text:
\begin{document}
\author{Ovidiu Munteanu and Mu-Tao Wang}
\title{The curvature of gradient Ricci solitons}
\date{}

\begin{abstract}
We study integral and pointwise bounds on the curvature of gradient
shrinking Ricci solitons. As applications, we discuss gap and compactness
results for gradient shrinkers.
\end{abstract}

\maketitle

\section{Introduction}

Our goal in this paper is to obtain further information about the curvature
of gradient shrinking Ricci solitons. This is important for a better
understanding and ultimately for the classification of these manifolds. The
classification of gradient shrinkers is known in dimensions 2 and 3, and
assuming locally conformally flatness, in all dimensions $n\geq 4$ (see \cite
{P, NW, CWZ, PW, Zh, MS, M}). Many of the techniques used in these works
required some control of the Ricci curvature. For example, in \cite{PW}
gradient shrinking Ricci solitons which are locally conformally flat were
classified assuming an integral condition on the Ricci tensor. This
condition and other integral estimates of the curvature were later proved in
\cite{MS}.

Without making the strong assumption of being conformally flat, it is
natural to ask whether similar estimates are true for the Riemann curvature
tensor. In this paper we are able to prove pointwise estimates on the
Riemann curvature, assuming in addition that the Ricci curvature is bounded.
We will show that any gradient shrinking Ricci soliton with bounded Ricci
curvature has Riemann curvature tensor growing at most polynomially in the
distance function. We note that by Shi's local derivative estimates we can
then obtain growth estimates on all derivatives of the curvature. This, in
particular, proves weighted $L^2$ estimates for the Riemann curvature tensor
and its covariant derivatives.

We point out that for self shrinkers of the mean curvature flow Colding and
Minicozzi \cite{CM} were able to prove weighted $L^2$ estimates for the
second fundamental form, assuming the mean curvature is positive. These
estimates were instrumental in the classification of stable shrinkers. Our
estimates can be viewed as parallel to theirs, however the classification of
gradient Ricci solitons is still a major open question in the field.

A gradient shrinking Ricci soliton is a Riemannian manifold $\left(
M,g\right) $ for which there exists a potential function $f$ such that
\begin{equation}  \label{soliton_eq}
R_{ij}+f_{ij}=\frac 12g_{ij}.
\end{equation}

It can be shown directly from the equation that the quantity $\left| \nabla
f\right| ^2+R-f$ is constant on $M$ hence we can normalize $f$ such that
\begin{equation*}
\left| \nabla f\right| ^2+R=f.
\end{equation*}

Let us denote with $\left| Rc\right| $ and $\left| Rm\right| $ the norms of
the Ricci and Riemann tensors, respectively. We now state the main result of
this paper.

\begin{theorem}
Let $\left( M,g,f\right) $ be a gradient shrinking Ricci soliton with
bounded Ricci curvature. Then the Riemann curvature tensor grows at most
polynomially in the distance function i.e.
\begin{equation*}
\left| Rm\right| \left( x\right) \leq C\left( r\left( x\right) +1\right) ^a,
\end{equation*}
for some constant $a>0.$
\end{theorem}

We apply this theorem to prove a gap result for gradient shrinkers. We show
that if the Ricci curvature is small enough everywhere on $M$ then the
soliton is isometric to the Gaussian soliton $\left( \mathbb{R}^n,dx^2,\frac
14\left| x\right| ^2\right) .$

\begin{corollary}
Let $\left( M,g,f\right) $ be a gradient shrinking Ricci soliton. If $\left|
Rc\right| \leq \frac 1{100n^2}$ on $M$ then $M$ is isometric to the Gaussian
soliton.
\end{corollary}

Yokota \cite{Y} has obtained a gap theorem for gradient shrinking Ricci
solitons, namely he showed that if $\left( 4\pi \right) ^{-\frac
n2}\int_Me^{-f}>1-\varepsilon _n$ for some $\varepsilon _n$ depending on $n$
then $M$ is isometric to the Gaussian soliton.

Another application of our main theorem is in relation to compactness
results for Ricci solitons. This topic has been studied recently in both
compact (\cite{CS, W, Z}) and noncompact (\cite{HM}) settings. We recall a
recent result for complete noncompact shrinkers, due to Haslhofer and
M\"{u}ller \cite{HM}. Let $\left( M_i,g_i,\bar{f}_i\right) $ a sequence of
gradient shrinkers with the potentials $\bar{f}_i$ normalized such that $
\left( 4\pi \right) ^{-\frac n2}\int_{M_i}e^{-\bar{f}_i}=1$. Consider $z_i$
a point where $\bar{f}_i$ attains its minimum. Assume that Perelman's
entropy $\mu _i$ has a uniform bound from below i.e. there exists $\bar{\mu}$
such that
\begin{equation*}
\mu _i:=\left( 4\pi \right) ^{-\frac n2}\int_{M_i}\left( \left| \nabla \bar{f
}_i\right| ^2+R_{g_i}+\bar{f}_i-n\right) e^{-\bar{f}_i}\geq \bar{\mu}.
\end{equation*}
If, moreover, for any $i$ and $r>0$
\begin{equation*}
\int_{B_{z_i}\left( r\right) }\left| Rm_{g_i}\right| ^{n/2}\leq E\left(
r\right) ,
\end{equation*}
then a subsequence of $\left( M_i,g_i,\bar{f}_i,z_i\right) $ converges to an
orbifold gradient shrinker in the pointed Cheeger-Gromov sense.

As a consequence of our main theorem, we can show the following result. Here
it is not important how we normalize the function $f,$ but we need to take $
x_0$ a point where $f$ assumes its minimum value on $M.$ Moreover, here we
also need to take $n\geq 6.$ Notice however that in dimensions 2 and 3
shrinking solitons are completely classified and in dimension 4, Haslhofer
and M\"{u}ller have proved that $\int_{B_{z_i}\left( r\right) }\left|
Rm_{g_i}\right| ^2\leq E\left( r\right) $ using the Gauss-Bonnet theorem.
However, it seems that their argument is more special and works only in
lower dimensions.

\begin{corollary}
Let $\left( M,g,f\right) $ be a gradient shrinker with $dimM=n\geq 6$ and $
\left| Rc\right| \leq K$ on $M$. There exists a $r_0$ depending on $n$ and $K
$ such that if $\int_{B_{x_0}\left( r_0\right) }\left| Rm\right| ^{n/2}\leq L
$ for a minimum point $x_0$ of $f$, then for any $r>0$ we have
\begin{equation*}
\int_{B_{x_0}\left( r\right) }\left| Rm\right| ^{n/2}\leq E\left( r\right) ,
\end{equation*}
where $E$ depends on $n,K$ and $L$.
\end{corollary}

In particular, this Corollary and the main theorem in \cite{HM} implies
compactness of shrinkers assuming Ricci curvature bounds and only local
curvature bounds.

\begin{corollary}
Let $\left( M_i,g_i,\bar{f}_i\right) $ be a sequence of gradient shrinking
Ricci solitons normalized by $\left( 4\pi \right) ^{-\frac n2}\int_{M_i}e^{-
\bar{f}_i}=1.$ Assume $\mu _i\geq \bar{\mu}$ and that we have a uniform
bound on the Ricci curvature $\left| Rc_{g_i}\right| \leq K$. Then there
exists $r_0$ depending on $n$ and $K$ such that if
\begin{equation*}
\sup_i\int_{B_{z_i}\left( r_0\right) }\left| Rm_{g_i}\right| ^{n/2}
\end{equation*}
is finite for $z_i$ a minimum point of $\bar{f}_i$, then a subsequence of $
\left( M_i,g_i,\bar{f}_i,z_i\right) $ converges to an orbifold gradient
shrinker in the pointed Cheeger-Gromov sense.
\end{corollary}

\section{Proof of the curvature estimate}

The idea of the proof of Theorem 1 is the following.

From the Ricci soliton equation \eqref{soliton_eq}, we can estimate $\Delta
_f\left| Rm\right| ^2\geq -c\left| Rm\right| ^3,$ where $\Delta _f=\Delta
-\nabla f\cdot \nabla .$ It is natural therefore to attempt to use Moser
iteration for this problem. Since Ricci is bounded below, the Sobolev
constant is uniformly bounded on arbitrary balls of fixed radius $=1.$
Therefore it is known that the Moser iteration will work if we can control
the $L^p$ norm (for $p>n/2$) of $\left| Rm\right| $ on any ball of radius
one. This is quite technical and it is done in the second Lemma below.

To get $L^p$ estimates, we use the Ricci soliton equation and that the Ricci
curvature is bounded. At the core of our estimates is a formula that relates
the divergence of the Riemann curvature tensor of a Ricci soliton to the
gradient of its Ricci curvature.

