sense that for such nonlocal problem  one can easily compare conditions \eqref{estLiang2002} and \eqref{estNC1p} graphically.
This comparison is given on Figure \ref{nc1pFig1} where we depict three sets of admissible values of $\alpha_1 \in \C$ for $t_1=1$.
\begin{figure}[ht]\label{nc1pFig1}
\vspace*{-3pt}
\centering\includegraphics[width=0.95\linewidth]{nc1p_alpha_t=1f}

\vspace*{-3pt}
\caption{Admissible values of parameter $\alpha_1 \in \C$ from nonlocal condition of problem  \eqref{exNCP1p} obtained by: \cbox{gray80} -- estimate  \eqref{estLiang2002} ($\theta=\pi/4$), \cbox{gray60} --  estimate \eqref{estNC1p} ($\theta=\pi/4$) and  \cbox{gray40} by estimate \eqref{estNC1p} ($\theta=\pi/6$). Spectral parameter $\rho=1$ everywhere.}
\end{figure}
Observe that for the operator $A$ with spectral parameters $(1,\pi/4)$ ($\rho=1$, $\theta=\pi/4$) the set of admissible $\alpha_1$ obtained from  \eqref{estNC1p} (interior of the region coloured in \cbox{gray60}) contains in itself as a subset the admissible set obtained by  \eqref{estLiang2002} (coloured in \cbox{gray80}). This set remains the same for the whole family of sectorial operator coefficients with some fixed $\rho$ and $\forall \theta \in [0, \pi/2]$ since \eqref{estLiang2002} are independent of $\theta$.
While in reality the admissible set grows larger when we make $\theta$ smaller.  Check for example the corresponding set for the case  $\theta=\pi/6$ obtained using \eqref{estNC1p} which is coloured in \cbox{gray40} on Figure \ref{nc1pFig1}.  In the limiting case of $\theta=0$ when $A$ is self-adjoint this set becomes equal to $\C \backslash (-\infty; -e^{t_1\rho})$.

As partial case of \eqref{exNCP1p} one can consider the problem
with $A = -\frac{d^2}{dx^2} + \pi^2 - 1,$ $D(A)=\left\{v(x) : v \in H^2, \ v(0) = v(1)=0 \right\}$, $\rho = 1$ and \eqref{bp1:nc} in the form $u(0)+2e u(1) = 3\sin(\pi x)$.
Even thought the solution to such problem exists and has the form $u(x,t) = e^{-t} \sin(\pi x)$ condition \eqref{estLiang2002} fails.
Meanwhile the approach described above shows that solution to the mentioned problem exists $\forall \alpha_1 \in \C \backslash (-\infty; -e)$.
\end{example}

To convince the reader that situation complicates when the nonlocal condition consists of more than one value of unknown at the given times we study a problem with two-point nonlocal condition.

\begin{example}\label{exBS}
Let us consider the problem
	\begin{equation}\label{exNCP2p}
  \begin{array}{l}
    u'_t+Au=f(t), \quad t \in [0,T],\\[1em]
    u(0)+\alpha_1 u(t_1) + \alpha_2 u(t_2)=u_0,\quad 0<t_1<t_2\leq T.
  \end{array}
\end{equation}

Such nonlocal condition yields the following $B(z)$:
$$
B(z) = 1 +\alpha_1e^{-zt_1}+\alpha_2e^{-zt_2},
$$
whence it is clear that a closed form representation of $\Ker{B(z)}$ is not available in general.

The function $B(z)$ remains entire for any fixed $\alpha_1,\alpha_2$, $t_1,t_2$.
So its roots can be accurately approximated numerically using various methods \cite{Henrici1974} (Newtons method and its modifications, gradient methods,  numerical methods based on the argument principle and numerical quadratures, etc.) which implementations are available as a part of many modern mathematical programs (Octave, Maxima, Matlab, Maple).

By fixing the values
\begin{equation}\label{par1exNCP2p}
\alpha_1=-0.13,\  \alpha_2=3,\  t_1=1/2,\  t_2=1,
\end{equation}
we  get
$$
B(z) = 1 -0.13 e^{-\frac{z}{2}}+3e^{-z}.
$$
If we additionally assume that operator $A$ has the spectral parameters  $(0, \theta)$ it becomes obvious that condition \eqref{estLiang2002} is inapplicable in such situation ($0.13+3 >1$).
The approximate calculation of zeros of $B(z)$ carried by Maple package or, to be more exact, the function \verb#Analytic# (being the implementation of modified Newton's method) gives us  :
$$
\Ker(B(z)) = -2.09255541146 + 4\pi i,
$$
here all given digits are significant.
Combining this information, Theorem \ref{thmNCExist} and the fact that the spectrum of $A$ lies in a right half-plane of $\C$ we conclude that the generalized solution to problem \eqref{exNCP2p} exists for any $\theta \in [0, \pi/2]$.
\end{example}
All in all, the performed numerical analysis will always allow us to clarify the existence of a solution to \eqref{bp1} as long as the nonlocal parameters from \eqref{bp1:nc} are fixed.
For many application of \eqref{bp} with $n>2$ this is not enough as one still would like to have some a priory information about the admissible parameters set rather than simply check the existence of solution for some fixed values  of nonlocal parameters.

This often happens in applications to control theory where one must guarantee the solution's existence for a certain submanifold in the space of parameter values.

In the remaining part of the work we propose the technique how to estimate $\Ker{B(z)}$ by means of some well known bounds on roots of polynomial.

\section{Zeros of $B(z)$ and equivalent problem for polynomial}\label{sec2}

At first we assume that all $t_k$ from nonlocal condition \eqref{bp1:nc} are rational numbers.
This assumption in itself is quite adequate in practice because the computer representation of $t_k$ rely on a fixed size mantissa \cite{ieeeKahan1987}.
Every  $t_k$ admits the representation
\begin{equation*}
t_k= \frac{\lambda_k}{\mu_k}, \quad \lambda_k \in \mathbb{Z},\quad  \mu_k \in \mathbb{N},
\end{equation*}
Next we set $c_k=\frac{Q \lambda_k}{\mu_k}$, with  $Q=\mbox{LCM}(\mu_1,\mu_2, \ldots, \mu_n)$. The function
$B(z)$ is periodic with period $2 \pi Q i$. Thus, using the arguments from Example \ref{exNCP1p} the set $\mathbb{C} \backslash \Sigma$ can be safely reduced to $D_{Q} \backslash  \Omega_Q$
$$
\Omega_Q=\Sigma(A) \cap D_{Q},
$$
here $D_{Q}$ is a strip around the real axis with the width $2\pi Q$ (See Figure \ref{picQSPhi}, a)).

