text:
\section{\label{sec: Examples}Examples}

In⋅the examples below we assume that $\Omega$ is a bounded $C^{1,1}$\nobreakdash-do\-main in~$\R^{d}$. For⋅each⋅$1\le k\le n$,
let $\Gamma_{k}$ either be a $C^{1,1}$\nobreakdash-submanifold
of⋅$\partial\Omega$ with boundary (possibly disconnected) or coincide
with⋅$\emptyset$ or~$\partial\Omega$. By⋅$W_{q}^{1}\rbr{\Omega;\Gamma_{k}}$,
we denote the $W_{q}^{1}$\nobreakdash-closure of smooth functions
on⋅$\Omega$ that vanish near~$\Gamma_{k}$. It⋅is⋅known  that
if $1<q_{0}<q_{1}<\infty$, then $W_{q_{\theta}}^{1}\rbr{\Omega;\Gamma_{k}}$
is the complex interpolation space between⋅$W_{q_{0}}^{1}\rbr{\Omega;\Gamma_{k}}$
and⋅$W_{q_{1}}^{1}\rbr{\Omega;\Gamma_{k}}$, where⋅$1/q_{\theta}=\rbr{1-\theta}/q_{0}+\theta/q_{1}$
and⋅$\theta\in\rbr{0,1}$. We set⋅$\set W_{q}^{1}\rbr{\Omega;\C^{n}}=\oplus_{k=1}^{n}W_{q}^{1}\rbr{\Omega;\Gamma_{k}}$.

\subsection{Strongly elliptic operators}

Let⋅$p=2$, and let $\set H^{1}\rbr{\Omega;\C^{n}}=\set W_{2}^{1}\rbr{\Omega;\C^{n}}$.
Suppose that the operator~$\op A^{\varepsilon}$ is weakly coercive
 uniformly in⋅$\varepsilon$ for⋅$\varepsilon$ sufficiently small,
that is, there are $\varepsilon_{0}\in(0,1]$ and $c_{A}>0$ and⋅$C_{A}<\infty$
so that for all⋅$\varepsilon\in\set E=(0,\varepsilon_{0}]$
\begin{equation}
\Re\rbr{\op A^{\varepsilon}u,u}_{\Omega}+C_{A}\norm u_{2,\Omega}^{2}\ge c_{A}\norm{Du}_{2,\Omega}^{2},\qquad u\in\set H^{1}\rbr{\Omega;\C^{n}}.\label{est: A^e is elliptic}
\end{equation}
With⋅this assumption, $\op A^{\varepsilon}$ becomes strongly elliptic,
which means that the function~$A$ satisfies the Legendre–Hadamard
condition
\begin{equation}
\Re\abr{A\rbr{\wc,\wc}\hairspace\xi\otimes\eta,\xi\otimes\eta}\ge c_{A}\abs{\xi}^{2}\abs{\eta}^{2},\qquad\xi\in\R^{d},\eta\in\C^{n}\label{est: Legendre-Hadamard condition}
\end{equation}
(see⋅Lemma~\ref{lem: Strong coercivity of A(x)} below). What is
more,  a simple calculation based on boundedness and coercivity of⋅$\op A^{\varepsilon}$
shows that if $\varepsilon\in\set E$, then $\op A^{\varepsilon}$
is an $m$-sectorial operator with sector
\[
\set S=\bigcbr{z\in\C\colon\abs{\Im z}\le c_{A}^{-1}\norm A_{L_{\infty}}\rbr{\Re z+C_{A}}}
\]
independent of⋅$\varepsilon$, and therefore (\ref{est: Original operator | Norm of (A=001D4B-=0003BC)=00207B=0000B9})
holds for⋅$p=2$ and any⋅$\varepsilon\in\set E$ as long as~$\mu\notin\set S$.
It⋅then follows from Shneiberg's stability theorem~\cite{Shneiberg:1974}
(see also~\cite[Section~17.2]{Agranovich:2013}) that the estimate~(\ref{est: Original operator | Norm of (A=001D4B-=0003BC)=00207B=0000B9})
is valid for any⋅$p$ satisfying
\[
\bigabs{\tan\tfrac{\pi}{2}\bigrbr{\tfrac{1}{2}-\tfrac{1}{p}}}<\inf_{\varepsilon\in\set E}\Bigrbr{\,\inf_{q\in\rbr{1,\infty}}\norm{\op A_{\mu}^{\varepsilon}}_{\set W_{q}^{1}\to\set W_{q}^{-1}}^{-1}\norm{\rbr{\op A_{\mu}^{\varepsilon}}^{-1}}_{\set H^{-1}\to\set H^{1}}^{-1}\!}.
\]
Using⋅(\ref{est: Original operator | A=001D4B is bounded}) and⋅(\ref{est: A^e is elliptic}),
one can easily see that the right-hand side here is  bounded below
by a positive constant~$C_{\mu}$, depending only on⋅$\mu$ and the
ellipticity constants~$c_{A}$, $C_{A}$ and~$\norm A_{C}$. Thus,
we have⋅(\ref{est: Original operator | Norm of (A=001D4B-=0003BC)=00207B=0000B9})
for all⋅$p\in\set P_{\mu}=\rbr{p_{\mu}^{\dual},p_{\mu}}$, where
$p_{\mu}$ solves
\[
\tan\tfrac{\pi}{2}\bigrbr{\tfrac{1}{2}-\tfrac{1}{p_{\mu}}}=C_{\mu}.
\]

