text:
\begin{document}
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\begin{center}
{\bf {\large Weakly singular integral inequalities and global solutions for fractional differential equations of  Riemann-Liouville type}}
\end{center}
\title{}
\maketitle

\bigskip

\begin{center}
Tao Zhu\\
\bigskip
Department of Mathematics and Physics, Nanjing
Institute of Technology,\\
Nanjing, 211100, P. R. China \\
\end{center}

\baselineskip=20pt

\bigskip

\bigskip
\begin{abstract}
In this paper, we obtain some new results about weakly singular integral inequalities. These inequalities are used
to discuss the global existence and uniqueness results for fractional differential equations of  Riemann-Liouville type. Some examples are provided to illustrate the applicability of our main results.\\
Keywords: Weakly singular integral inequalities; Fixed point theorem;
Fractional differential equations; Riemann-Liouville fractional derivative.\\
MSC2010: 39B62; 34A08; 26A33.
\end{abstract}
\bigskip

\section{Introduction}
\bigskip
In 1981, Henry [6] studied the following linear integral
inequality
\begin{equation}
u(t)\leq a(t)+b\int_{0}^{t}(t-s)^{\beta-1}u(s)ds.
\end{equation}
Another version of a weakly singular result of Henry [6] is the following
\begin{equation}
u(t)\leq
at^{\alpha-1}+b\int_{0}^{t}(t-s)^{\beta-1}s^{\gamma-1}u(s)ds.
\end{equation}
Since weakly singular integral inequalities are a well-known tool in proving the existence, uniqueness and stability of integral equations, evolution equations and fractional differential equations, many scholars have began to study weakly singular integral inequalities and obtained a number of versions of weakly singular integral inequalities (for example, see [2,4,5,6,8,9,11,15,17,18,19,20]). Recently, Medve\v{d} [8,9] studied the following nonlinear integral inequality
\begin{equation}
u(t)\leq
a(t)+\int_{0}^{t}(t-s)^{\beta-1}l(s)\omega(u(s))ds,
\qquad t\in[0,T) (0<T\leq+\infty).
\end{equation}
Zhu [19] used the same method as in [8,9] to improve the proof of the inequality (1.3).
By a new method, Zhu [20] investigated the following integral inequality
\begin{equation}
u(t)\leq a(t)+b(t)(\int_{0}^{t}l^{p}(s)u^{p}(s)ds)^{\frac{1}{p}},
\qquad t\in[0,\infty).
\end{equation}
The inequality (1.4) was first studied by Willett in [16] by using the Minkowski inequality.
The aim of this paper is to continue the research in this field.
We first deal with the following nonlinear integral
inequality
\begin{equation}
u(t)\leq a(t)+b(t)(\int_{0}^{t}l(s)\omega(u(s))ds)^{\frac{1}{p}},
\qquad t\in[0,T).
\end{equation}
As far as I know, there have been few papers to study the integral inequality (1.5). Then using the integral inequality (1.5), we study the following weakly singular integral inequality
\begin{equation}
u(t)\leq
a(t)+b(t)\int_{0}^{t}(t-s)^{\beta-1}l(s)\omega(u(s))ds,
\qquad t \in[0,T),
\end{equation}
and the weakly singular integral inequality
\begin{equation}
u(t)\leq
at^{-\alpha}+bt^{-\delta}\int_{0}^{t}(t-s)^{\beta-1}f(s,u(s))ds,
\qquad t\in(0,T),
\end{equation}
where $a, b\geq 0$, $\alpha>\delta\geq 0$ and $0<\beta<1$.

As applications of our results, we present the existence and uniqueness results
of the following fractional differential equation
\begin{equation}
\begin{cases}
D_{r}^{\beta}x(t)=f(t,x(t))  \qquad \beta \in(0,1),\\
\lim_{t\rightarrow 0^{+}}t^{1-\beta}x(t)=x_{0},
\end{cases}
\end{equation}
where $D_{r}^{\beta}$ is the Riemann-Liouville fractional
derivative. In the recent years, many scholars have devoted to the study of the fractional differential
equation (1.8). For example, Trif [14] studied the global existence of solutions for the fractional differential
equations (1.8)  when
$f(t,x)\leq p(t)|x|+q(t)$, where $p\in C_{\alpha}(0,+\infty)$ and
$q\in C_{1-\beta}(0,+\infty)$ with $0\leq\alpha<\beta$ and
$2\beta-\alpha>1$. Webb [15] obtained the existence results of the fractional differential equation (1.8) under the assumption of nonnegative function $f(t,x)=t^{-\gamma}g(t,x)$, where $g(t,x)\leq M(1+x)$, $M>0$ and $0\leq\gamma<\beta$. Zhu [20] presented the existence results
of the fractional differential equation (1.8) when $|f(t, x)|\leq l(t)|x|+k(t)$ with $t^{\beta-1}l(t)\in C(0,+\infty)\bigcap L_{Loc}^{p}[0,+\infty)$ and
$k(t)\in C(0,+\infty)\bigcap L_{Loc}^{p}[0,+\infty)$ ($p>\frac{1}{\beta}$).

In this paper, by fixed point theorem and weakly singular integral inequality, we prove the existence of global solutions of the fractional differential
equation (1.8) when $|f(t, x)|\leq l(t)\omega(t^{1-\beta}|x|)$, where $l(t)\in C(0,+\infty)$ with $t^{1-\beta}l(t)\in L^{p}_{Loc}[0,+\infty)$ ($p>\frac{1}{\beta}$) and nonnegative
nondecreasing function $\omega\in C[0,+\infty)$ with $\lim_{t\rightarrow+\infty}\frac{t}{\omega(t)}=K (0<K\leq+\infty)$. Applying the above conclusion, we can easily obtain the following results. If $|f(t,x)|\leq l(t)|x|^{\gamma}+k(t)$, where $t^{(1-\gamma)(1-\beta)}l(t)\in
C(0,+\infty)\bigcap L^{p}_{Loc}[0,+\infty)$ and $t^{1-\beta}k(t)\in
C(0,+\infty)\bigcap L^{p}_{Loc}[0,+\infty)$$(0<\gamma\leq1, p>\frac{1}{\beta})$, then the fractional differential equation (1.8) has at least one
global solution in $C_{1-\beta}(0,+\infty)$. We also prove a unique solution of the fractional differential
equation (1.8) when $|f(t,x)-f(t,y)|\leq l(t)|x-y|$, where $l(t)\in
C(0,+\infty)\bigcap L^{p}_{Loc}[0,+\infty)$ and $t^{1-\beta}|f(t,0)|\in
C(0,+\infty)\bigcap L^{p}_{Loc}[0,+\infty)$$(p>\frac{1}{\beta})$. Our results improve and generalize some of the results in [14,15,20]. Finally, we will give some examples to illustrate  the applicability of our results.

\section{weakly singular integral inequalities}
In this section, we first study the nonlinear integral
inequality (1.5). Then we discuss the weakly singular integral inequalities (1.6) and (1.7) by the integral
inequality (1.5).

\begin{thm}
Let $p\geq 1$, let $a(t), b(t)\in C[0,T)$ $(0<T\leq+\infty)$ be nonnegative, nondecreasing
functions, let $l(t)\in C(0,T)\bigcap L^{1}_{Loc}[0,T)$ be a nonnegative function, and let $\omega\in C[0,+\infty)$ be a nondecreasing, nonnegative function.
Assume that $u(t)$ is a
continuous and nonnegative function on $[0,T)$ with
\begin{equation} u(t)\leq
a(t)+b(t)(\int_{0}^{t}l(s)\omega(u(s))ds)^{\frac{1}{p}},
\qquad t \in [0,T).
\end{equation}
Then
\begin{equation}
\begin{split}
u(t)&\leq
2^{1-\frac{1}{p}}\left(\Omega^{-1}\left(\Omega(a^{p}(t))+b^{p}(t)\int_{0}^{t}l(s)ds\right)\right)^{\frac{1}{p}},
\qquad t\in[0,T_{1}],
\end{split}
\end{equation}
where $\Omega(x)=\int_{1}^{x}\frac{1}{\mu(t)}dt$,
$\mu(t)=\omega(2^{1-\frac{1}{p}}t^{\frac{1}{p}})$,
$\Omega^{-1}$ is the inverse of $\Omega$, and $T_{1}\in (0,T)$ is
such that $\Omega(a^{p}(t))+b^{p}(t)\int_{0}^{t}l(s)ds\in
\mathrm{Dom}(\Omega^{-1})$ for all $t\in [0,T_{1}]$.
\end{thm}
\begin{proof}Fix any $T_{0}\in[0,T_{1}]$. Then for $t\in[0,T_{0}]$, from (2.1), we have
\begin{equation}
u(t)\leq a(T_{0})+b(T_{0})(\int_{0}^{t}l(s)\omega(u(s))ds)^{\frac{1}{p}}.
\end{equation}
Let $\nu(t)=b^{p}(T_{0})\int_{0}^{t}l(s)\omega(u(s))ds$,
then we can get
\begin{equation}
\begin{split}
\nu^{'}(t)&= b^{p}(T_{0})l(t)\omega(u(t))\\
&\leq b^{p}(T_{0})l(t)\omega(a(T_{0})+\nu^{\frac{1}{p}}(t))\\
&\leq b^{p}(T_{0})l(t)\omega(2^{1-\frac{1}{p}}(a^{p}(T_{0})+\nu(t))^{\frac{1}{p}})\\
&= b^{p}(T_{0})l(t)\mu(a^{p}(T_{0})+\nu(t)).\\
\end{split}
\end{equation}
This yields
\begin{equation}
\frac{(a^{p}(T_{0}))^{'}+\nu^{'}(t)}{\mu(a^{p}(T_{0})+\nu(t))}=\frac{\nu^{'}(t)}{\mu(a^{p}(T_{0})+\nu(t))}\leq b^{p}(T_{0})l(t),\\
\end{equation}
or
\begin{equation}
\begin{split}
\frac{d}{dt}\Omega(a^{p}(T_{0})+\nu(t))&\leq b^{p}(T_{0})l(t).
\end{split}
\end{equation}
Integrating the inequality (2.6) from $0$ to $t\in[0,T_{0}]$, we obtain
\begin{equation}
\begin{split}
\Omega(a^{p}(T_{0})+\nu(t))&\leq \Omega(a^{p}(T_{0}))+\int_{0}^{t}b^{p}(T_{0})l(s)ds,
\end{split}
\end{equation}
then
\begin{equation}
\begin{split}
a^{p}(T_{0})+\nu(t)&\leq
\Omega^{-1}\left(\Omega(a^{p}(T_{0}))+b^{p}(T_{0})\int_{0}^{t}l(s)ds\right), \qquad
t\in[0,T_{0}].
\end{split}
\end{equation}
So
\begin{equation}
\begin{split}
a^{p}(T_{0})+\nu(T_{0})&\leq
\Omega^{-1}\left(\Omega(a^{p}(T_{0}))+b^{p}(T_{0})\int_{0}^{T_{0}}l(s)ds\right).
\end{split}
\end{equation}
Then, from (2.1) and (2.9), we know
\begin{equation}
\begin{split}
u(T_{0})&\leq a(T_{0})+\nu^{\frac{1}{p}}(T_{0})\\
&\leq 2^{1-\frac{1}{p}}(a^{p}(T_{0})+\nu(T_{0}))^{\frac{1}{p}}\\
&\leq 2^{1-\frac{1}{p}}\left(\Omega^{-1}\left(\Omega(a^{p}(T_{0}))+b^{p}(T_{0})\int_{0}^{T_{0}}l(s)ds\right)\right)^{\frac{1}{p}}.\\
\end{split}
\end{equation}
Now replace $T_{0}$ by $t$ in the inequality (2.10), we obtain the following
result
\begin{equation}
u(t)\leq 2^{1-\frac{1}{p}}\left(\Omega^{-1}\left(\Omega(a^{p}(t)+b^{p}(t)\int_{0}^{t}l(s)ds\right)\right)^{\frac{1}{p}},  \qquad
t\in[0,T_{1}],
\end{equation}
where $T_{1}\in(0,T)$ is such that
$\Omega(a^{p}(t))+b^{p}(t)\int_{0}^{t}l(s)ds\in\mathrm{Dom}(\Omega^{-1})$ for all $t\in
[0,T_{1}]$.
\end{proof}

As a consequence of Theorem 2.1, we can obtain the following result.
\begin{cor}
Let $p\geq 1$ and $0<\gamma\leq1$, let $a(t)$, $b(t)\in C[0,+\infty)$ be nondecreasing, nonnegative functions, let $l(t)\in C(0,+\infty)\bigcap L^{1}_{Loc}[0,+\infty)$ be a nonnegative
function, and $u(t)$ be a
continuous, nonnegative function on $[0,+\infty)$ with
\begin{equation} u(t)\leq
a(t)+b(t)(\int_{0}^{t}l(s)u^{p\gamma}(s)ds)^{\frac{1}{p}},
\qquad t \in [0,+\infty).
\end{equation}
Then the following assertions holds:\\
If $\gamma=1$, then we have
\begin{equation}
\begin{split}
u(t)&\leq
2^{1-\frac{1}{p}}a(t)\exp\left(\frac{2^{p-1}b^{p}(t)}{p}\int_{0}^{t}l(s)ds\right),
\qquad t\in[0,+\infty).
\end{split}
\end{equation}
If $0<\gamma<1$, then we get
\begin{equation}
\begin{split}
u(t)&\leq
2^{1-\frac{1}{p}}\left(a^{p(1-\gamma)}(t)+(1-\gamma)2^{(p-1)\gamma}b^{p}(t)\int_{0}^{t}l(s)ds\right)^{\frac{1}{p(1-\gamma)}},
\qquad t\in[0,+\infty).
\end{split}
\end{equation}
\end{cor}