Now we would like to make use of Theorem \ref{thmNCExist}. For that one needs to check whether $\Ker{B(z)} \subset D_{Q} \backslash  \Omega_Q$.  This problem is just as difficult as the corresponding problem for $\C\backslash \Sigma$.

A mapping
\begin{equation}
\label{bp_transf1}
\varphi(z)=\exp(-z/Q)
\end{equation}
transforms  \eqref{zerosExp} into the following form
\begin{equation}
		\label{zerosPol}
		P(z)=1+\suml_{k=1}^{n}\alpha_k z^{c_k}.
\end{equation}
It is well known \cite{Henrici1974}, that \eqref{zerosExp} is one-to-one conformal mapping of $\Omega_Q$  onto  $\Phi$ (see Figure \ref{picQSPhi} b)).
\begin{figure}[ht]
\vspace*{-3pt}
\begin{tabular}{lr}
		\begin{overpic}[width=0.49\linewidth]
		{inscribecircle_t=0_th=pi_4}
			\put(15,95){\textbf{a)}}
		\end{overpic}&
		\begin{overpic}[width=0.49\linewidth]
		{circumcircle_t=0_th=pi_4}
			\put(15,95){\textbf{b)}}
		\end{overpic}
				\\

\end{tabular}
\vspace*{-3pt}
\caption{The regions $\Sigma$, $D_Q$ and their images for $A$ with spectral parameters $\theta=\pi/4$, $\rho=0$ ($Q=1$):
 \textbf{a)} intersection of spectrum  $\Sigma$  and the set $D_Q$;
 \textbf{b)} the region $\Phi$ and its encompassing circle. Points $c_{1/2},b,o_1$ are the preimages of $C_{1/2},B,O_1$.}
\label{picQSPhi}
\end{figure}
By using it we achieved two goals. First of all the selected mapping transforms the entire function $B(z)$ into polynomial $P(z)$ with real coefficients (provided that all $\alpha_i$ are real). Secondly this mapping reduces the zeros finding problem for the exterior of $\Sigma$ to the same problem for the exterior of a bounded set $\Phi \subset \C$.
Or speaking more precisely the conditions guarantying that all roots of $P(z)$ lie outside $\Phi$ or equivalently $\Ker{B(z)} \subset D_{Q} \backslash  \Omega_Q$ would be necessary and sufficient to prove that existence and uniquenesses of solution \eqref{bp1IntRed}.
The majority of results related to such conditions for polynomials are devoted to the situation when a circle is considered in place of  $\Phi$
(to review existent results in that field see \cite{Milovanovic2000a, Milovanovic2000}, as well as \cite{Henrici1974, Kravanja2000}).

That is why we first encircle $\Phi$ and then use readily available zero-free conditions for that circle. Such approach will make the resulting conditions only sufficient for all $\theta \in [0,\pi/2)$ except for the limiting case $\theta = \pi/2$ when $\Phi$ is a circle by construction.

For any given operator $A$ with spectral parameters $(\rho,\theta)$ the boundary of $\Phi$ can be parametrized as follows
$$
\partial \Phi = \left\{\exp\left(\frac{-Z(x)}{Q}\right): x\in[0,+\infty]\right\},
$$
where $Z(x)$ is a parametrization of $\partial\Omega_Q$:
$$
Z(x)=\rho+x+i \cdot\left\{
\begin{array}{ll}
	x \tan{\theta},&\quad x\tan\theta< Q\pi,\\
	Q\pi,&\quad x\tan\theta\geq Q\pi.\\
\end{array}
\right.
$$
A closer look at the expression for $\partial \Phi$ unveils that a vertical linear diameter of $\Phi$ is proportional to the magnitude of spectral angle, and the horizontal diameter of $\Phi$ is reversely proportional to $\rho$.
This observation suggests us to describe the encompassing circle as a circumcircle of a triangle with the vertices
$$
B=\max_{z \in \partial \Phi}\Re(z)+0i=\exp(-\rho/Q),
$$
and  $C_{1/2} \in \partial \Phi $ which are symmetric with respect to the real axis.
The coordinates of $C_{1}$ are chosen to maximize the
distance $|O_1-B|$ under the constrain $|O_1-B|^2=|O_1-C_i|^2$,  here $O_1$ is a circumcentre of $\triangle_{BC_1C_2}$.
Using the definition of $Z(x)$, \eqref{bp_transf1} and some basic facts from calculus we reduce the mentioned maximization problem to the following equation
\begin{equation}\label{eq_maxdist}

\begin{split}
\exp\left(-\frac{2x}{Q}\right)\left[\cos\left(\frac{x\tan\theta}{Q}\right)-\tan(\theta)\sin\left(\frac{x\tan\theta}{Q}\right)\right]+
\\
 +\cos\left(\frac{x\tan\theta}{Q}\right)+\tan(\theta)\sin\left(\frac{x\tan\theta}{Q}\right)
=  2\exp\left(-\frac{x}{Q}\right).
\end{split}

\end{equation}
It has a positive solution for $Q\in \N$ and $\forall \theta \in [0,\pi/2]$. Assume that $x_d$ is a solution of \eqref{eq_maxdist}, then
$$
C_{1/2}=\varphi(\rho+x_d \pm i x_d \tan\theta),
\quad
O_1=
 \frac{\varphi(2\rho)-\Re(C_1)^2-\Im(C_1)^2}{2\left(\varphi(\rho)-\Re(C_1)\right)},
$$
while the radius of circumcircle $r=\varphi(\rho)-O_1$ (The picture of $\Phi$, its encompassing circle along with their inverse images  are shown in Figure \ref{picQSPhi}  ).