Let $\set P=\cup_{\mu\notin\set S}\set P_{\mu}$. We show that
(\ref{est: Original operator | Norm of (A=001D4B-=0003BC)=00207B=0000B9})
holds, in fact, for any⋅$p\in\set P$ and⋅$\mu\notin\set S$ uniformly
in~$\varepsilon\in\set E$. Indeed, suppose that $\mu,\nu\notin\set S$
and choose⋅$p\in\rbr{2,p_{\nu}}$, so that $\rbr{\op A_{\nu}^{\varepsilon}}^{-1}$
maps⋅$\set W_{p}^{-1}\rbr{\Omega;\C^{n}}$ to⋅$\set W_{p}^{1}\rbr{\Omega;\C^{n}}$
continuously. From⋅the Sobolev embedding theorem, we know that $L_{2}\rbr{\Omega}$
is  embedded in⋅$W_{q^{\dual}}^{1}\rbr{\Omega}^{*}$ for⋅$q\in\sbr{2,2^{*}}$,
and in particular in⋅$\set W_{2^{*}\meet p}^{-1}\rbr{\Omega;\C^{n}}$
(see⋅(\ref{est: Norm of pi})). Hence, the first resolvent identity
\begin{equation}
\rbr{\op A_{\mu}^{\varepsilon}}^{-1}=\rbr{\op A_{\nu}^{\varepsilon}}^{-1}+\rbr{\mu-\nu}\rbr{\op A_{\nu}^{\varepsilon}}^{-1}\rbr{\op A_{\mu}^{\varepsilon}}^{-1}\label{eq: 1st resolvent id}
\end{equation}
yields that $\rbr{\op A_{\mu}^{\varepsilon}}^{-1}$ is bounded from⋅$\set W_{2^{*}\meet p}^{-1}\rbr{\Omega;\C^{n}}$
to⋅$\set W_{2^{*}\meet p}^{1}\rbr{\Omega;\C^{n}}$. Repeating this
procedure finitely many times, if need be, we conclude that the operator~$\rbr{\op A_{\mu}^{\varepsilon}}^{-1}$
is bounded from⋅$\set W_{p}^{-1}\rbr{\Omega;\C^{n}}$ to⋅$\set W_{p}^{1}\rbr{\Omega;\C^{n}}$
as well.
\begin{rem}
No⋅necessary and sufficient algebraic condition for⋅$A$ to assure⋅(\ref{est: A^e is elliptic})
is known. A⋅simpler condition not involving⋅$\varepsilon$ and still
implying the weak coercivity  on⋅$\mathring{H}^{1}\rbr{\Omega}^{n}$
is that for some⋅$c>0$ and all~$x\in\Omega$
\begin{equation}
\Re\rbr{A\rbr{x,\wc}\hairspace Du,Du}_{\R^{d}}\ge c\norm{Du}_{2,\R^{d}}^{2},\qquad u\in H^{1}\rbr{\R^{d}}^{n}.\label{est: Sufficient condition for A to be coercive}
\end{equation}
That this hypothesis suffices can be seen by noticing that (\ref{est: Sufficient condition for A to be coercive})
is invariant under dilation and therefore remains true with⋅$A\rbr{x,\varepsilon^{-1}y}$
in place of~$A\rbr{x,y}$. Since $A$ is uniformly continuous in
the first variable, a localization argument then leads to⋅(\ref{est: A^e is elliptic}),
with $c_{A}<c$, $C_{A}>0$ and⋅$\set H^{1}\rbr{\Omega;\C^{n}}=\mathring{H}^{1}\rbr{\Omega}^{n}$.

To⋅give an example of⋅$A$ satisfying the strong coercivity condition
on⋅$\mathring{H}^{1}\rbr{\Omega}^{n}$ (i.e., with⋅$C_{A}=0$), take
a matrix first-order differential operator~$b\rbr D$ with symbol
\[
\xi\mapsto b\rbr{\xi}=\sum_{k=1}^{d}b_{k}\xi_{k},
\]
where~$b_{k}\in\C^{m\times n}$. Suppose that the symbol has the
property that $\rank b\rbr{\xi}=n$ for any⋅$\xi\in\R^{d}\setminus\cbr 0$,
or, equivalently, that there is⋅$c_{b}>0$ such that
\[
b\rbr{\xi}^{*}b\rbr{\xi}\ge c_{b}\abs{\xi}^{2},\qquad\xi\in\R^{d}.
\]
Extending $u\in\mathring{H}^{1}\rbr{\Omega}^{n}$ by zero outside⋅$\Omega$
and applying the Fourier transform, we see that the operator~$b\rbr D^{*}b\rbr D$
is strongly coercive on⋅$\mathring{H}^{1}\rbr{\Omega}^{n}$:
\begin{equation}
\norm{b\rbr D\hairspace u}_{2,\Omega}^{2}\ge c_{b}\norm{Du}_{2,\Omega}^{2},\qquad u\in\mathring{H}^{1}\rbr{\Omega}^{n}.\label{est: Coercivity of b(D)*b(D) on Ho}
\end{equation}
Let $g\in C^{0,1}\rbr{\wb{\Omega};\wt L_{\infty}\rbr{\cell}}^{m\times m}$
with⋅$\Re g$ uniformly positive definite and let⋅$A_{kl}=b_{k}^{*}gb_{l}$.
Then, by⋅(\ref{est: Coercivity of b(D)*b(D) on Ho}),
\[
\begin{aligned}\Re\rbr{A^{\varepsilon}Du,Du}_{\Omega} & =\Re\rbr{g^{\varepsilon}b\rbr D\hairspace u,b\rbr D\hairspace u}_{\Omega}\\
 & \ge c_{b}\norm{\rbr{\Re g}^{-1}}_{L_{\infty}}^{-1}\norm{Du}_{2,\Omega}^{2}
\end{aligned}
\]
for⋅all⋅$u\in\mathring{H}^{1}\rbr{\Omega}^{n}$. Purely periodic
operators of this type were studied, e.g., in⋅\cite{PSu:2012} and⋅\cite{Su:2013-1}.