Now, we begin to study the  weakly singular integral inequalities (1.6) and (1.7) by the above Theorem 2.1.
\begin{thm}
Let $\beta\in(0,1)$, $p>\frac{1}{\beta}$ and $q=\frac{p}{p-1}$, let $a(t), b(t)\in C[0,T)$ be nondecreasing, nonnegative functions, let $l(t)\in C(0,T)\bigcap L^{p}_{Loc}[0,T)$ be a nonnegative
function, let $\omega\in C[0,+\infty)$ be a nondecreasing, nonnegative function, Assume that $u(t)$ is a
continuous, nonnegative function on $[0,T)$ with
\begin{equation}
u(t)\leq
a(t)+b(t)\int_{0}^{t}(t-s)^{\beta-1}l(s)\omega(u(s))ds,
\qquad t\in[0,T).
\end{equation}
Then
\begin{equation}
u(t)\leq 2^{1-\frac{1}{p}}\left(\Omega^{-1}\left(\Omega(a^{p}(t))+c^{p}(t)\int_{0}^{t}l^{p}(s)ds\right)\right)^{\frac{1}{p}},  \qquad
t\in[0,T_{1}],
\end{equation}
where $c(t)=\frac{t^{\beta-1+\frac{1}{q}}b(t)}{(q(\beta-1)+1)^{\frac{1}{q}}}$, $\Omega(x)=\int_{1}^{x}\frac{1}{\mu(t)}dt$,
$\mu(t)=\omega^{p}(2^{1-\frac{1}{p}}t^{\frac{1}{p}})$,
$\Omega^{-1}$ is the inverse of $\Omega$, and $T_{1}\in(0,T)$ is such that
$\Omega(a^{p}(t))+c^{p}(t)\int_{0}^{t}l^{p}(s)ds\in\mathrm{Dom}(\Omega^{-1})$ for all $t\in
[0,T_{1}]$.
\end{thm}
\begin{proof}From (2.15), since $p>\frac{1}{\beta}$ and $\frac{1}{p}+\frac{1}{q}=1$, then $q(\beta-1)+1>0$, using the
H\"{o}lder inequality, we know
\begin{equation}
\begin{split}
u(t)&\leq a(t)+b(t)(\int_{0}^{t}(t-s)^{q(\beta-1)}ds)^{\frac{1}{q}}(\int_{0}^{t}(l(s)\omega(u(s)))^{p}ds)^{\frac{1}{p}}\\
&\leq a(t)+\frac{t^{\beta-1+\frac{1}{q}}b(t)}{(q(\beta-1)+1)^{\frac{1}{q}}}(\int_{0}^{t}l^{p}(s)\omega^{p}(u(s))ds)^{\frac{1}{p}}.\\
\end{split}
\end{equation}
Then, by Theorem 2.1, we get
\begin{equation}
u(t)\leq 2^{1-\frac{1}{p}}\left(\Omega^{-1}\left(\Omega(a^{p}(t))+c^{p}(t)\int_{0}^{t}l^{p}(s)ds\right)\right)^{\frac{1}{p}},  \qquad
t\in[0,T_{1}],
\end{equation}
where $T_{1}\in(0,T)$ is such that
$\Omega(a^{p}(t))+c^{p}(t)\int_{0}^{t}l^{p}(s)ds\in\mathrm{Dom}(\Omega^{-1})$ for all $t\in
[0,T_{1}]$.
Thus, we complete the proof of Theorem 2.3.
\end{proof}

\begin{thm}
Let $a, b\geq 0$, $\alpha>\delta\geq 0$ and $0<\beta<1$, $p>\frac{1}{\beta}$ and $q=\frac{p}{p-1}$, $f:(0,T)\times\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function and
$$|f(t,x)|\leq l(t)\omega(t^{\alpha}|x|),$$
where nonnegative
function $l(t)\in C(0,T)\bigcap L^{p}_{Loc}[0,T)$ and nondecreasing, nonnegative function $\omega\in C[0,+\infty)$, and let $t^{\alpha}u(t)$ be a
continuous, nonnegative function on $[0,T)$ with
\begin{equation}
u(t)\leq
at^{-\alpha}+bt^{-\delta}\int_{0}^{t}(t-s)^{\beta-1}f(s,u(s))ds,
\qquad t\in(0,T).
\end{equation}
Then
\begin{equation}
u(t)\leq 2^{1-\frac{1}{p}}t^{-\alpha}\left(\Omega^{-1}\left(\Omega(a^{p})+c^{p}(t)\int_{0}^{t}l^{p}(s)ds\right)\right)^{\frac{1}{p}},  \qquad
t\in(0,T_{1}],
\end{equation}
where $c(t)=\frac{bt^{\alpha-\delta+\beta-1+\frac{1}{q}}}{(q(\beta-1)+1)^{\frac{1}{q}}}$, $\Omega(x)=\int_{1}^{x}\frac{1}{\mu(t)}dt$,
$\mu(t)=\omega^{p}(2^{1-\frac{1}{p}}t^{\frac{1}{p}})$,
$\Omega^{-1}$ is the inverse of $\Omega$, and $T_{1}\in(0,T)$ is such that
$\Omega(a^{p})+c^{p}(t)\int_{0}^{t}l^{p}(s)ds\in\mathrm{Dom}(\Omega^{-1})$ for all $t\in
[0,T_{1}]$.
\end{thm}
\begin{proof}
Let $v(t)=t^{\alpha}u(t)$, then we get
\begin{equation}
\begin{split}
v(t)&\leq
a+bt^{\alpha-\delta}\int_{0}^{t}(t-s)^{\beta-1}f(s,s^{-\alpha}v(s))ds\\
&\leq
a+bt^{\alpha-\delta}\int_{0}^{t}(t-s)^{\beta-1}l(s)\omega(v(s))ds\\
&\leq
a+\frac{bt^{\alpha-\delta+\beta-1+\frac{1}{q}}}{(q(\beta-1)+1)^{\frac{1}{q}}}(\int_{0}^{t}l^{p}(s)\omega^{p}(u(s))ds)^{\frac{1}{p}}.\\
\end{split}
\end{equation}
Using Theorem 2.1, we obtain the inequality (2.20) and complete the proof.
\end{proof}

In Theorem 2.4, if $\alpha=1-\beta$ and $\delta=0$, then we obtain the following conclusion under the weaker condition $t^{1-\beta}l(t)\in C(0,T)\bigcap L^{p}_{Loc}[0,T)$.
\begin{thm}
Let $a\geq 0$ and $0<\beta<1$, $p>\frac{1}{\beta}$ and $q=\frac{p}{p-1}$, let $b(t)$ be a nondecreasing, nonnegative continuous function on $[0,T)$, $f:(0,T)\times\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function and
$$|f(t,x)|\leq l(t)\omega(t^{1-\beta}|x|),$$
where nonnegative
function $t^{1-\beta}l(t)\in C(0,T)\bigcap L^{p}_{Loc}[0,T)$ and nondecreasing, nonnegative function $\omega\in C[0,+\infty)$, and let $t^{1-\beta}u(t)$ be a
continuous, nonnegative function on $[0,T)$ with
\begin{equation}
u(t)\leq
at^{\beta-1}+b(t)\int_{0}^{t}(t-s)^{\beta-1}f(s,u(s))ds,
\qquad t\in(0,T).
\end{equation}
Then
\begin{equation}
u(t)\leq
2^{1-\frac{1}{p}}t^{\beta-1}\left(\Omega^{-1}\left(\Omega(a^{p})+c^{p}(t)\int_{0}^{t}s^{p(1-\beta)}l^{p}(s)ds\right)\right)^{\frac{1}{p}}, \qquad t\in(0,T_{1}],
\end{equation}
where $c(t)=\frac{2^{\frac{1}{q}}b(t)t^{\beta-1+\frac{1}{q}}}{(q\beta-q+1)^{\frac{1}{q}}}$, $\Omega(x)=\int_{1}^{x}\frac{1}{\mu(t)}dt$,
$\mu(t)=\omega^{p}(2^{1-\frac{1}{p}}t^{\frac{1}{p}})$,
$\Omega^{-1}$ is the inverse of $\Omega$, and $T_{1}\in (0,T)$ is
such that $\Omega(a^{p})+c^{p}(t)\int_{0}^{t}s^{p(1-\beta)}l^{p}(s)ds\in
\mathrm{Dom}(\Omega^{-1})$ for all $t\in [0,T_{1}]$.
\end{thm}
\begin{proof}
Let $v(t)=t^{1-\beta}u(t)$, from (2.22) and $0<q(1-\beta)<1$, using the
H\"{o}lder inequality, we know
\begin{equation}
\begin{split}
v(t)&\leq
a+t^{1-\beta}b(t)\int_{0}^{t}(t-s)^{\beta-1}f(s,s^{\beta-1}v(s))ds\\
&\leq a+t^{1-\beta}b(t)\int_{0}^{t}(t-s)^{\beta-1}l(s)\omega(v(s))ds\\
&= a+b(t)\int_{0}^{t}(\frac{1}{t-s}+\frac{1}{s})^{1-\beta}s^{1-\beta}l(s)\omega(v(s))ds\\
&\leq a+b(t)(\int_{0}^{t}(\frac{1}{t-s}+\frac{1}{s})^{q(1-\beta)}ds)^{\frac{1}{q}}(\int_{0}^{t}s^{p(1-\beta)}l^{p}(s)\omega^{p}(v(s))ds)^{\frac{1}{p}}\\
&\leq a+b(t)(\int_{0}^{t}(t-s)^{q(\beta-1)}+s^{q(\beta-1)}ds)^{\frac{1}{q}}(\int_{0}^{t}s^{p(1-\beta)}l^{p}(s)\omega^{p}(v(s))ds)^{\frac{1}{p}}\\
&= a+\frac{2^{\frac{1}{q}}b(t)t^{\beta-1+\frac{1}{q}}}{(q\beta-q+1)^{\frac{1}{q}}}(\int_{0}^{t}s^{p(1-\beta)}l^{p}(s)\omega^{p}(v(s))ds)^{\frac{1}{p}}.\\
\end{split}
\end{equation}
By Theorem 2.1, then we have
\begin{equation}
v(t)\leq
2^{1-\frac{1}{p}}\left(\Omega^{-1}\left(\Omega(a^{p})+c^{p}(t)\int_{0}^{t}s^{p(1-\beta)}l^{p}(s)ds\right)\right)^{\frac{1}{p}}, \qquad t\in[0,T_{1}].
\end{equation}
Thus, we complete the proof.
\end{proof}

As a consequence of Theorem 2.5, we can get the following conclusion.
\begin{cor}
Let $a\geq 0$, $0<\gamma\leq1$ and $0<\beta<1$, $p>\frac{1}{\beta}$ and $q=\frac{p}{p-1}$, let $b(t)$ be a nondecreasing, nonnegative continuous function on $[0,+\infty)$, let $l(t)$ be a nonnegative
function with $t^{(1-\gamma)(1-\beta)}l(t)\in C(0,+\infty)\bigcap L^{p}_{Loc}[0,+\infty)$, and let $t^{1-\beta}u(t)$ be a
continuous, nonnegative function on $[0,+\infty)$ with
\begin{equation}
u(t)\leq
at^{\beta-1}+b(t)\int_{0}^{t}(t-s)^{\beta-1}l(s)u^{\gamma}(s)ds,
\qquad t\in(0,+\infty).
\end{equation}
Then the following assertions holds:\\
If $\gamma=1$, we can obtain
\begin{equation}
u(t)\leq
2^{1-\frac{1}{p}}at^{\beta-1}\exp\left(\frac{2^{p-1}c^{p}(t)}{p}\int_{0}^{t}l^{p}(s)ds\right), \qquad t\in(0,+\infty),
\end{equation}
where $c(t)=\frac{2^{\frac{1}{q}}b(t)t^{\beta-1+\frac{1}{q}}}{(q\beta-q+1)^{\frac{1}{q}}}$.\\
If $0<\gamma<1$, then we can obtain
\begin{equation}
u(t)\leq
2^{1-\frac{1}{p}}t^{\beta-1}\left(a^{p(1-\gamma)}+(1-\gamma)2^{(p-1)\gamma}c^{p}(t)\int_{0}^{t}s^{p(1-\gamma)(1-\beta)}l^{p}(s)ds\right)^{\frac{1}{p(1-\gamma)}}, \qquad t\in(0,+\infty),
\end{equation}
where $c(t)$ is defined as in (2.27).
\end{cor}

\begin{rem}
Willett [16] and Zhu [20] studied the inequality (2.1) for the special case $\omega(t)=t^{p}$. Medve\v{d} [8,Theorem 1], Medve\v{d} [9,Theorem 2] and Zhu [19, Theorem 2.7] studied the nonlinear integral inequality (2.15) under the assumptions that $a(t)\in C^{1}[0,T)$  is a nondecreasing, nonnegative function and $b(t)$ is a constant. In Theorem 2.3, we only suppose that $a(t), b(t)$ are nondecreasing, nonnegative continuous functions on $[0,T)$. Hence, our results improve and generalize Theorem 1 in [8],  Theorem 2 in [9] and Theorem 2.7 in [19]. By a reduction to the classical Gronwall inequality, Webb [15] studied the inequality (2.19) when $f(t,u(t))=t^{-\gamma}u(t)$. In [20], Zhu proved the inequality (2.19) when $f(t,u(t))=l(t)u(t)$. In Theorem 2.4, we study the inequality (2.19) when $|f(t,x)|\leq l(t)\omega(t^{\alpha}|x|)$, where $l(t)\in C(0,T)\bigcap L^{p}_{Loc}[0,T)$. For the special $\alpha=1-\beta$ and $\delta=0$, we study the inequality (2.22) under the weaker condition $t^{1-\beta}l(t)\in C(0,T)\bigcap L^{p}_{Loc}[0,T)$. Theorem 2.4, Theorem 2.5 and Corollary 2.6 generalize Theorem 3.9 in [15] and Theorem 3.4 in [20].
\end{rem}

\begin{exa}
Suppose that $t^{\frac{1}{2}}u(t)$ is a
continuous, nonnegative function and satisfies the following inequality
\begin{equation}
u(t)\leq
t^{\frac{-1}{2}}+t^{\frac{-1}{3}}\int_{0}^{t}(t-s)^{-\frac{1}{3}}s^{\frac{-1}{12}}u^{\frac{1}{2}}(s)ds.
\end{equation}
\end{exa}
Let $p=2$. Using Theorem 2.4, we know $l(t)=t^{\frac{-1}{3}}$, $\mu(t)=2^{\frac{1}{2}}t^{\frac{1}{2}}$ and $\Omega(x)=2^{\frac{1}{2}}(x^{\frac{1}{2}}-1)$, then we obtain
\begin{equation}
u(t)\leq2^{\frac{1}{2}}t^{\frac{-1}{2}}+9t^{\frac{1}{2}},  \qquad
t\in(0,+\infty).
\end{equation}

\begin{exa}
Suppose that $t^{\frac{1}{3}}u(t)$ is a
continuous, nonnegative function and satisfies the inequality
\begin{equation}
u(t)\leq
t^{-\frac{1}{3}}+\int_{0}^{t}(t-s)^{-\frac{1}{3}}s^{\frac{-1}{2}}u^{\frac{1}{2}}(s)ds.
\end{equation}
\end{exa}
Let $p=2$. By Corollary 2.6, we get
\begin{equation}
u(t)\leq 2^{\frac{1}{2}}t^{\frac{-1}{3}}+18t^{\frac{1}{3}},  \qquad
t\in(0,+\infty).
\end{equation}
\begin{rem}
From (2.31), we know $t^{\frac{-1}{2}}u^{\frac{1}{2}}(t)=t^{\frac{-2}{3}}(t^{\frac{1}{3}}u(t))^{\frac{1}{2}}$ and $t^{\frac{-2}{3}}\notin L^{p}_{Loc}[0,T)$$(p>\frac{3}{2})$, hence we can not obtain the estimate of the inequality (2.31) by Theorem 2.4. From Example 2.8 and Example 2.9,
we know that Theorem 2.5 is better than Theorem 2.4 for the special $\alpha=1-\beta$ and $\delta=0$.
\end{rem}

\section{global solutions for fractional differential equation}
In this section, we give the existence and uniqueness results for the fractional differential equation (1.8). we first introduce some notations, definitions, and preliminary facts which are used throughout this section. For more details about fractional calculus, we refer the reader to the
texts ([3,7,10,12]).