We first prove some Lemmas.
Everywhere in this section $\left( M,g,f\right) $
is a gradient shrinking Ricci soliton with bounded Ricci curvature.
We use
the notation:
\begin{eqnarray*}
\left| Rc\right| ^2 &=&\sum \left| R_{ij}\right| ^2, \\
\left| \nabla Rc\right| ^2 &=&\sum \left| \nabla _kR_{ij}\right| ^2, \\
\left| Rm\right| ^2 &=&\sum \left| R_{ijkl}\right| ^2, \\
\left| \nabla Rm\right| ^2 &=&\sum \left| \nabla _hR_{ijkl}\right| ^2.
\end{eqnarray*}

We recall some basic identities for shrinking Ricci solitons, which are
essential in the proof. First, taking trace of the soliton equation we get $
R+\Delta f=\frac n2.$ As we have mentioned above, using the Bianchi
identities and normalizing $f$ we get $\left| \nabla f\right| ^2+R=f.$ This
normalization of $f$ will be assumed throughout the paper. Other formulas
that follow from Bianchi and Ricci identities and the soliton equation are
\begin{eqnarray*}
\nabla _iR &=&2R_{ij}f_j, \\
\nabla _kR_{jk} &=&R_{jk}f_k, \\
\nabla _lR_{ijkl} &=&R_{ijkl}f_l, \\
\nabla _jR_{ki}-\nabla _iR_{kj} &=&R_{ijkl}f_l.
\end{eqnarray*}
We will not prove these here since they are quite standard, see e.g. \cite
{ENM}.

\begin{lemma}
We have:
\begin{eqnarray}
\left| \nabla Rc\right| ^2 &\leq &\frac 12\Delta \left| Rc\right| ^2-\frac
12\nabla f\cdot \nabla \left| Rc\right| ^2+c\left| Rm\right|   \label{f1} \\
\left| \nabla Rm\right| ^2 &\leq &\frac 12\Delta \left| Rm\right| ^2-\frac
12\nabla f\cdot \nabla \left| Rm\right| ^2+c\left| Rm\right| ^3  \label{f2}
\end{eqnarray}
\end{lemma}

Proof. It is known (see \cite{PW}) that
\begin{eqnarray*}
\Delta R_{ij} &=&\nabla f\cdot \nabla R_{ij}+R_{ij}-2R_{ikjh}R_{kh} \\
\Delta Rm &=&\nabla f\cdot \nabla Rm+Rm-2\left( Rm^2+Rm^{\#}\right) .
\end{eqnarray*}
The Lemma follows immediately from here. $\blacksquare $

We now prove the following estimate, which is of independent interest.

\begin{lemma}
For any $p\geq 2$ there exist positive constants $C$ and $a$ such that
\begin{equation*}
\int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\leq C.
\end{equation*}
In particular, there exist positive constants $C$ and $a$ such that for any $
x\in M$ we have:
\begin{equation*}
\int_{B_x\left( 1\right) }\left| Rm\right| ^p\leq C\left( r\left( x\right)
+1\right) ^{2a}.
\end{equation*}
\end{lemma}

Proof. Let us denote by $\rho :=2\sqrt{f}$ and for $r>>1$ let
\begin{equation*}
D\left( r\right) :=\left\{ x\in M:\;\rho \left( x\right) \leq r\right\} .
\end{equation*}
Notice that $D\left( r\right) $ is always compact, in fact approximates well
the geodesic ball of radius $r$ when $r$ is large. Here we recall that $f$
has the following asymptotics, see \cite{CZ}:
\begin{equation*}
\left( \frac 12r\left( x\right) -c\right) ^2\leq f\left( x\right) \leq
\left( \frac 12r\left( x\right) +c\right) ^2,\;\;\text{for}\;\;r\left(
x\right) \geq r_0.
\end{equation*}
Here and below we denote by $r\left( x\right) $ the distance from $x$ to a
fixed point $x_0\in M.$ Let us point out moreover that if $x_0$ is chosen to
be a minimum point of $f$ then $c$ and $r_0$ will depend only on $n,$ see
\cite{HM}.

We take the following cut-off
\begin{equation*}
\phi =\left\{
\begin{array}{c}
\frac 1{r^2}\left( \frac 14r^2-f\left( x\right) \right) \\
0
\end{array}
\right.
\begin{array}{l}
\text{if} \\
\text{if}
\end{array}
\begin{array}{l}
x\in D\left( r\right) \\
x\in M\backslash D\left( r\right)
\end{array}
\end{equation*}

Since the Ricci curvature is bounded on $M$, let us set
\begin{equation*}
K=\sup_M\left| Rc\right| .
\end{equation*}

Let $a$ be a fixed number to be determined later, depending on $n,K$ and $p.$
Consider also $q$ a large enough integer, $q\geq 2p+1$. We will discuss
first the case when $p\geq 3,$ the case $2\leq p<3$ will follow immediately
by H\"{o}lder's inequality.

We have, integrating by parts, that
\begin{gather}
a\int_M\left| Rm\right| ^p\left| \nabla f\right| ^2\left( f+1\right)
^{-a-1}\phi ^q=-\int_M\left| Rm\right| ^p\nabla f\cdot \nabla \left(
f+1\right) ^{-a}\phi ^q  \label{a} \\
=\int_M\left| Rm\right| ^p\left( \Delta f\right) \left( f+1\right) ^{-a}\phi
^q+\int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\nabla f\cdot \nabla \phi
^q  \notag \\
+\int_M\left( \nabla \left| Rm\right| ^p\cdot \nabla f\right) \left(
f+1\right) ^{-a}\phi ^q.  \notag
\end{gather}
Let us explain why we take these functions in (\ref{a}). We take $\left|
\nabla f\right| ^2$ because we want to use integration by parts and the
symmetries of the Riemann curvature tensor (which are implied by the soliton
equation). For shrinking solitons the factor $e^{-f}$ seems much more
convenient than $\left( f+1\right) ^{-a},$ however the former factor only
gives exponential growth control. So to prove weighted $L^2$ estimates for
the Riemann curvature it is easier to start from $\int_M\left| Rm\right|
^p\left| \nabla f\right| ^2e^{-f}\phi ^q$ and carry all the estimates below,
with some useful simplifications, but in order to prove Lemma 2 we need to
use the less natural weight $\left( f+1\right) ^{-a},$ at the expense of
more complicated computations.

Everywhere in this section $c$ will denote a constant that depends on $n$, $
p $, $q$ and $K$ but not on $a$. We will use $C_1,$ $C_2,...$ etc. to denote
finite constants that have a more complicated dependence, such as on $
\sup_\Omega \left| Rm\right| $ over some compact set $\Omega .$ However, we
stress that all the constants $c$ or $C_1,C_2...$ are independent of $r.$

We now check that:
\begin{equation*}
a\left| \nabla f\right| ^2\left( f+1\right) ^{-a-1}-\Delta f\left(
f+1\right) ^{-a}=\left( a\frac{f-R}{f+1}-\left( \frac n2-R\right) \right)
\left( f+1\right) ^{-a}.
\end{equation*}
Furthermore, since $R\geq 0$ on any gradient Ricci soliton ( \cite{C, Ca})
we see that there exists a constant $r_1$ depending on $n,K$ and $a$ (e.g., $
r_1=\sqrt{8a\left( K+1\right) }$ ) such that on $M\backslash D\left(
r_1\right) $ we have
\begin{equation*}
a\left| \nabla f\right| ^2\left( f+1\right) ^{-a-1}-\Delta f\left(
f+1\right) ^{-a}\geq \left( a-n\right) \left( f+1\right) ^{-a}.
\end{equation*}

Notice also that by the choice of cut-off we have:
\begin{equation*}
\nabla f\cdot \nabla \phi ^q=-\frac q{r^2}\phi ^{q-1}\left| \nabla f\right|
^2\leq 0.
\end{equation*}
Using these simple estimates in (\ref{a}) it follows that
\begin{equation}
\left( a-n\right) \int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\phi
^q\leq \int_M\left( \nabla \left| Rm\right| ^p\cdot \nabla f\right) \;\left(
f+1\right) ^{-a}\phi ^q+C_1.  \label{b}
\end{equation}
Here we have set
\begin{equation*}
C_1:=\int_{D\left( r_1\right) }\left( -a\left| \nabla f\right| ^2\left(
f+1\right) ^{-1}+\Delta f+a-n\right) \left| Rm\right| ^p\left( f+1\right)
^{-a}\phi ^q.
\end{equation*}

Let us compute, using the Bianchi identities:
\begin{eqnarray*}
\nabla \left| Rm\right| ^2\cdot \nabla f &=&2f_h\left( \nabla
_hR_{ijkl}\right) R_{ijkl} \\
&=&4f_h\left( \nabla _lR_{ijkh}\right) R_{ijkl}
\end{eqnarray*}
Therefore, using this in the right hand side of (\ref{b}) we get
\begin{gather*}
\int_M\left( \nabla \left| Rm\right| ^p\cdot \nabla f\right) \;\left(
f+1\right) ^{-a}\phi ^q \\
=2p\int_Mf_h\left( \nabla _lR_{ijkh}\right) R_{ijkl}\left| Rm\right|
^{p-2}\left( f+1\right) ^{-a}\phi ^q \\
=-2p\int_MR_{ijkh}\nabla _l\left( f_hR_{ijkl}\left| Rm\right| ^{p-2}\left(
f+1\right) ^{-a}\phi ^q\right)  \\
=-2p\int_MR_{ijkh}f_{hl}R_{ijkl}\left| Rm\right| ^{p-2}\left( f+1\right)
^{-a}\phi ^q \\
-2p\int_MR_{ijkh}f_h\left( \nabla _lR_{ijkl}\right) \left| Rm\right|
^{p-2}\left( f+1\right) ^{-a}\phi ^q \\
-2p\int_MR_{ijkh}f_hR_{ijkl}\left( \nabla _l\left| Rm\right| ^{p-2}\right)
\left( f+1\right) ^{-a}\phi ^q \\
+2ap\int_MR_{ijkh}f_hR_{ijkl}f_l\left| Rm\right| ^{p-2}\left( f+1\right)
^{-a-1}\phi ^q \\
-2pq\int_MR_{ijkh}f_hR_{ijkl}\phi _l\left| Rm\right| ^{p-2}\left( f+1\right)
^{-a}\phi ^{q-1} \\
=I+II+III+IV+V.
\end{gather*}
It is easy to see that since the Ricci curvature is bounded,
\begin{equation*}
I=-2p\int_MR_{ijkh}f_{hl}R_{ijkl}\left| Rm\right| ^{p-2}\left( f+1\right)
^{-a}\phi ^q\leq c\int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\phi ^q.
\end{equation*}
Furthermore, using that for a gradient shrinker we have (see \cite{ENM, MS})
\begin{equation*}
\nabla _lR_{ijkl}=R_{ijkl}f_l
\end{equation*}
we see that
\begin{equation*}
II+IV=-2p\int_MR_{ijkh}f_hR_{ijkl}f_l\left| Rm\right| ^{p-2}\left( 1-\frac
a{f+1}\right) \left( f+1\right) ^{-a}\phi ^q\leq C_2,
\end{equation*}
where we have set
\begin{equation*}
C_2:=2p\int_{D\left( 2\sqrt{a-1}\right) }\left| R_{ijkh}f_h\right| ^2\left|
Rm\right| ^{p-2}\left( \frac a{f+1}-1\right) \left( f+1\right) ^{-a}\phi ^q.
\end{equation*}
The estimate above follows because if $f\left( x\right) +1>a$ then the
integral is negative. Let us point out however that in fact each of $\left|
II\right| $ and $\left| IV\right| $ can be estimated by a similar argument
as in the proof of inequality (\ref{f}) below. Clearly, since $a$ is
independent of $r$, so are the constants $C_1$ and $C_2$ obtained so far.