For⋅coercivity on⋅$H^{1}\rbr{\Omega}^{n}$, one needs to require a
stronger condition on the symbol, namely, that $\rank b\rbr{\xi}=n$
for any⋅$\xi\in\C^{d}\setminus\cbr 0$, not just⋅$\xi\in\R^{d}\setminus\cbr 0$,
implying that
\begin{equation}
\norm{b\rbr D\hairspace u}_{2,\Omega}^{2}\ge c_{b}\norm{Du}_{2,\Omega}^{2}-C_{b}\norm u_{2,\Omega}^{2},\qquad u\in H^{1}\rbr{\Omega}^{n},\label{est: Coercivity of b(D)*b(D) on H}
\end{equation}
with⋅$c_{b}>0$, see~\cite[Section~3.7, Theorem~7.8]{Necas:2012}.
Then, obviously, for any⋅$u\in H^{1}\rbr{\Omega}^{n}$
\[
\begin{aligned}\Re\rbr{A^{\varepsilon}Du,Du}_{\Omega} & =\Re\rbr{g^{\varepsilon}b\rbr D\hairspace u,b\rbr D\hairspace u}_{\Omega}\\
 & \ge\norm{\rbr{\Re g}^{-1}}_{L_{\infty}}^{-1}\bigrbr{c_{b}\norm{Du}_{2,\Omega}^{2}-C_{b}\norm u_{2,\Omega}^{2}},
\end{aligned}
\]
where⋅$A$ and⋅$g$ are as above. Such operators in the purely periodic
setting appeared in~\cite{Su:2013-2}.

\end{rem}
Now we turn to the cell problem and the effective operator. The⋅first
thing that we need to check is that the cell problem~(\ref{def: N})
has a unique solution for which⋅(\ref{assu: N is Lipschitz}) holds.
Lemma~\ref{lm: A(x)} contains a sufficient condition to conclude
these, and we will see in a moment that the operator~$\op A\rbr x$
does indeed meet the hypothesis of that lemma.
\begin{lem}
\label{lem: Strong coercivity of A(x)}Assume that~\textup{(\ref{est: A^e is elliptic})}
holds. Then for any⋅$x\in\Omega$
\begin{equation}
\Re\rbr{\op A\rbr x\hairspace u,u}_{\cell}\ge c_{A}\norm{Du}_{2,\cell}^{2},\qquad u\in\wt H^{1}\rbr{\cell}^{n}.\label{est: Coercivity of D*A(x,=0022C5)D on Q}
\end{equation}

\end{lem}
\begin{proof}
Fix $\smash[t]{u^{\rbr{\varepsilon}}}=\varepsilon u^{\varepsilon}\varphi$
with $u\in\wt C^{1}\rbr{\cell}^{n}$ and~$\varphi\in C_{c}^{\infty}\rbr{\Omega}$.
We substitute $\smash[t]{u^{\rbr{\varepsilon}}}$ into (\ref{est: A^e is elliptic})
and let $\varepsilon$ tend to~0. Then, because $\smash[t]{u^{\rbr{\varepsilon}}}$
and⋅$\smash[t]{Du^{\rbr{\varepsilon}}}-\rbr{Du}^{\varepsilon}\varphi$
converge in⋅$L_{2}$ to~$0$,
\[
\lim_{\varepsilon\to0}\Re\int_{\Omega}\abr{A^{\varepsilon}\rbr x\hairspace\rbr{Du}^{\varepsilon}\rbr x,\rbr{Du}^{\varepsilon}\rbr x}\abs{\varphi\rbr x}^{2}\dd x\ge\lim_{\varepsilon\to0}c_{A}\int_{\Omega}\abs{\rbr{Du}^{\varepsilon}\rbr x}^{2}\abs{\varphi\rbr x}^{2}\dd x.
\]
It⋅is well known  that if⋅$f\in C_{c}\rbr{\R^{d};\wt L_{\infty}\rbr{\cell}}$,
then
\[
\lim_{\varepsilon\to0}\int_{\R^{d}}f^{\varepsilon}\rbr x\dd x=\int_{\R^{d}}\int_{\cell}f\rbr{x,y}\dd x\dd y
\]
(see, e.g.,~\cite[Lemmas~5.5 and~5.6]{Al:1992}). As⋅a⋅result,
\[
\Re\int_{\Omega}\int_{\cell}\abr{A\rbr{x,y}\hairspace Du\rbr y,Du\rbr y}\abs{\varphi\rbr x}^{2}\dd x\dd y\ge c_{A}\int_{\Omega}\int_{\cell}\abs{Du\rbr y}^{2}\abs{\varphi\rbr x}^{2}\dd x\dd y.
\]
But $\varphi$ is an arbitrary function in $C_{c}^{\infty}\rbr{\Omega}$
and $A$ is uniformly continuous in the first variable,⋅so
\[
\Re\int_{\cell}\abr{A\rbr{x,y}\hairspace Du\rbr y,Du\rbr y}\dd y\ge c_{A}\int_{\cell}\abs{Du\rbr y}^{2}\dd y
\]
for⋅all⋅$x\in\Omega$, as required.
\end{proof}
We have shown that, for any⋅$x\in\Omega$, the operator~$\op A\rbr x$
is an isomorphism of⋅$\wt H^{1}\rbr{\cell}^{n}\!/\C$ onto⋅$\wt H^{-1}\rbr{\cell}^{n}$
and that the ellipticity constants of⋅$\op A\rbr x$ are better that
those of⋅$\op A^{\varepsilon}$ (cf.⋅(\ref{est: A^e is elliptic})
with~(\ref{est: Coercivity of D*A(x,=0022C5)D on Q})). Then the
Shneiberg stability theorem yields that $\op A\rbr x$ is  an isomorphism
of⋅$\wt W_{p}^{1}\rbr{\cell}^{n}\!/\C$ onto⋅$\wt W_{p}^{-1}\rbr{\cell}^{n}$
for any~$p\in\set P$. Thus, the hypothesis of Lemma~\ref{lm: A(x)}
 is verified.