Let $\alpha\in (0,1)$, we denote $C_{\alpha}(0,T]=\{x(t):
x(t)\in C(0,T]$ and $\lim_{t\rightarrow 0^{+}}t^{\alpha}x(t)$
exists $\}$. Let $\|x\|_{\alpha}=\sup_{0<t\leq T}t^{\alpha}|x(t)|$, then $C_{\alpha}(0,T]$ endowed with the norm $\|\cdot\|_{\alpha}$ is a Banach space. We denote $C_{\alpha}(0,+\infty)=\{x(t):
x(t)\in C(0,+\infty)$ and $\lim_{t\rightarrow 0^{+}}t^{\alpha}x(t)$
exists $\}$.

\begin{defn}\label{defn: 3.1}[3]
Let $\beta\in(0,1)$, The operator $I^{\beta}$, defined on $L^{1}[0,T]$ by
$$I^{\beta}\varphi(t)=\frac{1}{\Gamma(\beta)}\int_{0}^{t}\frac{\varphi(s)}{(t-s)^{1-\beta}}ds, \qquad t\in[0,T], $$
is called the Riemann-Liouville fractional integral operator of
order $\beta$.
\end{defn}

\begin{defn}\label{defn: 3.2}[3]
Let $\beta\in(0,1)$, The operator $D_{r}^{\beta}$, defined by
$$D_{r}^{\beta}\varphi(t)=\frac{d}{dt}I^{1-\beta}\varphi(t)=\frac{1}{\Gamma(1-\beta)}\frac{d}{dt}\int_{0}^{t}\frac{\varphi(s)}{(t-s)^{\beta}}ds,\qquad a.e.t\in[0,T],$$
where $I^{1-\beta}\varphi(t)$ is an absolutely continuous function, is called the Riemann-Liouville fractional differential operator of order $\beta$.
\end{defn}

\begin{thm}\label{thm: 3.3}[13]
Let $S$ be a convex subset of a Banach space $E$ and assume $0\in S$. Let $F: S\rightarrow S$ be a continuous and compact map, and let the set $\{s\in S :x=\lambda Fx$ for some $\lambda\in(0,1)\}$ be bounded. Then $F$ has at least one fixed point in $S$.
\end{thm}

\begin{lem}
Let $0<\mu<1$, then
$$\varphi(t)=\frac{t^{\mu}-1}{(t-1)^{\mu}}$$
is a nondecreasing function on $(1,+\infty)$ and $\lim_{t\rightarrow 1^{+}}\varphi(t)=0$.
\end{lem}
\begin{proof}
We know
$$\varphi{'}(t)=\frac{\mu(t-1)^{\mu-1}[1-t^{\mu-1}]}{(t-1)^{2\mu}},$$
then $\varphi{'}(t)>0$ for $t\in(1,+\infty)$. Hence $\varphi(t)$ is a nondecreasing function on $(1,+\infty)$. Next, we get
$$\lim_{t\rightarrow 1^{+}}\varphi(t)=\lim_{t\rightarrow 1^{+}}\frac{(t-1)^{1-\mu}}{t^{1-\mu}}=0.$$
Thus, we complete the proof.
\end{proof}

\begin{lem}
Suppose that $s^{1-\beta}\rho(s)\in L^{p}[0,1]$, then
\begin{equation}
\left|\int_{0}^{t}(\frac{t}{t-s})^{1-\beta}\rho(s)ds\right|\leq \frac{2^{\frac{1}{q}}t^{\beta-1+\frac{1}{q}}}{(q\beta-q+1)^{\frac{1}{q}}}\left(\int_{0}^{t}s^{p(1-\beta)}|\rho(s)|^{p}ds\right)^{\frac{1}{p}}
\end{equation}
for $t\in[0,1]$, and if $0<t_{1}\leq t_{2}\leq1$, then
\begin{equation}
\begin{split}
&\left|\int_{0}^{t_{2}}(\frac{t_{2}}{t_{2}-s})^{1-\beta}\rho(s)ds-\int_{0}^{t_{1}}(\frac{t_{1}}{t_{1}-s})^{1-\beta}\rho(s)ds\right|\\
&\leq\frac{2^{\frac{1}{q}}(t_{2}-t_{1})^{\beta-1+\frac{1}{q}}}{(q\beta-q+1)^{\frac{1}{q}}}\left(\int_{t_{1}}^{t_{2}}s^{p(1-\beta)}|\rho(s)|^{p}ds\right)^{\frac{1}{p}}\\
&+\left(\frac{(t_{2}-t_{1})^{1+q(\beta-1)}+t_{1}^{1+q(\beta-1)}-t_{2}^{1+q(\beta-1)}}{q\beta-q+1}\right)^{\frac{1}{q}}\left(\int_{0}^{t_{1}}s^{p(1-\beta)}|\rho(s)|^{p}ds\right)^{\frac{1}{p}}.\\
\end{split}
\end{equation}
where $\beta\in(0,1)$, $p>\frac{1}{\beta}$ and $q=\frac{p}{p-1}$.
\end{lem}
\begin{proof}
By the H\"{o}lder inequality and $0<q(1-\beta)<1$, we have
\begin{equation}
\begin{split}
\left|\int_{0}^{t}(\frac{t}{t-s})^{1-\beta}\rho(s)ds\right|&\leq
\int_{0}^{t}(\frac{1}{t-s}+\frac{1}{s})^{1-\beta}s^{1-\beta}|\rho(s)|ds\\
&\leq \left(\int_{0}^{t}(\frac{1}{t-s}+\frac{1}{s})^{q(1-\beta)}ds\right)^{\frac{1}{q}}\left(\int_{0}^{t}s^{p(1-\beta)}|\rho(s)|^{p}ds\right)^{\frac{1}{p}}\\
&\leq \left(\int_{0}^{t}(t-s)^{q(\beta-1)}+s^{q(\beta-1)}ds\right)^{\frac{1}{q}}\left(\int_{0}^{t}s^{p(1-\beta)}|\rho(s)|^{p}ds\right)^{\frac{1}{p}}\\
&=\frac{2^{\frac{1}{q}}t^{\beta-1+\frac{1}{q}}}{(q\beta-q+1)^{\frac{1}{q}}}\left(\int_{0}^{t}s^{p(1-\beta)}|\rho(s)|^{p}ds\right)^{\frac{1}{p}}.\\
\end{split}
\end{equation}
Thus, we obtain the inequality (3.1).

Now, let us prove the inequality (3.2), from (3.2), we get
\begin{eqnarray*}
\lefteqn{\left|\int_{0}^{t_{2}}(\frac{t_{2}}{t_{2}-s})^{1-\beta}\rho(s)ds-\int_{0}^{t_{1}}(\frac{t_{1}}{t_{1}-s})^{1-\beta}\rho(s)ds\right|}\\
&\leq&\left|\int_{t_{1}}^{t_{2}}(\frac{t_{2}}{t_{2}-s})^{1-\beta}\rho(s)ds\right|+\left|\int_{0}^{t_{1}}\left[(\frac{t_{2}}{t_{2}-s})^{1-\beta}-(\frac{t_{1}}{t_{1}-s})^{1-\beta}\right]\rho(s)ds\right|.\\
\end{eqnarray*}
Similarly with the proof of the inequality (3.3), we obtain
\begin{equation}
\left|\int_{t_{1}}^{t_{2}}(\frac{t_{2}}{t_{2}-s})^{1-\beta}\rho(s)ds\right|\leq\frac{2^{\frac{1}{q}}(t_{2}-t_{1})^{\beta-1+\frac{1}{q}}}{(q\beta-q+1)^{\frac{1}{q}}}\left(\int_{t_{1}}^{t_{2}}s^{p(1-\beta)}|\rho(s)|^{p}ds\right)^{\frac{1}{p}}.\\
\end{equation}
From Lemma 3.4, we get
\begin{equation}
(\frac{x}{x-1})^{1-\beta}-(\frac{1}{x-1})^{1-\beta}\leq(\frac{y}{y-1})^{1-\beta}-(\frac{1}{y-1})^{1-\beta}
\end{equation}
for $1<x\leq y<+\infty$. Let $x=\frac{t_{1}}{s}$ and $y=\frac{t_{2}}{s}$ in (3.5), then
\begin{equation}
(\frac{t_{1}}{t_{1}-s})^{1-\beta}-(\frac{s}{t_{1}-s})^{1-\beta}\leq(\frac{t_{2}}{t_{2}-s})^{1-\beta}-(\frac{s}{t_{2}-s})^{1-\beta},
\end{equation}
where $0<s< t_{1}\leq t_{2}\leq1$. Then we know
\begin{equation}
\begin{split}
&\int_{0}^{t_{1}}\left|(\frac{t_{2}}{t_{2}-s})^{1-\beta}-(\frac{t_{1}}{t_{1}-s})^{1-\beta}\right||\rho(s)|ds
\leq\int_{0}^{t_{1}}\left[(\frac{s}{t_{1}-s})^{1-\beta}-(\frac{s}{t_{2}-s})^{1-\beta}\right]|\rho(s)|ds\\
&\leq\left(\int_{0}^{t_{1}}\left[(\frac{1}{t_{1}-s})^{1-\beta}-(\frac{1}{t_{2}-s})^{1-\beta}\right]^{q}ds\right)^{\frac{1}{q}}\left(\int_{0}^{t_{1}}s^{p(1-\beta)}|\rho(s)|^{p}ds\right)^{\frac{1}{p}}\\
&\leq\left(\int_{0}^{t_{1}}(\frac{1}{t_{1}-s})^{q(1-\beta)}-(\frac{1}{t_{2}-s})^{q(1-\beta)}ds\right)^{\frac{1}{q}}\left(\int_{0}^{t_{1}}s^{p(1-\beta)}|\rho(s)|^{p}ds\right)^{\frac{1}{p}}\\
&\leq\left(\frac{(t_{2}-t_{1})^{1+q(\beta-1)}+t_{1}^{1+q(\beta-1)}-t_{2}^{1+q(\beta-1)}}{q\beta-q+1}\right)^{\frac{1}{q}}\left(\int_{0}^{t_{1}}s^{p(1-\beta)}|\rho(s)|^{p}ds\right)^{\frac{1}{p}}.\\
\end{split}
\end{equation}
By the inequality (3.4) and (3.7), then the inequality (3.2) holds for all $0<t_{1}\leq t_{2}\leq1$.
\end{proof}

\begin{thm}
Let $p>\frac{1}{\beta}$ and $q=\frac{p}{p-1}$. Suppose  $f: (0,T]\times\mathbb{R}\rightarrow\mathbb{R}$ is a continuous
function, and there exist nonnegative function $l(t)$ with $t^{1-\beta}l(t)\in
C(0,T]\bigcap L^{p}[0,T]$ and nonnegative
nondecreasing function $\omega\in C[0,+\infty)$ such that
$$|f(t, x)|\leq l(t)\omega(t^{1-\beta}|x|)$$
for all $(t,x)\in (0,T]\times\mathbb{R}$. Then the fractional differential equation
(1.8) has at least one solution in $C_{1-\beta}(0,T]$ provided that
$$\Omega(|x_{0}|^{p})+c^{p}(t)\int_{0}^{t}s^{p(1-\beta)}l^{p}(s)ds\in\mathrm{Dom}(\Omega^{-1})$$
for all $t\in[0,T]$, where $c(t)=\frac{2^{\frac{1}{q}}t^{\beta-1+\frac{1}{q}}}{\Gamma(\beta)(q(\beta-1)+1)^{\frac{1}{q}}}$, $\Omega(x)=\int_{1}^{x}\frac{1}{\mu(t)}dt$,
$\mu(t)=\omega^{p}(2^{1-\frac{1}{p}}t^{\frac{1}{p}})$,
$\Omega^{-1}$ is the inverse of $\Omega$.
\end{thm}
\begin{proof}
Step 1. Since $|f(t, x(t))|\leq l(t)\omega(t^{1-\beta}|x(t)|)$, $x(t)\in C_{1-\beta}(0,T]$ and $t^{1-\beta}l(t)\in
C(0,T]\bigcap L^{p}[0,T]$, then we know $f(t, x(t))\in L^{1}[0,T]$. By Theorem 6.2 of [1], we know that $x(t)\in C_{1-\beta}(0,T]$ satisfies the fractional differential equation (1.8) if and only if it satisfies the following Volterra integral equation
\begin{equation}
x(t)=x_{0}t^{\beta-1}+\frac{1}{\Gamma(\beta)}\int_{0}^{t}(t-s)^{\beta-1}f(s,x(s))ds,\qquad
t\in(0, T].
\end{equation}

Step 2. Let $G: C_{1-\beta}(0,T]\rightarrow C_{1-\beta}(0,T]$ be the
operator defined by
\begin{equation}
(Gx)(t)=x_{0}t^{\beta-1}+\frac{1}{\Gamma(\beta)}\int_{0}^{t}(t-s)^{\beta-1}f(s,x(s))ds.
\end{equation}
We will prove that $G$ is a compact operator. To see this, let $\Omega\in C_{1-\beta}(0,T]$ be bounded and $\|x\|_{1-\beta}\leq M$ for each $x\in \Omega$,
we will show that $t^{1-\beta}G(\Omega)$ is uniformly bounded and
equicontinuous on $[0,T]$. Firstly, we prove that $t^{1-\beta}G(\Omega)$ is uniformly bounded. Let $x\in \Omega$, then we have
\begin{equation}
\begin{split}
|t^{1-\beta}Gx(t)|
&\leq|x_{0}|+\frac{t^{1-\beta}}{\Gamma(\beta)}|\int_{0}^{t}(t-s)^{\beta-1}f(s,x(s))ds|\\
&\leq |x_{0}|+\frac{1}{\Gamma(\beta)}(\int_{0}^{t}(\frac{1}{t-s}+\frac{1}{s})^{q(1-\beta)}ds)^{\frac{1}{q}}(\int_{0}^{t}s^{p(1-\beta)}l^{p}(s)\omega^{p}(s^{1-\beta}|x(s)|)ds)^{\frac{1}{p}}\\
&\leq |x_{0}|+\frac{2^{\frac{1}{q}}\omega(M)t^{\beta-1+\frac{1}{q}}}{\Gamma(\beta)(q(\beta-1)+1)^{\frac{1}{q}}}(\int_{0}^{t}s^{p(1-\beta)}l^{p}(s)ds)^{\frac{1}{p}}\\
&\leq |x_{0}|+\frac{2^{\frac{1}{q}}\omega(M)T^{\beta-1+\frac{1}{q}}}{\Gamma(\beta)(q(\beta-1)+1)^{\frac{1}{q}}}(\int_{0}^{T}s^{p(1-\beta)}l^{p}(s)ds)^{\frac{1}{p}}.\\
\end{split}
\end{equation}
This proves that the set that $t^{1-\beta}G(\Omega)$ is uniformly bounded. Secondly, we prove that $t^{1-\beta}G(\Omega)$ is an equicontinuous family. For any $x\in\Omega$, let $0\leq t_{1}<t_{2}\leq T$,
we get
\begin{equation}
\begin{split}
|t_{2}^{1-\beta}Gx(t_{2})-t_{1}^{1-\beta}Gx(t_{1})|
&=\frac{1}{\Gamma(\beta)}|\int_{0}^{t_{2}}(\frac{t_{2}}{t_{2}-s})^{1-\beta}f(s,x(s))ds-\int_{0}^{t_{1}}(\frac{t_{1}}{t_{1}-s})^{1-\beta}f(s,x(s))ds|.\\
\end{split}
\end{equation}
Since $|f(t, x(t))|\leq l(t)\omega(t^{1-\beta}|x(t)|)\leq l(t)\omega(M)$ and $t^{1-\beta}l(t)\in
C(0,T]\bigcap L^{p}[0,T]$. By Lemma 3.5, we know that the right-hand side of the above
inequality (3.11) tends to zero as $t_{2}\rightarrow t_{1}$. Therefore, $t^{1-\beta}G(\Omega)$ is an equicontinuous family.