We use the above estimates in (\ref{b}), and get that
\begin{gather}
\left( a-c\right) \int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\phi ^q
\label{c} \\
\leq -2p\int_MR_{ijkh}f_hR_{ijkl}\left( \nabla _l\left| Rm\right|
^{p-2}\right) \left( f+1\right) ^{-a}\phi ^q  \notag \\
+\frac{2pq}{r^2}\int_M\left| R_{ijkh}f_h\right| ^2\left| Rm\right|
^{p-2}\left( f+1\right) ^{-a}\phi ^{q-1}+C_1+C_2.  \notag
\end{gather}
Recall that for Ricci solitons we have (see e.g. \cite{ENM, MS})
\begin{equation*}
R_{ijkh}f_h=\nabla _jR_{ik}-\nabla _iR_{kj},
\end{equation*}
so that we can estimate the first term in (\ref{c}) by
\begin{gather}
-2p\int_MR_{ijkh}f_hR_{ijkl}\left( \nabla _l\left| Rm\right| ^{p-2}\right)
\left( f+1\right) ^{-a}\phi ^q  \label{d} \\
\leq c\int_M\left| \nabla Rc\right| \left| \nabla Rm\right| \left| Rm\right|
^{p-2}\left( f+1\right) ^{-a}\phi ^q  \notag \\
\leq c\int_M\left| \nabla Rc\right| ^2\left| Rm\right| ^{p-1}\left(
f+1\right) ^{-a}\phi ^q+c\int_M\left| \nabla Rm\right| ^2\left| Rm\right|
^{p-3}\left( f+1\right) ^{-a}\phi ^q.  \notag
\end{gather}
We now work on the second term in (\ref{c}), and we will use below some
interpolations which are important throughout the rest of the proof.

\begin{gather}
\frac 1{r^2}\int_MR_{ijkh}f_hR_{ijkl}f_l\left| Rm\right| ^{p-2}\left(
f+1\right) ^{-a}\phi ^{q-1}  \notag \\
=\frac 2{r^2}\int_M\left( \nabla _jR_{ik}\right) R_{ijkl}f_l\left| Rm\right|
^{p-2}\left( f+1\right) ^{-a}\phi ^{q-1}  \notag \\
=-\frac 2{r^2}\int_MR_{ik}\nabla _j\left( R_{ijkl}f_l\left| Rm\right|
^{p-2}\left( f+1\right) ^{-a}\phi ^{q-1}\right)  \notag \\
=-\frac 2{r^2}\int_MR_{ik}f_{lj}R_{ijkl}\left| Rm\right| ^{p-2}\left(
f+1\right) ^{-a}\phi ^{q-1}  \notag \\
-\frac 2{r^2}\int_MR_{ik}f_l\left( \nabla _jR_{ijkl}\right) \left| Rm\right|
^{p-2}\left( f+1\right) ^{-a}\phi ^{q-1}  \notag \\
-\frac 2{r^2}\int_MR_{ik}R_{ijkl}f_l\left( \nabla _j\left| Rm\right|
^{p-2}\right) \left( f+1\right) ^{-a}\phi ^{q-1}  \notag \\
+\frac{2a}{r^2}\int_MR_{ik}R_{ijkl}f_lf_j\left| Rm\right| ^{p-2}\left(
f+1\right) ^{-a-1}\phi ^{q-1}  \notag \\
-\frac{2(q-1)}{r^2}\int_MR_{ik}R_{ijkl}f_l\phi _j\left| Rm\right|
^{p-2}\left( f+1\right) ^{-a}\phi ^{q-2}  \notag \\
=I+II+III+IV+V.  \label{e}
\end{gather}
Since the Ricci curvature is bounded, we get
\begin{eqnarray*}
I &=&-\frac 2{r^2}\int_MR_{ik}f_{lj}R_{ijkl}\left| Rm\right| ^{p-2}\left(
f+1\right) ^{-a}\phi ^{q-1} \\
&\leq &\frac c{r^2}\int_M\left| Rm\right| ^{p-1}\left( f+1\right) ^{-a}\phi
^{q-1}.
\end{eqnarray*}
Using again that $\nabla _lR_{ijkl}=R_{ijkl}f_l$ and that $\left| \nabla
f\right| ^2\leq f\leq \frac 14r^2$ on $D\left( r\right) ,$ it follows
\begin{equation*}
II\leq c\int_M\left| Rm\right| ^{p-1}\left( f+1\right) ^{-a}\phi ^{q-1}.
\end{equation*}
Furthermore,
\begin{eqnarray*}
III &=&-\frac 2{r^2}\int_MR_{ik}R_{ijkl}f_l\left( \nabla _j\left| Rm\right|
^{p-2}\right) \left( f+1\right) ^{-a}\phi ^{q-1} \\
&\leq &c\int_M\left| \nabla Rm\right| \left| Rm\right| ^{p-2}\left(
f+1\right) ^{-a}\phi ^{q-1} \\
&\leq &c\int_M\left| \nabla Rm\right| ^2\left| Rm\right| ^{p-3}\left(
f+1\right) ^{-a}\phi ^q+c\int_M\left| Rm\right| ^{p-1}\left( f+1\right)
^{-a}\phi ^{q-2}.
\end{eqnarray*}
Similarly,
\begin{eqnarray*}
IV &=&\frac{2a}{r^2}\int_MR_{ik}R_{ijkl}f_lf_j\left| Rm\right| ^{p-2}\left(
f+1\right) ^{-a-1}\phi ^{q-1} \\
&\leq &\frac{ca}{r^2}\int_M\left| Rm\right| ^{p-1}\left( f+1\right)
^{-a}\phi ^{q-1} \\
&\leq &c\int_M\left| Rm\right| ^{p-1}\left( f+1\right) ^{-a}\phi ^{q-1},
\end{eqnarray*}
by taking $r\geq \sqrt{a}.$ Finally, we also have
\begin{equation*}
V\leq \frac c{r^2}\int_M\left| Rm\right| ^{p-1}\left( f+1\right) ^{-a}\phi
^{q-2}.
\end{equation*}
Using these estimates in (\ref{e}) we get
\begin{gather}
\frac{2pq}{r^2}\int_M\left| R_{ijkh}f_h\right| ^2\left| Rm\right|
^{p-2}\left( f+1\right) ^{-a}\phi ^{q-1}  \label{f} \\
\leq c\int_M\left| \nabla Rm\right| ^2\left| Rm\right| ^{p-3}\left(
f+1\right) ^{-a}\phi ^q  \notag \\
+c\int_M\left| Rm\right| ^{p-1}\left( f+1\right) ^{-a}\phi ^{q-2}.  \notag
\end{gather}
Notice moreover that we can interpolate, using Young's inequality:
\begin{gather}
\int_M\left| Rm\right| ^{p-1}\left( f+1\right) ^{-a}\phi ^{q-2}\leq
\int_M\left| Rm\right| ^{p-1}\phi ^{q\frac{p-1}p}\phi ^{\frac qp-2}\left(
f+1\right) ^{-a}  \notag \\
\leq \varepsilon \int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\phi
^q+c\left( \varepsilon \right) \int_M\left( f+1\right) ^{-a}\phi ^{q-2p}.
\label{g}
\end{gather}
We want to use this in (\ref{f}), hence here we can take $\varepsilon =1.$
Moreover, to guarantee that $\phi ^{q-2p}$ is well defined, we take $q\geq
2p+1$.

Plug (\ref{g}) in (\ref{f}) and then use (\ref{f}) and (\ref{d}) in (\ref{c}
); it results that

\begin{gather}
\left( a-c\right) \int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\phi
^q\leq c\int_M\left| \nabla Rc\right| ^2\left| Rm\right| ^{p-1}\left(
f+1\right) ^{-a}\phi ^q  \label{h} \\
+c\int_M\left| \nabla Rm\right| ^2\left| Rm\right| ^{p-3}\left( f+1\right)
^{-a}\phi ^q+C_1+C_2+C_0.  \notag
\end{gather}
We have denoted with
\begin{equation*}
C_0:=c\int_M\left( f+1\right) ^{-a},
\end{equation*}
and observe that taking $q=2p+1$ and taking $a$ such that $a>\frac n2+1$
then $C_0$ is a finite constant, independent of $r$. Indeed it is known that
the volume growth of $M$ is polynomial, see \cite{CZ}.

