# Product of exponentials concentrates
around the exponential of the sum

Michael Anshelevich, Austin Pritchett  Department of Mathematics, Texas A&M
University, College Station, TX 77843-3368<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

For two matrices $A$ and $B$, and large $n$, we show that most products of $n$
factors of $e^{A/n}$ and $n$ factors of $e^{B/n}$ are close to $e^{A+B}$. This
extends the Lie-Trotter formula. The elementary proof is based on the relation
between words and lattice paths, asymptotics of binomial coefficients, and
matrix inequalities. The result holds for more than two matrices.

###### 2010 Mathematics Subject Classification:

Primary 15A16; Secondary 05A16

This work was supported in part by a Simons Foundation Collaboration Grant.

##  1\. Introduction.

Matrix products do not commute. One familiar consequence is that in general,

$e^{A}e^{B}\neq e^{A+B}.$

(Here for a square matrix $A$, the expression $e^{A}$ can be defined, for
example, using the power series expansion of the exponential function.)
However, a vestige of the “product of exponentials is the exponential of the
sum” property remains, as long as we take the factors in a very special
_alternating_ order.

######  Theorem (Lie-Trotter product formula).

Let $A$ and $B$ be complex square matrices. Then

$\lim_{n\rightarrow\infty}\left(e^{A/n}e^{B/n}\right)^{n}\rightarrow e^{A+B},$

where the convergence is with respect to any matrix norm.

This result goes back to Sophus Lie, see [1] or Proposition 16(b) below for an
elementary proof. Clearly, if we take $n$ factors $e^{A/n}$ and $n$ factors
$e^{B/n}$ but multiply them in a different order, the result will not always
converge to $e^{A+B}$. For example,

$\left(e^{A/n}\right)^{n}\left(e^{B/n}\right)^{n}=e^{A}e^{B},\quad\left(e^{B/n}\right)^{n}\left(e^{A/n}\right)^{n}=e^{B}e^{A}.$

The reader is invited to try out plotting all of such products for their
preferred choices of (real) matrices at
https://austinpritchett.shinyapps.io/nexpm_visualization/

Nevertheless, in this article we show that, for large $n$, the overwhelming
majority of products of $n$ factors $e^{A/n}$ and $n$ factors $e^{B/n}$ will
be close to $a^{A+B}$. In other words, such products _concentrate_ around
$e^{A+B}$. To give a precise formulation, we introduce some notation.

######  Definition 1.

Denote by $\mathcal{W}_{n}$ the set of all words in $A$ and $B$ which contain
exactly $n$ $A$’s and $n$ $B$’s. Denote by $w[i]$ the $i$’th letter in $w$.

######  Theorem 2.

Let $A$ and $B$ be complex square matrices. Consider all $\binom{2n}{n}$
products of $e^{A/n}$ and $e^{B/n}$ of the form $\prod_{i=1}^{2n}e^{w[i]/n}$
for $w\in\mathcal{W}_{n}$. Among these products, the proportion of those which
differ from $e^{A+B}$ in norm by less than

$\sqrt{\frac{\ln n}{n}}$

goes to $1$ as $n\rightarrow\infty$.

Along the way to the proof of this result, we discuss several metrics on the
space of words, which are interesting in their own right.

This expanded version also contains an appendix, which does not appear in the
published version. In it, we provide several figures illustrating possible
shapes of the set of products.

##  2\. Words and paths.

We define three metrics on the set of words $\mathcal{W}_{n}$.

######  Definition 3.

Let $w$ be a word. A _swap_ is an interchange of two neighboring letters in
$w$. The _swap distance_ $\operatorname{\rho_{\mathit{swap}}}(w,v)$ between
two words $w,v\in\mathcal{W}_{n}$ is the minimal number of swaps needed to
transform $w$ into $v$.

This metric may remind some readers of the bubble-sort algorithm.

######  Example 4.

We may swap

$AABB\mapsto ABAB\mapsto ABBA\mapsto BABA\mapsto BBAA.$

It is not hard to check that this is the minimal number of swaps needed, so

$\operatorname{\rho_{\mathit{swap}}}(AABB,BBAA)=4.$

To define the other two metrics, it is convenient to represent a word by a
_lattice path_. To be able to consider words of different length on equal
footing, our lattice paths will be normalized.

######  Definition 5.

A lattice path connects the origin $(0,0)$ to the point $(1,1)$ by a path
consisting of $n$ horizontal and $n$ vertical segments of length $1/n$.

We may identify words in $\mathcal{W}_{n}$ with such paths. For a word $w$,
denote

$w_{A}[j]=\\#\left\\{i\leq j:w[i]=A\right\\}$

the number of $A$’s among the first $j$ letters, and the same for $w_{B}[j]$.
Then the path corresponding to $w$ consists of points

$\left\\{\frac{1}{n}(w_{A}[j],w_{B}[j]):1\leq j\leq 2n\right\\}$

connected by straight line segments, with $A$ corresponding to a horizontal
step, and $B$ to a vertical step. See Figure 1.

Figure 1. The path corresponding to the word $AABBBBAAABAABBBAABAB$.

######  Definition 6.

For two words $w,v\in\mathcal{W}_{n}$, define the distance $\rho_{1}(w,v)$ to
be the (unsigned) area of the region between the paths. See Figure 2.

Figure 2. $\rho_{1}(w,v)=\frac{24}{100}$, while
$\rho_{\infty}(w,v)=\frac{4}{10}$. In the second plot,
$\left|w_{A}[5]-v_{A}[5]\right|=2$.

######  Lemma 7.

We can express $\rho_{1}(w,v)$ directly in terms of the words $w,v$ as
follows:

$\begin{split}\rho_{1}\left(w,v\right)&=\frac{1}{2n^{2}}\sum_{j=1}^{2n}\Bigl{|}\Bigl{(}w_{A}[j]-w_{B}[j]\Bigr{)}-\Bigl{(}v_{A}[j]-v_{B}[j]\Bigr{)}\Bigr{|}\\\
&=\frac{1}{n^{2}}\sum_{j=1}^{2n}\Bigl{|}w_{A}[j]-v_{A}[j]\Bigr{|}\\\
&=\frac{1}{2n^{2}}\sum_{j=1}^{2n}\left(\Bigl{|}w_{A}[j]-v_{A}[j]\Bigr{|}+\Bigl{|}w_{B}[j]-v_{B}[j]\Bigr{|}\right).\end{split}$

Here the first representation compares the excess of the number of $A$’s over
the number of $B$’s in $w$ and $v$.

###### Proof.

To obtain the second expression, we slice the region between the paths into
NW-SE diagonal regions. For each $j$,

(1) $w_{A}[j]+w_{B}[j]=v_{A}[j]+v_{B}[j]=j,$

and there are exactly $\Bigl{|}w_{A}[j]-v_{A}[j]\Bigr{|}$ squares located on
the diagonal between the points $\frac{1}{n}(w_{A}[j],w_{B}[j])$ and
$\frac{1}{n}(v_{A}[j],v_{B}[j])$. See Figure 2. For the first and third
expressions, again using the identity (1),

$\begin{split}\Bigl{|}\Bigl{(}w_{A}[j]-w_{B}[j]\Bigr{)}-\Bigl{(}v_{A}[j]-v_{B}[j]\Bigr{)}\Bigr{|}&=2\Bigl{|}w_{A}[j]-v_{A}[j]\Bigr{|}\\\
&=\Bigl{|}w_{A}[j]-v_{A}[j]\Bigr{|}+\Bigl{|}w_{B}[j]-v_{B}[j]\Bigr{|}.\qed\end{split}$

######  Definition 8.

The third distance we will consider is

$\begin{split}\rho_{\infty}\left(w,v\right)&=\frac{1}{n}\max_{1\leq j\leq
2n}\Bigl{|}\Bigl{(}w_{A}[j]-w_{B}[j]\Bigr{)}-\Bigl{(}v_{A}[j]-v_{B}[j]\Bigr{)}\Bigr{|}\\\
&=\frac{2}{n}\max_{1\leq j\leq n}\Bigl{|}w_{A}[j]-v_{A}[j]\Bigr{|}\\\
&=\frac{1}{n}\max_{1\leq j\leq
2n}\left(\Bigl{|}w_{A}[j]-v_{A}[j]\Bigr{|}+\Bigl{|}w_{B}[j]-v_{B}[j]\Bigr{|}\right).\end{split}$

It can interpreted as the maximal difference between the corresponding points
on the paths as measured in the NW-SE direction, with appropriate
normalization.

Clearly

(2) $\rho_{1}\leq\rho_{\infty}.$

We now observe that the swap metric and the path metric are related in a
simple way.

######  Theorem 9.

$\rho_{1}(w,v)=\dfrac{1}{n^{2}}\operatorname{\rho_{\mathit{swap}}}(w,v)$.

###### Proof.

Each swap of neighboring letters changes the area between the paths by
$\dfrac{1}{n^{2}}$. So
$\rho_{1}(w,v)\leq\dfrac{1}{n^{2}}\operatorname{\rho_{\mathit{swap}}}(w,v)$.
On the other hand, unless the words are equal, we can find an $A$ followed by
a $B$ such that at that point in the word, one word has more $A$’s than the
other one. Then swapping these $A$ and $B$ decreases $\rho_{1}$. So one can
always transform a word $w$ into $v$ by $n^{2}\rho_{1}(w,v)$ swaps. ∎

######  Proposition 10.

Let $M\in\left\\{1,\ldots,n\right\\}$. The number of words
$w\in\mathcal{W}_{n}$ for which the $\rho_{\infty}$ distance to the _standard
word_ $\operatorname{\overline{\mathrm{w}}}=ABAB\ldots AB\in\mathcal{W}_{n}$
is at least $M/n$ is at most $2\binom{2n}{n-M+1}$.