\section{Sufficient conditions for existence of solution}\label{sec3}

Let us review what we have done so far. Starting from Theorem \ref{thmNCExist} we reduce the problem of clarifying whether $\Ker{B(z)} \subset \C \backslash \Sigma$  to the corresponding problem for zeros of the polynomial $P(z)$ lying in the exterior of the circle $|z-O_1|\leq r$. Conditions guarantying such layout of roots \cite{Milovanovic2000,Milovanovic2000a} are obtained, as a rule, from equivalent conditions for the interior of the circle. That is why most of the related results are formulated for the circle with centre at origin. Some of them in addition operate with the unit circle only. To accommodate this observation we introduce two alternative forms of \eqref{zerosPol}:
with the given circle transformed to the unit circle centered at the origin
		\begin{equation}
			P_1(z')= P(O_1+rz')=\suml_{k=0}^{c_n} \alpha'_k z'^k,
		\label{zerosPol1}
		\end{equation}
and with the given circle transformed to the circle centered at the origin
		\begin{equation}
			P_2(z'')=P(O_1+z'')=\suml_{k=0}^{c_n} \alpha''_k z''^k.
		\label{zerosPol2}
		\end{equation}
At this point we need to make use of several results estimating the radius of zero-free circle in terms of the polynomial coefficients. First of these results is a so-called SchurCohn test \cite{Henrici1974}. It establishes the necessary and sufficient conditions for the roots of polynomial to lie in the region $|z|>1$.

\begin{definition}
Given $P^\star(z)=z^n\overline{P(1/\overline{z})}$ we define  the Schur transform $T$ of polynomial $P(z)$ by
$$
\begin{array}{rl}
TP(z):=&\overline{\alpha_0}P(z)-\alpha_n P^\star(z)\\
=& \suml_{k=0}^{n-1} (\overline{\alpha_0} \alpha_k - \alpha_n \overline{\alpha_{n-k}})z^k.
\end{array}
$$
\end{definition}

\begin{theorem}[{\cite{Henrici1974}}]\label{tShur}
Let $P(z)$ is a polynomial of degree $n>0$. All zeros of  $P(z)$ lie in the exterior of the circle $|z| \leq 1$ if and only if for all $k=1,2, \ldots, n$
\begin{equation}
	\gamma_k > 0,
	\label{shurCrit}
\end{equation}
where $\gamma_k := T^k P(0)$ and $T^k P= T(T^{k-1} P)$.
\end{theorem}

\begin{corollary}
If the coefficients  $\alpha_k$ satisfy
$$
\alpha_0\geq \alpha_{1}\geq \ldots \geq \alpha_{n-1} \geq \alpha_n > 0,
$$
then all zeros of $P(z)$ lie outside the circle {$|z|\leq 1$}.
\end{corollary}

The application Schur-Cohn test to $P_1(z)$ produces a system of $c_n$-th innequalities for the coefficients of nonlocal condition \eqref{bp1:nc} which is sufficient for the solution's existence.
Next four Lemmas are more convenient than Theorem \ref{tShur} from the computational standpoint since the number of the produced inequalities are independent on the polynomial degree.

{
\begin{lemma}\label{lem_zpol_1}
All zeros of $P(z)$  lie in th region
$$
|z|\geq \frac{|\alpha_0|}{|\alpha_0|+M},
$$
where $M=\max\limits_{1\leq k \leq n}|\alpha_k|$.\\
\end{lemma}

\begin{lemma}\label{lem_zpol_2}
All zeros of $P(z)$ satisfy the inequality
$$
|z|\geq \frac{|\alpha_0|}{\left[|\alpha_0|+M^q \right]^{1/q}},\quad M=\left(\sum\limits_{k=1}^n |\alpha_k|^p\right)^{1/p},\quad
p,\ q \in \mathbb{R}_+,\quad
\frac{1}{p}+\frac{1}{q}=1
$$

\end{lemma}
}{
Next estimate is due to M. Fujiwara \cite{Fujiwara1916}.
It is an optimal homogeneous estimate  in the space of polynomials \cite{Batra2004}:
\begin{lemma}\label{lem_zpol_3}
All zeros of  $P(z)$ belong to the region
$$
|z|\geq \frac{1}{2}\min\limits_{\alpha_i\neq 0 }\left\{\left|\frac{\alpha_0}{\alpha_1}\right|, \left|\frac{\alpha_0}{\alpha_2}\right|^{1/2}, \ldots, \left|\frac{\alpha_0}{\alpha_{n-1}}\right|^{1/(n-1)}, \left|\frac{2\alpha_0}{\alpha_n}\right|^{1/n}\right\}.
$$
\end{lemma}
}
The last of the estimates given here was proved by H.~Linden.
This estimate in its original form gives bounds on the real and imaginary part of zeros separately. It has been adapted to fit within the framework studied here.

\begin{lemma}\label{lem_zpol_4}
All  zeros of $P(z)$ belong to the region  $|z|\geq \max\{V_1^{-1},V_2^{-1}\}$, where
$$
V_1=\cos{\frac{\pi}{n+1}}+\frac{|\alpha_n|}{2|\alpha_0|}\left(\left|\frac{\alpha_1}{\alpha_n}\right|+\sqrt{1+\suml_{k=1}^{n-1}\left|\frac{\alpha_k}{\alpha_n}\right|^2}\right),
$$
$$
V_2=\frac{1}{2}\left(\left|\frac{\alpha_1}{\alpha_0}\right|+\cos\frac{\pi}{n}\right)+
\frac{1}{2}\left[\left(\left|\frac{\alpha_1}{\alpha_0}\right|-\cos\frac{\pi}{n}\right)^2+
\left(1+\left|\frac{\alpha_n}{\alpha_0}\right|\sqrt{1+\suml_{k=2}^{n-1}\left|\frac{\alpha_k}{\alpha_n}\right|^2}\right)^2\right]^{1/2}.
$$
\end{lemma}

By combining the estimates given by Theorem \ref{tShur} or Lemmas \ref{lem_zpol_1} -- \ref{lem_zpol_4} with Theorem \ref{thmNCExist} we obtain new sufficient conditions for the existence and uniquenesses of the solution to \eqref{bp1}.