We finish the proof by estimating each of the two terms in the right hand
side of (\ref{h}). Start with the first, which by (\ref{f1}) we have:
\begin{gather}
2\int_M\left| \nabla Rc\right| ^2\left| Rm\right| ^{p-1}\left( f+1\right)
^{-a}\phi ^q\leq \int_M\left( \Delta \left| Rc\right| ^2\right) \left|
Rm\right| ^{p-1}\left( f+1\right) ^{-a}\phi ^q  \label{i} \\
-\int_M\left( \nabla f\cdot \nabla \left| Rc\right| ^2\right) \left|
Rm\right| ^{p-1}\left( f+1\right) ^{-a}\phi ^q+c\int_M\left| Rm\right|
^p\left( f+1\right) ^{-a}\phi ^q  \notag
\end{gather}
Let us observe that
\begin{gather*}
\int_M\left( \Delta \left| Rc\right| ^2\right) \left| Rm\right| ^{p-1}\left(
f+1\right) ^{-a}\phi ^q \\
=-\int_M\nabla \left| Rc\right| ^2\cdot \nabla \left( \left| Rm\right|
^{p-1}\left( f+1\right) ^{-a}\phi ^q\right) \\
=-\int_M\nabla \left| Rc\right| ^2\cdot \nabla \left| Rm\right| ^{p-1}\left(
f+1\right) ^{-a}\phi ^q \\
+a\int_M\left( \nabla \left| Rc\right| ^2\cdot \nabla f\right) \left|
Rm\right| ^{p-1}\left( f+1\right) ^{-a-1}\phi ^q \\
-q\int_M\left( \nabla \left| Rc\right| ^2\cdot \nabla \phi \right) \left|
Rm\right| ^{p-1}\left( f+1\right) ^{-a}\phi ^{q-1} \\
\leq c\int_M\left| \nabla Rc\right| \left| \nabla Rm\right| \left| Rm\right|
^{p-2}\left( f+1\right) ^{-a}\phi ^q \\
+ca\int_M\left| \nabla Rc\right| \left| Rm\right| ^{p-1}\left( f+1\right)
^{-a}\phi ^q \\
+\frac cr\int_M\left| \nabla Rc\right| \left| Rm\right| ^{p-1}\left(
f+1\right) ^{-a}\phi ^{q-1} \\
\leq c\int_M\left| \nabla Rc\right| \left| \nabla Rm\right| \left| Rm\right|
^{p-2}\left( f+1\right) ^{-a}\phi ^q \\
+ca\int_M\left| \nabla Rc\right| \left| Rm\right| ^{p-1}\left( f+1\right)
^{-a}\phi ^{q-1}.
\end{gather*}
Furthermore, let us use that
\begin{eqnarray*}
&&c\int_M\left| \nabla Rc\right| \left| \nabla Rm\right| \left| Rm\right|
^{p-2}\left( f+1\right) ^{-a}\phi ^q \\
&\leq &\frac 14\int_M\left| \nabla Rc\right| ^2\left| Rm\right| ^{p-1}\left(
f+1\right) ^{-a}\phi ^q \\
&&+c\int_M\left| \nabla Rm\right| ^2\left| Rm\right| ^{p-3}\left( f+1\right)
^{-a}\phi ^q,
\end{eqnarray*}
and, similarly,
\begin{eqnarray*}
&&ca\int_M\left| \nabla Rc\right| \left| Rm\right| ^{p-1}\left( f+1\right)
^{-a}\phi ^{q-1} \\
&\leq &\frac 14\int_M\left| \nabla Rc\right| ^2\left| Rm\right| ^{p-1}\left(
f+1\right) ^{-a}\phi ^q \\
&&+ca^2\int_M\left| Rm\right| ^{p-1}\left( f+1\right) ^{-a}\phi ^{q-2}.
\end{eqnarray*}
We conclude from above that
\begin{eqnarray*}
&&\int_M\left( \Delta \left| Rc\right| ^2\right) \left| Rm\right|
^{p-1}\left( f+1\right) ^{-a}\phi ^q \\
&\leq &\frac 12\int_M\left| \nabla Rc\right| ^2\left| Rm\right| ^{p-1}\left(
f+1\right) ^{-a}\phi ^q+c\int_M\left| \nabla Rm\right| ^2\left| Rm\right|
^{p-3}\left( f+1\right) ^{-a}\phi ^q \\
&&+ca^2\int_M\left| Rm\right| ^{p-1}\left( f+1\right) ^{-a}\phi ^{q-2}.
\end{eqnarray*}
Moreover,

\begin{gather*}
-\int_M\left( \nabla f\cdot \nabla \left| Rc\right| ^2\right) \left|
Rm\right| ^{p-1}\left( f+1\right) ^{-a}\phi ^q \\
\leq c\int_M\left| \nabla Rc\right| \left| Rm\right| ^{p-1}\left| \nabla
f\right| \left( f+1\right) ^{-a}\phi ^q \\
\leq \frac 12\int_M\left| \nabla Rc\right| ^2\left| Rm\right| ^{p-1}\left(
f+1\right) ^{-a}\phi ^q+c\int_M\left| Rm\right| ^{p-1}\left| \nabla f\right|
^2\left( f+1\right) ^{-a}\phi ^q.
\end{gather*}
We can use the same idea as in (\ref{g}) to bound:
\begin{eqnarray*}
&&\int_M\left| Rm\right| ^{p-1}\left| \nabla f\right| ^2\left( f+1\right)
^{-a}\phi ^q \\
&\leq &c\int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\phi
^q+c\int_M\left( f+1\right) ^{-a}\left| \nabla f\right| ^{2p}\phi ^q.
\end{eqnarray*}
Once again, we can take $a$ large enough e.g. $a>\frac n2+p+1$ so that
\begin{equation*}
\int_M\left( f+1\right) ^{-a}\left| \nabla f\right| ^{2p}<\int_M\left(
f+1\right) ^{-a+p}<\infty .
\end{equation*}
We also use (\ref{g}) for $\varepsilon =\frac 1{a^2}$ to get
\begin{eqnarray*}
&&ca^2\int_M\left| Rm\right| ^{p-1}\left( f+1\right) ^{-a}\phi ^{q-2} \\
&\leq &c\int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\phi
^q+ca^{2p}\int_M\left( f+1\right) ^{-a}\phi ^{q-2p}.
\end{eqnarray*}
Therefore, plugging all in (\ref{i}) gives
\begin{gather*}
\int_M\left| \nabla Rc\right| ^2\left| Rm\right| ^{p-1}\left( f+1\right)
^{-a}\phi ^q\leq c\int_M\left| \nabla Rm\right| ^2\left| Rm\right|
^{p-3}\left( f+1\right) ^{-a}\phi ^q \\
+c\int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\phi
^q+ca^{2p}\int_M\left( f+1\right) ^{-a+p},
\end{gather*}
which by (\ref{h}) yields

\begin{equation}
\left( a-c\right) \int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\phi
^q\leq c\int_M\left| \nabla Rm\right| ^2\left| Rm\right| ^{p-3}\left(
f+1\right) ^{-a}\phi ^q+C,  \label{k}
\end{equation}
where
\begin{eqnarray*}
C &:&=C_1+C_2+C_3, \\
C_3 &:&=ca^{2p}\int_M\left( f+1\right) ^{-a+p}.
\end{eqnarray*}
Finally, let us use (\ref{f2}) to check that
\begin{gather}
2\int_M\left| \nabla Rm\right| ^2\left| Rm\right| ^{p-3}\left( f+1\right)
^{-a}\phi ^q\leq \int_M\left( \Delta \left| Rm\right| ^2\right) \left|
Rm\right| ^{p-3}\left( f+1\right) ^{-a}\phi ^q  \label{l} \\
-\int_M\left( \nabla f\cdot \nabla \left| Rm\right| ^2\right) \left|
Rm\right| ^{p-3}\left( f+1\right) ^{-a}\phi ^q+c\int_M\left| Rm\right|
^p\left( f+1\right) ^{-a}\phi ^q.  \notag
\end{gather}
The first term in the formula above is
\begin{gather*}
\int_M\left( \Delta \left| Rm\right| ^2\right) \left| Rm\right| ^{p-3}\left(
f+1\right) ^{-a}\phi ^q \\
=-\int_M\left( \nabla \left| Rm\right| ^2\cdot \nabla \left| Rm\right|
^{p-3}\right) \left( f+1\right) ^{-a}\phi ^q \\
+a\int_M\left( \nabla \left| Rm\right| ^2\cdot \nabla f\right) \left|
Rm\right| ^{p-3}\left( f+1\right) ^{-a-1}\phi ^q \\
-q\int_M\left( \nabla \left| Rm\right| ^2\cdot \nabla \phi \right) \left|
Rm\right| ^{p-3}\left( f+1\right) ^{-a}\phi ^{q-1}.
\end{gather*}
Choosing $p\geq 3$ guarantees that

\begin{equation*}
-\int_M\left( \nabla \left| Rm\right| ^2\cdot \nabla \left| Rm\right|
^{p-3}\right) \left( f+1\right) ^{-a}\phi ^q\leq 0.
\end{equation*}
On the other hand,
\begin{gather*}
a\int_M\left( \nabla \left| Rm\right| ^2\cdot \nabla f\right) \left|
Rm\right| ^{p-3}\left( f+1\right) ^{-a-1}\phi ^q \\
\leq ac\int_M\left| \nabla Rm\right| \left| Rm\right| ^{p-2}\left(
f+1\right) ^{-a}\phi ^q \\
\leq \frac 14\int_M\left| \nabla Rm\right| ^2\left| Rm\right| ^{p-3}\left(
f+1\right) ^{-a}\phi ^q+ca^2\int_M\left| Rm\right| ^{p-1}\left( f+1\right)
^{-a}\phi ^q.
\end{gather*}
Similarly we find:

\begin{gather*}
-q\int_M\left( \nabla \left| Rm\right| ^2\cdot \nabla \phi \right) \left|
Rm\right| ^{p-3}\left( f+1\right) ^{-a}\phi ^{q-1} \\
\leq \frac 14\int_M\left| \nabla Rm\right| ^2\left| Rm\right| ^{p-3}\left(
f+1\right) ^{-a}\phi ^q \\
+c\int_M\left| Rm\right| ^{p-1}\left( f+1\right) ^{-a}\phi ^{q-2}.
\end{gather*}
Notice moreover that
\begin{gather*}
-\int_M\left( \nabla f\cdot \nabla \left| Rm\right| ^2\right) \left|
Rm\right| ^{p-3}\left( f+1\right) ^{-a}\phi ^q \\
\leq 2\int_M\left| \nabla Rm\right| \left| \nabla f\right| \left| Rm\right|
^{p-2}\left( f+1\right) ^{-a}\phi ^q \\
\leq \frac 12\int_M\left| \nabla Rm\right| ^2\left| Rm\right| ^{p-3}\left(
f+1\right) ^{-a}\phi ^q+c\int_M\left| Rm\right| ^{p-1}\left| \nabla f\right|
^2\left( f+1\right) ^{-a}\phi ^q.
\end{gather*}
Using these estimates in (\ref{l}) we get:
\begin{gather*}
\int_M\left| \nabla Rm\right| ^2\left| Rm\right| ^{p-3}\left( f+1\right)
^{-a}\phi ^q\leq c\int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\phi ^q \\
+ca^2\int_M\left| Rm\right| ^{p-1}\left( f+1\right) ^{-a+1}\phi ^{q-2}.
\end{gather*}
From interpolation as in formula (\ref{g}), for $\varepsilon =\frac 1{a^2},$
it follows:
\begin{eqnarray*}
&&ca^2\int_M\left| Rm\right| ^{p-1}\left( f+1\right) ^{-a+1}\phi ^{q-2} \\
&\leq &c\int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\phi
^q+ca^{2p}\int_M\left( f+1\right) ^{-a+p}\phi ^{q-2p}.
\end{eqnarray*}
Therefore, we have proved:
\begin{equation*}
\int_M\left| \nabla Rm\right| ^2\left| Rm\right| ^{p-3}\left( f+1\right)
^{-a}\phi ^q\leq c\int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\phi
^q+ca^{2p}\int_M\left( f+1\right) ^{-a+p}.
\end{equation*}
This, by (\ref{k}) implies that
\begin{equation*}
\left( a-c\right) \int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\phi
^q\leq C.
\end{equation*}
Recall that $c$ is a constant depending only on $n,p$ and $K$ while $a$ is a
sufficiently large arbitrary number. This shows that there exists $a$
depending on $n,p$ and $K$ such that
\begin{equation*}
\int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\phi ^q\leq C.
\end{equation*}
To conclude the proof of the Lemma notice that if
\begin{equation*}
\rho \left( x\right) \leq \frac 12r,
\end{equation*}
then
\begin{equation*}
\phi \left( x\right) \geq \frac 3{16},
\end{equation*}
hence this shows that
\begin{equation*}
\int_{D\left( \frac 12r\right) }\left| Rm\right| ^p\left( f+1\right)
^{-a}\leq C.
\end{equation*}
Since $C$ is independent of $r,$ making $r\rightarrow \infty $ we get
\begin{equation*}
\int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\leq C.
\end{equation*}
Moreover, from $C=C_1+C_2+C_3$ and the expressions for these constants we
see that in fact we have the estimate
\begin{equation*}
\int_M\left| Rm\right| ^p\left( f+1\right) ^{-a}\leq c\int_{D\left(
r_0\right) }\left| Rm\right| ^p\left( f+1\right) ^{-a}+c\int_M\left(
f+1\right) ^{-a+p},
\end{equation*}
where $r_0$ is a fixed number, depending on $n,p$ and $K.$

This proves the Lemma in the case $p\geq 3$. We can interpolate as in (\ref
{g}) to get the claim for any $p\geq 2.$ The claim that $\int_{B_x\left(
1\right) }\left| Rm\right| ^p\leq C\left( r\left( x\right) +1\right) ^{2a}$
follows immediately from the asymptotics of $f,$ see above.  $\blacksquare $

We are now ready to finish the proof of the Theorem. We will be brief here,
since this part of the proof is standard.

From (\ref{f2}) we infer that
\begin{eqnarray*}
\Delta \left| Rm\right| ^2 &\geq &2\left| \nabla Rm\right| ^2+\nabla f\cdot
\nabla \left| Rm\right| ^2-c\left| Rm\right| ^3 \\
&\geq &-c\left( \left| Rm\right| +\left| \nabla f\right| ^2\right) \left|
Rm\right| ^2.
\end{eqnarray*}
Therefore, if we denote
\begin{equation*}
u:=c\left( \left| Rm\right| +\left| \nabla f\right| ^2\right) ,
\end{equation*}
then
\begin{equation*}
\Delta \left| Rm\right| ^2\geq -u\left| Rm\right| ^2.
\end{equation*}
Since we assumed the Ricci curvature is bounded below, there exists a
uniform bound on the Sobolev constant of the ball $B_x\left( 1\right) .$
More exactly, for any $\varphi $ with support in $B_x\left( 1\right) $ we
have:
\begin{equation*}
\left( \int_{B_x\left( 1\right) }\varphi ^{\frac{2n}{n-2}}\right) ^{\frac{n-2
}n}\leq C_S\int_{B_x\left( 1\right) }\left( \left| \nabla \varphi \right|
^2+\varphi ^2\right) .
\end{equation*}
Indeed, we can use the Sobolev inequality in \cite{SC} and notice that we
can control the volume of $B_x\left( 1\right) $ from below uniformly in $x$
since $M$ satisfies a log Sobolev inequality and the Ricci curvature is
bounded, see \cite{CN}. The constant $C_S$ depends only on $n$, the Ricci
curvature lower bound on $M$ and on $\inf_{x\in M}vol\left( B_x\left(
1\right) \right) .$ As proved by Carillo and Ni in \cite{CN}, $vol\left(
B_x\left( 1\right) \right) \geq k>0,$ where $k$ depends on $K=\sup_M\left|
Rc\right| $ and on Perelman's $\mu $ invariant. We remark that with our
normalization of $f,$ the $\mu $ invariant can be computed as $\mu =\log
\left( \left( 4\pi \right) ^{\frac n2}\int_Me^{-f}\right) <\infty .$

Then the standard Moser iteration, see \cite{L}, implies that
\begin{equation*}
\left| Rm\right| ^2\left( x\right) \leq A\int_{B_x\left( 1\right) }\left|
Rm\right| ^2,
\end{equation*}
where
\begin{equation*}
A:=\bar{C}\left( \int_{B_x\left( 1\right) }u^n+1\right) ,
\end{equation*}
for a constant $\bar{C}$ depending only on $n$ and $C_S.$ Since we showed in
Lemma 2 that $\int_{B_x\left( 1\right) }u^p$ grows at most polynomially in $
r\left( x\right) ,$ the theorem follows from here.$\blacksquare $

\section{Gap and compactness theorems}

In this section we prove the gap theorem and the compactness theorem of
shrinking Ricci solitons, based on the estimates proved above. Since
everywhere in this section the Ricci curvature is bounded, we can apply
Theorem 1 to see that $\left| Rm\right| \left( x\right) \leq C\left(
1+r\left( x\right) \right) ^{2a}.$ We follow a similar argument as in
Theorem 1, this time using the weight $e^{-f}$ and paying more attention to
the dependence on the Ricci curvature bound. The computations will be
simpler, and we will use many times the identities
\begin{eqnarray*}
\nabla _l\left( R_{ijkl}e^{-f}\right) &=&0, \\
\nabla _j\left( R_{ij}e^{-f}\right) &=&0.
\end{eqnarray*}
Clearly, we do not need to use a cut-off here, since all the curvature terms
will be integrable with respect to $e^{-f},$ by Theorem 1. As in the
previous section, we take
\begin{equation*}
K=\sup_M\left| Rc\right| .
\end{equation*}
We also assume that $K>0,$ since otherwise there is nothing to prove. We use
the notation $\Delta _f=\Delta -\nabla f\cdot \nabla $ and note that $\Delta
_f$ is self adjoint with respect to the weighted volume $e^{-f}dv.$

Since
\begin{equation*}
\Delta _f\left( f\right) =\Delta f-\left| \nabla f\right| ^2=\frac n2-f,
\end{equation*}
it follows that
\begin{gather}
\int_M\left| Rm\right| ^p\left( f-\frac n2\right) e^{-f}=-\int_M\left|
Rm\right| ^p\Delta _f\left( f\right) e^{-f}=\int_M\nabla f\cdot \nabla
\left| Rm\right| ^pe^{-f}  \label{1} \\
=p\int_M\nabla _hR_{ijkl}R_{ijkl}f_h\left| Rm\right|
^{p-2}e^{-f}=2p\int_M\nabla _lR_{ijkh}R_{ijkl}f_h\left| Rm\right|
^{p-2}e^{-f}  \notag \\
=-2p\int_MR_{ijkh}f_{hl}R_{ijkl}\left| Rm\right|
^{p-2}e^{-f}-2p\int_MR_{ijkh}f_hR_{ijkl}\nabla _l\left( \left| Rm\right|
^{p-2}\right) e^{-f}.  \notag
\end{gather}
As in Theorem 1, we take $p\geq 3.$ Using the soliton equation and the Ricci
curvature bound $\left| Rc\right| \leq K,$ we find that
\begin{eqnarray*}
R_{ijkh}f_{hl}R_{ijkl} &=&\frac 12\left| Rm\right| ^2-R_{ijkh}R_{ijkl}R_{hl}
\\
&\geq &\left( \frac 12-K\right) \left| Rm\right| ^2.
\end{eqnarray*}
We can also estimate
\begin{gather*}
-2p\int_MR_{ijkh}f_hR_{ijkl}\nabla _l\left( \left| Rm\right| ^{p-2}\right)
e^{-f}=-4p\int_M\left( \nabla _jR_{ik}\right) R_{ijkl}\nabla _l\left( \left|
Rm\right| ^{p-2}\right) e^{-f} \\
\leq 4p\left( p-2\right) \int_M\left| \nabla Rc\right| \left| \nabla
Rm\right| \left| Rm\right| ^{p-2}e^{-f}.
\end{gather*}
Then (\ref{1}) shows that
\begin{equation}
\int_M\left( f-\frac n2+p\left( 1-2K\right) \right) \left| Rm\right|
^pe^{-f}\leq 4p^2\int_M\left| \nabla Rc\right| \left| \nabla Rm\right|
\left| Rm\right| ^{p-2}e^{-f}.  \label{2}
\end{equation}
We estimate the right hand side of (\ref{2}) as follows:
\begin{gather}
2\int_M\left| \nabla Rc\right| \left| \nabla Rm\right| \left| Rm\right|
^{p-2}e^{-f}\leq \frac 1{pK}\int_M\left| \nabla Rc\right| ^2\left| Rm\right|
^{p-1}e^{-f}  \label{2b} \\
+pK\int_M\left| \nabla Rm\right| ^2\left| Rm\right| ^{p-3}e^{-f}.  \notag
\end{gather}
Furthermore, we have
\begin{eqnarray*}
\Delta _f\left| Rc\right| ^2 &=&2\left| \nabla Rc\right| ^2+2\left|
Rc\right| ^2-4R_{ikjh}R_{ij}R_{kh} \\
&\geq &2\left| \nabla Rc\right| ^2-4K^2\left| Rm\right| .
\end{eqnarray*}
Consequently,
\begin{gather*}
\int_M\left| \nabla Rc\right| ^2\left| Rm\right| ^{p-1}e^{-f}\leq \frac
12\int_M\left( \Delta _f\left| Rc\right| ^2\right) \left| Rm\right|
^{p-1}e^{-f}+2K^2\int_M\left| Rm\right| ^pe^{-f} \\
=-\frac 12\int_M\nabla \left| Rc\right| ^2\cdot \nabla \left| Rm\right|
^{p-1}e^{-f}+2K^2\int_M\left| Rm\right| ^pe^{-f} \\
\leq \left( p-1\right) K\int_M\left| \nabla Rc\right| \left| \nabla
Rm\right| \left| Rm\right| ^{p-2}e^{-f}+2K^2\int_M\left| Rm\right| ^pe^{-f}.
\end{gather*}
Using this in (\ref{2b}) we get
\begin{equation*}
\int_M\left| \nabla Rc\right| \left| \nabla Rm\right| \left| Rm\right|
^{p-2}e^{-f}\leq \frac{2K}p\int_M\left| Rm\right| ^pe^{-f}+pK\int_M\left|
\nabla Rm\right| ^2\left| Rm\right| ^{p-3}e^{-f},
\end{equation*}
which, after plugging into (\ref{2}), yields
\begin{equation}
\int_M\left( f-\frac n2+p\left( 1-10K\right) \right) \left| Rm\right|
^pe^{-f}\leq 4p^3K\int_M\left| \nabla Rm\right| ^2\left| Rm\right|
^{p-3}e^{-f}.  \label{3}
\end{equation}
Finally, we have:
\begin{equation*}
2\int_M\left| \nabla Rm\right| ^2\left| Rm\right| ^{p-3}e^{-f}\leq
\int_M\left( \Delta _f\left| Rm\right| ^2\right) \left| Rm\right|
^{p-3}e^{-f}+20\int_M\left| Rm\right| ^pe^{-f}.
\end{equation*}
Indeed, to see this one only has to check the details of the proof of (\ref
{f2}). Integrating by parts and using that $p\geq 3$ we find
\begin{equation*}
\int_M\left( \Delta _f\left| Rm\right| ^2\right) \left| Rm\right|
^{p-3}e^{-f}=-\int_M\left( \nabla \left| Rm\right| ^2\cdot \nabla \left|
Rm\right| ^{p-3}\right) e^{-f}\leq 0.
\end{equation*}
Therefore, from (\ref{3}) we conclude that for $p\geq 3$ we have:
\begin{equation}
\int_M\left( f-\frac n2+p\left( 1-50p^2K\right) \right) \left| Rm\right|
^pe^{-f}\leq 0.  \label{4}
\end{equation}
We are now ready to prove the Corollaries.