As⋅for the effective operator, one can prove that, for any⋅$\mu\notin\set S$,
the inverse for⋅$\op A_{\mu}^{\varepsilon}$ converges in the weak
operator topology and then the limit is an isomorphism of⋅$\set H^{-1}\rbr{\Omega;\C^{n}}$
onto⋅$\set H^{1}\rbr{\Omega;\C^{n}}$, which is, in fact, the inverse
for⋅$\op A_{\mu}^{0}$, see~\cite[Lemma~6.2]{Tartar:2010}. We conclude
that $\rbr{\op A_{\mu}^{0}}^{-1}$ is also an isomorphism as a mapping
from⋅$\set W_{p}^{-1}\rbr{\Omega;\C^{n}}$ to⋅$\set W_{p}^{1}\rbr{\Omega;\C^{n}}$
for any⋅$p\in\set P$, because it is the limit in the weak operator
topology. In⋅the case of the Dirichlet and Neumann problems, this,
and the fact that $A^{0}$ is Lipschitz, yields the assumption~(\ref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})
with⋅$s=1$ and any⋅$p\in\set P$ (see, e.g.,~\cite[Chapter~4]{McL:2000}).
In⋅the case of the mixed Dirichlet–Neumann problem, we assume additionally
that $A^{0}$ is self-adjoint; then it follows from~\cite{Savare:1997}
that (\ref{est: Norm of (A=002070-=0003BC)=00207B=0000B9}) holds
for⋅$s=1/2$ and⋅$p=2$.

Of⋅course, all these results are true for the dual counterparts with
the same range of⋅$p$, because $c_{A^{\dual}}=c_{A}$, $C_{A^{\dual}}=C_{A}$
and~$\norm{A^{\dual}}_{L_{\infty}}=\norm A_{L_{\infty}}$.

It⋅remains to discuss the interior energy estimate~(\ref{est: Local enery estimate}).
Let⋅$p\in\set P\cap[2,\infty)$. Applying the functional~$\op A_{\mu}^{\varepsilon}u$
to⋅$\abs{\chi}^{2}u$, where⋅$\chi\in C_{c}^{0,1}\rbr{\Omega}$, and
using⋅(\ref{est: A^e is elliptic}), we arrive at the well-known Caccioppoli
inequality:
\[
\norm{\chi Du}_{2,\supp\chi}\lesssim\norm u_{2,\supp\chi}+\norm{\chi\op A_{\mu}^{\varepsilon}u}_{-1,2,\Omega}^{*},\qquad u\in\set H^{1}\rbr{\Omega;\C^{n}}.
\]
Therefore, Lemma~\ref{lem: Original operator | Interior energy estimate}
with⋅$q=2$ yields~(\ref{est: Local enery estimate}).

To⋅summarize, assume that $\op A^{\varepsilon}$ satisfies the coercivity
condition~(\ref{est: A^e is elliptic}). Then for Dirichlet or Neumann
boundary conditions the global results (see Theorem~\ref{thm: Convergence and Approximation with 1st corrector}–Theorem~\ref{thm: Convergence with e}) are valid with⋅$s=s^{\dual}=1$ and⋅$p\in\set P$ and the local
results (see Theorem~\ref{thm: Interior approximation with 1st corrector}–Corollary~\ref{cor: Interior convergence for fractional derivative})
are valid with⋅$s=s^{\dual}=1$ and~$p\in\set P\cap[2,\infty)$;
for mixed Dirichlet–Neumann boundary conditions all these results
hold true with⋅$s=s^{\dual}=1/2$ and⋅$p=2$, provided that $A^{0}$
is self-adjoint.
\begin{rem}
The⋅constants~$p_{\mu}$, and hence the interval~$\set P$, can
be expressed explicitly. We note that generally one would not expect
$\set P$ to be too wide. In⋅fact,  it must shrink to⋅$\cbr 2$
as the ellipticity of the family~$\op A^{\varepsilon}$ becomes ``bad''
(that is, the ratio~$c_{A}^{-1}\norm A_{L_{\infty}}$ grows), see~\cite{Mey:1963}.
In⋅the⋅next subsection we provide an example where $p$ may be chosen
 arbitrary large.
\end{rem}

\subsection{Strongly elliptic operators with $\boldsymbol{\protect\VMO}$\protect\nobreakdash-coefficients}

Throughout this subsection, we restrict our attention to the Dirichlet
and Neumann problems. Let $\op A^{\varepsilon}$ be as in the previous
subsection. Assume further that $A\in L_{\infty}\rbr{\Omega;\VMO\rbr{\R^{d}}}$,
meaning that $\sup_{x\in\Omega}\eta_{A\rbr{x,\wc}}\rbr r\to0$ as⋅$r\to0$.

Using the reflection technique, we extend $A$ to be a function belonging
to both $C_{c}^{0,1}\rbr{\R^{d};\wt L_{\infty}\rbr{\cell}}$ and~$L_{\infty}\rbr{\R^{d};\VMO\rbr{\R^{d}}}$.
Notice that $A^{\varepsilon}$ is then a $\VMO$\nobreakdash-function.
Indeed, $A^{\varepsilon}$ obviously belongs to the space~$\BMO\rbr{\R^{d}}$,
with $\norm{A^{\varepsilon}}_{\BMO}\le2\norm A_{L_{\infty}}$. Next,
after dilation, we may suppose that~$\varepsilon=1$. Given an⋅$\epsilon>0$
small, there is $r>0$ such that $\osc_{A}\rbr r<\epsilon/3$ and
$\eta_{A\rbr{x,\wc}}\rbr r<\epsilon/3$. Then, since
\[
\begin{aligned}\dashint_{B_{R}\rbr{x_{0}}}\abs{A^{1}\rbr x-m_{B_{R}\rbr{x_{0}}}\rbr{A^{1}}}\dd x & \le\dashint_{B_{R}\rbr{x_{0}}}\abs{A\rbr{x_{0},x}-m_{B_{R}\rbr{x_{0}}}\rbr{A\rbr{x_{0},\wc}}}\dd x\\
 & \quad+2\dashint_{B_{R}\rbr{x_{0}}}\abs{A\rbr{x,x}-A\rbr{x_{0},x}}\dd x,
\end{aligned}
\]
we have
\[
\eta_{A^{1}}\rbr r\le\sup_{x_{0}\in\R^{d}}\eta_{A\rbr{x_{0},\wc}}\rbr r+2\osc_{A}\rbr r<\epsilon,
\]
and the claim follows.