\begin{theorem}\label{thmSufCond}
Assume that  $A$, $f(t)$ and $u_0$ satisfy the conditions of Theorem \ref{thmNCExist}.
The generalized solution \eqref{bp1IntRed} of nonlocal problem \eqref{bp1} exists if either of the following is true.
\begin{enumerate}
	\item $\exists\ z > 1$, which along with the coefficients of  $P(\varphi(\rho) z)$ from (\ref{zerosPol}) satisfies the conditions of Theorem \ref{tShur} or at least one of Lemmas \ref{lem_zpol_1} -- \ref{lem_zpol_4}.
	\item $\exists\ z > 1$, which along with the coefficients of $P_1(z)$ from (\ref{zerosPol1}) satisfies the conditions of Theorem \ref{tShur} or at least one of Lemmas \ref{lem_zpol_1} -- \ref{lem_zpol_4}.
	\item $\exists\ z > r$, which along with the coefficients of $P_2(z)$ from (\ref{zerosPol2}) satisfies the conditions of at least one of Lemmas \ref{lem_zpol_1} -- \ref{lem_zpol_4}.
\end{enumerate}
\end{theorem}

\begin{proof}
The results formulated in Theorem \ref{tShur} and Lemmas \ref{lem_zpol_1} -- \ref{lem_zpol_4} guaranty that as soon as at least one of three theorem's prepositions is valid all zeros of $B(z)$  will lie outside $\Phi$. Hence according to Theorem \ref{thmNCExist} the solution of \eqref{bp1} exists and is unique.
\end{proof}

Propositions 1-3 of Theorem  \ref{thmSufCond} are ordered in such a way that the first proposition deals with a circle obtained by putting $\theta = \pi/2$. It is therefore valid for $A$ with any $\theta \in (0;\pi/2]$.
The other two propositions use the parameters of encompassing circle defined above. These propositions will lead to more general sufficient conditions for $\theta <\pi/2$.
To illustrate these facts we apply different proposition stated in Theorem \ref{thmSufCond} to some concrete examples of nonlocal conditions.

\begin{example}
Let us again consider nonlocal problem (\ref{bp1}) with operator coefficient $A$ ($\theta=\theta_0, \rho=0$) and the Bicadze-Samarskii--type nonlocal condition
\begin{equation}\label{eq:nc_ex2}
u(0)+\alpha_1 u(t_1)=\alpha_2 u(t_2).
\end{equation}
As we have mentioned in Example \ref{exBS} the set $\Ker(B(z))$ can not be found in a closed form.

Estimate (\ref{estLiang2002}) yields:
$$

|\alpha_1|+|\alpha_2|<1.
$$

Meanwhile the application of proposition 1 from Theorem \ref{thmSufCond} together with Schur-Cohn algorithm with $\theta_0=\pi/2$  lead us to system of inequalities:
\begin{equation}\label{eqintshura2}
\left\{
\begin{array}{l}
	|\alpha_2|<1,\\
	|1-\alpha_2^2|>|\alpha_1(1-\alpha_2)|.
\end{array}\right.
\end{equation}
Here we assumed that $t_1=1,\ t_2=2$ and $Q=1$.
Solution of \eqref{eqintshura2} are graphically compared to with set of pairs $(\alpha_1, \alpha_2)$ satisfying \eqref{estLiang2002} in Figure \ref{picShura2} a).
\begin{figure}[ht]
\begin{tabular}{c}
		\begin{overpic}[width=0.49\linewidth]
		{pic_tshura_n=2_filled}
			\put(5,65){\textbf{a)}}
			\put(95,38){\small $\alpha_1$}
			\put(51,65){\small $\alpha_2$}
		\end{overpic}
		\begin{overpic}[width=0.49\linewidth]
		{pic_tshura_n=6_filled}
			\put(5,65){\textbf{b)}}
			\put(95,38){\small $\alpha_1$}
			\put(51,65){\small $\alpha_2$}
		\end{overpic}
				\\
\end{tabular}
\caption{The admissible sets of $\alpha_{1,2} \in \R$ from \eqref{eq:nc_ex2} using condition \eqref{estLiang2002} (coloured in \cbox{gray80}) and  proposition 1 from Theorem \ref{thmSufCond} $\rho=0$, $\forall \theta_0 \in [0, \pi/2)$ (coloured in \cbox{gray40}), where:
\textbf{a)} $t_1=1, t_2=2$;
\textbf{b)} $t_1=3, t_2=8$.}
\label{picShura2}
\end{figure}
Our approach apparently gives more general conditions than \eqref{estLiang2002} or its particular case from \cite{Byszewski1992} even though we made sufficient conditions \eqref{eqintshura2} independent on $\theta_0$.

To compare different conditions quantitatively we can weight the areas of corresponding admissible parameters sets against each other.
In that sense the area of admissible parameters set obtained by application of Proposition 1, Theorem \ref{thmSufCond} for $t_1=1,\ t_2=2$ and $Q=1$ is twice bigger than the set obtained by the application of \eqref{estLiang2002}.
The number of inequalities in \eqref{eqintshura2} obtained from \eqref{shurCrit} depends on the ratio $t_2/t_1$ from the nonlocal condition. So it will grow if we increase $t_2/t_1$.

As a result the corresponding admissible parameters set defined by \eqref{shurCrit} is going to shrink in size and in the limit $t_2/t_1\rightarrow\infty$ will become equal to the set defined by \eqref{estLiang2002}.
To illustrate this behaviour we provide (see Figure \ref{picShura2} b) the same comparison of admissible parameter sets as in Figure \ref{picShura2} a) but for the case $t_1=3,t_2=8$. Recall that all these results are valid  for any sectorial operator $A$ with $\rho=0$.

Proposition 2 of Theorem \ref{thmSufCond} ought to be more advantageous for operator coefficients with some fixed $\theta<\pi/2$. Let us fix $\theta_0 =\pi/3$ and calculate the centre and the radius of circumcircle.
We get
$$
O_1 = 0.3950734246 , \quad
r = 0.6049265754.
$$
By substituting  these parameters into \eqref{zerosPol1} and \eqref{zerosPol2} we produce two alternative forms of $P(z)$
$$
\begin{array}{ll}
P_1(z') &\approx 0.37 \alpha_2 {z'}^2+(0.6 \alpha_1+0.48 \alpha_2) z'+1+0.4 \alpha_1+0.16 \alpha_2 , \\
P_2(z'') &\approx \alpha_2 {z''}^2+(\alpha_1+0.79 \alpha_2) {z''}+1+0.4 \alpha_1+0.16 \alpha_2
\end{array}
$$
for the given nonlocal condition.