\textit{Proof of Corollary 1.}

Let us take $p=n.$ We check from (\ref{4}) that if $K\leq \frac 1{100n^2}$
then
\begin{equation*}
\int_Mf\left| Rm\right| ^ne^{-f}\leq 0.
\end{equation*}
Recall that $f\,$is normalized such that $\left| \nabla f\right| ^2+R=f,$
and in particular, since any gradient shrinker has $R\geq 0$ (see \cite{C,
Ca}) it follows that $f\geq 0.$ Thus the above inequality implies that $M$
is flat i.e. $\left( M,g,f\right) $ is the Gaussian soliton $\left(
\mathbb{R}^n,dx^2,\frac 14\left| x\right| ^2\right) .\blacksquare $

\textit{Proof of Corollary 2.}

We take $p=\frac n2$ in (\ref{4}) to see that
\begin{equation}
\int_M\left( f-7n^3K\right) \left| Rm\right| ^{\frac n2}e^{-f}\leq 0.
\label{5}
\end{equation}
We fix $x_0$ a point where $f$ achieves its minimum on $M$. Then we have
(see \cite{CZ, HM})
\begin{equation}
\frac 14\left[ \left( d\left( x_0,x\right) -5n\right) _{+}\right] ^2\leq
f\left( x\right) \leq \frac 14\left( d\left( x_0,x\right) +\sqrt{2n}\right)
^2,  \label{7}
\end{equation}
where $a_{+}:=\max \left\{ 0,a\right\} .$

Let us set $r_0:=6n+\sqrt{28n^3K}.$ Using this in (\ref{5}) we see that:
\begin{gather*}
\int_{M\backslash B_{x_0}\left( r_0\right) }\left| Rm\right| ^{\frac
n2}e^{-f}\leq \int_{f\geq 7n^3K+1}\left| Rm\right| ^{\frac n2}e^{-f} \\
\leq \int_{f\geq 7n^3K+1}\left( f-7n^3K\right) \left| Rm\right| ^{\frac
n2}e^{-f}\leq 7n^3K\int_{f\leq 7n^3K+1}\left| Rm\right| ^{\frac n2}e^{-f} \\
\leq 7n^3K\int_{B_{x_0}\left( r_0\right) }\left| Rm\right| ^{\frac n2}\leq
7n^3KL.
\end{gather*}
Using again (\ref{7}) shows that for any $r>0$
\begin{eqnarray*}
\int_{B_{x_0}\left( r\right) }\left| Rm\right| ^{\frac n2} &\leq &E\left(
r\right) ,\;\;\text{for} \\
E\left( r\right) &:&=7n^3KLe^{\frac 14\left( r+\sqrt{2n}\right) ^2}.
\end{eqnarray*}

This proves Corollary 2. $\blacksquare $

config: arxiv


TITLE: Solving a simple Schrödinger equation with Fast Fourier Transforms
QUESTION [2 upvotes]: While trying to solve a stochastic Gross-Pitaevskii equation I have found a problem that can be tracked down to something buggy occurring in the simplest Schrodinger equation possible:
$$\partial_t \psi = i \partial^2_x \psi$$
This has the very simple solution of a plane wave:
$$\psi(t) = A e^{i k x - i\omega t}, \text{with } \omega = k^2$$
and my problem is that in some cases I cannot recover this very simple solution numerically. This is due to the fact that my code uses a very standard Fast Fourier Transform method to go to k-space, where the evolution is trivial, and then does the inverse FFT to go back to real space. This method naturally enforces periodic boundary conditions in your simulation grid since you are expanding your solution as a sum of periodic functions. What occurs is that, when the initial condition is a plane wave with a periodicity that does not match the domain size, the algorithm is not able to provide the correct solution.
I show here an example. The problem is clear in $|\psi(t)|$, which should remain constant:

The effect is less dramatic in the phase but still present. One can see some ripples that come from the bottom edges.

I'm not surprised by the fact that, if I enforce periodic boundary conditions, computing the evolution of a wave which is not periodic in the simulation domain yields problems. However, the method is very standard and used extensively, with packages like XMDS (xmds.org) employing it by default. Therefore, I am surprised by the fact that I did not find any mention of this method failing to solve such an extremely simple example.
My question is, is there something that I'm missing? Should I just get over it and assume that this is not a good method if I expect a solution of this kind? Does someone know a reference where this is documented?
Note: Cross-posted in https://math.stackexchange.com/questions/1612508/solving-a-simple-schrodinger-equation-with-fast-fourier-transforms, and also in https://scicomp.stackexchange.com/questions/21821/solving-a-simple-schroedinger-equation-with-fast-fourier-transforms.

REPLY [2 votes]: If I understand correctly, what you're doing amounts to:

Starting with initial data $\psi(x,0)$ that is represented as a truncated Fourier series.
Computing the exact (up to roundoff errors) time evolution of this initial condition on a periodic domain, using discrete Fourier transforms.

This is often referred to as a Fourier spectral method.  Note that no matter what initial data you wish impose, by representing it using a finite, equispaced set of point values and then solving with Fourier transforms, you are effectively representing it as a truncated Fourier series.  Therefore, "when the initial condition is a plane wave with a periodicity that does not match the domain size", by sampling that initial condition on your grid and imposing periodicity, you'll obtain an initial condition with a range of frequencies in it.  Your PDE is dispersive, so you'll see those different frequencies separate, as shown in your first plot.
You claim that "$|\psi(t)|$ should remain constant"; I think you mean to refer to $|\psi(x,t)|$.  But with a dispersive equation and an initial condition that is not monochromatic, this claim is not justified.
By the way, I believe that scicomp.SE is a more useful place to post questions like this.


text:
{\bf Problem.} For what value of $n$ is $5 \times 8 \times 2 \times n = 7!$?
{\bf Level.} Level 1
{\bf Type.} Counting & Probability
{\bf Solution.} To make the arithmetic a little easier, we can write out 7! as $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 = 7 \times 5 \times 3^2 \times 2^4$. And, $5 \times 8 \times 2 = 5 \times 2^4$. So, $n = \frac{7 \times 5 \times 3^2 \times 2^4}{5 \times 2^4} = 7 \times 3^2 = \boxed{63}$.

set_name: MATH


TITLE: A certain class of representations
QUESTION [0 upvotes]: Let $g$ be a non-identity element in a torsion-free amenable group, does there exist a finite-dimensional unitary representation $\pi$ with $\pi(g)\neq 1$?
(The word "finite-dimensional" was initially omitted: as mentioned in the comments the answer is a trivial "yes" then, by considering the left regular representation.)

REPLY [1 votes]: The buzzword to look for is "maximally almost periodic".  This is part of the theory of Bohr Compactifications.  Start with a (locally compact) group $G$.  I do not believe amenability and (especially) torsion freeness has much if any baring on what following.
The Bohr Compactification is the maximal compact group $K$ for which there is a dense range homomorphism $\theta:G\rightarrow K$.  Any finite-dimensional unitary rep of $G$ factors through $\theta$ (by the representation theory of compact groups).  It follows that finite-dimensional unitary reps separate the points of $G$ if and only if $\theta$ is injective.  By definition, this means that $G$ is "maximally almost periodic" (MAP).  One can also construct $K$ using almost periodic functions, and here $G$ is MAP if and only if the almost periodic functions separate the points of $G$.
The Freudenthai-Weil theorem says that for connected locally compact groups, $G$ is MAP if and only if $G=\mathbb R^n \times L$ for some compact $L$.  Chapter 16 of Dixmier's book on $C^\ast$-algebras is a good source for all of this.
Searching around MathOverFlow will find examples of groups which are not MAP.