As⋅a⋅result, if $\varepsilon\in\set E$ is fixed and⋅$\mu\notin\set S$,
the inverse of⋅$\op A_{\mu}^{\varepsilon}$ is a continuous map from⋅$\set W_{p}^{-1}\rbr{\Omega;\C^{n}}$
to⋅$\set W_{p}^{1}\rbr{\Omega;\C^{n}}$ for each⋅$p\in\rbr{1,\infty}$,
see~\cite{Shen:2018}. Hence, in order to prove⋅(\ref{est: Original operator | Norm of (A=001D4B-=0003BC)=00207B=0000B9}),
we need only show that its norm is uniformly bounded in~$\varepsilon$.
We do this by  treating⋅$\op A^{\varepsilon}$ as a local perturbation
of a purely periodic operator and then applying results for purely
periodic operators with rapidly oscillating coefficients.

First observe that if $B_{R}$ is a ball with center in⋅$\wb{\Omega}$
and radius~$R$, then, by⋅(\ref{est: A^e is elliptic}),
\[
\Re\rbr{A^{\varepsilon}\rbr{x_{0},\wc}\hairspace Dv,Dv}_{B_{R}\cap\Omega}\ge\rbr{c_{A}-\osc_{A}\rbr R}\hairspace\norm{Dv}_{2,B_{R}\cap\Omega}^{2}-C_{A}\norm v_{2,B_{R}\cap\Omega}^{2}
\]
for⋅all⋅$v$ in⋅$\set H^{1}\rbr{B_{R}\cap\Omega;\C^{n}}$, the space
of functions whose zero extensions to⋅$\Omega$ belong to⋅$\set H^{1}\rbr{\Omega;\C^{n}}$.
 It⋅follows that for⋅$R$ small enough, the operator~$D^{*}A^{\varepsilon}\rbr{x_{0},\wc}\hairspace D$
from⋅$\set H^{1}\rbr{B_{R}\cap\Omega;\C^{n}}$ to the dual space~$\set H^{-1}\rbr{B_{R}\cap\Omega;\C^{n}}$
is $m$\nobreakdash-sectorial, with sector
\[
\set S_{R}=\bigcbr{z\in\C\colon\abs{\Im z}\le\rbr{c_{A}-\osc_{A}\rbr R}^{-1}\norm A_{L_{\infty}}\rbr{\Re z+C_{A}}}
\]
converging pointwise to $\set S$ as⋅$R\to0$, that is, $\dist\rbr{z,\set S_{R}}\to\dist\rbr{z,\set S}$
for~$z\in\C$.

Now,  fix $\mu\not\in\set S$ and find $R_{0}>0$ such that $\osc_{A}\rbr R\le c_{A}/2$
and~$\mu\notin\set S_{R}$ as long as~$R\le R_{0}$. Let⋅$F\in C_{c}^{\infty}\rbr{\Omega}^{dn}$
and $u_{\varepsilon}=\rbr{\op A_{\mu}^{\varepsilon}}^{-1}D^{*}F$.
Take⋅$\chi\in C_{c}^{\infty}\rbr{B_{R}}$, $R\le R_{0}$, with the
properties that $0\le\chi\rbr x\le1$ and⋅$\chi=1$ on~$1/2B_{R}$.
Then $v_{\varepsilon}=\chi u_{\varepsilon}$  obviously satisfies
\[
D^{*}A^{\varepsilon}\rbr{x_{0},\wc}\hairspace Dv_{\varepsilon}-\mu v_{\varepsilon}=D^{*}\rbr{A^{\varepsilon}\rbr{x_{0},\wc}-A^{\varepsilon}}Dv_{\varepsilon}+g
\]
in⋅the sense of functionals on⋅$\set H^{1}\rbr{B_{R_{0}}\cap\Omega;\C^{n}}$,
where ${g=\chi D^{*}F+D^{*}\rbr{A^{\varepsilon}D\chi\cdot u_{\varepsilon}}}-\rbr{D\chi}^{*}\cdot A^{\varepsilon}Du_{\varepsilon}$.
This is a purely periodic problem, for which we know that the operator~$D^{*}A^{\varepsilon}\rbr{x_{0},\wc}\hairspace D-\mu$
is an isomorphism of⋅$\set W_{q}^{1}\rbr{B_{R_{0}}\cap\Omega;\C^{n}}$
onto⋅$\set W_{q}^{-1}\rbr{B_{R_{0}}\cap\Omega;\C^{n}}$ for any⋅$q\in\rbr{1,\infty}$,
with uniformly  bounded inverse, see~\cite{Shen:2018}. Assuming
 that⋅$p>2$ (the⋅other case will follow by duality), we immediately
find that
\[
\norm{Dv_{\varepsilon}}_{2^{*}\meet p,B_{R_{0}}\cap\Omega}\lesssim\osc_{A}\rbr R\hairspace\norm{Dv_{\varepsilon}}_{2^{*}\meet p,B_{R_{0}}\cap\Omega}+\norm g_{-1,2^{*}\meet p,B_{R_{0}}\cap\Omega}^{*},
\]
the⋅constant not depending on~$R$. Choosing $R$ sufficiently
small, we may absorb the first term on the right into the left-hand
side. Since
\[
\norm g_{-1,2^{*}\meet p,B_{R_{0}}\cap\Omega}^{*}\lesssim\norm F_{2^{*}\meet p,B_{R}\cap\Omega}+\norm{u_{\varepsilon}}_{1,2,B_{R}\cap\Omega}
\]
(we have used the Sobolev embedding theorem to estimate the⋅$L_{2^{*}\meet p}$\nobreakdash-norm
of⋅$u_{\varepsilon}$ and the⋅$\set W_{2^{*}\meet p}^{-1}$\nobreakdash-norm
of~$Du_{\varepsilon}$), it follows that
\[
\norm{Du_{\varepsilon}}_{2^{*}\meet p,1/2B_{R}\cap\Omega}\lesssim\norm F_{2^{*}\meet p,B_{R}\cap\Omega}+\norm{u_{\varepsilon}}_{1,2,B_{R}\cap\Omega}.
\]
Now, cover⋅$\Omega$ with balls of radius~$R$ to obtain
\[
\norm{Du_{\varepsilon}}_{2^{*}\meet p,\Omega}\lesssim\norm F_{2^{*}\meet p,\Omega}+\norm{u_{\varepsilon}}_{1,2,\Omega}\lesssim\norm F_{2^{*}\meet p,\Omega}.
\]
After⋅a⋅finite number of repetitions, if need be, we get
\[
\norm{Du_{\varepsilon}}_{p,\Omega}\lesssim\norm F_{p,\Omega}.
\]