###### Proof.

Note first that

$\begin{split}\rho_{\infty}(w,\operatorname{\overline{\mathrm{w}}})&=\frac{1}{n}\max_{1\leq
j\leq
2n}\Bigl{|}\Bigl{(}w_{A}[j]-w_{B}[j]\Bigr{)}-\Bigl{(}\operatorname{\overline{\mathrm{w}}}_{A}[j]-\operatorname{\overline{\mathrm{w}}}_{B}[j]\Bigr{)}\Bigr{|}\\\
&\leq\frac{1}{n}\max_{1\leq j\leq
2n}\Bigl{|}w_{A}[j]-w_{B}[j]\Bigr{|}+\frac{1}{n}.\end{split}$

Suppose $\rho_{\infty}(w,\operatorname{\overline{\mathrm{w}}})\geq M/n$, so
that

$\max_{1\leq j\leq 2n}\Bigl{|}w_{A}[j]-w_{B}[j]\Bigr{|}\geq M-1.$

We now apply the Reflection Principle, see [4] or Section 10.3 of [2]. Let $j$
be the smallest index such that

(3) $\Bigl{|}w_{A}[j]-w_{B}[j]\Bigr{|}=M-1;$

note that such a $j$ exists. Let $\tilde{w}$ be a word of length $2n$ (not
necessarily in $\mathcal{W}_{n}$) such that

  * •

for $i\leq j$, $\tilde{w}[i]=w[i]$,

  * •

for $i>j$, $\tilde{w}[i]=A$ if $w[i]=B$, and $\tilde{w}[i]=B$ if $w[i]=A$.

Since $w_{A}[j]=w_{B}[j]\pm(M-1)$, $\tilde{w}$ contains

$w_{A}[j]+(n-w_{B}[j])=n\pm(M-1)$

$A$’s. Conversely, because $j$ is the _smallest_ index satisfying equation
(3), this procedure can be reversed, and each word $v$ with $n\pm(M-1)$ $A$’s
arises as $\tilde{w}$ for a unique $w\in\mathcal{W}_{n}$. It remains to note
that the total number of words of length $2n$ with $n\pm(M-1)$ $A$’s is

$2\binom{2n}{n-M+1}.\qed$

######  Remark 11.

Recall the little-o, big-O, and asymptotic notation. For two positive
sequences $(a_{n})$ and $(b_{n})$, we write

  * •

$a_{n}=o(b_{n})$ if $a_{n}/b_{n}\rightarrow 0$

  * •

$a_{n}=O(b_{n})$ is $a_{n}/b_{n}$ is bounded

  * •

$a_{n}\sim b_{n}$ if $a_{n}/b_{n}\rightarrow 1$

######  Corollary 12.

Let $p(n)$ be a positive sequence such that $p(n)\rightarrow\infty$ and
$p(n)=o(n^{1/6})$. Then for large $n$, the proportion of words
$w\in\mathcal{W}_{n}$ for which

$\rho_{\infty}(w,\operatorname{\overline{\mathrm{w}}})\geq\frac{p(n)}{\sqrt{n}}$

is asymptotically at most $2e^{-p(n)^{2}}$.

###### Proof.

The desired proportion is at most

$2\frac{\binom{2n}{n-[\sqrt{n}p(n)]+1}}{\binom{2n}{n}},$

where $[\sqrt{n}p(n)]$ denotes the integer part. The asymptotics of this
expression can be found using Stirling’s formula, see equation (5.43) in [5].
∎

######  Remark 13.

$\rho_{\infty}$ is closely related to the notion of “span” from [3], where its
asymptotic expected value is computed (more precisely, “span” is the $\tau$
statistic in the final section below). Related analysis for paths which lie
entirely above the main diagonal is sometimes called the Sock Counting
Problem, for reasons we invite the reader to discover.

##  3\. Matrix estimates.

We now recall that $A,B$ are actually matrices in
$M_{d}(\mathbb{C})=\mathbb{C}^{d\times d}$. Denote by $\left\|{\cdot}\right\|$
some sub-multiplicative norm on this matrix space, such as the operator norm
or the Frobenius norm.

######  Lemma 14.

For every word $w\in\mathcal{W}_{n}$

$\left\|{e^{w[1]/n}e^{w[2]/n}\cdots e^{w[2n]/n}}\right\|\leq
e^{\left\|{A}\right\|+\left\|{B}\right\|},$

with a uniform bound which does not depend on the word or on $n$.

###### Proof.

Since the norm is sub-multiplicative,

$\left\|{e^{C}}\right\|=\left\|{\sum_{k=0}^{\infty}\frac{1}{k!}C^{k}}\right\|\leq\sum_{k=0}^{\infty}\frac{1}{k!}\left\|{C}\right\|^{k}=e^{\left\|{C}\right\|}.$

Therefore

$\begin{split}\left\|{e^{w[1]/n}e^{w[2]/n}\cdots
e^{w[2n]/n}}\right\|&\leq\prod_{i=1}^{2n}\left\|{e^{w[i]/n}}\right\|\leq\prod_{i=1}^{2n}e^{\left\|{w[i]}\right\|/n}\\\
&=e^{\sum_{i=1}^{2n}\left\|{w[i]}\right\|/n}=e^{\left\|{A}\right\|+\left\|{B}\right\|}.\qed\end{split}$

The following estimates can be improved (with a longer argument), but suffice
for our purposes.

######  Lemma 15.

For large $n$,

$\left\|{e^{A/n}e^{B/n}-e^{(A+B)/n}}\right\|\leq\frac{1}{n^{2}}\left\|{AB-
BA}\right\|$

and

$\left\|{e^{A/n}e^{B/n}-e^{B/n}e^{A/n}}\right\|\leq\frac{2}{n^{2}}\left\|{AB-
BA}\right\|.$

###### Proof.

If $AB=BA$, the result is immediate, so we assume that $AB\neq BA$. Then

$\begin{split}&\left\|{e^{A/n}e^{B/n}-e^{(A+B)/n}-\frac{1}{2n^{2}}(AB-
BA)}\right\|\\\
&\quad=\left\|{\sum_{k=0}^{\infty}\frac{1}{n^{k}}\sum_{\ell=0}^{k}\frac{1}{\ell!(k-\ell)!}A^{\ell}B^{k-\ell}-\sum_{k=0}^{\infty}\frac{1}{n^{k}}\frac{1}{k!}(A+B)^{k}-\frac{1}{2n^{2}}(AB-
BA)}\right\|\\\
&\quad=\left\|{\sum_{k=3}^{\infty}\frac{1}{n^{k}}\frac{1}{k!}\left(\sum_{\ell=0}^{k}\binom{k}{\ell}A^{\ell}B^{k-\ell}-(A+B)^{k}\right)}\right\|\\\
&\quad\leq
2\sum_{k=3}^{\infty}\frac{1}{n^{k}}\frac{1}{k!}(\left\|{A}\right\|+\left\|{B}\right\|)^{k}\\\
&\quad\leq\frac{2}{n^{3}}e^{\left\|{A}\right\|+\left\|{B}\right\|}.\end{split}$

Therefore

$\left\|{e^{A/n}e^{B/n}-e^{(A+B)/n}}\right\|\leq\frac{1}{2n^{2}}\left\|{AB-
BA}\right\|+\frac{2}{n^{3}}e^{\left\|{A}\right\|+\left\|{B}\right\|}\leq\frac{1}{n^{2}}\left\|{AB-
BA}\right\|$

for large $n$. The second estimate follows from the first. ∎

######  Proposition 16.

  1. (a)

Swapping two neighboring letters in a word changes the corresponding product
by $O(1/n^{2})$. More precisely,

$\Bigl{\|}e^{w[1]/n}\cdots e^{w[i]/n}e^{w[i+1]/n}\cdots e^{w[2n]/n}\\\
-e^{w[1]/n}\cdots e^{w[i+1]/n}e^{w[i]/n}\cdots
e^{w[2n]/n}\Bigr{\|}\leq\frac{2}{n^{2}}\left\|{AB-
BA}\right\|e^{\left\|{A}\right\|+\left\|{B}\right\|}.$

  2. (b)

The Lie-Trotter formula:

$\left\|{\left(e^{A/n}e^{B/n}\right)^{n}-e^{A+B}}\right\|\leq\frac{1}{n}\left\|{AB-
BA}\right\|e^{\left\|{A}\right\|+\left\|{B}\right\|}.$

###### Proof.

For (a), using the two preceding lemmas,

$\begin{split}&\Bigl{\|}e^{w[1]/n}\cdots e^{w[i]/n}e^{w[i+1]/n}\cdots
e^{w[2n]/n}\\\ &\qquad-e^{w[1]/n}\cdots e^{w[i+1]/n}e^{w[i]/n}\cdots
e^{w[2n]/n}\Bigr{\|}\\\ &\quad=\Bigl{\|}e^{w[1]/n}\cdots e^{w[i-1]/n}\\\
&\quad\qquad\times\Bigl{(}e^{w[i]/n}e^{w[i+1]/n}-e^{w[i+1]/n}e^{w[i]/n}\Bigr{)}e^{w[i+2]/n}\cdots
e^{w[2n]/n}\Bigr{\|}\\\ &\quad\leq
e^{\left\|{A}\right\|+\left\|{B}\right\|}\left\|{e^{A/n}e^{B/n}-e^{B/n}e^{A/n}}\right\|\\\
&\quad\leq\frac{2}{n^{2}}e^{\left\|{A}\right\|+\left\|{B}\right\|}\left\|{AB-
BA}\right\|.\end{split}$

Similarly, for (b),

$\begin{split}\left\|{\left(e^{A/n}e^{B/n}\right)^{n}-e^{A+B}}\right\|&=\left\|{\left(e^{A/n}e^{B/n}\right)^{n}-\left(e^{(A+B)/n}\right)^{n}}\right\|\\\
&\leq
n\left\|{e^{A/n}e^{B/n}-e^{(A+B)/n}}\right\|e^{\left\|{A}\right\|+\left\|{B}\right\|}\\\
&\leq\frac{1}{n}\left\|{AB-
BA}\right\|e^{\left\|{A}\right\|+\left\|{B}\right\|}.\qed\end{split}$

######  Theorem 17.

For fixed matrices $A,B$, the map
$F:(\mathcal{W}_{n},\rho_{1})\rightarrow(M_{d}(\mathbb{C}),\left\|{\cdot}\right\|)$
given by

$F(w)=e^{w[1]/n}e^{w[2]/n}\cdots e^{w[2n]/n}$

is Lipschitz continuous, with the Lipschitz constant independent of $n$.