Step 3. We now show that $G$ is continuous, that is $x_{n}\rightarrow x$ (that is $x_{n}(t)\rightarrow x(t)$ uniformly in $t$) implies $Gx_{n}\rightarrow Gx$. Since $x_{n}\rightarrow x$ in $C_{1-\beta}(0,T]$, then there exists
$r>0$ such that $\|x_{n}\|_{1-\beta}\leq r$ and $\|x\|_{1-\beta}\leq r$.
For every $s\in(0, T]$, we have
$$f(s,x_{n}(s))\rightarrow f(s,x(s)).$$
Using the assumption of $f$, we get
\begin{equation}
\begin{split}
(\frac{t}{t-s})^{1-\beta}|f(s,x_{n}(s))-f(s,x(s))|
&\leq
2\omega(r)(\frac{t}{t-s})^{1-\beta}l(s).\\
\end{split}
\end{equation}
Since $t^{1-\beta}l(t)\in C(0,T]\bigcap L^{p}[0,T]$, using (3.1) in Lemma 3.5, then we know that the function $$s\rightarrow
2\omega(r)(\frac{t}{t-s})^{1-\beta}l(s)$$ is integrable for $s\in
(0,t)$. By means of the Lebesgue dominated convergence theorem yields
$$|\int_{0}^{t}(\frac{t}{t-s})^{1-\beta}[f(s, x_{n}(s))-f(s, x(s))]ds|\rightarrow 0$$
as $n\rightarrow +\infty$. Therefore $t^{1-\beta}Gx_{n}(t)\rightarrow t^{1-\beta}Gx(t)$
pointwise on $[0,T]$ as $n\rightarrow +\infty$. With the fact that $G$ is compact, we get $G: C_{1-\beta}(0,T]\rightarrow C_{1-\beta}(0,T]$ is continuous.

Step 4. We shall prove that the set $\Lambda=\{x \in
C_{1-\beta}(0,T]: x=\lambda Gx$ for some $0<\lambda<1\}$ is
bounded. Indeed, let $x\in\Lambda$, then
\begin{equation}
\begin{split}
|x(t)|&\leq
|x_{0}|t^{\beta-1}+\frac{1}{\Gamma(\beta)}\int_{0}^{t}(t-s)^{\beta-1}|f(s,x(s))|ds\\
&\leq
|x_{0}|t^{\beta-1}+\frac{1}{\Gamma(\beta)}\int_{0}^{t}(t-s)^{\beta-1}l(s)\omega(s^{1-\beta}|x(s)|)ds.\\
\end{split}
\end{equation}
By Theorem 2.5, we obtain
\begin{equation}
x(t)\leq 2^{1-\frac{1}{p}}t^{\beta-1}\left(\Omega^{-1}\left(\Omega(|x_{0}|^{p})+c^{p}(t)\int_{0}^{t}s^{p(1-\beta)}l^{p}(s)ds\right)\right)^{\frac{1}{p}},  \qquad
t\in[0,T],
\end{equation}
and
\begin{equation}
\|x\|_{1-\beta}\leq 2^{1-\frac{1}{p}}\left(\Omega^{-1}\left(\Omega(|x_{0}|^{p})+c^{p}(T)\int_{0}^{T}s^{p(1-\beta)}l^{p}(s)ds\right)\right)^{\frac{1}{p}}.
\end{equation}
Then, the set $\Lambda$ is bounded.

Finally, applying fixed point theorem 3.3, the operator
$G$ has a fixed point $x(t)\in C_{1-\beta}(0,T]$ which is the
solution of the equation (1.8).
\end{proof}

Now, we investigate the existence of global solutions of the fractional differential
equation (1.8).
\begin{thm}
Let $p>\frac{1}{\beta}$ and $q=\frac{p}{p-1}$. Suppose  $f: (0,+\infty)\times\mathbb{R}\rightarrow\mathbb{R}$ is a continuous
function, and there exist nonnegative function $l(t)\in C(0,+\infty)$ with $t^{1-\beta}l(t)\in L^{p}_{Loc}[0,+\infty)$ and nonnegative
nondecreasing function $\omega\in C[0,+\infty)$ with $\lim_{t\rightarrow+\infty}\frac{t}{\omega(t)}=K (0<K\leq+\infty)$ such that
$$|f(t, x)|\leq l(t)\omega(t^{1-\beta}|x|)$$
for all $(t,x)\in (0,+\infty)\times\mathbb{R}$. Then the fractional differential equation
(1.8) has at least one global solution in $C_{1-\beta}(0,+\infty)$.
\end{thm}
\begin{proof}
We know
\begin{equation}
\lim_{t\rightarrow+\infty}\frac{t}{\mu(t)}=\lim_{t\rightarrow+\infty}\frac{t}{\omega^{p}(2^{1-\frac{1}{p}}t^{\frac{1}{p}})}=2^{1-p}K^{p}.
\end{equation}
Since $\int_{0}^{+\infty}\frac{1}{t}dt$ is divergent, from (3.16), we get that $\int_{0}^{+\infty}\frac{1}{\mu(t)}dt$ is also divergent. Then we get $[0,+\infty)\in\mathrm{Dom}(\Omega^{-1})$ and $\Omega(|x_{0}|^{p})+c^{p}(t)\int_{0}^{t}s^{p(1-\beta)}l^{p}(s)ds\in\mathrm{Dom}(\Omega^{-1})$
for every $t\in[0,+\infty)$, where $\Omega(x)$ and $c(t)$ are defined as in Theorem 3.6.

For any $T>0$, from Theorem 3.6, We know that the equation (1.8) has at least one
solution in $C_{1-\beta}(0, T]$. Since $T$ can be chosen arbitrarily large, then the equation (1.8)
has at least one global solution in $C_{1-\beta}(0,+\infty)$. Thus, we complete the proof of Theorem 3.7.
\end{proof}

From Theorem 3.7, we can immediately obtain the following conclusion.
\begin{cor}
Let $0<\gamma\leq1$, $p>\frac{1}{\beta}$ and $q=\frac{p}{p-1}$. Suppose  $f: (0,+\infty)\times\mathbb{R}\rightarrow\mathbb{R}$ is a continuous
function, and there exist nonnegative functions $l(t)$ and $k(t)$ such that
$$|f(t, x)|\leq l(t)|x|^{\gamma}+k(t)$$
for all $(t,x)\in (0,+\infty)\times\mathbb{R}$, where $t^{(1-\gamma)(1-\beta)}l(t)\in
C(0,+\infty)\bigcap L^{p}_{Loc}[0,+\infty)$ and $t^{1-\beta}k(t)\in
C(0,+\infty)\bigcap L^{p}_{Loc}[0,+\infty)$. Then the fractional differential equation
(1.8) has at least one solution in $C_{1-\beta}(0,+\infty)$.
\end{cor}
\begin{proof}
Since
\begin{equation}
\begin{split}
|f(t,x)|&\leq t^{\gamma(\beta-1)}l(t)(t^{1-\beta}|x|)^{\gamma}+k(t)\\
&\leq\left(t^{\gamma(\beta-1)}l(t)+k(t)\right)\left((t^{1-\beta}|x|)^{\gamma}+1\right),\\
\end{split}
\end{equation}
then we know
$$t^{1-\beta}\left(t^{\gamma(\beta-1)}l(t)+k(t)\right)=t^{(1-\gamma)(\beta-1)}l(t)+t^{1-\beta}k(t)\in L_{Loc}^{p}[0,+\infty),$$
and if $0<\gamma<1$, then
\begin{equation}
\lim_{t\rightarrow+\infty}\frac{t}{t^{\gamma}+1}=+\infty,
\end{equation}
if $\gamma=1$, then
\begin{equation}
\lim_{t\rightarrow+\infty}\frac{t}{t^{\gamma}+1}=1.
\end{equation}
Applying Theorem 3.7, we know that fractional differential equation
(1.8) has at least one solution in $C_{1-\beta}(0,+\infty)$. Thus the proof is
complete.
\end{proof}

\begin{rem}
In Theorem 4.2 of [14], Trif presented the existence result when
$f(t,x)\leq p(t)|x|+q(t)$, where $p\in C_{\alpha}(0,+\infty)$ and
$q\in C_{1-\beta}(0,+\infty)$ with $0\leq\alpha<\beta$ and
$2\beta-\alpha>1$.

Webb [15, Theorem 4.11] proved the existence results of the equation (1.8) when nonnegative function $f(t,x)=t^{-\gamma}g(t,x)$, where $g(t,x)\leq M(1+x)$, $M>0$ and $0\leq\gamma<\beta$.

In Theorem 4.3 of [20], Zhu proved that the fractional differential equation (1.8) has at least one global solution in $C_{1-\beta}(0,+\infty)$ when $|f(t, x)|\leq l(t)|x|+k(t)$, where $t^{\beta-1}l(t)\in C(0,+\infty)\bigcap L_{Loc}^{p}[0,+\infty)$ and
$k(t)\in C(0,+\infty)\bigcap L_{Loc}^{p}[0,+\infty)$
$(p>\frac{1}{\beta}, \beta\in(0,1))$.

From Corollary 3.8, we can easily obtain that our
results include and generalize Theorem 4.2 in [14], Theorem 4.11 in [15] and Theorem 4.3 in [20]. We notice that our hypothesis in Corollary 3.8 is weaker than that imposed by Zhu [20, Theorem 4.3].
\end{rem}

\begin{thm}
Let $p>\frac{1}{\beta}$ and $q=\frac{p}{p-1}$. If $f: (0,+\infty)\times\mathbb{R}\rightarrow\mathbb{R}$ is a continuous
function and
$$|f(t,x)-f(t,y)|\leq l(t)|x-y|$$ for all $x, y\in \mathbb{R}$ and
$t\in(0,+\infty)$, where $l(t)\in C(0,+\infty)\bigcap
L_{Loc}^{p}[0,+\infty)$ and $t^{1-\beta}|f(t,0)|\in L_{Loc}^{p}[0,+\infty)$. Then the equation (1.8)
has a unique solution on $(0,+\infty)$.
\end{thm}
\begin{proof}
We know
\begin{equation}
\begin{split}
|f(t,x)|&\leq|f(t,x)-f(t,0)|+|f(t,0)|\\
&\leq l(t)|x|+|f(t,0)|.\\
\end{split}
\end{equation}
Since $l(t)\in C(0,+\infty)\bigcap
L_{Loc}^{p}[0,+\infty)$ and $t^{1-\beta}|f(t,0)|\in L_{Loc}^{p}[0,+\infty)$, applying Corollary 3.8, we know that the fractional differential equation
(1.8) has at least one solution in $C_{1-\beta}(0,+\infty)$. We suppose that $x_{1}(t)$,
$x_{2}(t)$ are two global solutions of the equation (1.8). Then
\begin{equation}
\begin{split}
|x_{1}(t)-x_{2}(t)|&=\left|\frac{1}{\Gamma(\beta)}\int_{0}^{t}(t-s)^{\beta-1}(f(s,x_{1}(s))-f(s,x_{2}(s)))ds\right|\\
&\leq\frac{1}{\Gamma(\beta)}\int_{0}^{t}(t-s)^{\beta-1}l(s)|x_{1}(s)-x_{2}(s)|ds.\\
\end{split}
\end{equation}
By (2.27) in Corollary 2.6, we can get $x_{1}(t)=x_{2}(t)$. Thus the proof is
complete.
\end{proof}

\begin{rem}
In [14, Theorem 4.1], Trif presented the uniqueness result of the fractional differential equation (1.8) when
$f(t,x)=p(t)x+q(t)$, where $p\in C_{\alpha}(0,+\infty)$ and
$q\in C_{1-\beta}(0,+\infty)$ with $0\leq\alpha<\beta$.

Webb [15, Theorem 4.11] proved the uniqueness result of the equation (1.8) when $f(t,x)=t^{-\gamma}g(t,x)$, where $g(t,x)\leq M(1+x)$ and $|g(t,x)-g(t,y)|\leq L|x-y|$, $M>0$, $L>0$ and $0\leq\gamma<\beta$.

In [20, Theorem 4.4], Zhu proved that the fractional differential equation (1.8) has a unique solution in $C_{1-\beta}(0,+\infty)$ when $|f(t, x)-f(t, y)|\leq l(t)|x-y|$, where $t^{\beta-1}l(t)\in C(0,+\infty)\bigcap L_{Loc}^{p}[0,+\infty)$ and
$|f(t,0)|\in C(0,+\infty)\bigcap L_{Loc}^{p}[0,+\infty)$
$(p>\frac{1}{\beta})$.