Next, the hypothesis of Lemma~\ref{lm: A(x)} is satisfied, because
of Lemma~\ref{lem: Strong coercivity of A(x)} and the fact that
$A\rbr{x,\wc}\in\VMO\rbr{\R^{d}}$ with $\VMO$\nobreakdash-modulus
bounded uniformly in~$x$. Finally,  (\ref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})
 and⋅(\ref{est: Local enery estimate}) hold for, respectively, $s=1$
and any⋅$p\in\rbr{1,\infty}$ and any⋅$p\in[2,\infty)$, as indicated
previously.

Summarizing, if $A\in L_{\infty}\rbr{\Omega;\VMO\rbr{\R^{d}}}$ satisfies
the coercivity condition~(\ref{est: A^e is elliptic}), then the
global results (see Theorem~\ref{thm: Convergence and Approximation with 1st corrector}–Theorem~\ref{thm: Convergence with e})
are valid with⋅$s=s^{\dual}=1$ and⋅$p\in\rbr{1,\infty}$ and the
local results (see Theorem~\ref{thm: Interior approximation with 1st corrector}–Corollary~\ref{cor: Interior convergence for fractional derivative})
are valid with⋅$s=s^{\dual}=1$ and~$p\in[2,\infty)$.

config: arxiv


text:
\section{Sequence of Differences on Generalized Pentagonal Numbers}
Tags: Generalized Pentagonal Numbers

\begin{theorem}
Recall the [[Definition:Generalized Pentagonal Number|generalised pentagonal numbers]] $GP_n$ for $n = 0, 1, 2, \ldots$
Consider the [[Definition:Integer Sequence|sequence]] defined as $\Delta_n = GP_{n + 1} - GP_n$:
:$1, 1, 3, 2, 5, 3, 7, 4, 9, 5, 11, 6, 13, 7, 15, 8, \ldots$
{{OEIS|A026741}}
Then:
:The values of $\Delta_n$ for [[Definition:Odd Integer|odd $n$]] consist of the [[Definition:Odd Integer|odd numbers]]
:The values of $\Delta_n$ for [[Definition:Even Integer|even $n$]] consist of the [[Definition:Natural Numbers|natural numbers]].
\end{theorem}

\begin{proof}
Recall the definition of the [[Definition:Generalized Pentagonal Number/Definition 2|generalised pentagonal numbers]] $GP_n$ for $n = 0, 1, 2, \ldots$
:$GP_n = \begin{cases} \dfrac {m \paren {3 m + 1} } 2 & : n = 2 m \\
\dfrac {m \paren {3 m - 1} } 2 & : n = 2 m - 1 \end{cases}$
for $n = 0, 1, 2, \ldots$
Hence:
{{begin-eqn}}
{{eqn | l = \Delta_{2 n - 1}
      | r = GP_{2 n} - GP_{2 n - 1}
      | c =
}}
{{eqn | r = \dfrac {n \paren {3 n + 1} } 2 - \dfrac {n \paren {3 n - 1} } 2
      | c =
}}
{{eqn | r = \dfrac {3 n^2 + n - 3 n^2 + n} 2
      | c =
}}
{{eqn | r = \dfrac {2 n} 2
      | c =
}}
{{eqn | r = n
      | c =
}}
{{end-eqn}}
which defines the [[Definition:Sequence|sequence]] of [[Definition:Natural Number|natural numbers]].
Then:
{{begin-eqn}}
{{eqn | l = \Delta_{2 n}
      | r = GP_{2 n + 1} - GP_{2 n}
      | c =
}}
{{eqn | r = GP_{2 \paren {n + 1} - 1} - GP_{2 n}
      | c =
}}
{{eqn | r = \dfrac {\paren {n + 1} \paren {3 \paren {n + 1} - 1} } 2 - \dfrac {n \paren {3 n + 1} } 2
      | c =
}}
{{eqn | r = \dfrac {\paren {3 n^2 + 6 n + 3} - \paren {n + 1} - 3 n^2 - n} 2
      | c =
}}
{{eqn | r = \dfrac {4 n + 2} 2
      | c =
}}
{{eqn | r = 2 n + 1
      | c =
}}
{{end-eqn}}
which defines the [[Definition:Sequence|sequence]] of [[Definition:Odd Integer|odd numbers]].
{{qed}}
\end{proof}

config: wiki


TITLE: Connectedness of the set
QUESTION [0 upvotes]: I'm new to Real Analysis, so I'm struggling with proofs. What is this set $$E = \Big([−1, 1] \times \{0\}\Big) ∪ \Big((−\infty, −1] \times \mathbb R\Big) \cup \Big([1, \infty) \times \mathbb R\Big)$$ and how to prove that its connected?