###### Proof.

By Proposition 16(a) and Theorem 9, for any two words $w,v\in\mathcal{W}_{n}$,

(4)
$\begin{split}\left\|{\prod_{i=1}^{2n}e^{w[i]/n}-\prod_{i=1}^{2n}e^{v[i]/n}}\right\|&\leq\frac{\operatorname{\rho_{\mathit{swap}}}(w,v)}{n^{2}}2\left\|{AB-
BA}\right\|e^{\left\|{A}\right\|+\left\|{B}\right\|}\\\
&=\rho_{1}(w,v)2\left\|{AB-
BA}\right\|e^{\left\|{A}\right\|+\left\|{B}\right\|}.\end{split}$

So the map $F$ is Lipschitz continuous, with the constant depending only on
$A$ and $B$. ∎

######  Remark 18.

One can identify the lattice paths discussed above with non-decreasing step
functions from $[0,1]$ to $[0,1]$ which (for some $n$) take values in
$\left\\{\frac{k}{n}:0\leq k\leq n\right\\}$ and are constant on the intervals
in the uniform partition of $[0,1]$ into $n$ subintervals. It is easy to check
that the closure of this space, with respect to the metric $\rho_{1}$, is the
space of _all_ increasing functions from $[0,1]$ to $[0,1]$. By Theorem 17,
the map $F$ extends continuously to a map from the space of all such
increasing functions (with the $\rho_{1}$ metric) to $M_{d}(\mathbb{C})$.

##  4\. The main result.

###### Proof of Theorem 2.

Fix $c>0$. Applying Corollary 12 with $p(n)=\sqrt{c\ln n}$, the proportion of
words $w\in\mathcal{W}_{n}$ for which

$\rho_{\infty}(w,\operatorname{\overline{\mathrm{w}}})\geq\sqrt{c\frac{\ln
n}{n}}$

is asymptotically at most

$2e^{-c\ln n}=\frac{2}{n^{c}}.$

On the other hand, by inequality (2), for $w$ with

$\rho_{\infty}(w,\operatorname{\overline{\mathrm{w}}})<\sqrt{c\frac{\ln
n}{n}}$

we also have

$\rho_{1}(w,\operatorname{\overline{\mathrm{w}}})<\sqrt{c\frac{\ln n}{n}}.$

By equation (4), for such $w$,

$\left\|{F(w)-F(\operatorname{\overline{\mathrm{w}}})}\right\|\leq\sqrt{c\frac{\ln
n}{n}}2\left\|{AB-BA}\right\|e^{\left\|{A}\right\|+\left\|{B}\right\|}.$

Finally, by Proposition 16(b), for such $w$,

(5) $\begin{split}\left\|{F(w)-e^{A+B}}\right\|&\leq\left(\sqrt{c\frac{\ln
n}{n}}+\frac{1}{2n}\right)2\left\|{AB-
BA}\right\|e^{\left\|{A}\right\|+\left\|{B}\right\|}\\\ &\leq
4\sqrt{c\frac{\ln n}{n}}\left\|{AB-
BA}\right\|e^{\left\|{A}\right\|+\left\|{B}\right\|}\end{split}$

for large $n$.

If $AB=BA$, then $F(w)=e^{A+B}$ for each $w\in\mathcal{W}_{n}$. If $AB\neq
BA$, set

$c=\frac{1}{\left(4\left\|{AB-
BA}\right\|e^{\left\|{A}\right\|+\left\|{B}\right\|}\right)^{2}}.$

It follows that the proportion of words $w\in\mathcal{W}_{n}$ with
$\left\|{F(w)-e^{A+B}}\right\|<\sqrt{\frac{\ln n}{n}}$ is at least

$1-\frac{2}{n^{c}},$

and so goes to one as $n\rightarrow\infty$. ∎

For readers familiar with probability theory, we can state a cleaner result.
We will need the following device.

######  Lemma (Borel-Cantelli).

Denote by $P(E)$ the probability of an event. If a family of events
$\left\\{E_{n}:n\in\mathbb{N}\right\\}$ has the property that the series
$\sum_{n=1}^{\infty}P(E_{n})<\infty$, then almost surely, an element $x$ lies
in at most finitely many $E_{n}$’s.

######  Corollary 19.

Let $\mathcal{W}_{n}$ and $F$ be as before. Put on $\mathcal{W}_{n}$ the
uniform measure, so that each word has probability $\dfrac{1}{\binom{2n}{n}}$.
Let $\mathcal{W}=\prod_{n=1}^{\infty}\mathcal{W}_{n}$ be the collection of all
sequences of words of progressively longer length, with the usual product
probability measure. Then for $\mathbf{w}=(w_{1},w_{2},\ldots)\in\mathcal{W}$,
almost surely with respect to this product measure,

$F(w_{n})\rightarrow e^{A+B}$

in the matrix norm as $n\rightarrow\infty$.

###### Proof.

In the proof of Theorem 2, take $c>1$. Since the series $\sum\frac{1}{n^{c}}$
converges, combining equation (5) and the Borel-Cantelli lemma, almost surely

$\left\|{F(w_{n})-e^{A+B}}\right\|\leq 4\sqrt{c\frac{\ln n}{n}}\left\|{AB-
BA}\right\|e^{\left\|{A}\right\|+\left\|{B}\right\|}$

for all but finitely many $n$. This implies that $F(w_{n})\rightarrow
e^{A+B}$. ∎

######  Remark 20.

We don’t need the full power of Theorem 2 to prove the preceding corollary.
Indeed, all we need is that for any $\varepsilon>0$,
$\left\|{F(w_{n})-e^{A+B}}\right\|<\varepsilon$ for all but finitely many
terms. This corresponds to the asymptotics of the binomial coefficient
$\binom{2n}{n-[n\varepsilon]}$ for $\varepsilon<1$. These asymptotics
(describing the large rather than moderate deviations from the mean) are both
easier and better known, namely

$\binom{2n}{n-[n\varepsilon]}\sim\sqrt{1-\varepsilon^{2}}e^{-H(\varepsilon)n}\binom{2n}{n},$

see for example Section 5.3 in [5]. Here

$H(\varepsilon)=(1+\varepsilon)\ln(1+\varepsilon)+(1-\varepsilon)\ln(1-\varepsilon).$

Since the function $x\ln x$ is concave up, $H(\varepsilon)\geq 0$. So the
series $\sum_{n}e^{-H(\varepsilon)n}$ converges, and the Borel-Cantelli lemma
still implies the result.

##  5\. The case of several matrices.

Similar results hold if instead of $A$ and $B$, we start with an $N$-tuple of
matrices $A_{1},\ldots,A_{N}\in M_{d}(\mathbb{C})$. Several parts of the
argument require modification, while others are almost the same (and so are
only outlined).

######  Definition 21.

Let $\mathcal{W}_{n}^{(N)}$ be the collection of all words of length $Nn$
containing exactly $n$ of each $A_{j}$, $1\leq j\leq N$. Define the standard
word $\operatorname{\overline{\mathrm{w}}}$ to be the word $A_{1}A_{2}\ldots
A_{N}$ repeated $n$ times. Define the swap distance exactly as before,

$\rho_{1}\left(w,v\right)=\frac{1}{N^{2}n^{2}}\sum_{j=1}^{Nn}\sum_{k=1}^{N}\Bigl{|}w_{A_{k}}[j]-v_{A_{k}}[j]\Bigr{|},$

and

$\rho_{\infty}\left(w,v\right)=\frac{2}{n}\max_{\begin{subarray}{c}1\leq k\leq
N\\\ 1\leq j\leq Nn\end{subarray}}\Bigl{|}w_{A_{k}}[j]-v_{A_{k}}[j]\Bigr{|}.$

We also define $F$ by exactly the same formula as in Theorem 17.

######  Example 22.

$\operatorname{\rho_{\mathit{swap}}}$ and $\rho_{1}$ no longer determine each
other. Indeed, omitting the normalization factor,

$\rho_{1}(ACB,BCA)=2+2+0=4\quad\text{while}\quad\operatorname{\rho_{\mathit{swap}}}(ACB,BCA)=3.$

On the other hand,

$\rho_{1}(ABC,BCA)=2+1+1=4\quad\text{while}\quad\operatorname{\rho_{\mathit{swap}}}(ABC,BCA)=2.$

However, all we really need is an inequality between
$\operatorname{\rho_{\mathit{swap}}}$ and $\rho_{\infty}$, which still holds.

######  Proposition 23.

$\operatorname{\rho_{\mathit{swap}}}\leq\frac{1}{2}N^{2}n^{2}\rho_{\infty}$.

###### Proof.

Suppose the $i$’th position is the first one where $w$ and $v$ differ, and
$w[i]=A_{k}$. Then the next $A_{k}$ appears in $v$ no later than
$N\frac{n}{2}\rho_{\infty}(w,v)$ positions away. So no more than $(Nn)\cdot
N\frac{n}{2}\rho_{\infty}(w,v)$ swaps are necessary to transform $v$ into $w$.
∎

Counting exactly the number of words which lie within a given $\rho_{\infty}$
distance from the standard word is a difficult question, see Section 10.17 in
[2]. For our needs, the following slightly different estimate suffices.

######  Definition 24.

Let $w\in W_{n}^{(N)}$. Denote

$\tau(w)=\frac{1}{n}\max_{\begin{subarray}{c}1\leq k,\ell\leq N\\\ 1\leq j\leq
Nn\end{subarray}}\Bigl{|}w_{A_{k}}[j]-w_{A_{\ell}}[j]\Bigr{|}.$

Just like $\rho_{\infty}(w,\operatorname{\overline{\mathrm{w}}})$, $\tau(w)$
measures how far the path corresponding to $w$ is from the straight path
connecting the origin to $(n,n,\ldots,n)$. In fact,

######  Lemma 25.

For any $w\in W_{n}^{(N)}$,

$\frac{1}{2}\rho_{\infty}(w,\operatorname{\overline{\mathrm{w}}})\leq\tau(w)+\frac{1}{n}.$

###### Proof.

Note that
$\operatorname{\overline{\mathrm{w}}}_{A_{k}}[j]=\left[\frac{j+N-k}{N}\right]$
and $\sum_{\ell=1}^{N}w_{A_{\ell}}[j]=j$. So

$\begin{split}\Bigl{|}w_{A_{k}}[j]-\operatorname{\overline{\mathrm{w}}}_{A_{k}}[j]\Bigr{|}&\leq\left|w_{A_{k}}[j]-\frac{j}{N}\right|+\left|\left[\frac{j+N-k}{N}\right]-\frac{j}{N}\right|\\\
&\leq\frac{1}{N}\sum_{\ell=1}^{N}\Bigl{|}w_{A_{k}}[j]-w_{A_{\ell}}[j]\Bigr{|}+1\\\
&\leq\max_{k,\ell}\Bigl{|}w_{A_{k}}[j]-w_{A_{\ell}}[j]\Bigr{|}+1.\qed\end{split}$

######  Proposition 26.

  1. (a)

The number of words $w\in\mathcal{W}_{n}^{(N)}$ for which the $\rho_{\infty}$
distance to the standard word is at least $(2M+2)/n$ is at most
$2N^{2}\binom{Nn}{n-M,n+M,n,\ldots,n}$.