TITLE: What increases faster: $(\log n)^2$ or $n^{1/3} + \log n$
QUESTION [0 upvotes]: What increases faster: $(\log n)^2$ or $n^{1/3} + \log n$ and why? And also, what increases faster $\log n$ or $n^{x}$, where $x$ is a random positive constant number?

REPLY [3 votes]: Recall that:

$$
f(n)=O(g(n)) \iff \text{For some } C\ge0,\quad\lim_{n\to\infty} \dfrac{f(n)}{g(n)} = C
$$

Hence, $n^{1/3}+\log n$ grows faster (that is, we have $(\log n)^2 = O(n^{1/3}+\log n)$), since by L'Hôpital's Rule, we have:
$$ \begin{align*}
\lim_{n\to\infty} \dfrac{(\log n)^2}{n^{1/3}+\log n} &= \lim_{n\to\infty} \dfrac{\dfrac{2\log n}{n}}{\dfrac{1}{3}n^{-2/3}+\dfrac{1}{n}}\\
&= \lim_{n\to\infty} \dfrac{2\log n}{\dfrac{1}{3}n^{1/3}+1}\\
&= \lim_{n\to\infty} \dfrac{\dfrac{2}{n}}{\dfrac{1}{9}n^{-2/3}}\\
&= \lim_{n\to\infty} \dfrac{2}{\dfrac{1}{9}n^{1/3}}\\
&= 18\lim_{n\to\infty} \dfrac{1}{n^{1/3}}\\
&= 18(0)=0\ge0
\end{align*} $$

In general, notice that for any constant $x>0$, $n^x$ will grow faster than a logarithm (that is, $\log n = O(n^x)$), since:
$$
\lim_{n\to\infty} \dfrac{\log n}{n^x} = \lim_{n\to\infty} \dfrac{\dfrac{1}{n}}{xn^{x-1}}= \lim_{n\to\infty} \dfrac{1}{xn^x}= \dfrac{1}{x}\lim_{n\to\infty} \dfrac{1}{n^x}= \dfrac{1}{x}(0)=0\ge0
$$


TITLE: Proving a Conditional Probability Theorem (General Addition Rule)
QUESTION [1 upvotes]: If $B$ is an event with $P(B) > 0$, prove that the set function $Q(A) = P(A|B)$
satisfies the axioms for a probability measure. ,
$$P(A ∪ C|B) = P(A|B) + P(C|B) − P(A ∩ C|B)$$
What i tried
Basically the question is asking to prove the above Probability axiom using the fact that $Q(A) = P(A|B)$ .
$P(A ∩ C|B)=-P(A ∪ C|B)+P(A)+P(C|B)$
Substituting this to the Probability axiom it becomes
$$P(A ∪ C|B) = P(A|B) + P(C|B)+P(A ∪ C|B)-P(A)-P(C|B)$$
Simplifying it it becomes
$$P(A|B) =P(A)$$
This shows that events $A$ and $B$ are independent of each other. Hence showing $Q(A) = P(A|B)$ Is my method correct? Could anyone explain . Thanks

REPLY [2 votes]: Between the Question and some of the Comments, there seems to
be some confusion that we need to dispel from the start:
First, if $A$ and $B$ are independent, then $P(A|B) = P(A)$, but
there seems no reason to assume that.
Second, in most formulations of probability, the first equation you
write is a theorem, not an axiom.

I will begin by showing that $Q$ satisfies some of the $axioms$
of probability. (There are a few gaps, which I hope you will have
no trouble filling in.)
1) Show $Q(S) = 1$, where $S$ denotes the space on which $P$ and $Q$ are defined.
$$Q(S) = P(S|B) = P(S \cap B)/P(B) = P(B)/P(B) = 1.$$
2) Show $Q(A) \ge 0$, where $A$ is an event in $S$.
$$Q(A) = P(A \cap B)/P(B) \ge 0,$$
because $P(A \cap B) \ge 0$ and $P(B) > 0,$
3) If $A_1$ and $A_2$ are disjoint events in $S$, then
$Q(A_1 \cup A_2) = Q(A_1)+Q(A_2).$
$$Q(A_1 \cup A_2) = \frac{P((A_1 \cup A_2)\cap B)}{P(B)}
=\frac{P((A_1 \cap B) \cup (A_2 \cap B))}{P(B)}\\
= \frac{P(A_1 \cap B) + P(A_2 \cap B)}{P(B)} = Q(A_1) + Q(A_2),$$
because $A_1 \cap B$ and $A_2 \cap B$ must be disjoint.
Notes:
(a) I used $A_1$ and $A_2$ instead of $A$ and $C$ because I think
there is potential distraction owing to the alphabetical order
of the letters. [In order to keep sight of event $B$, some
books would write $Q$ as $P_B,$ but I did not do that.]
(b) Depending on the level of your course, there may be a fourth
axiom involving an infinite sequence of mutually disjoint  events $A_i$ and an
infinite sum of probabilities. The proof for $Q$ is similar to that
of axiom (3).
(c) Depending on the level of your course, there may be
a definition of 'events' as subsets of $S$, which are assigned
probabilities.

Now for the specific $theorem$ about $Q$ you mentioned in your
problem.
If $A_1$ and $A_2$ are not necessarily disjoint, it follows that $$Q(A_1) + Q(A_2) = Q(A_1)+Q(A_2)-Q(A_1 \cap A_2).$$
If we have proved $Q$ obeys all of the probability axioms,
this is just the same as the proof that
$$P(A_1) + P(A_2) = P(A_1)+P(A_2)-P(A_1 \cap A_2).$$
If you want to prove this relationship directly from the
definition of $Q$ as a conditional probability given $B$,
then the proof is very similar to the one shown for axiom (3).


TITLE: Find a matrix $P$ that orthogonally diagonalizes $I-uu^T$ if $u=\pmatrix{1\\-1\\1}$.
QUESTION [0 upvotes]: Find a matrix $P$ that orthogonally diagonalizes $I-uu^T$ if $u=\pmatrix{1\\-1\\1}$.
Let matrix $A=I-uu^T=\pmatrix{0&1&-1\\1&0&1\\-1&1&0}$.
I have followed the usual procedure and obtained matrix $P=\pmatrix{\frac{1}{\sqrt3}&-\frac{1}{\sqrt6}&\frac{1}{\sqrt2}\\-\frac{1}{\sqrt3}&\frac{1}{\sqrt6}&\frac{1}{\sqrt2}\\\frac{1}{\sqrt3}&\frac{2}{\sqrt6}&0}$.
In the midst of solving for $P^TAP$, I realized it got really tedious and long-winded. Is there is a better method to go about obtaining the diagonal matrix $D$ instead of computing it directly like so? I can't help but feel like I have missed something.

REPLY [1 votes]: Assuming that you actually wanted the matrix $P$ as stated in the title and first sentence of your question, here are some short cuts.
Every eigenvector $v$ of a matrix $A$ is also an eigenvector of $I-A$: $$(I-A)v=v-A v=v-\lambda v=(1-\lambda)v.$$ We can also see from this that if $\lambda$ is an eigenvalue of $A$, then $1-\lambda$ is an eigenvalue of $I-A$.
The matrix $uu^T$ has rank one—every column is a scalar multiple of $u$— and its kernel consists of all vectors $v$ for which $u^Tv=0$, i.e., the orthogonal complement of $u$. Thus, an orthonormal basis of $u^\perp$ together with $u/\|u\|$ is an orthonormal basis of $\mathbb R^3$ that consists of eigenvector of $I-uu^T$, so you can take as $P$ the matrix with these vectors as its columns. The eigenvalues of $uu^T$, incidentally, are $0$ and $u^Tu$. The former is obvious because the nullity of $uu^T$ is $2$, while the latter is easily found by rearranging $(uu^T)u$.
For this problem, you can take another short cut since you’re working in $\mathbb R^3$: if you take any vector $v$ that’s orthogonal to $u$ and then compute $u\times v$, the three vectors form an orthogonal basis and all that’s left to do is to normalize them. You can find by inspection that $(1,1,0)^T$ is orthogonal to $u$, and $(1,-1,1)^T\times(1,1,0)^T=(-1,1,2)^T$, from which a quick computation produces $$P=\left[\begin{array}{rcr}\frac1{\sqrt3}&\frac1{\sqrt2}&-\frac1{\sqrt6}\\-\frac1{\sqrt3}&\frac1{\sqrt2}&\frac1{\sqrt6}\\\frac1{\sqrt3}&0&\frac2{\sqrt6}\end{array}\right].$$ The matrix $D=P^{-1}(I-uu^T)P$ has the corresponding eigenvalues along its main diagonal and $u^Tu=3$, so $D=\operatorname{diag}(-2,1,1)$.


TITLE: Quotients of $\ell_\infty$ by separable subspaces
QUESTION [14 upvotes]: Given a (closed) separable subspace $M$ of $\ell_\infty$, I am interested in conditions implying that the quotient $\ell_\infty/M$ is isomorphic to a subspace of $\ell_\infty$.
It is not difficult to see that being $M$ reflexive is sufficient, and Bourgain proved that $\ell_\infty/c_0$ does not admit an equivalent strictly convex norm (while $\ell_\infty$ does). So $\ell_\infty/c_0$ is not isomorphic to a subspace of $\ell_\infty$.