REPLY [1 votes]: The union of connected sets is connected if the intersection is nonempty:
Let $\mathcal{F}$ be a family of connected sets with $\bigcap \mathcal{F} \neq \emptyset$.
Suppose we have separated sets $A,B \subset X$ such that $A \cup B = \bigcup \mathcal{F}$ and let $x \in \bigcap \mathcal{F}$.
Now suppose $x \in A$ and $B \neq \emptyset$.
Then we have $y \in B$, and as $B \subset \bigcup \mathcal{F}$, there exists $F \in \mathcal{F}$ such that $y \in F$.
But also note that $x \in F$, since $x \in \bigcap \mathcal{F}$, so $x \in F \cap A$.
Now I'll prove that if $F$ is connected and $F \subset A \cup B$ with $A$ and $B$ separated, then $F \subset A$ or $F \subset B$.
Write $F = (F \cap A) \cup (F \cap B)$.
Note that $F \cap A$ and $F \cap B$ are separated, because if $z \in \overline{F \cap A} \cap (F \cap B)$, then $z \in B$ and $z \in \overline{F \cap A} \subset \overline{A}$.
Contradiction, since $z \in \overline{A} \cap B$, but $A$ and $B$, by hypothesis, are separated.
But, if $F \cap A$ and $F \cap B$ are separated and $F$ is connected, by definition $F \subset F \cap A \implies F \subset A$ or $F \subset F \cap B \implies F \subset B$.
Finally, note that $x \in F \cap A$ and $y \in F \cap B$, so we can't have $F \subset A$ or $F \subset B$. Contradiction. Then $B$ must be $\emptyset$.
Now note that the point $(-1,0) \in [-1,1] \times \{0\} \cap (-\infty,-1] \times \mathbb{R}$ and the point $(1,0) \in [-1,1] \times \{0\} \cap [1,\infty) \times \mathbb{R}$.
Then we only have to prove that each component is connected.
But you can see that $[-1,1] \times \{0\}$ is homeomorphic to $[-1,1]$, just consider the projection $\pi : [-1,1] \times \{0\} \to [-1,1]$. Then use the fact that $I \subset \mathbb{R}$ is connected iff $I$ is an interval.
To prove that $(-\infty,-1] \times \mathbb{R}$ and $[1,\infty) \times \mathbb{R}$ is connected, you can use that they're path-connected and path-connected implies connected (proof here).
I'll make the case $[1,\infty) \times \mathbb{R}$:
Let $(a,b),(c,d) \in [1,\infty) \times \mathbb{R}$.
Then $f : [0,1] \to [1,\infty) \times \mathbb{R}$ given by $f(t) = (1-t)(a,b) + t(c,d)$ is a continuous path connecting both points ($f$ is a sum and product of continuous functions).
We only have to show that $f([0,1]) \subset [1,\infty) \times \mathbb{R}$, ie, that $(1-t)a + tc \geq 1$ and $(1-t)b + td \in \mathbb{R}$.
The second is obvious, then I'll just prove $(1-t)a + tc \geq 1$.
If $a \leq c$, then $ta \leq tc \implies (1-t)a + ta \leq (1-t)a + tc$.
But $(1-t)a + ta = a \geq 1$, the result follows.
If $c \leq a$, then $(1-t)c \leq (1-t)a \implies (1-t)c + tc \leq (1-t)a + tc$.
But $(1-t)c + tc = c \geq 1$. The result also follows.

REPLY [0 votes]: $$E=([−1,1]×{0})∪((−∞,−1]×R)∪([1,∞)×R)$$
The first part is a line segment on the $x-axis$ connecting $-1$ to $1$.
The second part is part of the plane to the left of $x=-1$ and including the boundary line $x=-1$
The third part is part of the plane to the right of $x=1$ and including the boundary line $x=1$
The union is a connected set because you can join any point of the set to any other point via a path in the set.
you may have to pass through the bridge $[-1,1]$ if your points are in different sides of the bridge.


TITLE: Let $n = {1,2,3, ...}$, prove that $\left(1+\frac{1}{n}\right)^n$ is bounded above.
QUESTION [1 upvotes]: I know that the sequence tends to $e \approx 2.718281828...$, but I am having a hard time proving it. Does anyone have any hints for me?

REPLY [3 votes]: For an integer $k\geq 0$ by Bernoulli's inequality $$\left(1+\frac1n\right)^{-k}=\left(1-\frac1{n+1}\right)^k\geq 1-\frac{k}{n+1}.$$ In particular for $k=\lceil n/2 \rceil\leq (n+1)/2$ we have $$\left(1+\frac1n\right)^{-k} \geq \tfrac12$$ and by inverting both sides $$\left(1+\frac1n\right)^k \leq 2.$$ Now $n\leq 2k$ and so $$\left(1+\frac1n\right)^n\leq \left(1+\frac1n\right)^{2k} \leq 2^2=4.$$


TITLE: Nowhere-Dense-Sets Clarified
QUESTION [0 upvotes]: If A is a nowhere dense set, it means that "A closure" has an empty interior.
Canwe also say that "A" too has an empty interior?
I believe Yes, we can say so.
Because "A closure" is just the union of A and it's limit points. The set of limit points are always closed(since they are not open). Now when we talk about "A Closure's interior", we refer to A's interior. And so if we say A closure has an empty interior, it means "A Closure" sans the closed subset of A (which is the set of it's limit points) has an empty interior. In other words, A has an empty interior? Am I Right?

REPLY [0 votes]: If $A$ is nowhere dense, then, as $\operatorname{int}(A) \subseteq \operatorname{int}(\overline{A}) = \emptyset$, $A$ has empty interior.
But e.g. $([0,1] \cap \Bbb Q)$ has empty interior in $\Bbb R$, but is not nowhere dense, so it's quite a bit stronger than having empty interior.


TITLE: Can a function be differentiable while having a discontinuous derivative?
QUESTION [2 upvotes]: Recently I came across functions like $x^2\sin(1/x)$ and $x^3\sin(1/x)$ where the derivatives were discontinuous. Can there exist a function whose derivative is not conitnuous, and yet the function is differentiable? If yes, please provide some examples.

REPLY [1 votes]: The functions you mentioned are in fact differentiable, so you can use them as examples.


TITLE: Derive forward Euler method for two-variable function
QUESTION [2 upvotes]: I need to derive the forward Euler method for solving ODEs and I would like some comments on what I have so far; overdot denote the time derivative: $\dot x \overset{def}{=} dx/dt$.
Say we have $\dot x = f(t, x)$ and want to approximate the solution $x(t)$.
The initial conditions are $x(t_0, x_0)$ and the approximate solution $x_1$ after one time $h$ step is sought, $x_1 \approx x(t_0 + h, x_0)$. $x$ is expanded as a Taylor series in $t$ around $(t_0, x_0)$:
$$
x(t_0 + h, x_0) = x(t_0, x_0) + \dot x(t_0, x_0)h + \frac{1}{2} \ddot x(t_0, x_0)h^2 + \dots\\
%
= x(t_0, x_0) + f(t_0, x_0)h + \mathcal{O}(h^2).
$$
Discarding anything of order $\mathcal{O}(h^2)$ or higher we are left with the Euler solution:
$$
x_1 = x(t_0, x_0) + f(t_0, x_0)h.
$$
It's pretty close to the derivation found on wikipedia; however, wikipedia does not include $x_0$ in the Taylor expansion.
Any comments - is the above correct?
EDIT: typo.