  2. (b)

Let $p(n)=o(n^{1/6})$. Then the proportion of words $w\in\mathcal{W}_{n}$ for
which the $\rho_{\infty}$ distance to the standard word is at least
$\frac{p(n)}{\sqrt{n}}$ goes to zero as $n\rightarrow\infty$.

###### Proof.

For part (a), by the preceding lemma, it suffices to consider words with
$n\tau(w)\geq M$. Let $j$ be the smallest index such that for some $k,\ell$,

$\Bigl{|}w_{A_{k}}[j]-w_{A_{\ell}}[j]\Bigr{|}=M,$

and let $k$ and $\ell$ be the indices for which this occurs. Then applying the
reflection principle as in Proposition 10 just to the letters $A_{k}$ and
$A_{\ell}$ (keeping all the other letters in their places), the number of such
paths is at most the multinomial coefficient
$2\binom{Nn}{n-M,n+M,n,\ldots,n}$. Since there are $N^{2}$ choices for the
pair $(k,\ell)$, the result follows.

For part (b), we note that the ratio of multinomial coefficients

$\frac{\binom{Nn}{n-M,n+M,n,\ldots,n}}{\binom{Nn}{n,\ldots,n}}=\frac{n!n!}{(n-M)!(n+M)!}=\frac{\binom{2n}{n-M}}{\binom{2n}{n}}.$

So the direct application of Corollary 12 gives the result. ∎

######  Corollary 27.

Put on $\mathcal{W}_{n}^{(N)}$ the uniform measure, and let
$\mathcal{W}^{(N)}=\prod_{n=1}^{\infty}\mathcal{W}_{n}^{(N)}$, with the usual
product probability measure. Then for
$\mathbf{w}=(w_{1},w_{2},\ldots)\in\mathcal{W}^{(N)}$, almost surely with
respect to this product measure,

$F(w_{n})\rightarrow e^{A_{1}+\ldots+A_{N}}$

in the matrix norm as $n\rightarrow\infty$.

Acknowledgements. The first author would like to thank Matthew Junge, who
reminded him that words can be treated as random walks. The authors are
grateful to Harold Boas for numerous comments, which led to a substantial
improvement of the article (the remaining errors are, of course, our own), and
to the reviewers for a careful reading of the manuscript and helpful comments.

## References

  * [1] Herzog, G. (2014). A proof of Lie’s product formula. _Amer. Math. Monthly._ 121(3): 254–257.
  * [2] Krattenthaler, C. (2015). Lattice path enumeration. In: Bóna, M., ed. _Handbook of enumerative combinatorics_. Discrete Math. Appl. (Boca Raton). Boca Raton, FL: CRC Press, pp. 589–678.
  * [3] Panny, W., Prodinger, H. (1985). The expected height of paths for several notions of height. _Studia Sci. Math. Hungar._ 20(1-4): 119–132.
  * [4] Renault, M. (2008). Lost (and found) in translation: André’s actual method and its application to the generalized ballot problem. _Amer. Math. Monthly._ 115(4): 358–363.
  * [5] Spencer, S. (2014). _Asymptopia_. Student Mathematical Library, vol. 71. Providence, RI: American Mathematical Society. With Laura Florescu.

##  Appendix A The set of products.

We now return to the case of two matrices. The results earlier in the paper
are concerned with the asymptotic _density_ of the sets $F(\mathcal{W}_{n})$.
A quite different question is to give an asymptotic description of these sets
themselves, or perhaps of the closure

$\mathcal{S}=\overline{\bigcup_{n=1}^{\infty}F(\mathcal{W}_{n})}\subset
M_{d}(\mathbb{C}).$

It is clear that $\mathcal{S}$ is connected, closed, and bounded. It is also
easy to check that all of its elements have same determinant, and so lie in a
hypersurface. Beyond these elementary properties, we do not in general have a
good description of this set. Several examples are included in Figure 3. The
two-dimensional images may be hard to read, so the reader is invited to take
advantage of the three-dimensional functionality at
https://austinpritchett.shinyapps.io/nexpm_visualization/

Figure 3. The sets $F(\mathcal{W}_{8})$ for several choices of $A$ and $B$. We
plot the $(1,1),(1,2)$, and $(2,1)$ entries, and indicate the value of the
$(2.2)$ entry via the color.

We finish with an example where the set $\mathcal{S}$, and in fact the
function $F:\mathcal{W}_{n}\rightarrow M_{2}(\mathbb{C})$, can be described
completely. Denote by $E_{ij}$ the matrix with a $1$ in the $(i,j)$ position,
and $0$’s elsewhere.

######  Remark 28.

Recall that in Remark 18 we identified a word with a non-decreasing step
function from $[0,1]$ to $[0,1]$. Here is an explicit description of this
correspondence. For the $i$’th $A$ in $w$, denote by $h_{i}(w)$ the number of
$B$’s which have appeared in $w$ before it. Equivalently, the $i$’th $A$
appears in position $i+h_{i}(w)$ in $w$. Then the function
$L_{w}:[0,1]\rightarrow[0,1]$ corresponding to $w$ takes the value $h_{i}(w)$
on the interval $\left(\frac{i-1}{n},\frac{i}{n}\right)$. See Figure 1.

######  Theorem 29.

Let $A=E_{12}$ and $B=E_{11}$ in $M_{2}(\mathbb{C})$. Using the notation just
above,

  1. (a)

For $w\in\mathcal{W}_{n}$,

$F(w)=\begin{pmatrix}e&\frac{1}{n}\left(e^{h_{1}(w)/n}+\ldots+e^{h_{n}(w)/n}\right)\\\
0&1\end{pmatrix}.$

  2. (b)

For a general increasing function $L:[0,1]\rightarrow[0,1]$ as in Remark 18,

$F(L)=\begin{pmatrix}e&\int_{0}^{1}e^{L(x)}\,dx\\\ 0&1\end{pmatrix}.$

###### Proof.

It suffices to prove recursively that for any $1\leq j\leq 2n$,

$\prod_{i=1}^{j}e^{w[i]/n}=\begin{pmatrix}e^{w_{B}[j]/n}&\frac{1}{n}\sum_{i=1}^{w_{A}[j]}e^{h_{i}(w)/n}\\\
0&1\end{pmatrix}.$

First consider $j=1$. If the first letter of $w$ is $A$,

$e^{A/n}=\begin{pmatrix}1&\frac{1}{n}\\\
0&1\end{pmatrix}=\begin{pmatrix}e^{w_{B}[1]/n}&\frac{1}{n}\sum_{i=1}^{w_{A}[1]}e^{h_{i}(w)/n}\\\
0&1\end{pmatrix}$

since $h_{1}(w)=0$. If the first letter of $w$ is $B$,

$e^{B/n}=\begin{pmatrix}e^{1/n}&0\\\
0&1\end{pmatrix}=\begin{pmatrix}e^{w_{B}[1]/n}&\frac{1}{n}\sum_{i=1}^{w_{A}[1]}e^{h_{i}(w)/n}\\\
0&1\end{pmatrix}$

since the latter sum is empty. Now recursively, if $w[j+1]=A$,

$\begin{split}&\begin{pmatrix}e^{w_{B}[j]/n}&\frac{1}{n}\sum_{i=1}^{w_{A}[j]}e^{h_{i}(w)/n}\\\
0&1\end{pmatrix}\ \begin{pmatrix}1&\frac{1}{n}\\\ 0&1\end{pmatrix}\\\
&\qquad=\begin{pmatrix}e^{w_{B}[j]/n}&\frac{1}{n}\sum_{i=1}^{w_{A}[j]}e^{h_{i}(w)/n}+\frac{1}{n}e^{w_{B}[j]/n}\\\
0&1\end{pmatrix}\\\
&\qquad=\begin{pmatrix}e^{w_{B}[j+1]/n}&\frac{1}{n}\sum_{i=1}^{w_{A}[j+1]}e^{h_{i}(w)/n}\\\
0&1\end{pmatrix}\end{split}$

Indeed, since $w[j+1]$ is the $w_{A}[j+1]$’th $A$ in $w$, we have
$w_{B}[j+1]=w_{B}[j]=h_{i}(w)$ for $i=w_{A}[j+1]$. Similarly, if $w[j+1]=B$,

$\begin{split}&\begin{pmatrix}e^{w_{B}[j]/n}&\frac{1}{n}\sum_{i=1}^{w_{A}[j]}e^{h_{i}(w)/n}\\\
0&1\end{pmatrix}\ \begin{pmatrix}e^{1/n}&0\\\ 0&1\end{pmatrix}\\\
&\qquad=\begin{pmatrix}e^{(w_{B}[j]+1)/n}&\frac{1}{n}\sum_{i=1}^{w_{A}[j]}e^{h_{i}(w)/n}\\\
0&1\end{pmatrix}\\\
&\qquad=\begin{pmatrix}e^{w_{B}[j+1]/n}&\frac{1}{n}\sum_{i=1}^{w_{A}[j+1]}e^{h_{i}(w)/n}\\\
0&1\end{pmatrix}\end{split}$

since $w_{A}[j+1]=w_{A}[j]$.