From Theorem 3.10, we can easily get that Theorem 3.10 includes Theorem 4.1 in [14], Theorem 4.11 in [15] and Theorem 4.4 in [20]. We notice that our hypothesis in Theorem 3.10 is weaker than that imposed by Zhu [20, Theorem 4.4].
\end{rem}

\begin{exa}
\begin{equation}
\begin{cases}
D_{r}^{\frac{2}{3}}x(t)=t^{\frac{-3}{4}}x^{\frac{1}{2}}(t)+t^{\frac{-1}{2}}x(t),\\
\lim_{t\rightarrow 0^{+}}t^{\frac{1}{3}}x(t)=1.
\end{cases}
\end{equation}
\end{exa}
We know
\begin{equation}
t^{\frac{-3}{4}}x^{\frac{1}{2}}(t)+t^{\frac{-1}{2}}x(t)\leq (t^{\frac{-11}{12}}+t^{\frac{-5}{6}})
\left((t^{\frac{1}{3}}x(t))^{\frac{1}{2}}+t^{\frac{1}{3}}x(t)\right),
\end{equation}
let $l(t)=t^{\frac{-11}{12}}+t^{\frac{-5}{6}}$, then $t^{\frac{1}{3}}l(t)=t^{\frac{-7}{12}}+t^{\frac{-1}{2}}\in L^{p}_{Loc}[0,+\infty)$($p>\frac{3}{2}$), and
\begin{equation}
\lim_{t\rightarrow+\infty}\frac{t}{t^{\frac{1}{2}}+t}=1.
\end{equation}
Applying Theorem 3.7, we know that the equation (3.22) has at least one global solution in $C_{\frac{1}{3}}(0,+\infty)$.

\begin{exa}
\begin{equation}
\begin{cases}
D_{r}^{\frac{2}{3}}x(t)=t^{\frac{-2}{3}}\ln(1+x^{\frac{1}{2}}(t)),\\
\lim_{t\rightarrow 0^{+}}t^{\frac{1}{3}}x(t)=1.
\end{cases}
\end{equation}
\end{exa}
Using Corollary 3.8, we know
\begin{equation}
t^{\frac{-2}{3}}\ln(1+x^{\frac{1}{2}}(t))\leq t^{\frac{-2}{3}}x^{\frac{1}{2}}(t),
\end{equation}
and $t^{\frac{1}{6}}l(t)=t^{\frac{-1}{2}}\in C(0,+\infty)\bigcap L^{p}_{Loc}[0,+\infty)$($p>\frac{3}{2}$), then the equation (3.25) has at least one global solution on $(0,+\infty)$.

In Example 3.13, we know $$t^{\frac{-2}{3}}\ln(1+x^{\frac{1}{2}}(t))\leq\frac{t^{\frac{-2}{3}}(x(t)+1)}{2}.$$ Thus, Theorem 4.2 in [14], Theorem 4.11 in [15] or Theorem 4.3 in [20] can not be applied to Example 3.13 (see Remark 3.9).

\begin{exa}
\begin{equation}
\begin{cases}
D_{r}^{\frac{2}{3}}x(t)=t^{\frac{-1}{2}}\frac{x^{2}(t)}{1+x(t)}+t^{\frac{-3}{4}},\\
\lim_{t\rightarrow 0^{+}}t^{\frac{1}{3}}x(t)=1.
\end{cases}
\end{equation}
\end{exa}
We know $$|\frac{x^{2}}{1+x}-\frac{y^{2}}{1+y}|\leq|x-y|, \qquad  x,y\in[0,+\infty).$$
Using Theorem 3.10, since $t^{\frac{-1}{2}}\in C(0,+\infty)\bigcap
L_{Loc}^{p}[0,+\infty)$ and $t^{\frac{-5}{12}}\in C(0,+\infty)\bigcap
L_{Loc}^{p}[0,+\infty)$ $(p>\frac{3}{2})$, then the equation (3.27) has a unique solution on $(0,+\infty)$.

Since $t^{\frac{-3}{4}}\notin C_{\frac{1}{3}}(0,+\infty)$ and $t^{\frac{-3}{4}}\notin L^{p}[0,+\infty)$($p>\frac{3}{2}$), then Theorem 4.1 in [14], Theorem 4.11 in [15] or Theorem 4.4 in [20] can not be applied to Example 3.14 (see Remark 3.11).

\bigskip

\centerline{Acknowledgements}
The research was supported by Scientific Research Foundation of Nanjing
Institute of Technology(No: CKJB201508). I would like to express my sincere gratitude to the anonymous referee for his or her comments and suggestions.


TITLE: Inner products with spherical harmonics in quantum mechanics
QUESTION [1 upvotes]: Let $|l,m\rangle$ be a simultaneous eigenstate of operators $L^2$ and $L_z$ and we want to calculate $\langle l,m|\cos(\theta)|l,m'\rangle$ where $\theta$ is the angle $[0,\pi]$. It is true that in general $$\langle l,m|\cos(\theta)|l,m'\rangle=0 \tag{1}$$ for the same $l$? Why?
Does the fact that equation $(1)$ is valid have anything to do with parity?

REPLY [1 votes]: It is true in general, and $l$ doesn't have to be the same: the matrix element is zero because $\cos\theta$ doesn't depend on $\phi$, so the azimuthal wavefunction factorizes and we get a factor of $\delta_{m,m'}$.
More explicitly, the spherical harmonics can be written as $Y_{lm} = N_{lm} P_{lm}(\cos\theta) e^{im\phi}$, where $P_{lm}$ are the associated Legendre polynomials and $N_{lm}$ are normalization constants. Inserting this into the integral we get
$$\begin{align}
\langle l, m| \cos\theta |l', m'\rangle &= \int \left(N_{lm} P_{lm}(\cos\theta) e^{-im\phi}\right) \cos\theta \left( N_{l'm'} P_{l'm'}(\cos\theta) e^{im'\phi} \right) \cos\theta\, d\theta\, d\phi \\
&= N_{lm} N_{l'm'} \int P_{lm}(\cos\theta)P_{l'm'}(\cos\theta) \cos^2\theta\, d\theta \int e^{i(m'-m)\phi}\,d\phi
\end{align}$$
and the last integral is proportional to $\delta_{m,m'}$.


TITLE: Prove that the image of a a closed and bounded interval in $\mathbb{R}$ is a a closed and bounded interval in $\mathbb{R}$?
QUESTION [2 upvotes]: According to the excercise 7.22 of the book Topology by Franzosa:

Combining the Extreme Value Theorem and the Intermediate Value Theorem, prove the following theorem:
Let $[a, b]$ be a closed and bounded interval in $\mathbb{R}$, and assume that $f : [a, b] \rightarrow \mathbb{R}$  is continuous. Then the image of $f$ is a closed
and bounded interval in $\mathbb{R}$.

Of course we can prove the theorem by the fact that image of compact domain is compact if the function is continuous. But what about proving it by using only the Extreme Value Theorem and the Intermediate Value Theorem?
Thanks a lot.

REPLY [10 votes]: How's this:
Left $f:[a,b]\rightarrow\mathbb R$ be continuous. We know that $[a,b]$ is compact, so by the extreme value theorem, $f$ achieves both it's maximum (say, $d$) at a point $y\in [a,b]$ and it's minimum (say, $c$) at a point $x\in[a,b]$. Thus, $f([a,b])\subseteq[c,d]$. Without loss of generality, assume $x<y$. Let $z\in[c,d]$. Then by the intermediate value theorem, there exists $w$ such that $x<w<y$ and such that $f(w)=z$. Thus, $f([a,b])=[c,d].$


TITLE: Show that a single binomial probability can be written as the sum of two binomial probabilities
QUESTION [3 upvotes]: I am trying to prove that the sum of two independent binomial probabilities with the same parameter (probability of success in a single trial) is equivalent to a binomial probability with the same parameter , but an index (number of trials) and number of successes that is equivalent to the sum of the indexes from the first two binomial probabilities and the sum of the number of successes from the first two binomial probabilities.
I think the basic idea is that the binomial probability of an event can be broken up and calculated in two "chunks" which can be summed to find the desired binomial probability.
Sorry for not including any mathematical operators in this post; I am unfamiliar with how they are actually written in this type of a post.
Edit:
${n \choose x} \pi^x (1 - \pi)^{n-x} = {a \choose y} \pi^y (1 - \pi)^{a-y} + {b \choose z} \pi^z(1 - \pi)^{b-z}$ where $a+b=n$ and $y+z=x$

REPLY [1 votes]: Let $X$ be the number of successes when an experiment is repeated independently $m$  times, with probability of success $p$ each time. Now suppose the experiment is done again $n$ times.
Then $X$ has binomial distribution, parameters $m$ and $p$, while $Y$ is binomial with parameters $n$ and $p$.
It is clear that $X+Y$ is the number of successes when the experiment is repeated independently $m+n$ times. So $X+Y$ has binomial distribution with parameters $m+n$ and $p$.
Remark: One can also do the problem the hard way. Let $W=X+Y$. Then
$$\Pr(W=w)=\sum_{i=0}^w \Pr(X=i)\Pr(Y=w-i).$$
Using the usual binomial distribution formulas, we get
$$\Pr(W=w)=\sum_{i=0}^w\binom{m}{i}p^i(1-p)^{w-i}\binom{n}{w-i}p^{w-i}(1-p)^{n-(w-i)}.$$
Manipulation of the above formula yields, after a while,
$$\Pr(W=w)=\binom{m+n}{w}p^w(1-p)^{m+n-w}.$$


TITLE: If $F_1$ and $F_2$ are the antiderivatives of $f(x)$, then $F_1=F_2$?
QUESTION [0 upvotes]: Per question: True or false? If $F_1$ and $F_2$ are the antiderivatives of $f(x)$, then $F_1=F_2$.
If I had $f(x) = 2x$, and I did an anti-derivative by $\int 2x dx = x^2 +c$. Does this mean because of $c$, the $F_1$ cannot equal  $F_2$?

REPLY [0 votes]: The statement is equivalent to:
$$F_1'=F_2'=f$$
but as others have touched on, the equality is not maintained for a definite integral. What this means is that whilst:
$$\int_{x_1}^{x_2}F_1\,dx=\int_{x_1}^{x_2}F_2\,dx$$
$$F_1(x_2)-F_1(x_1)=F_2(x_2)-F_2(x_1)$$
this does not imply that either of the following are true:
$$F_1(x_1)=F_2(x_1)\\F_1(x_2)=F_2(x_2)$$
which is just the same as how:
$$2-1=100-99\nRightarrow 2=100$$

This is what it means for functions to differ by a constant, and how two functions can be antiderivatives of the same function


TITLE: Problem of Number Theory
QUESTION [3 upvotes]: Can you give a tip for this question?
For all $ n $ odd positive integer, let be
$ F(n) = \# \{ a($mod $ n): a^{n-1} \equiv 1($mod $n) \} $
that is the number of elements of the set.

Show that
$$ F(n) = \prod_{p|n} gcd(p-1,n-1) $$
Let be $ F_0(n) $ the numbers $ a($mod $n) $ such that $ a^n \equiv a($mod $n) $. Find a formula like the previous one,
Show that if $ F_0(n) $ < n, then $ F_0(n) \leq \frac{2}{3} n $. Show that if $ n \neq 6 $ and $ F_0(n) < n $ then $ F_0(n) \leq \frac{3}{5} n $

REPLY [1 votes]: I have skimmed over some steps; you can try filling in the details.
1) Write $n=\prod_{i=1}^{m}{p_i^{a_i}}$, where $p_i$ are odd primes.
2) There exists a primitive root $\pmod{p_i^{a_i}}$.
3) Using 2), conclude that the number of solutions $\pmod{p_i^{a_i}}$ of $a^k \equiv 1 \pmod{p_i^{a_i}}$ is $\gcd(\varphi(p_i^{a_i}), k)$.
4) Apply 4) for $k=n-1$ and Chinese Remainder Theorem to get $F(n)=\prod_{i=1}^{m}{\gcd(\varphi(p_i^{a_i}), n-1)}=\prod_{p\mid n}{\gcd(p-1, n-1)}$.
5) If we want to look at even $n$, then for $2^l \| n$, we get $F(n)=\alpha\prod_{p\mid n, 2 \nmid p}{\left(\gcd(p-1, n-1) \right)}$. where $\alpha$ is the number of solutions to $a^{n-1} \equiv 1\pmod {2^l}$. We get $a \equiv 1 \pmod{2^l}$. Thus $$F(n)=\prod_{p\mid n}{\left(\gcd(p-1, n-1) \right)}$$

For $F_0(n)$, a similar approach applies and we get $\gcd(\varphi(p_i^{a_i}), k)+1$ solutions in step $3$, so $$F_0(n)=\prod_{p\mid n}{\left(\gcd(p-1, n-1)+1 \right)}$$ The analysis of even $n$ is also similar, and we get the same formula.

Note $\gcd(p-1, n-1)+1 \leq p$ so $F_0(n) \leq \prod_{p \mid n}{p} \leq n$. If there is strict inequality, either $n$ is not squarefree or $\gcd(p-1, n-1)<p-1$ for some $p$.
In the former, if $q^2 \mid n$, $F_0(n) \leq \prod_{p \mid n}{p} \leq \frac{n}{q} \leq \frac{n}{2}$.
In the latter case clearly $p>2$, and we have $\gcd(p-1, n-1) \leq \frac{p-1}{2}$ so $\gcd(p-1, n-1) \leq \frac{p+1}{2} \leq \frac{2}{3}p$. Thus $F_0(n) \leq \frac{2n}{3}$.

Note that if $n$ not squarefree or $p \geq 5$ in case 2, we get $F_0(n) \leq \frac{3n}{5}$. Thus if $\frac{3n}{5}<F_0(n)<n$ then we must have $n$ squarefree and $p=3$ in case $2$, $\gcd(3-1, n-1)=1$, $\gcd(p-1, n-1)=p-1$ for $p \mid n, p \not =3$.
We have $\gcd(2, n-1)=1$ so $2 \mid n$. Thus $6 \mid n$
If $p \mid n$ for some $p>3$ then $\gcd(p-1, n-1)=p-1$ implies $n$ odd, a contradiction.
Thus $n$ is squarefree and a multiple of $6$, i.e. $n=6$.


TITLE: Investigate convergence of series
QUESTION [2 upvotes]: Formula represented on this image :
$$\sum_{n=1}^{\infty}\left(3^{\frac{1}{n}}-1\right)\sin \frac{\pi}{n}$$
I know that this is convergent but how can I show it?

REPLY [3 votes]: HINT
We have that
$$\left(3^{\frac{1}{n}}-1\right)\sin \frac{\pi}{n}=\left(3^{\frac{1}{n}}-1\right)\frac{\pi}{n}\frac{\sin \frac{\pi}{n}}{\frac{\pi}{n}} =\frac{\log 3}{n}\,\frac{e^\frac{\log 3}{n}-1}{\frac{\log 3}{n}}\,\frac{\pi}{n}\,\frac{\sin \frac{\pi}{n}}{\frac{\pi}{n}}$$
then use limit comparison test with $\sum \frac1{n^2}$.


TITLE: Reference for light interaction with many electron atoms
QUESTION [0 upvotes]: I am trying to understand light interaction with  many electron atoms.
I know that to calculate the bound states  we can use Hartree-Fock method but for light interaction with matter i do not know it this method works.
I am searching for references and open source programs.
Could anyone help me please?