Application of proposition 2 of Theorem \ref{thmSufCond} along with \eqref{shurCrit} (setting $t_1=1,\ t_2=2$ as before) gives us the set of admissible $(\alpha_1,\alpha_2)$ depicted in Figure \ref{picShura2Pzp1} a).
\begin{figure}[ht]
\begin{tabular}{lr}
		\begin{overpic}[width=0.47\linewidth]
		{pic_tshura_n=2_pzp_filled}
			\put(5,95){\textbf{a)}}
			\put(73,22){\small $\alpha_1$}
			\put(45,95){\small $\alpha_2$}
		\end{overpic}&
		\begin{overpic}[width=0.49\linewidth]
		{pic_tshura_n=2_pzp_z1_z2_filled}
			\put(5,95){\textbf{b)}}
			\put(95,45){\small $\alpha_1$}
			\put(52,95){\small $\alpha_2$}
		\end{overpic}
				\\

\end{tabular}
\caption{
Admissible values of parameters $\alpha_{1,2} \in \R$ from nonlocal condition \eqref{eq:nc_ex2}  $\rho=0$, $\theta = \pi/3$ obtained via estimate \eqref{estLiang2002} (coloured in \cbox{gray80}) and:
\textbf{a)}  proposition 2 from Theorem \ref{thmSufCond} and estimate \eqref{shurCrit} coloured in \cbox{gray40};
\textbf{b)} proposition 3 of Theorem \ref{thmSufCond} based on Lemma \ref{lem_zpol_1} (coloured in \cbox{gray60}) and Lemma \ref{lem_zpol_2} (coloured in \cbox{gray30}).
}
\label{picShura2Pzp1}
\end{figure}
\FloatBarrier
One can see that this set contains both admissible parameters sets obtained from proposition 1 of the same theorem and condition \eqref{estLiang2002}.
The area of the admissible parameters set resulting from Schur-Kohn test has grown considerably comparing to Figure \ref{picShura2}.
Similarly to Example \ref{ex1} this grow is caused by the usage of smaller $\theta_0=\pi/3$, because smaller $\theta_0$ leads to the region $\Phi$ with a smaller diameter.

All applications of Theorem \ref{thmSufCond} demonstrated hitherto are sensitive to the values $t_i$ from nonlocal condition. For situation where such sensitivity is unfavourable one may wish to use zero bounds from Lemmas \ref{lem_zpol_1} -- \ref{lem_zpol_4} instead of the Schur-Cohn test in propositions of Theorem \ref{thmSufCond}.
Graphical comparisons of the admissible parameters sets given by these Lemmas are presented on figures \ref{picShura2Pzp1} b) trough \ref{picShura2Pzp2} b).
To make graphical comparisons more straightforward we kept the boundary of the set obtained by application of proposition 2 from Theorem \ref{thmSufCond} (see figure \ref{picShura2Pzp1} a) on all mentioned figures.
\begin{figure}[ht]
\begin{tabular}{lr}
		\begin{overpic}[width=0.49\linewidth]
		{pic_tshura_n=2_pzp_z1_z3_filled}
			\put(5,95){\textbf{a)}}
			\put(95,45){\small $\alpha_1$}
			\put(52,95){\small $\alpha_2$}
		\end{overpic}&
		\begin{overpic}[width=0.49\linewidth]
		{pic_tshura_n=2_pzp_z2_z4_filled}
			\put(5,95){\textbf{b)}}
			\put(95,45){\small $\alpha_1$}
			\put(52,95){\small $\alpha_2$}
		\end{overpic}

				\\
\end{tabular}
\caption{
Admissible values of parameters $\alpha_{1,2} \in \R$ from nonlocal condition \eqref{eq:nc_ex2} $\rho=0, \theta=\pi/3$ obtained via estimate \eqref{estLiang2002} (coloured in \cbox{gray80}) and proposition 3 of Theorem \ref{thmSufCond} based on:
\textbf{a)} the estimate from Lemma \ref{lem_zpol_1} coloured in \cbox{gray60} and the estimate from Lemma \ref{lem_zpol_3} coloured in \cbox{gray40}
\textbf{b)} the estimate from Lemma \ref{lem_zpol_2} coloured in \cbox{gray60} and the estimate from Lemma \ref{lem_zpol_4} coloured in \cbox{gray40}
}
\label{picShura2Pzp2}
\end{figure}

One thing the reader would immediately note is that the dominance of necessary conditions presented in this work over the estimate \eqref{estLiang2002} is no longer absolute.
The sets obtained from Lemmas \ref{lem_zpol_1} -- \ref{lem_zpol_3} do not fully cover the set given by \eqref{estLiang2002}. But still any of these Lemmas performs better than condition \eqref{estLiang2002} in terms of the area in the parameters space.
The same is true for Lemma \ref{lem_zpol_4} which lead to the set having the area at least six times bigger than the area coloured in \cbox{gray80}.  In addition to that the condition given by Lemma \ref{lem_zpol_4} clearly generalize \eqref{estLiang2002}.
This dominance of Lemma \ref{lem_zpol_4} over the other necessary conditions may not always be valid because the mutual  interdependence of root estimates from  lemmas \ref{lem_zpol_1} -- \ref{lem_zpol_4} are not yet known.
Therefore, they are advised to be used jointly.

\end{example}
\FloatBarrier
A priori estimates for the zero-free circle mentioned in Lemmas \ref{lem_zpol_1} -- \ref{lem_zpol_4} have been derived by inversion of the corresponding polynomial zero bounds. These particular zero bounds have been selected among others from \cite{Milovanovic2000} \cite{Milovanovic2000a}  based on the numerical comparison of its performance for a number of nonlocal conditions.
More detailed discussion about various root finding methods and their potential in application to theorems \ref{thmNCExist} and \ref{thmSufCond} are given in \cite{PhdThSytnykEN}.
In this work it is also described how to generalize the results of section \ref{sec3} to the case when some of $t_i$ are irrational. All codes for the generation of figures, circle parameters calculation and numerical checks of the presented necessary conditions can be found at\newline \url{imath.kiev.ua/~sytnik/research/works/nonlocal-2014}.

\section{Conclusions and future work}
By exploiting the connection between nonlocal evolution problem \eqref{bp1} and its classical counterpart we derive the reduction operator representation \eqref{reprDelta}.
It enabled us to work out the conditions for existence of the mild solution to \eqref{bp1}.
Analogous existence analysis is possible for other evolution problems  as long as the exact representation of $B(z)$ similar to \eqref{zerosExp} is obtainable and one can characterize the set free from zeros of $B(z)$.  The nonlocal evolutional problems for the abstract time dependant Schro\"dinder equation and the abstract second order linear differential equation are both tractable by our approach. They will constitute the subject for our future analysis.
\selectlanguage{english}

config: arxiv


TITLE: How to write this expression in sigma notation?
QUESTION [0 upvotes]: Write this expression in sigma notation:
$$\frac 1{2k} + \frac 2{3k} + \frac 3{4k} + \ldots + \frac{k-1}{k^2} + \frac 1{k+1}.$$

REPLY [0 votes]: $$\sum\limits_{i=1}^k \dfrac i {(i+1)k}$$
(Note that when $i=k-1$ the summand simplifies to $\dfrac{k-1}{k^2}$,
and when $i=k$ the summand simplifies to $\dfrac1{k+1}$.)