Part (b) follows: the expression in part (a) is the Riemann sum for the
integral $\int_{0}^{1}e^{L(x)}\,dx$, and since the function $L$ is increasing,
it is Riemann integrable. ∎

######  Remark 30.

In the example above, $\mathcal{S}$ is a (one-dimensional) curve from
$e^{A}e^{B}$ to $e^{B}e^{A}$. There are several general situations where this
behavior also occurs. Denoting $[A,B]=AB-BA$ the commutator of $A$ and $B$,
these include

  * •

Quasi-commuting matrices [1] for which the commutator is non-zero but commutes
with both $A$ and $B$,

  * •

Matrices which satisfy $[A,B]=sB$,

  * •

$A=\begin{pmatrix}x&1\\\ 0&x\end{pmatrix},\quad B=\begin{pmatrix}a&0\\\
0&b\end{pmatrix}$

for $a\neq b$.

## References

  * [1] Neal H. McCoy, _On quasi-commutative matrices_ , Trans. Amer. Math. Soc. 36 (1934), no. 2, 327–340. MR 1501746


# Gradient estimates for a nonlinear diffusion equation on complete manifolds

Jia-Yong Wu  Department of Mathematics, East China Normal University,
Shanghai, China 200241<EMAIL_ADDRESS>
(Date: January 1, 2009.)

###### Abstract.

Let $(M,g)$ be a complete non-compact Riemannian manifold with the
$m$-dimensional Bakry-Émery Ricci curvature bounded below by a non-positive
constant. In this paper, we give a localized Hamilton-type gradient estimate
for the positive smooth bounded solutions to the following nonlinear diffusion
equation

$u_{t}=\Delta u-\nabla\phi\cdot\nabla u-au\log u-bu,$

where $\phi$ is a $C^{2}$ function, and $a\neq 0$ and $b$ are two real
constants. This work generalizes the results of Souplet and Zhang (Bull.
London Math. Soc., 38 (2006), pp. 1045-1053) and Wu (Preprint, 2008).

###### Key words and phrases:

local gradient estimate; nonlinear diffusion equation; Bakry-Émery Ricci
curvature

###### 2000 Mathematics Subject Classification:

Primary 58J35; Secondary 58J35, 58J05. Chinese Library Classification:
O175.26; O186.12

##  1\. Introduction

Let $(M,g)$ be an $n$-dimensional non-compact Riemannian manifold with the
$m$-dimensional Bakry-Émery Ricci curvature bounded below. Consider the
following diffusion equation:

(1.1) $u_{t}=\Delta u-\nabla\phi\cdot\nabla u-au\log u-bu$

in $B(x_{0},R)\times[t_{0}-T,t_{0}]\subset M\times(-\infty,\infty)$, where
$\phi$ is a $C^{2}$ function, and $a\neq 0$ and $b$ are two real constants.
Eq. (1.1) is closely linked with the gradient Ricci solitons, which are the
self-similar solutions to the Ricci flow introduced by Hamilton [3]. Ricci
solitons have inspired the entropy and Harnack estimates, the space-time
formulation of the Ricci flow, and the reduced distance and reduced volume.

Below we recall the definition of Ricci solitons (see also Chapter 4 of [4]).

######  Definition 1.1.

A Riemannian manifold $(M,g)$ is called a _gradient Ricci soliton_ if there
exists a smooth function $f:M\rightarrow\mathbb{R}$, sometimes called
_potential function_ , such that for some constant $c\in\mathbb{R}$, it
satisfies

(1.2) $Ric(g)+\nabla^{g}\nabla^{g}f=cg$

on $M$, where $Ric(g)$ is the Ricci curvature of manifold $M$ and
$\nabla^{g}\nabla^{g}f$ is the Hessian of $f$. A soliton is said to be
_shrinking_ , _steady_ or _expanding_ if the constant $c$ is respectively
positive, zero or negative.

Suppose that $(M,g)$ be a gradient Ricci soliton, and $c$, $f$ are described
in Definition A. Letting $u=e^{f}$, under some curvature assumptions, we can
derive from (1.2) that (cf. [5], Eq. (7))

(1.3) $\Delta u+2cu\log u=(A_{0}-nc)u,$

for some constant $A_{0}$. Eq. (1.3) is a nonlinear elliptic equation and a
special case of Eq. (1.1). For this kind of equations, Ma (see Theorem 1 in
[5]) obtained the following result.

Theorem A ([5]). _Let $(M,g)$ be a complete non-compact Riemannian manifold of
dimension $n\geq 3$ with Ricci curvature bounded below by the constant
$-K:=-K(2R)$, where $R>0$ and $K(2R)\geq 0$, in the metric ball $B_{2R}(p)$.
Let $u$ be a positive smooth solution to the elliptic equation_

(1.4) $\Delta u-au\log u=0$

_with $a>0$. Let $f=\log u$ and let $(f,2f)$ be the maximum among $f$ and
$2f$. Then there are two uniform positive constant $c_{1}$ and $c_{2}$ such
that_

(1.5) $\displaystyle|\nabla f|^{2}-a(f,2f)$ $\displaystyle\leq$
$\displaystyle\frac{n\Big{[}(n+2)c^{2}_{1}+(n-1)c^{2}_{1}(1+R\sqrt{K})+c_{2}\Big{]}}{R^{2}}+2n\Big{(}|a|+K\Big{)}$

_in $B_{R}(p)$._

Then Yang (see Theorem 1.1 in [6]) extended the above result and obtained the
following local gradient estimate for the nonlinear equation (1.1) with
$\phi\equiv c_{0}$, where $c_{0}$ is a fixed constant.

Theorem B ([6]). _Let $M$ be an $n$-dimensional complete non-compact
Riemannian Manifold. Suppose the Ricci curvature of $M$ is bounded below by
$-K:=-K(2R)$, where $R>0$ and $K(2R)\geq 0$, in the metric ball $B_{2R}(p)$.
If $u$ is a positive smooth solution to Eq. (1.1) with $\phi\equiv c_{0}$ on
$M\times[0,\infty)$ and $f=\log u$, then for any $\alpha>1$ and $0<\delta<1$_

(1.6) $\displaystyle|\nabla f|^{2}(x,t)-\alpha af(x,t)-\alpha b-\alpha
f_{t}(x,t)$ $\displaystyle\leq$ $\displaystyle\frac{n\alpha^{2}}{2\delta
t}+\frac{n\alpha^{2}}{2\delta}\Bigg{\\{}\frac{2\epsilon^{2}}{R^{2}}+\frac{\nu}{R^{2}}+\sigma+\frac{\epsilon^{2}}{R^{2}}(n-1)\left(1+R\sqrt{K(2R)}\right)$
$\displaystyle+\frac{K(2R)}{\alpha-1}+\frac{n\alpha^{2}\epsilon^{2}}{8(1-\delta)(\alpha-1)R^{2}}\Bigg{\\}}$

_in $B_{R}(p)\times(0,\infty)$, where $\epsilon>0$ and $\nu>0$ are some
constants and where $\sigma=a/2$ if $a>0$; $\sigma=-a$ if $a<0$._

Recently, the author (see Theorem 1.1 in [2]) used Souplet-Zhang’s method in
[1] and obtained a localized Hamilton-type gradient estimate for the positive
smooth bounded solutions of the equation (1.1) with $\phi\equiv c_{0}$.

Theorem C ([2]). _Let $(M,g)$ be an $n$-dimensional non-compact Riemannian
manifold with $Ric(M)\geq-K$ for some constant $K\geq 0$. Suppose that
$u(x,t)$ is a positive smooth solution to the parabolic equation (1.1) with
$\phi\equiv c_{0}$ in $Q_{R,T}\equiv B(x_{0},R)\times[t_{0}-T,t_{0}]\subset
M\times(-\infty,\infty)$. Let $f:=\log u$. We also assume that there exists
non-negative constants $\alpha$ and $\delta$ such that $\alpha-f\geq\delta>0$.
Then there exist three dimensional constants $\tilde{c}$, $c(\delta)$ and
$c(\alpha,\delta)$ such that_

(1.7) $\frac{|\nabla
u|}{u}\leq\left(\frac{\tilde{c}}{R}\beta{+}\frac{c(\alpha,\delta)}{R}{+}\frac{c(\delta)}{\sqrt{T}}{+}c(\delta)\left(|a|+K\right)^{1/2}\kern-2.0pt{+}c(\delta)|a|^{1/2}\beta^{1/2}\right)\left(\alpha{-}\frac{b}{a}{-}\log
u\right)$

_in $Q_{R/2,T/2}$, where $\beta:=\max\left\\{1,|\alpha/\delta-1|\right\\}$._

The purpose of this paper is to extend Theorem C to the general nonlinear
diffusion equation (1.1) via the $m$-dimensional Bakry-Émery Ricci curvature.

Let us first recall some facts about the $m$-dimensional Bakry-Émery Ricci
curvature (please see [7, 8, 9, 10] for more details). Given an
$n$-dimensional Riemannian manifold $(M,g)$ and a $C^{2}$ function $\phi$, we
may define a symmetric diffusion operator $L:=\Delta-\nabla\phi\cdot\nabla,$
which is the infinitesimal generator of the Dirichlet form

$\mathcal{E}(f,g)=\int_{M}(\nabla f,\nabla g)\mathrm{d}\mu,\,\,\,\forall
f,g\in C_{0}^{\infty}(M),$

where $\mu$ is an invariant measure of $L$ given by
$\mathrm{d}\mu=e^{-\phi}\mathrm{d}x.$ It is well-known that $L$ is self-
adjoint with respect to the weighted measure $\mathrm{d}\mu$.

The $\infty$-dimensional Bakry-Émery Ricci curvature $Ric(L)$ is defined by

$Ric(L):=Ric+Hess(\phi),$

where $Ric$ and $Hess$ denote the Ricci curvature of the metric $g$ and the
Hessian respectively. Following the notation used in [10], we also define the
$m$-dimensional Bakry-Émery Ricci curvature of $L$ on an $n$-dimensional
Riemaniann manifold as follows

$Ric_{m,n}(L):=Ric(L)-\frac{\nabla\phi\otimes\nabla\phi}{m-n},$

where $m:=\mathrm{dim}_{BE}(L)$ is called the Bakry-Émery dimension of $L$.
Note that the number $m$ is not necessarily to be an integer and $m\geq
n=\mathrm{dim}M$.