REPLY [1 votes]: When atoms and molecules interact with electromagnetic radiation, there could be some transitions between rotational, vibrational or electronic states. Whenever electronic transitions happens, the Coulomb interaction is changed due to the redistribution of the electrons.
Hartree-Fock deals with the stationary states of many-electron systems, and do not describe the response to a time-dependent perturbation, like the electric field of an EM wave. Thus, you can not calculate dynamical properties nor atomic or molecular spectra using only Hartree-Fock, you need Time-Dependent Hartree-Fock. However, the HF method does not consider Coulomb correlation, so it is better to use Time-Dependent Density Functional Theory with an appropriate exchange-correlation functional to calculate molecular spectra more quantitatively.
There are many quantum chemistry/solid-state programs that incorporate the TDDFT scheme. For instance, Gaussian16 program includes TDDFT and also TDHF, but the software is too expensive. Nonetheless, there are programs like NWChem and DIRAC that have implemented the TDDFT scheme and have open access.
To read about the fundamentals of TDHF or TDDFT, I would recommend this two books:

Methods of Molecular Quantum Mechanics by R. McWeeny. This book contains the basics of electronic structure from HF theory to intermolecular interactions, passing through multiconfiguration theories, perturbation theory, linear response theory, and propagator and equation of motion methods. Chapter 12 has some sections about TDHF.
Time-Dependent Density-Functional Theory by C. Ullrich. It is a book dedicated exclusively to TDDFT. It contains the basic formalism of TDDFT and linear response theory and further developments. Also, Appendix O have a list with another popular computer codes which include TDDFT.


TITLE: Need help simplifying this Boolean expression
QUESTION [1 upvotes]: From my textbook, by using distributive law, its able to simplify:
$[(p \land \lnot q) \lor (p \land q)] \land q$
To:
$[p \land (\lnot q \lor q)] \land q $
I don't know how to get to this step, and here is how I've tried by distributing the first expression :
$[(p \lor p) \land (p \lor q) \land (\lnot q \lor p) \land (\lnot q \lor q)] \land q$
If I try continue to expand this, it will become very complex. I think I'm doing this in the wrong way and I will appreciate it very much if anyone can explain how the textbook got the simplified answer.
Thanks

REPLY [5 votes]: $$[(\color{blue}{p} \land \lnot q) \lor (\color{blue}{p} \land q)] \land q \equiv [\color{blue}{p} \land (\color{red}{{\lnot q \lor q}})] \land q\tag{distributivity}$$
$$ \equiv (\color{blue}{p} \land \color{red} \top) \land q$$
$$\equiv \color{blue}{p} \land q$$
That is, the distributive law declares that $$[p\land (\lnot q \lor q)] \equiv [(p \land \lnot q) \lor (p \land q)]$$  That means that it is the same law when we write it: $$[(p \land \lnot q) \lor (p \land q)] \equiv [p\land (\lnot q \lor q)]$$

Addendum


TITLE: Regular and irregular points
QUESTION [4 upvotes]: In the following 2 examples, I am trying to find all the singular points of the given equations and determine whether each one is regular or irregular
I can determine the singular points of the equations but I am having some trouble determining if they are regular or irregular.
1)  $$x^2 (1-x^2)y'' + \frac {2}{x}y'+4y =0$$
The singular points of the equation are $x=\pm 1, 0$
The answer is that $0$ is irregular and $\pm 1$ regular but I am not sure why
2) $$xy'' + (1-x)y' +xy =0$$
The singular point is $0$ but why is this considered regular?
Do I have to consider the $p (x)$ term in each equation?

REPLY [3 votes]: The definition of wikipedia :
For the ordinary differential equation: $$f''(x)+p_{1}(x)f'(x)+p_{0}(x)f(x)=0.\,$$

Point $a$ is an ordinary point when functions $p_1(x)$ and $p_0(x)$ are analytic at $x = a$.
Point $a$ is a regular singular point if $p_1(x)$ has a pole up to order $1$ at $x = a$ and $p_0$ has a pole of order up to $2$ at $x = a$.
Otherwise, point a is an irregular singular point.

For your example 1)
$$x^2 (1-x^2)y'' + \frac {2}{x}y'+4y =0$$
we divide by the coefficient function of $y''$ to obtain:
$$y'' + \frac {2}{x^3 (1-x^2)}y'+\dfrac{4}{x^2 (1-x^2)}y =0.$$
The singular points are $x=0$ and $x=\pm 1$.  From the definition, we see that $x=0$ is an irregular singular point because it is a third order pole. $x=\pm 1$ are order $1$ poles of the coefficient function of $y'$ and also order $1$ poles of the coefficient function of $y$. Hence, we know that these are regular singular points.
Can you do the next example?


TITLE: What is a permutation pattern?
QUESTION [1 upvotes]: The wikipedia entry for permutation pattern gives this as an example:

For example, in the permutation π = 391867452, π1=3 and π9=2. A permutation π is said to contain the permutation σ if there exists a subsequence of (not necessarily consecutive) entries of π that has the same relative order as σ, and in this case σ is said to be a pattern of π, written σ ≤ π. Otherwise, π is said to avoid the permutation σ. For example, the permutation π = 391867452 contains the pattern σ = 51342, as can be seen in the highlighted subsequence of π = 391867452 (or π = 391867452 or π = 391867452). Each subsequence (91674, 91675, 91672) is called a copy, instance, or occurrence of σ. Since the permutation π = 391867452 contains no increasing subsequence of length four, π avoids 1234.

In that example, is a pattern of 51342 the same as a pattern of, say, 62453 or 92483?  I mean, is it the just the the increase/decrease as well as the relative magnitude of each digit with respect to every other digit in the sequence that makes a pattern? Or must a pattern be "reduced" so that the value of the lowest digit in the pattern is 1 and the difference between the two values in the "sort" of the pattern is no more than 1?

REPLY [3 votes]: Rather than thinking of a pattern as a single permutation, think of it as an equivalence class with respect to order isomorphism (i.e. if positions compare in the same way, then the entries in those positions compare in the same way, too). That equivalence class has a canonical representative where the $k$th smallest position/letter is $k$. In your example, $51342$, $62453$ and $92483$ are order-isomorphic, and the canonical representative of their equivalence class is $51342$.


TITLE: minimum value of dot product with planar vector and planar axes
QUESTION [0 upvotes]: I am writing some code and want to select a calculation to do based on "how much" of a direction vector is in the X and Z axis (which are the planar axes in Unity). If more of the direction vector is along the X axis, I want to compute Y(X); but if more of the vector is along the Z axis, I want to compute Y(Z). I am hopeful that doing this will avoid undefined behavior when the curve is entirely (or almost entirely) parallel to the XY plane or ZY plane. Here is what I wrote:
// Vector3.right is the x-axis
if (Mathf.Abs(Vector3.Dot(direction,Vector3.right)) >= Mathf.Sqrt(2f) )
{
    //Calculate Y(x)
}
else
{
    //Calculate Y(z)
}

Is this logical? In this situation, would the dot product always be greater than or equal to $\sqrt{2}$ with respect to either the X or Z axis? That is the question.

REPLY [1 votes]: In Mathematica:
If[Abs[#[[1]]]>Abs[#[[2]]],f[#],g[#]]&

You simply look at the $x$ component (#[[1]]) and the $y$ (#[[2]]) component of the vector and see which is larger.


TITLE: Prob. 9, Sec. 3.10 in Kreyszig's functional analysis book: The image of ann isometric non-unitary operator on a Hilbert space
QUESTION [2 upvotes]: Let $H$ be a Hilbert space, let $T \colon H \to H$ be a linear operator such that $T$ is isometric but not unitary.
Then how to show that the image $T[H]$ is a proper closed subspace of $H$?
My effort:
Since $T$ is a linear operator, $T[H]$ is a (vector) subspace of $H$.
Since $T$ is isometric, for each $x \in H$, we have
$$ \langle Tx, Tx \rangle = \Vert Tx \Vert^2 = \Vert x \Vert^2 = \langle x, x \rangle.$$
If $H$ were complex, then $T$ would be unitary if and only if $T$ were isometric and surjective.
But $T$ is not unitary. So $T$ is not surjective. Thus $T[H]$ is a proper subspace of $H$. How to tackle the case when $H$ is real?
How to show that $T[H]$ is closed in $H$?
After reading @Braindead's comment:
Suppose that $y \in \mathrm{cl}(T[H])$. Let $y_n \colon= T x_n$ be a sequence in $T[H]$ such that $y_n$ converges to the point $y$. Then $y_n$ is a Cauchy sequence in $H$. And since $T$ is isometric, the sequence $x_n$ is also Cauchy in $H$, which is complete. So $x_n$ converges to some point $x \in H$. Again using the isometric property of $T$, we can conclude that $T$ is continuous, so that $T x_n$ converges to $T x$. But $Tx_n = y_n$. So $Tx = y$; that is, $y \in T[H]$, showing that $T[H]$ is closed.
Is the above proof correct?
Using th so-called polarisation identities, we obtain, for all $x, y \in H$,
\begin{align*}
\langle Tx, Ty \rangle & = \Re \langle Tx, Ty \rangle + \iota \Im \langle Tx, Ty \rangle \\
&= \frac{1}{4} \left( \Vert Tx + Ty \Vert^2 - \Vert Tx - Ty \Vert^2 \right) + \frac{\iota}4 \left( \Vert Tx + \iota Ty \Vert^2 - \Vert Tx - \iota Ty \Vert^2 \right) \\
&= \frac{1}{4} \left( \Vert T (x + y) \Vert^2 - \Vert T ( x - y) \Vert^2 \right) + \frac{\iota}{4} \left( \Vert T (x + \iota y) \Vert^2 - \Vert T ( x - \iota y) \Vert^2 \right) \\
& = \frac{1}{4} \left( \Vert x + y \Vert^2 - \Vert x - y \Vert^2 \right) + \frac{\iota}4 \left( \Vert x + \iota y \Vert^2 - \Vert x - \iota y \Vert^2 \right) \ \mbox{ [ since $T$ is isometric ]} \\
&= \Re \langle x, y \rangle + \iota \Im \langle x,y \rangle \\
&= \langle x, y \rangle.
\end{align*}
If $H$ is real, there would be no terms involving the $\iota$.
Thus, irrespective of whether $H$ is real or complex, using the definition of the Hilbert adjoint operator $T^*$ of $T$, we have
$$\langle T^* Tx, y \rangle = \langle Tx, T y \rangle = \langle x, y \rangle \ \mbox{ for all } \ x, y \in H.$$
This implies that $$T^* T x = x \ \mbox{ for all } \ x \in H,$$
and so $$T^*T = I. \ \tag{equation 1}$$
Now if $T$ were also surjective (i.e. if $T[H]$ were all of $H$), then $T$ would be bijective (and so $T^{-1}$ would also exist) and in that eventuality, we would obtain
$$TT^* = TT^*I = TT^* (TT^{-1}) = T(T^*T)T^{-1} = T I T^{-1} = I,$$
which together with equation (1) would imply $T^* = T^{-1}$ and hence $T$ would be unitary, contrary to our hypothesis.

REPLY [0 votes]: In general, an isometric linear operator satisfies $T^*T = I$.
By polarization identity we have $\langle Tx , Ty \rangle = \langle x ,y \rangle$, then $\langle Tx,Ty \rangle = \langle x,T^*Ty \rangle $, ie., $\langle x,(T^*T-I)y\rangle = 0$ for all $x,y$. Then $T^*T = I$.
In your case, if $T$ is subjective, then $T^{-1}$ exists and
$$T^{-1} = I T^{-1} = (T^*T)T^{-1} = T^*$$


TITLE: Are there infinitely many rational pairs $(a,b)$ which satisfy given equation?
QUESTION [9 upvotes]: I saw on some facebook page this concrete example:
$1.2^2+0.6^2=1.2+0.6$
The question that immediately  arises is:

Are there infinitely many pairs $(a,b)$ of rational numbers such that we have $a^2+b^2=a+b$?

Thank you for your response.

REPLY [4 votes]: Real solutions:
The equation
$$
a^2 + b^2 = a + b
$$
can be transformed:
$$
a^2 + b^2 = a + b \iff \\
a^2 - a + b^2 -b = 0 \iff \\
(a - 1/2)^2 + (b - 1/2)^2 = 1/4 + 1/4 = 1/2 = (1/\sqrt{2})^2 \quad (*)
$$
The solutions form a circle with origin $(1/2, 1/2)$ and radius $1/\sqrt{2}$ in $\mathbb{R}^2$.
They are
$$
(a, b) = (a, b(a)) = \left(a, (1/2) \pm\sqrt{(1/\sqrt{2})^2 - (a - 1/2)^2}\right)
$$
for $a \in [1/2 - 1/\sqrt{2}, 1/2 + 1/\sqrt{2}] = [a_-, a_+]$.
Rational solutions:
Among all those infinite many real solutions $(a, b)$ we expect infinite many rational solutions, but my topology is too bad to give a good argument here. If we insert arbitrary rational $a$ not all $b = b(a)$ are rational.
I tried to find all rational $a$ for which $b = b(a)$ is rational, but ran against a wall. However an infinite subset of such rational $a$ is still infinite, so we try to pick an easy one.
An infinite rational subset of solutions:
The subset I came up with consists of those $a$ which have the form
$$
a_n = \frac{1}{2} + \frac{1}{2n} < a_+ \quad (n \in \mathbb{N})
$$
which gives
\begin{align}
b_n
&= \frac{1}{2} \pm \sqrt{\frac{1}{2} - \frac{1}{4n^2}} \\
&= \frac{1}{2} \pm \frac{\sqrt{2n^2 - 1}}{2n} \\
&= \frac{n \pm \sqrt{2n^2 - 1}}{2n}
\end{align}
Then $b_n$ is rational, if
$$
2n^2 -1 = m^2
$$
for some $m \in \mathbb{N} \cup \{ 0 \}$.
We can rewrite this as
$$
m^2 - 2 n^2 = -1 \quad (**)
$$
which is a Diophantine equation related to the Pell equation with constant $D = 2$,
see negative Pell equation.
Solution of the negative Pell equation:
One solution to $(**)$ is $m = 1$ and $n = 1$, from which we can generate infinite many more solutions:
For odd $k \in \mathbb{N}$ we have:
$$
-1 = (1^2 - 2\cdot 1^2)^k = (m^2 - 2n^2) \Rightarrow \\
-1 = (1 + \sqrt{2})^k (1-\sqrt{2})^k = (m + \sqrt{2} n)(m - \sqrt{2} n)
$$
which because of
$$
(1 + \sqrt{2})^k
= \sum_{i=0}^k \binom{k}{i} 1^i (\sqrt{2})^{n-i}
= c_1 \cdot 1 + c_2 \sqrt{2} \quad (c_1, c_2 \in \mathbb{N}) \\
(1 - \sqrt{2})^k
= \sum_{i=0}^k \binom{k}{i} 1^i (-\sqrt{2})^{n-i}
= d_1 \cdot 1 - d_2 \sqrt{2} \quad (d_1, d_2 \in \mathbb{N})
$$
gives
$$
m + \sqrt{2} n = (1 + \sqrt{2})^k \\
m - \sqrt{2} n = (1 - \sqrt{2})^k \\
$$
and
$$
m = \frac{(1 + \sqrt{2})^k + (1 - \sqrt{2})^k}{2} \\
n = \frac{(1 + \sqrt{2})^k - (1 - \sqrt{2})^k}{2 \sqrt{2}}
$$
Here are the first ten solutions:
k = 1,  m = 1,       n = 1
k = 3,  m = 7,       n = 5
k = 5,  m = 41,      n = 29
k = 7,  m = 239,     n = 169
k = 9,  m = 1393,    n = 985
k = 11, m = 8119,    n = 5741
k = 13, m = 47321,   n = 33461
k = 15, m = 275807,  n = 195025
k = 17, m = 1607521, n = 1136689
k = 19, m = 9369319, n = 6625109

Plugging the Pell solutions into the circle solution:
We have
\begin{align}
a_k
&= \frac{1}{2} + \frac{1}{2n} \\
&= \frac{1}{2} +
\frac{\sqrt{2}}{(1 + \sqrt{2})^k - (1 - \sqrt{2})^k}
\in \mathbb{Q}
\end{align}
and
\begin{align}
b_k
&= \frac{1}{2} \pm \frac{m}{2n} \\
&= \frac{1}{2} \pm
\frac{\sqrt{2}}{2}
\frac{(1 + \sqrt{2})^k + (1 - \sqrt{2})^k}
{(1 + \sqrt{2})^k - (1 - \sqrt{2})^k}
\in \mathbb{Q}
\end{align}
for any odd $k \in \mathbb{N}$.
Here is a visualization:

(Large version)
Featured are the real solutions (blue circle) and a few of the rational solutions $Q_k^{(+)} = (a_k, b_k^{(+)})$ (green points), $Q_k^{(-)} = (a_k, b_k^{(-)})$ (red points) for $k \in \{ 1,3,5 \}$. The other ones pile up too much to be distinguishable in the plot.