REPLY [15 votes]: While I do not have a complete answer to the OP’s question, I made enough observations that I think it is worthwhile to record them as an answer.
(1) If $X$ and $Y$ are isomorphic (closed) subspaces of $\ell_\infty$, then $\ell_\infty/X$ embeds into $\ell_\infty$ iff $\ell_\infty/Y$ embeds into $\ell_\infty$.
Indeed, it is clear that $\ell_\infty/X$ embeds into $\ell_\infty$ iff
$(\ell_\infty \oplus \ell_\infty)/(X \oplus \{0\})$ embeds into $\ell_\infty$, so without loss of generality we can assume that $\ell_\infty$ embeds into both $\ell_\infty/X$ and into $\ell_\infty/Y$.  We then get from [LR, [Theorem 3(i)] that every isomorphism from $X$ onto $Y$ extends to an automorphism of $\ell_\infty$.
(2) Suppose that $Y$ is a subspace of $\ell_\infty$ and $X$ is a subspace of $Y$.  Assume that $\ell_\infty/Y$ embeds into $\ell_\infty$. Then $\ell_\infty/X$ embeds into $\ell_\infty$ iff $Y/X$ embeds into $\ell_\infty$.
The “only if” part is clear because $Y/X$ embeds into $\ell_\infty/X$.  The other direction is an easy consequence of Lindentrauss’ observation that if $U\subset V$ and both $U$ and $V/U$ embed into $\ell_\infty$, then $V$ embeds into $\ell_\infty$.   To prove this observation, let $T$ be an isomorphism from $U$ into $\ell_\infty$ and  $S$ an isomorphism from $V/U$ into $\ell_\infty$.  The space $\ell_\infty$ is $1$-injective (this is immediate from the Hahn-Banach theorem), so $T$ extends to a bounded linear mapping (which we also denote by $T$) from $V$ into $\ell_\infty$.   Denoting the quotient map from $V$ to $V/U$ by $Q$, we see that $T \oplus SQ: x\mapsto (Tx, SQx)$ defines an isomorphism from $V$ into $\ell_\infty \oplus_\infty \ell_\infty \equiv \ell_\infty$.
(3)  Suppose that $X$ is a subspace of of $\ell_\infty$ and $X$ is isomorphic to $Y^*$ for some separable $Y$.  Then $\ell_\infty/X$ embeds into $\ell_\infty$.
Indeed, $Y$, being separable, is isomorphic to the quotient space $\ell_1/W$ for some subspace $W$ of $\ell_1$, and  $W^\perp$ in $\ell_\infty$ is isomorphic to $X$.  But $\ell_\infty/{W^\perp}$ is isometric to $W^*$, which embeds into $\ell_\infty$ because $W$, being separable, is a quotient of $\ell_1$.  Hence by (1), the space $\ell_\infty/X$ also embeds into $\ell_\infty$.
(4) Suppose that $X$ a subspace of of $\ell_\infty$ and $X$ embeds into a separable conjugate space.  Then $\ell_\infty/X$ embeds into $\ell_\infty$.
This follows from (3), (2), and the fact that every separable space embeds into $\ell_\infty$.
(5) $\ell_\infty/{c_0}$ does not embed into $\ell_\infty$.  Hence if  $X$ is a separable subspace of of $\ell_\infty$ and $X$ contains a subspace isomorphic to $c_0$, then $\ell_\infty/X$ does not embed into $\ell_\infty$.
The second statement follows from the first and (1), (2).  The first statement is another observation that I think is due to Lindenstrauss; namely, that $\ell_\infty/{c_0}$ contains an isometric copy of $c_0(\Gamma)$ with $|\Gamma|=2^{\aleph_0}$ (consider the image in $\ell_\infty/{c_0}$ of characteristic functions of $2^{\aleph_0}$ infinite subsets of the natural numbers all of whose pairwise intersections are finite).  The space $c_0(\Gamma)$ does not embed into $\ell_\infty$ because the unit vector basis for the space is a non separable set that has weakly compact closure, and every weakly compact subset of the dual to a separable space is separable.
(6) If $X$ is a subspace of $\ell_\infty$ that is isomorphic to $L_1(0,1)$, then $\ell_\infty/X$ embeds into $\ell_\infty$. Hence if $X$ is a subspace of $\ell_\infty$ that embeds into $L_1(0,1)$, then $\ell_\infty/X$ embeds into $\ell_\infty$.
The second statement follows from the first by what is now standard reasoning.  For the first statement, note that $L_1(0,1)$ embeds into $C[0,1]^*$ as a complemented subspace, and by (3), if $Y$ is a subspace of $\ell_\infty$ isomorphic to $C[0,1]^*$, then $\ell_\infty/Y$ embeds into $\ell_\infty$.  Now use (2) and (1).
[LR] J. Lindenstrauss and H. P. Rosenthal, Automorphisms in $c_0$, $\ell_1$, and $m$,  Israel J. Math. 7 (1969), 227–239.


TITLE: Summing two independent Markov Chains on a discrete-time and space environment
QUESTION [1 upvotes]: I have the following problem on my Stochastic Processes problem set:

Let $(X_n)_{n \in \mathbb{N}}$ and $(Y_n)_{n \in \mathbb{N}}$ be two independent Markov Chains taking values on $I_{X}$ and $I_{Y}$ respectively, both countable sets. Also, let $(Z_{n})_{n \in \mathbb{N}}$ be iid random variables taking values on $\{0,1\}$. Now, define $W_n = X_n \mathbb{1}_{\{Z = 0\}} + Y_n \mathbb{1}_{\{Z = 1\}}$. Is $W_n$ a Markov Chain?

I understand that I have to show that
$\mathbb{P}(W_{n+1} = j| W_{n} = i, W_{n-1} = i_{n-1},..., W_0 = i_0) = \mathbb{P}(W_{n+1} = j| W_{n} = i)$,
for any states $j,i,i_{n-1}, ..., i_{0} \in I_{X}\cup I_{Y}$. I am stuck right now, but I have the impression that this result is false as stated.
For example, if $I_X$ and $I_Y$ were disjoint sets, observing that $W_n = i$ for some $i \in I_X$ gives me some information about $X_n$, although no information about $Y_n$. But if $i \in I_{X}\cup I_{Y}$, I can get "confused" since I cannot attribute this realization to one or another variable.
On the other hand, $Z_{n}$ is such a simple process and both $X_{n}$ and $X_{n}$ are Markov Chains, so the dependence on the past is somewhat limited.
I have no idea how to answer this question. I do know, however, that the sum of independent Markov Chains is not necessarily another Markov Chain, but I don't see how this result can help me here.
Thanks a lot in advance!!

REPLY [1 votes]: You're right about the case of disjoint $I_X$ and $I_Y$, and that's all you need. The question isn't asking whether $W_n$ might be a Markov chain under some circumstances (the answer is clearly yes, e.g. if $Z_n\equiv0$), but whether $W_n$ is necessarily a Markov chain, given the assumptions, so your counterexample is enough to disprove this. If $W_n\in I_X$, then it gives you no information about $Y_{n-1}$, but unless $Z_n\equiv0$ or $Y_{n+1}$ is independent of previous states, the next state $W_{n+1}$ depends on $Y_{n-1}$.
Edit in response to the comment:
A concrete counterexample is given by $Z_n$ uniformly distributed on $\{0,1\}$ and $X_n$ and $Y_n$ symmetric random walks on disjoint copies $\mathbb Z_X$ and $\mathbb Z_Y$ of $\mathbb Z$ starting at the respective origins at $n=0$. Assume that $Z_1=1$ and $Z_2=0$, so $W_1=y_1\in\mathbb Z_Y$ and $W_2=x_2\in\mathbb Z_X$. Now
$$P(W_3=a\mid W_1=y_1,W_2=x_2)=
\begin{cases}
\frac14&a\in\{y_1,x_2+1_X,x_2-1_X\}\\
\frac18&a\in\{y_1+2_Y,y_1-2_Y\}\\
0&\text{otherwise}
\end{cases}
$$
whereas e.g. $P(W_3=3_Y\mid W_2=x_2)=\frac1{16}$.


TITLE: Is uncertainty uncertain?
QUESTION [1 upvotes]: I was thinking about uncertainty and wondered if uncertainty is inherently uncertain. I know in the uncertainty inequality for position and momentum, $\Delta x  \Delta p \ge \hbar/2$, the presence of the greater than or equal to sign implies that the quantities are not fully known. This leads me to ask: Is uncertainty uncertain? How accurate is our knowledge of how much we know about a system? How accurate can this knowledge be? Note that this does not ask if the uncertainty principle is just a limitation of our knowledge, but if there is uncertainty in uncertainty itself.

REPLY [0 votes]: We must understand that the $\ge$ sign does not show that the uncertainty is not fully known. While measuring $x$ and $p$, there would be errors, which are due to the instrument used... (eg : If we are using a geometry scale, it would be 1mm  and if the least count of our momentum measuring apparatus is say 1cm/s then $\Delta P$ is 1 cm/s).
Uncertainty principle says that the error would always be greater than $\frac{\hbar}{2}$, implying that no matter how accurate your apparatus may be, a particle always has a small wave nature and hence, its position would always be uncertain after a limit.
To conclude, there is no uncertainty in uncertainity. Heisenberg was clever enough to write uncertainty relation as      $\Delta x  \Delta p \ge \hbar/2$
instead of writing the maximum possible accuracy for $\Delta x  \Delta p = \hbar/2$


TITLE: How to calculate limit $\lim_{x\to\infty}\left(\lfloor x\rfloor+\frac{1}{2}\right)\ln x-\ln(\lfloor x\rfloor!)$
QUESTION [1 upvotes]: While uses plot on www.wolframalpha.com, i found a problem:
The function $$f(x)=\left(\lfloor x\rfloor+\frac{1}{2}\right)\ln x-\ln(\lfloor x\rfloor!)$$
approximate the function
$$g(x)=x-\frac{1}{2}\ln\pi$$
Can you prove it?

REPLY [2 votes]: Use Stirling's formula:
$$x!\sim \sqrt{2\pi x}\left( \frac{x}{e}\right)^x$$
$$\begin{align}
\lim_{x\to\infty} f(x)&=(x+\frac{1}2)\ln(x)-\frac{1}{2}\ln(2\pi x)-x\ln(x)+x\\
\\
\lim_{x\to\infty} f(x)&=x-\frac{1}{2}\ln(2\pi)
\end{align}$$
Note in your plot (and in your post), it should be  $g(x)=x-\frac{1}{2}\ln(2\pi)$
Here is the updated plot