The main result of this paper can be stated in the following:

######  Theorem 1.2.

Let $(M,g)$ be an n-dimensional non-compact Riemannian manifold with
$Ric_{m,n}(L)\geq-K$ for some constant $K\geq 0$. Suppose that $u(x,t)$ is a
positive smooth solution to the diffusion equation (1.1) in $Q_{R,T}\equiv
B(x_{0},R)\times[t_{0}-T,t_{0}]\subset M\times(-\infty,\infty)$. Let $f:=\log
u$. We also assume that there exists non-negative constants $\alpha$ and
$\delta$ such that $\alpha-f\geq\delta>0$. Then there exist three dimensional
constants $\tilde{c}$, $c(\delta)$ and $c(\alpha,\delta,m)$ such that

(1.8) $\frac{|\nabla
u|}{u}\leq\left(\frac{\tilde{c}}{R}\beta{+}\frac{c(\alpha,\delta,m)}{R}{+}\frac{c(\delta)}{\sqrt{T}}{+}c(\delta)\left(|a|+K\right)^{1/2}\kern-3.0pt{+}c(\delta)|a|^{1/2}\beta^{1/2}\right)\left(\alpha{-}\frac{b}{a}{-}\log
u\right)$

in $Q_{R/2,T/2}$, where $\beta:=\max\left\\{1,|\alpha/\delta-1|\right\\}$.

We make some remarks on the above theorem below.

######  Remark 1.3.

(i). In Theorem 1.2, it seems that the assumption $\alpha-f\geq\delta>0$ is
reasonable. Because from this assumption, we can get $u\leq
e^{\alpha-\delta}$. We say that this upper bound of $u$ can be achieved in
some setting. For example, from Corollary 1.2 in [6], we know that positive
smooth solutions to the elliptic equation (1.4) with $a<0$ have $u(x)\leq
e^{n/2}$ for all $x\in M$ provided the Ricci curvature of $M$ is non-negative.

(ii). Note that the theorem still holds if $m$-dimensional Bakry-Émery Ricci
curvature is replaced by $\infty$-dimensional Bakry-Émery Ricci curvature. In
fact this result can be obtained by (2.10) in Section 2.

(iii). Theorem 1.2 generalizes the above mentioned Theorem C. When we choose
$\phi\equiv c_{0}$, we return Theorem C. The proof of our main theorem is
based on Souplet-Zhang’s gradient estimate and the trick used in [2] with some
modifications.

In particular, if $u(x,t)\leq 1$ is a positive smooth solution to the
diffusion equation (1.1) with $a<0$, then we have a simple estimate.

######  Corollary 1.4.

Let $(M,g)$ be an n-dimensional non-compact Riemannian manifold with
$Ric_{m,n}(L)\geq-K$ for some constant $K\geq 0$. Suppose that $u(x,t)\leq 1$
is a positive smooth solution to the diffusion equation (1.1) with $a<0$ in
$Q_{R,T}\equiv B(x_{0},R)\times[t_{0}-T,t_{0}]\subset
M\times(-\infty,\infty)$. Then there exist two dimensional constants $c$ and
$c(m)$ such that

(1.9) $\frac{|\nabla
u|}{u}\leq\left(\frac{c(m)}{R}+\frac{c}{\sqrt{T}}+c\sqrt{K+|a|}\right)\left(1-\frac{b}{a}+\log{\frac{1}{u}}\right)$

in $Q_{R/2,T/2}$.

######  Remark 1.5.

We point out that our localized Hamilton-type gradient estimate can be also
regarded as the generalization of the result of Souplet-Zhang [1] for the heat
equation on complete manifolds. In fact, the above Corollary 1.4 is similar to
the result of Souplet-Zhang (see Theorem 1.1 of [1]). From the inequality
(4.4) below, we can conclude that if $\phi\equiv c_{0}$ and $a=0$, then our
result can be reduced to theirs.

The method of proving Theorem 1.2 is the gradient estimate, which is
originated by Yau [11] (see also Cheng-Yau [12]), and developed further by Li-
Yau [13], Li [14] and Negrin [15]. Then R. S. Hamilton [16] gave an elliptic
type gradient estimate for the heat equation. But this type of estimate is a
global result which requires the heat equation defined on closed manifolds.
Recently, a localized Hamilton-type gradient estimate was proved by Souplet
and Zhang [1], which can be viewed as a combination of Li-Yau’s Harnack
inequality [13] and Hamilton’s gradient estimate [16]. In this paper, we
obtain a localized Hamilton-type gradient estimate for a general diffusion
equation (1.1) as Souplet and Zhang in [1] did for the heat equation on
complete manifolds. To prove Theorem 1.2, we mainly follow the arguments of
Souplet-Zhang in [1], together with some facts about Bakry-Émery Ricci
curvature. Note that the diffusion equation (1.1) is nonlinear. So our case is
a little more complicated than theirs.

The structure of this paper is as follows. In Section 2, we will give a basic
lemma to prepare for proving Theorem 1.2. Section 3 is devoted to the proof of
Theorem 1.2. In Section 4, we will prove Corollary 1.4 in the case $0<u\leq 1$
with $a<0$.

##  2\. A basic lemma

In this section, we will prove the following lemma which is essential in the
derivation of the gradient estimate of the equation (1.1). Replacing $u$ by
$e^{-b/a}u$, we only need to consider positive smooth solutions of the
following diffusion equation:

(2.1) $u_{t}=\Delta u-\nabla\phi\cdot\nabla u-au\log u.$

Suppose that $u(x,t)$ is a positive smooth solution to the diffusion equation
(1.1) in $Q_{R,T}\equiv B(x_{0},R)\times[t_{0}-T,t_{0}]$. Define a smooth
function

$f(x,t):=\log u(x,t)$

in $Q_{R,T}$. By (2.1), we have

(2.2) $\left(L-\frac{\partial}{\partial t}\right)f+|\nabla f|^{2}-af=0.$

Then we have the following lemma, which is a generalization of the computation
carried out in [1, 2].

######  Lemma 2.1.

Let $(M,g)$ be an n-dimensional non-compact Riemannian manifold with
$Ric_{m,n}(L)\geq-K$ for some constant $K\geq 0$. Let $f(x,t)$ is a smooth
function defined on $Q_{R,T}$ satisfying the diffusion equation (2.2). We also
assume that there exist non-negative constants $\alpha$ and $\delta$ such that
$\alpha-f\geq\delta>0$. Then for all $(x,t)$ in $Q_{R,T}$ the function

(2.3) $\omega:=\left|\nabla\log(\alpha-f)\right|^{2}=\frac{|\nabla
f|^{2}}{(\alpha-f)^{2}}$

satisfies the following inequality

(2.4) $\displaystyle\left(L-\frac{\partial}{\partial t}\right)\omega$
$\displaystyle\geq$
$\displaystyle\frac{2(1-\alpha)+2f}{\alpha-f}\left\langle\nabla
f,\nabla\omega\right\rangle+2(\alpha-f)\omega^{2}+2(a-K)\omega+\frac{2af}{\alpha-f}\omega.$

###### Proof.

By (2.3), we have

(2.5)
$\displaystyle\omega_{j}=\frac{2f_{i}f_{ij}}{(\alpha-f)^{2}}+\frac{2f^{2}_{i}f_{j}}{(\alpha-f)^{3}},$

(2.6)
$\displaystyle\Delta\omega=\frac{2|f_{ij}|^{2}}{(\alpha-f)^{2}}+\frac{2f_{i}f_{ijj}}{(\alpha-f)^{2}}+\frac{8f_{i}f_{j}f_{ij}}{(\alpha-f)^{3}}+\frac{2f^{2}_{i}f_{jj}}{(\alpha-f)^{3}}+\frac{6f^{2}_{i}f^{2}_{j}}{(\alpha-f)^{4}}$

and

(2.7) $\displaystyle L\omega$ $\displaystyle=\Delta\omega-\phi_{j}\omega_{j}$
$\displaystyle=\frac{2|f_{ij}|^{2}}{(\alpha{-}f)^{2}}+\frac{2f_{i}f_{ijj}}{(\alpha{-}f)^{2}}+\frac{8f_{i}f_{j}f_{ij}}{(\alpha{-}f)^{3}}+\frac{2f^{2}_{i}f_{jj}}{(\alpha{-}f)^{3}}+\frac{6f^{4}_{i}}{(\alpha{-}f)^{4}}-\frac{2f_{ij}f_{i}\phi_{j}}{(\alpha{-}f)^{2}}-\frac{2f^{2}_{i}f_{j}\phi_{j}}{(\alpha{-}f)^{3}}$
$\displaystyle=\frac{2|f_{ij}|^{2}}{(\alpha{-}f)^{2}}+\frac{2f_{i}(Lf)_{i}}{(\alpha{-}f)^{2}}+\frac{2(R_{ij}+\phi_{ij})f_{i}f_{j}}{(\alpha{-}f)^{2}}+\frac{8f_{i}f_{j}f_{ij}}{(\alpha{-}f)^{3}}+\frac{2f^{2}_{i}\cdot
Lf}{(\alpha{-}f)^{3}}+\frac{6f^{4}_{i}}{(\alpha{-}f)^{4}},$

where $f_{i}:=\nabla_{i}f$ and $f_{ijj}:=\nabla_{j}\nabla_{j}\nabla_{i}f$,
etc. By (2.3) and (2.2), we also have

(2.8) $\displaystyle\omega_{t}$
$\displaystyle=\frac{2\nabla_{i}f\cdot\nabla_{i}\Big{[}Lf+|\nabla
f|^{2}-af\Big{]}}{(\alpha-f)^{2}}+\frac{2|\nabla f|^{2}\Big{[}Lf+|\nabla
f|^{2}-af\Big{]}}{(\alpha-f)^{3}}$ $\displaystyle=\frac{2\nabla f\nabla
Lf}{(\alpha-f)^{2}}+\frac{4f_{i}f_{j}f_{ij}}{(\alpha-f)^{2}}-\frac{2a|\nabla
f|^{2}}{(\alpha-f)^{2}}+\frac{2f^{2}_{i}Lf}{(\alpha-f)^{3}}+\frac{2|\nabla
f|^{4}}{(\alpha-f)^{3}}-\frac{2af|\nabla f|^{2}}{(\alpha-f)^{3}}.$