TITLE: Asymptotic behaviour of an integral involving error function
QUESTION [0 upvotes]: I have the following parametric integral
$$F(a)=\int_0^a e^{-x^2\operatorname{erf}(x)}dx$$
where $\operatorname{erf}(x)$ is the error function.
I'm interested in asymptotic behaviour of this integral for the cases $a\to 0$ and $a\to \infty$.
Could someone clarify to me how to obtain an asymptotic expansion of $F(a)$ ?
Thanks a lot.

REPLY [2 votes]: For the case where $a$ is small, you can approximate
$$e^{-x^2 \text{erf}(x)}$$ by a simple $[n,n]$ Padé approximant $P_n$ built around $x=0$. For example
$$P_3=\frac{1+\frac 13 x^2 -\frac 1 {\sqrt \pi}x^3}{1+\frac 13 x^2 +\frac 1 {\sqrt \pi}x^3 }$$ whose error is $\frac {x^7} {45\sqrt \pi}$ which is easy to integrate using partial fraction decomposition; the denominator has only one real root and we shall face one logarithm and an arctangent.
For example, for $a=0.5$, numerical integration gives $0.48390392$ while the approximation of the integrand by $P_3$ gives $0.48389086$.


TITLE: Conditional Gambler ruin problem
QUESTION [1 upvotes]: A gambler repeatedly plays a game where in each round, he wins a dollar with
probability 1/3 and loses a dollar with probability 2/3. His strategy is “quit
when he is ahead by 2 dollars”, though some suspect he is a gambling addict anyway. Suppose that he starts with a million dollars. Show that the probability that he’ll ever be ahead by $2 is less than 1/4.
Is it possible to solve it without Markov chain and with a very basic reasoning.

REPLY [4 votes]: Let $P(i)$ denote the probability he'll get to $\$1,000,002$ if he starts from $\$i$.  We know that $$P(0)=0\;\;\;P(1,000,002)=1$$  You are asked to find $P(1,000,000)$.
Imagine he has $\$i$ and places a bet.  He either gets to state $(i+1)$ or to state $(i-1)$ and we have:  $$P(i) = \frac 13 P(i+1)+\frac 23 P(i-1)$$ It is easier to start near the "ruin state".  We see, for example, that $$P(1) = \frac 13 P(2)+\frac 23 P(0)=\frac 13 P(2)\;\;\;\Rightarrow\;\;\;P(2)=3P(1).$$  In a similar way we see that $$P(3) = 7P(1)\;\;\mbox{and}\;\;\;P(4)=15P(1)$$
We are lead to conjecture that $$P(n) = \left(2^{n}-1\right)P(1)$$
And this is easily confirmed by induction.
To finish, we compute $P(1)$ from the other boundary value:  $$P(1,000,002)=1=(2^{1,000,002}-1)P(1)$$
Whence $$P(1) = \frac {1}{(2^{1,000,002}-1)}$$
We now see the answer:  $$P(1,000,000)= \frac{2^{1,000,000}-1}{2^{1,000,002}-1}$$
From which the desired inequality follows at once.
We note that $4$ times the numerator is $2^{1,000,002}-4$ so $4P(1,000,000)=1-\frac{3}{2^{1,000,002}-1}$.  We may be below $\frac 14$ here, but not by much.


TITLE: PDF in book gives impossible answers
QUESTION [0 upvotes]: So I have a problem in my book with the following pdf.
$f(x) = \frac32 \cdot \frac{1}{x^2}$ for $1≤x≤3$
Otherwise $f(x) = 0$
For any pdf to be valid all probabilities need to be less than one but this obviously yields 150% chance at 1. I made a cdf from this
$F(x) = -\frac{1.5}{x}$ for $1≤x≤3$
and it performs as expected but I was marked wrong by my professor when I used the cdf to find $P[1≤x≤2.5] = .9$ and $P[1.5≤x≤2.5] = .4$. Could someone explain why this answer is incorrect? I don't think this integration is wrong. Even if I integrate from negative infinity to y that still yields the same answer.

REPLY [2 votes]: Your cdf is wrong.  For $1 \le x \le 3$ you should have
$$F(x) = \int_{-\infty}^x f(t)\; dt = \int_1^x \frac{3}{2t^2} \; dt = \frac{3}{2} - \frac{3}{2x}$$
while $F(x) = 0$ for $x < 1$ and $F(x) = 1$ for $x > 3$.


TITLE: How many degree $k$, monic polynomials factor completely in $\mathbb{Z}/p\mathbb{Z}$?
QUESTION [0 upvotes]: Suppose that we have the finite field $\mathbb{Z}/p\mathbb{Z}$, and we are considering adjoining the roots of degree $k$ monic polynomials to the field (whose coefficients are taken from this field).  So we have a total of $p^k$ possible polynomials that could factor to some degree.
Is there a formula for how many of these polynomials completely factor into linear factors?
I found this answer on Stack Exchange concerning irreducible polynomials.

REPLY [3 votes]: There is a bijection: $$\{\text{monic polynomials }f \text{ of degree $k$ which split completely}\}\to\{\text{multisets of size $k$, consisting of elements of }\mathbb Z/p\},$$
sending $f$ to the set of solutions to $f(x)=0$, which all lie in $\mathbb Z/p$ due to our assumption that $f$ splits into linear factors. In the other direction, take a multiset $S$ to the polynomial $\prod_{\alpha\in S}(x-\alpha)$. [Here, a multiset is a set with multiplicity.]
The number of such multisets is ${k+p-1\choose p-1}$—imagine having $k$ objects in a line, and putting $p-1$ dividers into them to make $p$ possibly empty groups. We can instead think about $k+p$ objects, and putting $p-1$ dividers, where we require the groups to be nonempty. There are $k+p-1$ places to put the dividers, so the total number is ${k+p-1\choose p-1}$.
Alternatively, consider the coefficient of $x^k$ in the generating function $(1+x+x^2+\dots)^p=\frac1{(1-x)^p}$. The formula for this is proved here.


TITLE: Let $H \leq G$: prove that $K=\{x\in G:\exists n \in\Bbb{N},\,x^n\in H\}$ is a subgroup of $G$?
QUESTION [1 upvotes]: $H$ is a subgroup of an abelian group $G$. We define
$$K = \{ x \in G : \exists n \in \Bbb{N}, x^n \in H\}.$$
How to prove that $K$ is a subgroup of $G$ ?

REPLY [2 votes]: First do the case $H=0$, then $K$ is known as the torsion subgroup of $G$. In general, $K$ is the torsion subgroup of $G/H$ pulled back along the projection $G \twoheadrightarrow G/H$.

REPLY [2 votes]: It is the set of all elements $x\in G$ such that
$$
\overline{x}=xH
$$
has finite order in $G/H$.
In other words, if we denote
$$
\pi:G\longrightarrow G/H
$$
the canonical surjection, and if we set $L$ to be the set of all finite order elements in $G/H$, then
$$
K=\pi^{-1}(L).
$$
So it suffices to check that $\pi$ is a group homomorphism and that $L$ is a subgroup of $G/H$. Both are easy.
This is far from being the fastest way, but that's how I see it.


text:
\section{Linear convergence of first-order methods}

For PG, it has been shown that linear convergence can be achieved for non-strongly convex objectives under various relaxed conditions \cite{luo1993error, drusvyatskiy2018error, karimi2016linear}. In particular, \eqref{eq:std-form-proj-grad-desc} can be shown to satisfy several such conditions and therefore guarantees linear convergence under PG. One notion of central significance in many such results is the \textit{Hoffman constant}. For $A\in \RR^{d\times r }$ and polyhedral set $\cX\subseteq \RR^d$, denote the (relative) Hoffman constant of $A$ w.r.t. $\cX$ as $H_\cX(A)$, which is the smallest $H>0$ such that, for any $z\in {\rm range}(A)$, $\cS = \{x: Ax=z \}$, it holds that $\|x - \Pi_{\cX\cap \cS}(x)\| \leq H \|Ax-z\|$ for all $x\in \cX$ (the Hoffman inequality). Note that $H_\cX(A)$ is always well-defined, finite, and depends only on $A$ and $\cX$ \cite{hoffman1952approximate, pena2020new}.

In the optimization context, the intuition is as follows: the set of optimal solutions $\cX^*$ can often be expressed as $\cX^* = \cX \cap \cS$, which means the distance to optimality can be bounded by a constant multiple of the residual $\|Ax-z\|$. Therefore, the Hoffman inequality can be viewed as an EB condition. Further details on this definition, as well as its connection to the classical explicit expression, can be found in Appendix \ref{app:char-hoffman-const}.

Using the Hoffman inequality, we have the following linear convergence guarantee. It is a special case of \cite[Theorem 5]{karimi2016linear}. A proof is in Appendix \ref{app:proof-h(Ax)+poly_ind-conv}, which is essentially the same as \cite[Appendix F]{karimi2016linear}. However, our proof is slightly refined to allow a weakened condition, provide more details on the relevant constants and conclude iterate convergence.
\begin{theorem}
	For \eqref{eq:std-form-proj-grad-desc} with $q = 0$, PG with a constant stepsize $\gamma_t = \frac{1}{L \|A\|^2}$ generates $x^t$ such that $ \frac{\mu}{2 H_\cX(A)} \|x^t - \Pi_{\cX^*}(x^t)\|^2 \leq f(x^t) - f^* \leq \left( 1 - \frac{\mu}{\max\{ \mu, LH_\cX(A)^2 \|A\|^2  \}} \right)^t \left(f(x^0) - f^* \right) $ for all $t$.
	 \label{thm:h(Ax)+poly_ind}
\end{theorem}

For the case of $q\neq 0$, we have the following instead, which relies heavily on the boundedness of $\cX$. The proof is in Appendix \ref{app:proof-h(Ax)+<q,x>+poly_ind-conv}. The key is to make use of the polyhedral structure of the set of optimal solutions and establish a QG condition similar to the one in \cite[Lemma 2.3]{beck2017linearly}. Here, we slightly refine the constants using the monotonicity of $f(x^t)$.
We note that linear convergence with different rates can also be established through invoking different error bound conditions that also hold for \eqref{eq:std-form-proj-grad-desc} (see, e.g., \cite{wang2014iteration, drusvyatskiy2018error}).

\begin{theorem}
	 There exists unique $z^*\in \RR^r$ and $t^*\in \RR$ such that $Ax^* = z^*$ and $\langle q, x^*\rangle = t^*$ for any $x^* \in \cX^*$. Furthermore, PG for \eqref{eq:std-form-proj-grad-desc} with constant stepsize $\gamma_t = \frac{1}{L \|A\|^2}$, starting from $x^0\in \cX$, generates $x^t$ such that $ \frac{1}{\kappa} \|x^t - \Pi_{\cX^*}(x^t)\| \leq f(x^t) - f^* \leq C \rho^t $, where\newline $\rho = 1 - \frac{1}{\max\{ 1, 2\kappa L \|A\|^2\}}$, $\kappa = H_\cX(A)^2 \left( C + 2 G D_A + \frac{2(G^2 + 1)}{\mu} \right)$,

	 $C = f(x^0) - f^*$, $G = \|\nabla h(z^*)\|$, $D_A = \sup_{x, x'\in \cX} \|A(x - x')\|$.
	 \label{thm:h(Ax)+<q,x>-lin-conv}
\end{theorem}

For FW, the classical sublinear convergence holds. Specifically, for \eqref{eq:std-form-proj-grad-desc}, using FW with static stepsizes or exact linesearch, that is, $\gamma_t = \frac{2}{2+t}$ or  $\gamma_t \in \argmin_{\gamma\in [0,1]} f(x^t + \gamma (w^t - x^t))$, it holds that $f(x^t) - f^* \leq \frac{{\rm Diam}(\cX)^2 L\|A\|^2}{k+2}$ for all $t$ (see, e.g., \cite[Theorem 1]{jaggi2013revisiting} and \cite[Theorem 2.3]{clarkson2010coresets}). The advantage of FW is that it can maintain highly sparse iterates when $\cX$ has sparse vertices.
In fact, if we start with $x^0$ being a vertex, then $x^t$ is a convex combination of $(t+1)$ vertices. For ME convex programs such as \eqref{eq:eisenberg-gale-primal}, the constraint set $\cX$ is a product of simplexes $\Delta_n$. Therefore, any vertex of $\cX$ has the form $x\in \{0,1\}^{n\times m}$, where each $(x_{1j}, \dots, x_{nj})$, $i\in [m]$ is a unit vector. Hence, $x^t$ is a linear combination of $(t+1)$ vertices and has at most $(t+1)m$ nonzeros.
There have been recent theories on linear convergence of an away-step variant of FW for \eqref{eq:std-form-proj-grad-desc} \cite{beck2017linearly, lacoste2015global}, in which a succinct ``vertex representation'' \cite[\S 3.2]{beck2017first} must be maintained through an additional, potentially costly reduction procedure. Initial trials suggest that it does not work well for our specific convex programs and are therefore not considered further.

\paragraph{PG with Linesearch.} Linesearch is often essential for numerical efficiency of gradient-based methods, especially when the Lipschitz constant cannot be computed.
Here, we consider a modified version of backtracking linesearch for PG (see, e.g., \cite[\S 10.4.2]{beck2017first} and \cite[\S 1.4.3]{beck2009gradient}). Specifically, given increment factor $\alpha\geq 1$, decrement factor $\beta\in (0,1)$, and stepsize upper bound $\Gamma >0$, at each iteration, set the (candidate) stepsize $\gamma_t$ to either the previous one $\gamma_{t-1}$ or $\min\{\Gamma, \alpha \gamma_{t-1}\}$, depending on whether backtracking occurs at all in the previous iteration. Then, perform the usual backtracking with factor $\beta$ until a sufficient decrease condition is met. denote the linesearch subroutine as $(x^{t+1}, \gamma_t, k_t) \leftarrow \mathcal{LS}_{\alpha, \beta, \Gamma}(x^t, \gamma_{t-1}, k_{t-1})$, where $k_{t-1}$ is the number of backtracking steps in the previous iteration, $\gamma_{-1} = \Gamma$, $k_{-1}=0$.