TITLE: Solve the following (logarithmic) function for x
QUESTION [0 upvotes]: $x^{log_{2}x}+16x^{-log_{2}x} = 17$

Looks horrible, I started by removing the exponents:
$e^{ln(x)*log_{2}x}+16e^{-ln(x)*log_{2}x}=17$ | ln()
$ln(x)*log_{2}x-16ln(x)*log_{2}x=ln(17)$
$ln(x)*log_{2}x(1-16)=ln(17)$
$-15ln(x)*log_{2}x=ln(17)$ |:(-15)
$ln(x)*log_{2}x=-\frac{ln(17)}{15}$
And here is the dead end for me, tried several different things I cannot write them all here but nothing lead to a solution. At least it looks better than at the beginning... :D
Is this even the right way to go or there are easier things? And is it possible to somehow
change $ln(x)*log_{2}x$ to something else, something which looks better?

REPLY [3 votes]: $$x^{ log_{ 2 }x }+\frac { 16 }{ x^{ log_{ 2 }x } } =17\\ { x }^{ 2\log _{ 2 }{ x }  }-17x^{ log_{ 2 }x }+16=0\\ \left( x^{ log_{ 2 }x }-16 \right) \left( x^{ log_{ 2 }x }-1 \right) =0\\ x^{ log_{ 2 }x }=16\Rightarrow \log _{ 2 }{ x^{ log_{ 2 }x } } =\log _{ 2 }{ 16 } \Rightarrow { \left( \log _{ 2 }{ x }  \right)  }^{ 2 }=4\Rightarrow \log _{ 2 }{ x } =\pm 2\Rightarrow { x }_{ 1 }=4,{ x }_{ 2 }=\frac { 1 }{ 4 } \\ x^{ log_{ 2 }x }=1\Rightarrow \log _{ 2 }{ x^{ log_{ 2 }x } } =\log _{ 2 }{ 1 } \Rightarrow { \left( \log _{ 2 }{ x }  \right)  }^{ 2 }=0\Rightarrow { x }_{ 3 }=1$$

REPLY [2 votes]: let $$y=x^{log_2x}$$
your equation becomes, $$y+\frac{16}y=17$$
solve it, you get two solutions: 1 and 16.
Now it becomes less horrible, $$x^{log_2x}=1 ~or~ 16 $$
This leads to solution $x =1$, $2^2$ and $2^{-2}$


TITLE: Divide 500 into certain group so that all no's 1 to 500 can be found.
QUESTION [0 upvotes]: 500 coins are there.
Divide 500 coins into certain bags such that any rupees from 1 to 500 can be found by the combination of the bag's coins.
What are the minimum nos of bags ?

REPLY [0 votes]: It seems to me you are not allowed to subtract the numbers, just add them.  In that case, having the bags with powers of $2$ is optimal.  If you are allowed to subtract them as well, powers of $3$.  The last bag won't follow the pattern, and provides flexibility to the rest.


TITLE: Weierstrass Approximation - Proof that Integral Exists
QUESTION [0 upvotes]: Trying to answer the following question;
Let $f(x)$ be a continuous real valued function on $[0,4]$. Given any $\epsilon>0$ prove there is a polynomial $p(x)$ such that $$\int_0^4|f(x)-p(x)|^2dx<\epsilon.$$
It is very clear how to show this for $$\int_0^4|f(x)-p(x)|dx,$$ (which I've worked on here: Does integration preserve uniform convergence of sequence? (Weierstrass Approximation Theorem)) but I'm not sure how to deal with the square.

REPLY [1 votes]: Note that if $|f(x)-p(x)|<\epsilon,$ then $|f(x)-p(x)|^2<\epsilon^2.$ So, it just comes down to fiddling with epsilons to remove the square. Spoiler below:

 Fix $\epsilon>0.$ By WAT, there exists a polynomial $p$ so that $|f(x)-p(x)|<\sqrt{\epsilon}/2$ for all $x\in [0,4]$. Now, $$\int\limits_0^4|f(x)-p(x)|^2\, dx\leq \sup_{x\in [0,4]} |f(x)-p(x)|^2 \int\limits_0^4 \, dx<\epsilon.$$


TITLE: An alternative to continued fraction and applications
QUESTION [8 upvotes]: This post is inspired by the Numberphile video 2.920050977316, advertising the paper A Prime-Representing Constant by Dylan Fridman, Juli Garbulsky, Bruno Glecer, James Grime and Massi Tron Florentin, involving an alternative to continued fractions. The goal of this post is to discuss the relevance of this alternative by asking whether it can prove the irrationality of numbers for which it was unknown before.
Let us first recall the notion of continued fraction. For a given number $\alpha>0$, consider the recurrence relation $u_0 = \alpha$ and $$ u_{n+1} = \begin{cases}
(u_n - \lfloor u_n \rfloor)^{-1} & \text{ if } u_n \neq \lfloor u_n \rfloor \\
 0  & \text{ otherwise }
\end{cases}$$
and let $a_n = \lfloor u_n \rfloor $. Then $$\alpha = a_0 + \frac{1}{a_1+\frac{1}{a_2+\frac{1}{\ddots}}}$$ denoted $[a_0; a_1, a_2, \dotsc]$. It is rational if and only if $a_n = 0$ for $n$ large enough. So it is a great tool to prove the irrationality of some numbers. For example, $\phi = [1;1,1, \dotsc]$ is the golden ratio, because $(\phi-1)^{-1}=\phi$.
Let $p_n$ be the $n$th prime, then we can consider the irrational number $[p_1;p_2,p_3, \dots] = 2.31303673643\ldots$ (A064442), which then compresses the data of all the prime numbers, in a more natural and efficient way than just taking $2.\mathbf{3}5\mathbf{7}11\mathbf{13}17\mathbf{19}\ldots$.
The paper mentioned above provides another interesting way to compress the primes numbers, which uses Bertrand’s postulate, i.e. $p_n < p_{n+1} < 2p_n$. This way is a kind of alternative to continued fractions. For a given number $\beta \ge 2$, consider the recurrence relation $u_1=\beta$ and $$u_{n+1} = \lfloor u_n \rfloor (u_n - \lfloor u_n \rfloor + 1).$$ Let $a_n= \lfloor u_n \rfloor $. Then $a_n \le a_{n+1} < 2a_n$ and the mentioned paper proves that $$\beta = \sum_{n=1}^{\infty}\frac{a_{n+1}-1}{\prod_{i=1}^{n-1}a_i}$$ denoted, let’s say, $(a_1,a_2,a_3, \dots )$.
By the mentioned paper:
Theorem 1: Let $(a_n)$ be a sequence of positive integers such that:

$a_n < a_{n+1} < 2a_n$,
$\frac{a_{n+1}}{a_n} \to 1$

then $\beta := (a_1,a_2,a_3, \dots ) := \sum_{n=1}^{\infty}\frac{a_{n+1}-1}{\prod_{i=1}^{n-1}a_i}$ is irrational.
It follows that the number $(p_1,p_2,p_3,\dots) = 2.920050977316\ldots$ is irrational.
Question: Can Theorem 1 be proved by some previously known methods?
Remark: The first point of Theorem 1 can be relaxed to $a_n \le a_{n+1} < 2a_n$, when $(a_n)$ is not eventually-constant.
For a given non-constant polynomial $P \in \mathbb{Z}[X]$ with a positive leading term and $P(n) \neq 0$ for all $n \in \mathbb{N}_{\ge 1}$, consider $a_n=P(n)$. Then it is easy to deduce from Theorem 1 that the number $e_P\mathrel{:=}(a_1,a_2, \dotsc )$ is irrational. For example, take $P(X)=X^k$, with $k \in \mathbb{N}_{\ge 1}$, then $$e_k:= \sum_{n=1}^{\infty} \frac{(n+1)^k-1}{n!^k}$$ is irrational. Note that $e_1 = e$ is Euler’s number.
The following result applies for an alternative proof of the irrationality of $e_k$ for all $k$, and of $e_P$ for many $P$ (not all), but not for $(p_1,p_2,p_3, \dots)$
Theorem 2: Let $(a_n)$ be a sequence of positive integers such that:

$a_n \le a_{n+1} < 2a_n$,
$\forall k \in \mathbb{N}_{\ge 1}$, $\exists m$ such that $k$ divides $a_m$,

then $\beta := (a_1,a_2,a_3, \dots ) := \sum_{n=1}^{\infty}\frac{a_{n+1}-1}{\prod_{i=1}^{n-1}a_i}$ is irrational.
proof: Assume that $\beta = \frac{p}{q}$. By assumption, there is $m$
such that $q$ divides $a_m$. By the mentioned paper, if $u_1=\beta$ and $u_{n+1} = \lfloor u_n \rfloor (u_n - \lfloor u_n \rfloor + 1)$, then
$a_n= \lfloor u_n \rfloor $. It is easy to see that $u_n$ can always be written with a denominator equal to $q$ (possibly not simplified). It follows that $u_{m+1}=a_m(u_m-a_m+1)$ and that $a_m u_m$ is an integer. So $u_{m+1}$ is an integer. It follows that for all $n>m$ then $u_n=u_{m+1}$, and so $a_n=a_{m+1}$. But the second point of Theorem 2 implies that $a_n \to \infty$, contradiction. $\square$
The following example will show that the condition $\frac{a_{n+1}}{a_n} \to 1$ is not necessary for the irrationality.
Consider $a_n=\lfloor \frac{3^n}{2^n} \rfloor + r_n$, with $0 \le r_n < n$ such that $n$ divides $a_n$. Adjust the sequence for $n$ small so that the first point of Theorem 2 holds. Then $\beta$ is irrational whereas $\frac{a_{n+1}}{a_n} \to \frac{3}{2} \neq 1$.
Bonus question: What is a necessary and sufficent condition for irrationality?
Joel Moreira suggested in this comment that it may be rational if and only if $(a_n)$ is eventually-constant. See the new post Do these rational sequences always reach an integer? dedicated to this question.
FYI, it is easy to compute that $$\pi = (3,  3,  4,  5,  5,  7,  10,  10,  13,  17,  31,  35,  67,  123,  223,  305,  414,  822,  1550,  2224, ...) $$

REPLY [3 votes]: I am sorry If the comment is misleading, and welcome to point out any mistakes in the following proof. This is a clarification of the previous comment.
And this is only a proof of the irrationality of $e_k$.
And the proof strategy is an imitation of Fourier's proof of the irrationality of Euler's number $e$.

if $\forall n=\mathbb{N}^{*} \quad n$, $n$ suffice large,
$$
\left(n!\right) \cdot a \notin \mathbb{Z} \quad \text { then } a \notin \mathbb{Q} \hspace{1cm}(1)
$$
WLOG, in the following calculation we don't distinguish $x,y$ if $x-y\in \mathbb{Z}$. And we write $x=y+\mathbb{Z}$ iff $x-y\in \mathbb{Z}$.
$\begin{aligned} m ! e_{k} +\mathbb{Z}&=\sum_{n \geq m+1} \frac{(n+1)^{k}-1}{(m+1) \cdots(n-1) n)^{k}}+\mathbb{Z} \\ &=\sum_{n \geqslant m+2} \frac{(n+1)^{k}-1}{((n-1) \cdots(n-1) n)^{k}}+\frac{(m+2)^{k}-1}{(m+1)^{k}}+\mathbb{Z}\\
&=\sum_{n \geq m+2} \frac{(n+1)^{k}-1}{((m+1) \cdots(n-1) n)^{k}}+\sum_{i=1}^{k-1} \frac{C_{k}^{i} \cdot(m+1)^{i}}{(m+1)^{k}}+1 +\mathbb{Z}\\
&=\sum_{n \geqslant m+2} \frac{(n+1)^{k}-1}{( m+1) \cdots(n-1) n)^{k}}+\sum_{i=1}^{k-1} \frac{C_{k}^{i}}{(m+1)^{i}}+\mathbb{Z}\hspace{1cm}(*)
\end{aligned}$
In fact in $(*)$ we have $\sum_{n \geqslant m+2} \frac{(n+1)^{k}-1}{((m+1) \cdots(n-1) n)^{k}}= O(\frac{1}{m^{k}})$, $\sum_{i=1}^{k-1} \frac{C_{k}^{i}}{(m+1)^{i}}=O(\frac{1}{m})$.
Now take $m$ suffice large, in fact $m=10000\cdot k^{100}$ is ok,
then
$$0< \sum_{n \geqslant m+2} \frac{(n+1)^{k}-1}{((m+1) \cdots(n-1) n)^{k}}+\sum_{i=1}^{k-1} \frac{C_{k}^{i}}{(m+1)^{i}}< 1$$
So $(*)\neq \mathbb{Z}$, so $(1)$ is true, $ e_{k}$ is not rational.