Combining (2.7) with (2.8), we can get

(2.9) $\displaystyle\left(L-\frac{\partial}{\partial t}\right)\omega$
$\displaystyle=\frac{2|f_{ij}|^{2}}{(\alpha-f)^{2}}+\frac{2(R_{ij}+\phi_{ij})f_{i}f_{j}}{(\alpha-f)^{2}}+\frac{8f_{i}f_{j}f_{ij}}{(\alpha-f)^{3}}+\frac{6f^{4}_{i}}{(\alpha-f)^{4}}$
$\displaystyle\,\,\,\,\,\,-\frac{4f_{i}f_{j}f_{ij}}{(\alpha-f)^{2}}-\frac{2f^{4}_{i}}{(\alpha-f)^{3}}+\frac{2af^{2}_{i}}{(\alpha-f)^{2}}+\frac{2aff^{2}_{i}}{(\alpha-f)^{3}}.$

Noting that $Ric_{m,n}(L)\geq-K$ for some constant $K\geq 0$, we have

(2.10) $(R_{ij}+\phi_{ij})f_{i}f_{j}\geq\frac{|\nabla\phi\cdot\nabla
f|^{2}}{m-n}-K|\nabla f|^{2}\geq-K|\nabla f|^{2}.$

By (2.5), we have

(2.11)
$\displaystyle\omega_{j}f_{j}=\frac{2f_{i}f_{j}f_{ij}}{(\alpha-f)^{2}}+\frac{2f^{2}_{i}f^{2}_{j}}{(\alpha-f)^{3}},$

and consequently,

(2.12) $\displaystyle
0=-2\omega_{j}f_{j}+\frac{4f_{i}f_{j}f_{ij}}{(\alpha-f)^{2}}+\frac{4f^{4}_{i}}{(\alpha-f)^{3}},$
(2.13) $\displaystyle
0=\frac{1}{\alpha-f}\left[2\omega_{j}f_{j}-\frac{4f^{4}_{i}}{(\alpha-f)^{3}}\right]-\frac{4f_{i}f_{j}f_{ij}}{(\alpha-f)^{3}}.$

Substituting (2.10) into (2.9) and then adding (2.9) with (2.12) and (2.13),
we can get

(2.14) $\displaystyle\left(L-\frac{\partial}{\partial t}\right)\omega$
$\displaystyle\geq\frac{2|f_{ij}|^{2}}{(\alpha-f)^{2}}-\frac{2K|\nabla
f|^{2}}{(\alpha-f)^{2}}+\frac{4f_{i}f_{j}f_{ij}}{(\alpha-f)^{3}}+\frac{2f^{4}_{i}}{(\alpha-f)^{4}}+\frac{2f^{4}_{i}}{(\alpha-f)^{3}}$
$\displaystyle\,\,\,\,\,\,+\frac{2(1-\alpha)+2f}{\alpha-f}f_{i}\omega_{i}+\frac{2af^{2}_{i}}{(\alpha-f)^{2}}+\frac{2aff^{2}_{i}}{(\alpha-f)^{3}}.$

Note that $\alpha-f\geq\delta>0$ implies

$\displaystyle\frac{2|f_{ij}|^{2}}{(\alpha-f)^{2}}+\frac{4f_{i}f_{j}f_{ij}}{(\alpha-f)^{3}}+\frac{2f^{4}_{i}}{(\alpha-f)^{4}}\geq
0.$

This, together with (2.14), yields the desired estimate (LABEL:lemmaequ3).

∎

##  3\. Proof of Theorem 1.2

In this section, we will use Lemma 2.1 and the localization technique of
Souplet-Zhang [1] to give the elliptic type gradient estimates on the positive
and bounded smooth solutions of the diffusion equation (1.1).

###### Proof.

First we give the well-known cut-off function by Li-Yau [13] (see also [1]) as
follows. We caution the reader that the calculation is not the same as that in
[13] due to the difference of the first-order term.

Let $\psi=\psi(x,t)$ be a smooth cut-off function supported in $Q_{R,T}$
satisfying the following properties:

  1. (1)

$\psi=\psi(d(x,x_{0}),t)\equiv\psi(r,t)$; $\psi(x,t)=1$ in $Q_{R/2,T/2}$,
$0\leq\psi\leq 1$;

  1. (2)

$\psi$ is decreasing as a radial function in the spatial variables;

  1. (3)

$\frac{|\partial_{r}\psi|}{\psi^{\epsilon}}\leq\frac{C_{\epsilon}}{R}$,
$\frac{|\partial^{2}_{r}\psi|}{\psi^{\epsilon}}\leq\frac{C_{\epsilon}}{R^{2}}$,
when $0<\epsilon<1$;

  1. (4)

$\frac{|\partial_{t}\psi|}{\psi^{1/2}}\leq\frac{C}{T}$.

From Lemma 2.1, by a straight forward calculation, we have

(3.1) $\displaystyle L(\psi\omega)-\frac{2(1-\alpha)+2f}{\alpha-f}\nabla
f\cdot\nabla(\psi\omega)-2\frac{\nabla\psi}{\psi}\cdot\nabla(\psi\omega)-(\psi\omega)_{t}$
$\displaystyle\geq$ $\displaystyle
2\psi(\alpha-f)\omega^{2}-\left[\frac{2(1-\alpha)+2f}{\alpha-f}\nabla
f\cdot\nabla\psi\right]\omega-2\frac{|\nabla\psi|^{2}}{\psi}\omega$
$\displaystyle+(L\psi)\omega-\psi_{t}\omega+2(a-K)\psi\omega+2\frac{af}{\alpha-f}\psi\omega.$

Let $(x_{1},t_{1})$ be a point where $\psi\omega$ achieves the maximum. By Li-
Yau [13], without loss of generality we assume that $x_{1}$ is not in the cut-
locus of $M$. Then at this point, we have

$\displaystyle L(\psi\omega)\leq 0,\,\,\,\,\,\,(\psi\omega)_{t}\geq
0,\,\,\,\,\,\,\nabla(\psi\omega)=0.$

Hence at $(x_{1},t_{1})$, by (LABEL:lemdx3), we get

(3.2) $\displaystyle 2\psi(\alpha-f)\omega^{2}(x_{1},t_{1})$
$\displaystyle\leq\Bigg{\\{}\left[\frac{2(1-\alpha)+2f}{\alpha-f}\nabla
f\cdot\nabla\psi\right]\omega+2\frac{|\nabla\psi|^{2}}{\psi}\omega-(L\psi)\omega$
$\displaystyle\,\,\,\,\,\,\,\,\,+\psi_{t}\omega-2(a-K)\psi\omega-2\frac{af}{\alpha-f}\psi\omega\Bigg{\\}}(x_{1},t_{1}).$

In the following, we will introduce the upper bounds for each term of the
right-hand side (RHS) of (3.2). Following similar arguments of Souplet-Zhang
([1], pp. 1050-1051), we have the estimates of the first term of the BHS of
(3.2)

(3.3) $\displaystyle\left[\frac{2f}{\alpha-f}\nabla
f\cdot\nabla\psi\right]\omega$ $\displaystyle\leq$ $\displaystyle
2|f|\cdot|\nabla\psi|\cdot\omega^{3/2}=2\left[\psi(\alpha-f)\omega^{2}\right]^{3/4}\cdot\frac{|f|\cdot|\nabla\psi|}{[\psi(\alpha-f)]^{3/4}}$
$\displaystyle\leq$
$\displaystyle\psi(\alpha-f)\omega^{2}+\tilde{c}\frac{(f|\nabla\psi|)^{4}}{[\psi(\alpha-f)]^{3}}\leq\psi(\alpha-f)\omega^{2}+\tilde{c}\frac{f^{4}}{R^{4}(\alpha-f)^{3}}$

and

(3.4) $\displaystyle\left[\frac{2(1-\alpha)}{\alpha-f}\nabla
f\cdot\nabla\psi\right]\omega$ $\displaystyle\leq$ $\displaystyle
2|1-\alpha||\nabla\psi|\omega^{3/2}=(\psi\omega^{2})^{3/4}\cdot\frac{2|1-\alpha||\nabla\psi|}{\psi^{3/4}}$
$\displaystyle\leq$
$\displaystyle\frac{\delta}{12}\psi\omega^{2}+c(\alpha,\delta)\left(\frac{|\nabla\psi|}{\psi^{3/4}}\right)^{4}\leq\frac{\delta}{12}\psi\omega^{2}+\frac{c(\alpha,\delta)}{R^{4}}.$

For the second term of the RHS of (3.2), we have

(3.5) $\displaystyle 2\frac{|\nabla\psi|^{2}}{\psi}\omega$
$\displaystyle=2\psi^{1/2}\omega\cdot\frac{|\nabla\psi|^{2}}{\psi^{3/2}}\leq\frac{\delta}{12}\psi\omega^{2}+c(\delta)\left(\frac{|\nabla\psi|^{2}}{\psi^{3/2}}\right)^{2}$
$\displaystyle\leq\frac{\delta}{12}\psi\omega^{2}+\frac{c(\delta)}{R^{4}}.$

For the third term of the RHS of (3.2), since $Ric_{m,n}(L)\geq-K$, by the
generalized Laplacian comparison theorem (see [9] or [10]),

$Lr\leq(m-1)\sqrt{K}\coth(\sqrt{K}r).$

Consequently, we have

(3.6) $\displaystyle-(L\psi)\omega$
$\displaystyle=-\left[(\partial_{r}\psi)Lr+(\partial^{2}_{r}\psi)\cdot|\nabla
r|^{2}\right]\omega$
$\displaystyle\leq-\left[\partial_{r}\psi(m-1)\sqrt{K}\coth(\sqrt{K}r)+\partial^{2}_{r}\psi\right]\omega$
$\displaystyle\leq-\left[\partial_{r}\psi(m-1)\left(\frac{1}{r}+\sqrt{K}\right)+\partial^{2}_{r}\psi\right]\omega$
$\displaystyle\leq\left[|\partial^{2}_{r}\psi|+2(m-1)\frac{|\partial_{r}\psi|}{R}+(m-1)\sqrt{K}|\partial_{r}\psi|\right]\omega$
$\displaystyle\leq\psi^{1/2}\omega\frac{|\partial^{2}_{r}\psi|}{\psi^{1/2}}+\psi^{1/2}\omega
2(m-1)\frac{|\partial_{r}\psi|}{R\psi^{1/2}}+\psi^{1/2}\omega(m-1)\frac{\sqrt{K}|\partial_{r}\psi|}{\psi^{1/2}}$
$\displaystyle\leq\frac{\delta}{12}\psi\omega^{2}+c(\delta,m)\left[\left(\frac{|\partial^{2}_{r}\psi|}{\psi^{1/2}}\right)^{2}+\left(\frac{|\partial_{r}\psi|}{R\psi^{1/2}}\right)^{2}+\left(\frac{\sqrt{K}|\partial_{r}\psi|}{\psi^{1/2}}\right)^{2}\right]$
$\displaystyle\leq\frac{\delta}{12}\psi\omega^{2}+\frac{c(\delta,m)}{R^{4}}+\frac{c(\delta,m)K}{R^{2}}.$