REPLY [0 votes]: The notation $x(t;t_0,x_0)$  (or similar, for instance $\phi(t;t_0,x_0)$ to use some greek) is used for the flow of the differential equation, i.e., the most general form of a solution.
In an initial value problem, the initial point $(t_0,x_0)$ is fixed, so there is little use in repeating the parameter in the notation of the solution function.
However, in the analysis of the Euler method, one uses all solutions $x(t;t_k,x_k)$ to the initial value problems starting in the iteration point. The long notation then is useful when one estimates how big the gaps are between neighboring trajectories, how the evolution of the gaps is bounded in time and how the single gaps sum up to the global error of the method.


TITLE: $\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}$ using residues
QUESTION [5 upvotes]: I'm trying to find $\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}dx$ where $\alpha>0$ is real.  My approach was to take an integral along the real line from $1/R$ to $R$, around the circle counterclockwise to $-R$, along the real line to $-1/R$, and then around the circle clockwise to $1/R$.  I have encountered 2 problems with this:

This path encloses one pole, at $z=\alpha i$.  I found the residue at $z=\alpha i$ to be $\frac{\ln(\alpha)+i\pi/2}{2\alpha i}$.  However, this gives me that $\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}dx=\frac{\pi(\ln(\alpha)+i\pi/2)}{2\alpha}$.  Since I have a real function integrated over the real line, there cannot be an imaginary part.  Where did I go wrong?  (Also, doing a few examples, the correct answer seems to be $\frac{\pi\ln(\alpha)}{2\alpha}$, the same as I have but without the imaginary part.)
At first chose my path so instead of going all the way around the upper semicircle, it only went 3/4 of the way around, as I wanted to avoid anything that might go wrong with the discontinuity of $\log(x)$ at the negative real axis.  When I do this, though, I get a different answer than before (the denominator of the fraction is $\alpha(1-e^{3\pi i/4})$ instead of $2\alpha$.  What am I doing wrong that gives me different answers?

REPLY [1 votes]: You were on the right track.  We have
$$\begin{align}
\int_{-\infty}^0 \frac{\log x}{x^2+\alpha^2}dx+\int_{0}^{\infty} \frac{\log x}{x^2+\alpha^2}dx&=2\pi i \left(\frac{\log (\alpha)+i\pi/2}{2i\alpha}\right)\\\\
&=\frac{\pi\log \alpha}{\alpha}+i\frac{\pi^2}{2\alpha}
\end{align} \tag 1$$
We change variables on the first integral on the left-hand side of $(1)$ by letting $x\to -x$.  Now, being careful to evaluate $\log (-x)=\log (x)+i\pi$ reveals
$$2\int_{0}^{\infty} \frac{\log x}{x^2+\alpha^2}dx+i\pi\int_0^{\infty}\frac{1}{x^2+\alpha^2}=\frac{\pi\log \alpha}{\alpha}+i\frac{\pi^2}{2\alpha}\tag 2$$
whereby equating real and imaginary parts yields of $(2)$
$$\bbox[5px,border:2px solid #C0A000]{\int_{0}^{\infty} \frac{\log x}{x^2+\alpha^2}dx=\frac{\pi\log \alpha}{2\alpha}}$$
and
$$\bbox[5px,border:2px solid #C0A000]{\int_0^{\infty}\frac{1}{x^2+\alpha^2}=\frac{\pi}{2\alpha}}$$
which agree with the answers reported by @MarkoRiedel and obtained with a semi-circular contour rather than a keyhole contour!


TITLE: Extremal graphs definition
QUESTION [2 upvotes]: I'm trying to understand extremal graph theory using Diestel's book. I have two problems in it's introduction. First the definition of extremality for graph $H$ and $n$ vertices is somewhat confusing. As the first point, I want to verify whether my understanding is right.
My understanding of $ex(n,H)$ is as follows:
given $n$ vertices, if you add $ex(n,H)$ number of edges among these vertices, then, no matter which vertices you connect with these edges, an $H$ subgraph appears inside this graph. And if you lose any one of these $ex(n,H)$ number of edges, the $H$ subgraph disappears.
This is different from being edge maximal in respect to not having an $H$ subgraph because the edge maximality could depend on the current arrangement of the edges in the graph.
Is this correct?
That description in the book is confusing enough (specially being a non-native speaker) but what adds to the confusion is the following figure (which is going to be my second problem):

My understanding of $P^3$ is that its a path of three vertices, which already appears in the graph in the right. But the book claims that this graph is edge maximal with $P^3 \nsubseteq G$. Am I confused what $P^3$ means?
Any help would be appreciated.
EDIT: the book I'm referring to is reinhard diestel's book on graph theory: 2

REPLY [2 votes]: The usual definition of $ex(n,H)$ is:

there exists a graph with $n$ vertices, $ex(n,H)$ edges, and without a copy of $H$, and
every graph with $n$ vertices and $ex(n,H)+1$ edges contains a copy of $H$.

Any graph which satisfies the first bullet point is then called extremal (with respect to the property of containing $H$).


TITLE: Does all unary language belongs to P questions?
QUESTION [1 upvotes]: It's a simple and maybe silly question. Does all unary language aka belongs to {1*} a P language. It seems I can always check the input unary string in linear time to see if there exists one in the language.

REPLY [3 votes]: The answer is no. Take a non recursive subset $S$ of $\Bbb{N}$. Then the language $\{1^n \mid n \in S\}$ will not even be decidable.