TITLE: Given $f :(a,b)\to\mathbb{R}$ is a monotone increasing function bounded above, show that $\lim_{x\to b^-} f(x)$ exists
QUESTION [2 upvotes]: Let $f :(a,b)\to\mathbb{R}$ be a monotone increasing function bounded above. Show that $\lim_{x\to b^-} f(x)$ exists.
Okay, so what I've done is shown below. Apparently it's not enough because I've just shown that $f(x_n)$ converges. Could anyone give any hints as to how I can complete the proof?
Let $x_n=b-\dfrac{1}{n},n\in\mathbb{N}$. Then since $\forall n\in\mathbb{N},x_n<b$ and $x_{n+1}=b-\dfrac{1}{n+1}>b-\dfrac{1}{n}=x_n$, $x_n$ is monotone increasing and bounded above. Thus, by the Monotone Convergence Theorem, it has a limit $\lim_{n\to\infty}x_n=b^-$. Assume that $f :(a,b)\to \mathbb{R}$ is monotone increasing and bounded above by some $L\in\mathbb{R}$. Since $f(x)$ is monotone increasing, we know that $f(x_n)<L \forall n\in\mathbb{N}$. But here I'm stuck. How do I show precisely that the given limit exists?

REPLY [1 votes]: Application of some sequence.
Let $x_n = b -1/n$. Then $x _n \to b^-$. Since $f$ is increasing and bounded above, so is the sequence $(f(x_n))_1^\infty$. Therefore  it converges. Let the limit be $A$. By definition, for each $\varepsilon > 0$, there is some $N \in \Bbb N^*$ s.t. $\vert f(x_n) - A\vert < \varepsilon$ whenever $n \geqslant N$. Therefore for $\underline {\delta = b - x_N}$, whenever $\underline{x_N = b -\delta < x < b}$, there is an $M \in \Bbb N^*$ that $M \geqslant N$ and $x_M \leqslant x \leqslant x_{M+1}$. Then by monotonicity, $f(x_M) \leqslant f(x) \leqslant f(x_{M+1})$, and
$$
\underline {- \varepsilon <}\ f(x_M) - A \leqslant \underline {f(x) - A} \leqslant f(x_{M+1}) - A\ \underline { < \varepsilon}\ ,
$$
i.e.
$$
\lim_{ x \to b^-} f(x) = A.
$$


TITLE: Supposed solution for the Poisson equation fulfills Laplace equation instead.
QUESTION [0 upvotes]: It seems basic, but I never understood properly. Suppose a spherical symmetric mass distribution, and a point outside the sphere. The gravitational field at the point can be calculated by summing all mass elements:$$\Phi = \int -\frac{Gdm}{|\mathbf r - \mathbf r'|} = \int -\frac{G\rho}{|\mathbf r - \mathbf r'|}dV$$ It is also the standard solution for the Poisson equation according to Wikipedia, where $f = -4\pi G\rho$.
Solving the integral, we get $$\Phi = -\frac{Gm}{r}$$ where m is the mass of the sphere and $r$ the distance to its center.
But the Laplacian of this solution is zero. It fulfills the Laplace equation (where $f = 0$) and not the Poisson's (where $f = -4\pi G\rho$).
I think that it can the mathematical translation for the fact that if all mass were at a singularity (the center), the field would be the same. In this case $\rho(\mathbf r) = 0$ everywhere. But anyway there is the problem of the singularity where $\rho(\mathbf 0) = \infty$.
How to deal with this anomaly?

REPLY [0 votes]: What do we know? We have a body, WLOG centered at the origin, with radius $R$, say. For simplicity, let's assume the body is small and dense enough (i.e, Mercury) for us to consider the mass distribution to be totally even. In other words,
$$\rho(r)=\begin{cases}\rho &r\leq R \\0 & r>R \end{cases}$$
As we are aware, the potential $U$ satisfies
$$\Delta U=4\pi G\rho$$
Outside the body, the density is zero, which reduces this to
$$\Delta U_{\text{out}}=0$$
I'll let you check on your own that this is solved by
$$U_{\text{out}}(r)=b_1+\frac{b_2}{r}$$
On the other hand, inside the body, the density is nonzero, so we have
$$\Delta U_{\text{in}}=4\pi G\rho$$
Once again, I will let you check for yourself that this is solved by
$$U_{\text{in}}(r)=a_1+\frac{a_2}{r}+\frac{2\pi G\rho r^2}{3}$$
As of now, we don't have enough information to uniquely specify the potential. We apply some physical conditions, imposing

$U$ is well-defined at the origin
And in particular $U(0)=0$. In physical terms, this means that it takes no work to move a particle at the origin, to the origin.
$U_{\text{in}}(R)=U_{\text{out}}(R)$, in other words, the inside and outside potential agree on the body's surface

You'll notice that these are only three conditions, but we have four unknown constants. This is fine - the thing we are truly interested in is the graviational field, $\boldsymbol g$ which is related to the potential via
$$\boldsymbol g=-\nabla U$$
Because of this differentiation, we have some freedom up to shifting by a constant. So, we can, without loss of generality, decide to set $b_1=0$. The first two conditions impose $a_1=a_2=0$ so this reduces to
$$U_{\text{in}}(r)=\frac{2\pi G\rho r^2}{3} \\ U_{\text{out}}(r)=\frac{b_2}{r}$$
Applying the matching condition, you can deduce
$$b_2=\frac{2\pi G\rho R^3}{3}$$
And so
$$\boxed{U(r)=\begin{cases}\frac{2\pi G\rho r^2}{3} & r\leq R \\ \frac{2\pi G\rho R^2}{3}\frac{R}{r}& r>R \end{cases}}$$