Now we estimate the fourth term:

(3.7) $\displaystyle|\psi_{t}|\omega$
$\displaystyle=\psi^{1/2}\omega\frac{|\psi_{t}|}{\psi^{1/2}}\leq\frac{\delta}{12}\left(\psi^{1/2}\omega\right)^{2}+c(\delta)\left(\frac{|\psi_{t}|}{\psi^{1/2}}\right)^{2}$
$\displaystyle\leq\frac{\delta}{12}\psi\omega^{2}+\frac{c(\delta)}{T^{2}}.$

Notice that we have used Young’s inequality below in obtaining
(LABEL:term1)-(3.7):

$ab\leq\frac{a^{p}}{p}+\frac{b^{q}}{q},\,\,\,\,\,\,\forall\,\,\,p,q>0\,\,\,\mathrm{with}\,\,\,\frac{1}{p}+\frac{1}{q}=1.$

Finally, we estimate the last two terms:

(3.8) $\displaystyle-2(a-K)\psi\omega\leq
2(|a|+K)\psi\omega\leq\frac{\delta}{12}\psi\omega^{2}+c(\delta)(|a|+K)^{2};$

and

(3.9) $-2\frac{af}{\alpha-f}\psi\omega\leq
2\frac{|a|\cdot|f|}{\alpha-f}\psi\omega\leq\frac{\delta}{12}\psi\omega^{2}+c(\delta)a^{2}\frac{f^{2}}{(\alpha-f)^{2}}.$

Substituting (LABEL:term1)-(3.9) to the RHS of (3.2) at $(x_{1},t_{1})$, we
get

(3.10) $\displaystyle 2\psi(\alpha-f)\omega^{2}$
$\displaystyle\leq\psi(\alpha-f)\omega^{2}+\frac{\tilde{c}f^{4}}{R^{4}(\alpha-f)^{3}}+\frac{\delta}{2}\psi\omega^{2}+\frac{c(\alpha,\delta)}{R^{4}}+\frac{c(\delta)}{R^{4}}+\frac{c(\delta,m)}{R^{4}}$
$\displaystyle\,\,\,\,\,\,+\frac{c(\delta,m)K}{R^{2}}+\frac{c(\delta)}{T^{2}}+c(\delta)(|a|+K)^{2}+c(\delta)a^{2}\frac{f^{2}}{(\alpha-f)^{2}}.$

Recall that $\alpha-f\geq\delta>0$, (3.10) implies

(3.11) $\displaystyle\psi\omega^{2}(x_{1},t_{1})$
$\displaystyle\leq\tilde{c}\frac{f^{4}}{R^{4}(\alpha-f)^{4}}+\frac{1}{2}\psi\omega^{2}(x_{1},t_{1})+\frac{c(\alpha,\delta)}{R^{4}}+\frac{c(\delta,m)}{R^{4}}+\frac{c(\delta,m)K}{R^{2}}$
$\displaystyle\,\,\,\,\,\,+\frac{c(\delta)}{T^{2}}+c(\delta)(|a|+K)^{2}+c(\delta)a^{2}\frac{f^{2}}{(\alpha-f)^{2}}.$

Furthermore, we need to estimate the RHS of (3.11). If $f\leq 0$ and
$\alpha\geq 0$, then we have

(3.12) $\displaystyle\frac{f^{4}}{(\alpha-f)^{4}}\leq
1,\,\,\,\,\,\,\,\,\,\,\,\,\frac{f^{2}}{(\alpha-f)^{2}}\leq 1;$

if $f>0$, by the assumption $\alpha-f\geq\delta>0$, we know that

(3.13)
$\displaystyle\frac{f^{4}}{(\alpha-f)^{4}}\leq\frac{(\alpha-\delta)^{4}}{\delta^{4}}=\left(\frac{\alpha}{\delta}-1\right)^{4},\,\,\,\,\,\,\,\,\,\,\,\,\frac{f^{2}}{(\alpha-f)^{2}}\leq\left(\frac{\alpha}{\delta}-1\right)^{2}.$

Plugging (3.12) (or (3.13)) into (3.11), we obtain

(3.14)
$\displaystyle(\psi\omega^{2})(x_{1},t_{1})\leq\frac{\tilde{c}\beta^{4}+c(\alpha,\delta,m)}{R^{4}}+\frac{c(\delta,m)K}{R^{2}}+\frac{c(\delta)}{T^{2}}+c(\delta)(|a|+K)^{2}+c(\delta)a^{2}\beta^{2},$

where $\beta:=\max\left\\{1,|\alpha/\delta-1|\right\\}$.

The above inequality implies, for all $(x,t)$ in $Q_{R,T}$

(3.15) $\displaystyle(\psi^{2}\omega^{2})(x,t)$
$\displaystyle\leq\psi^{2}(x_{1},t_{1})\omega^{2}(x_{1},t_{1})\leq\psi(x_{1},t_{1})\omega^{2}(x_{1},t_{1})$
$\displaystyle\leq\frac{\tilde{c}\beta^{4}+c(\alpha,\delta,m)}{R^{4}}+\frac{c(\delta,m)K}{R^{2}}+\frac{c(\delta)}{T^{2}}+c(\delta)(|a|+K)^{2}+c(\delta)a^{2}\beta^{2}.$

Note that $\psi(x,t)=1$ in $Q_{R/2,T/2}$ and $\omega={|\nabla
f|^{2}}/{(\alpha-f)^{2}}$. Therefore we have

(3.16) $\displaystyle\frac{|\nabla
f|}{\alpha-f}\leq\left(\frac{\tilde{c}\beta^{4}+c(\alpha,\delta,m)}{R^{4}}+\frac{c(\delta,m)K}{R^{2}}+\frac{c(\delta)}{T^{2}}+c(\delta)(|a|+K)^{2}+c(\delta)a^{2}\beta^{2}\right)^{1/4}.$

Since $f=\log u$, we get the following estimate for Eq. (2.1)

(3.17) $\displaystyle\frac{|\nabla
u|}{u}\leq\left(\frac{\tilde{c}\beta^{4}+c(\alpha,\delta,m)}{R^{4}}+\frac{c(\delta)}{T^{2}}+c(\delta)(|a|+K)^{2}+c(\delta)a^{2}\beta^{2}\right)^{1/4}\Big{(}\alpha-\log
u\Big{)}.$

Replacing $u$ by $e^{b/a}u$ gives the desired estimate (1.8). This completes
the proof of Theorem 1.2. ∎

##  4\. Proof of Corollary 1.4

###### Proof.

The proof is similar to that of Theorem 1.2. We still use the technique of a
cut-off function in a local neighborhood of Riemannian manifolds. For $0<u\leq
1$, we let $f=\log u$. Then $f\leq 0$. Set

$\omega:=|\nabla\log(1-f)|^{2}=\frac{|\nabla f|^{2}}{(1-f)^{2}}.$

By Lemma 2.1, we have

(4.1) $\displaystyle\left(L-\frac{\partial}{\partial
t}\right)\omega\geq\frac{2f}{1-f}\left\langle\nabla
f,\nabla\omega\right\rangle+2(1-f)\omega^{2}-2(|a|+K)\omega.$

We define a smooth cut-off function $\psi=\psi(x,t)$ in the same way as
Section 3. Follow all steps in the last section (see also pp. 1050-1051 in
[1]), we can easily get the following inequality

(4.2) $\displaystyle 2(1-f)\psi\omega^{2}$
$\displaystyle\leq(1-f)\psi\omega^{2}+\frac{cf^{4}}{R^{4}(1-f)^{3}}+\frac{\psi\omega^{2}}{2}+\frac{c}{R^{4}}$
$\displaystyle\,\,\,\,\,\,+\frac{c(m)}{R^{4}}+\frac{c(m)K}{R^{2}}+\frac{c}{T^{2}}+c(|a|+K)^{2},$

where we used similar estimates (LABEL:term1)-(3.9) with the difference that
these estimates do not contain the parameter $\delta$. Using the same method
as that in proving Theorem 1.2, for all $(x,t)$ in $Q_{R/2,T/2}$ we can get

(4.3) $\displaystyle\omega^{2}(x,t)$
$\displaystyle\leq\frac{c(m)}{R^{4}}+\frac{c(m)K}{R^{2}}+\frac{c}{T^{2}}+c(|a|+K)^{2}$
$\displaystyle\leq\frac{c(m)}{R^{4}}+\frac{c(m)}{R^{2}}(|a|+K)+\frac{c}{T^{2}}+c(|a|+K)^{2}$
$\displaystyle\leq\frac{c(m)}{R^{4}}+\frac{c}{T^{2}}+c(|a|+K)^{2}.$

Again, using the same argument in the proof of Theorem 1.2 gives

(4.4) $\frac{|\nabla
f|}{1-f}\leq\frac{c(m)}{R}+\frac{c}{\sqrt{T}}+c\sqrt{K+|a|},$

where $c$ is a constant depending only on $n$, $c(m)$ is a constant depending
only on $n$ and $m$.

Since $f=\log u$, we get

(4.5) $\frac{|\nabla
u|}{u}\leq\left(\frac{c(m)}{R}+\frac{c}{\sqrt{T}}+c\sqrt{K+|a|}\right)\cdot\left(1+\log{\frac{1}{u}}\right).$

At last, replacing $u$ by $e^{b/a}u$ above yields (1.9). ∎

## Acknowledgment

The author would like to thank Professor Yu Zheng for his helpful suggestions
on this problem, and for his encouragement. He would also like to thank the
referees for useful suggestions. This work is partially supported by the
NSFC10871069.

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