More details, as well as the proof of the following convergence result (in fact, a more general one that holds for the proximal gradient method), can be found in Appendix \ref{app:pg-ls-details}.
Let $L^\cX_f = \sup_{x,y\in \cX,\, x\neq y}\frac{\|\nabla f(x) - \nabla f(y)\|}{x-y} \leq L \|A\|^2$ be the (effective) Lipschitz constant of $\nabla f$ on $\cX$.
Note that, in addition to ensure linear convergence, linesearch can potentially improve the constant in the rate, since $L_f^\cX$ can be much smaller than $L\|A\|^2$.
\begin{theorem}
	PG with linesearch $\mathcal{LS}_{\alpha,\beta, \Gamma}$ for \eqref{eq:std-form-proj-grad-desc} with $q=0$ and $q\neq 0$ converges linearly with rate $1 - \frac{\mu }{\max\{\mu, H^2/\Gamma, H^2 L_f^\cX / \beta \}}$ (where $H = H_\cX(A)$) and $1 - \frac{1}{\max \{1, 2\kappa/\Gamma, 2\kappa L_f^\cX/\beta \}}$, respectively.  For both, the number of projections $\Pi_\cX$ per iteration is at most $1 + \frac{\log \frac{\Gamma}{\tilde{\gamma}}}{\log \frac{1}{\beta}}$, where $\tilde{\gamma} :=  \min\left\{ \Gamma, \beta/L^\cX_f\right\}$.
	\label{thm:pg-ls-conv}
\end{theorem}

config: arxiv


TITLE: plotting an fft function
QUESTION [1 upvotes]: need to plot the folowing:
$$X(j\omega)=\cal F\{{2\over\pi\lambda}\}*\cal F\{sinc(2w_m\lambda)\}-\cal F\{{4\over\pi\lambda}\}*\cal F\{sinc(w_m\lambda)\}$$
as a plot of $|X(j\omega)|$ for $|\omega|<10\pi$
the code I've wrote is:
%% define
t = -4:1/100:4;
w= -10*pi:1/100:10*pi;
wm=3*pi;
z=t-(pi/wm);
a=1./(pi*z);
b=sinc(2*wm*z);
c=sinc(wm*z);
Aa=fft(a);
Bb=fft(b);
Cc=fft(c);
Dd=fft(a);
A1=fftshift(Aa);
B1=fftshift(Aa);
C1=fftshift(Aa);
D1=fftshift(Aa);
x=2*conv(A1,B1)-4*conv(D1,C1);
mag=abs(x);
%%plot
figure;
plot(w,mag,'k','LineWidth',2);
title('Q1: |X(jw)| - frequency domain');
xlabel('w [hz]','FontSize',12);
ylabel('|X(jw)|','FontSize',12);
legend([{'|X(jw)|'}]);

I know my problem is with the w-axis, but I have no idea on how to fix it.

REPLY [0 votes]: You have two arrays of different dimensions in your plot.
After running your code, I am told that mag is 1 by 1601 and w is 1 by 6000 something.

Fix 1 quick and dirty:
w = linspace(-10*pi, 10*pi, 1601);

A better solution is to define w after mag since w isn't used until plotting. Then you can do
Fix 2 (a) and (b):
w = -10*pi:length(mag):10*pi;

or
w = linspace(-10*pi, 10*pi, length(mag));

This way if you change anything in the code that affects the length of mag, then w will change accordingly.


TITLE: Vanishing in etale motivic cohomology
QUESTION [2 upvotes]: As far as vanishing is concerned, the usual motivic cohomology has the following two properties (for a smooth scheme $X$ over a field):

$H^{p,q}(X, \mathbb Z) = 0$, if $p > q + dim(X)$; and
$H^{p,q}(X, \mathbb Z) = 0$, if $q<0$.

Are the analogous properties true for etale (or Lichtenbaum) motivic cohomology?  The only statement of this sort that I have heard of is the statement of the motivic Hilbert 90 ($H_L^{n+1,n}(Spec F, \mathbb Z)=0$, for a field $F$).  However, it is very hard prove and is a crucial step in the proof of the Bloch-Kato conjecture by Voevodsky.

REPLY [4 votes]: The easiest counterexample is $p=2, q=0$. In this case the usual motivic cohomology vanishes, and if $X$ is normal
$$H^2_{et}(X,\mathbf{Z})=H^1_{et}(X,\mathbf{Q}/\mathbf{Z})= \mathrm{Hom}(\pi_1^{et}(X),\mathbf{Q}/\mathbf{Z})$$
which does not vanish even for fields (which are not separably closed).
Another example is $p=3, q=1$. In this case you get the Brauer group which doesn't vanish in general.


TITLE: Where does the identity $\tau(nm) = \sum_{d \mid (m, n)} \mu(d) \tau(n/d) \tau(m/d)$ come from?
QUESTION [3 upvotes]: I was reading an article which states (without reference) a recurrence relation for a character sum associated to a quadratic extension of number fields. It is a bit complicated to state, but the analogue for the trivial extension $\mathbb Q / \mathbb Q$ and the trivial character would be that
$$\tau(nm) = \sum_{d \mid (m, n)} \mu(d) \tau(n/d) \tau(m/d)$$
where $\tau$ denotes the divisor counting function. This seems to be true. I quickly checked this when $mn$ has $1$ or $2$ prime divisors; it then reduces to the identities
$$a+b+1 = (a+1)(b+1) - ab$$
resp.
$$\begin{align*}(a+b+1)(c+d+1)
& = (a+1)(b+1)(c+1)(d+1) \\
& - ab(c+1)(d+1) \\
& - (a+1)(b+1)cd \\
& + abcd
\end{align*}$$
Where does this identity (if true) come from? Can it be understood by comparing the coefficients of two Dirichlet series? Modular forms? It is reminiscent of the recurrence relation for Hecke operators.

REPLY [1 votes]: For the proof of your identity let
$$n =N \prod_j p_j^{a_j+1}, \qquad m =M \prod_j p_j^{b_j+1}, \qquad \gcd(N,M)=\gcd(NM,\prod_j p_j)=1$$
Then $$\tau(nm) = \tau(N)\tau(M) \prod_j \tau(p_j^{a_j+b_j+2})=  \tau(N)\tau(M) \prod_j  ( \tau(p_j^{a_j+1})\tau(p_j^{b_j+1})- \tau(p_j^{a_j})\tau(p_j^{b_j}))$$
$$=\tau(N)\tau(M) \prod_{p_j^r \| (n,m)} \sum_{s=0}^r \mu(p_j^s)\tau(p_j^{a_j+1-s})\tau(p_j^{b_j+1-s})$$
$$ = \tau(N)\tau(M)\sum_{d |(n,m)} \mu(d) \tau(\frac{n}{Nd})\tau(\frac{m}{Md})$$ $$=\sum_{d |(n,m)} \mu(d) \tau(n/d)\tau(m/d)$$


TITLE: Why did the babylonians approximated the square root of two the way they did?
QUESTION [1 upvotes]: I was researching about the Newton-Raphson method and came across http://www.sosmath.com/calculus/diff/der07/der07.html. On the third line of the page near the end of the line it told us to consider $2/x_1$:

Consequently $3/2>\sqrt{2}$. If we now consider $2/x_1=4/3$, its square $16/9$ is of course smaller than $2$, so $2/x_1<\sqrt{2}$.

Why is that? Why did they divide $2$ by $x_1$ and not divide $1$ by $x_1$?

REPLY [0 votes]: Since $x_1$ is close to $\sqrt2$, taking $2\over x_1$ will get you another number close to $\sqrt2$ but on the other side from $x_1$. This way you can take the average of $x_1$ and the new number to approximate $\sqrt2$.


TITLE: Find the formula $\int_0^1f(x)dx\approx A_0f(0)+A_1f(1)$ that is exact for all functions of the form $f(x)=ae^x+b\cos(\pi x/2)$.
QUESTION [0 upvotes]: Find the formula $\int_0^1f(x)dx\approx A_0f(0)+A_1f(1)$ that is exact for all functions of the form $f(x)=ae^x+b\cos(\pi x/2)$.
I'm not sure how to go about this. Any solutions/hints are greatly appreciated.

REPLY [2 votes]: $$\int_0^1[ae^x+b\cos(\pi x/2)]{\rm d}x=a(e-1)+\frac{2b}{\pi}=\underbrace{1}_{A_1}\underbrace{(ae)}_{f(1)}+\underbrace{\frac{2b}{(a+b)\pi}}_{A_0}\underbrace{(a+b)}_{f(0)}$$


TITLE: Expectation of minimum of two uniform distribution
QUESTION [0 upvotes]: $x_1$ and $x_2$ are uniformly distributed in $[0,1]$ with correlation $\rho$.
What is $E[min(x_1,x_2)]$?
I know in a simple case when $x_1$ and $x_2$ are independent, then I can get the cdf of $min(x_1,x_2)$ first and then break the joint distribution down. Next get the pdf and then the expectation of $min(x_1,x_2)$.
But I have no idea how to break the cdf if they are no longer independent.

REPLY [1 votes]: HINT
The formula for the CDF of the minimum of two iid variables $$ P(X_{min} \le x) = 1 - P(X_{min}>x) = 1-P(X>x)^2$$ generalizes to $$ P(X_{min} \le x) = 1 - P(X_{min}>x) =  1 - P(X_1>x, X_2>x) $$ in terms of the joint distribution of $(X_1,X_2).$
EDIT
Actually, come to think of it, is the problem even well-defined? "Two uniform variables with correlation $\rho$" does not define a unique joint distribution. So unless the distribution of the min is somehow independent of everything except the correlation, I don't know that this is doable. Someone correct me if I'm wrong. Does it only depend on $\rho$ (or is there some obvious implied choice of the joint distribution like bivariate normal in the case with two normals?)
EDIT 2
For example, if $U_1\sim U(0,1)$ and conditional on $U_1,$ $U_2 = U_1$ with probability $1/2$ and $1-U_1$ with probability $1/2,$ then $U_1$ and $U_2$ are uncorrelated $U(0,1)$'s. We can compute $E(\min(U_1,U_2)) = 3/8.$ However, if $U_1$ and $U_2$ are independent $U(0,1)$'s (also uncorrelated of course), then $E(\min(U_1,U_2)) = 1/3.$ So the correlation is not enough to determine this quantity.


TITLE: Truncated Euler's sigma function
QUESTION [1 upvotes]: Does the arithmetical function
$$
s(n) := \sum_{k|n} \min\left\{k,\frac{n}{k}\right\}
$$
have well-studied properties? Can anybody help me by pointing out some studies about these functions?
As motivation, I am studying about the number of solutions of some Diophantine equations and try to come up with some identities involving arithmetic functions. One of the equations gives result involving this function.
Thank you.

REPLY [0 votes]: Let $d_1,\ldots,d_r$ be the positive divisors of $n$, in increasing order, hence $d_1=1$, $d_r=r$, and $n=d_id_{r+1-i}$ for all $i$. It is well-known that $r$ is odd if and only if $n$ is square. Therefore, if $n$ is a square then $r$ is odd and
$$
s(n)=d_{\frac{r+1}{2}}+2\left(d_1+\cdots+d_{\frac{r-1}{2}}\right)
$$
Otherwise $n$ is not square, so $n$ is even and
$$
s(n)=2\left(d_1+\cdots+d_{\frac{r}{2}}\right).
$$
To answer your question, which properties of $s$ you have in mind?


TITLE: Subgroups of $S_6$
QUESTION [8 upvotes]: What are all the subgroups of $S_6$ up to isomorphism? I have been able to find a few listings but not one I can be sure is complete.

REPLY [7 votes]: The GroupProps Subwiki is a decent resource for information about small groups (up to isomorphism) and their subgroups.
Drilling down a level we learn that $S_6$ is unique among the symmetric groups in having a different number of conjugacy classes (56) from automorphism classes (37) of subgroups.  That is, there are subgroups of $S_6$ which are isomorphic under an outer automorphism of $S_6$ but not under an inner automorphism (conjugacy).
Unfortunately the latter Wiki page has a very incomplete "Table classifying subgroups up to conjugacy", listing only 4 of the 56 conjugacy classes.  Of course equivalence up to isomorphism is even coarser than equivalence up to automorphism (of the whole group $S_6$).  A paper at arxiv.org lists 29 such subgroups up to isomorphism (in an applied context):
$$ D_6, D_3, D_2, Z_6, Z_3, Z_2, D_3\times D_3, Z_3×Z_3, (Z_3\times Z_3)⋊ Z_2,
D_3 \times Z_3, Z_2 \times Z_2 \times Z_2, S_3 ≀ Z_2, Z_4, Z_5, D_4, Z_2 \times Z_4, D_4, D_5, A_4, D_4 \times Z_2, Z_5 ⋊ Z_4,
A_4 \times Z_2, S_4, (Z_3 \times Z_3) ⋊ Z_4, S_4 \times Z_2, A_5, S_5, A_6, S_6 $$
and references a PDF that lists by order/index the 56 conjugacy classes of $S_6$ subgroups from this Web site.  There is an error in this list, as $D_4$ appears twice and the trivial group $Z_1$ is omitted.
For comparison with Jack Schmidt's Answer, the $Z_n$ here are the cyclic groups denoted $C_n$ by GAP. As Derek Holt points out, the notation here for dihedral group of order $2n$ is $D_n$, where GAP uses $D_{2n}$.  There's a redundancy in that $D_3 \cong S_3$, yet here both are used.


TITLE: Holomorphic function
QUESTION [2 upvotes]: Let $f(z)$ be a holomorphic  function  that  maps  the  unit  disk  to  the  unit  disk. Prove that $$|f^{(5)}(0)| \leq 120.$$  I use some concrete example it seems that this statement work out but i do not know how to come up with a formal proof help any one

REPLY [2 votes]: Maybe overly pedantic, but we can't really integrate over $|z| = 1$ without some justification, since $f$ is only assumed to be holomorphic on $|z| < 1$. It's not that big of a deal though, since we can integrate over the circle $|z| = r$ for any $r < 1$ to get:
$$
|f^{(5)}(0)| = \bigg| \frac{5!}{2\pi i}\int_{|z|=r} \frac{f(z)}{z^6}\,dz \bigg| \le \frac{120}{2\pi} \cdot 2\pi r \cdot \frac{1}{r^6} = \frac{120}{r^5}
$$
since $|f(z)| < 1$ on $|z|=r$ by assumption. The above inequality holds for all $r$, so by letting $r \to 1^-$, it follows that $|f^{5}(0)| \le 120$.


TITLE: If $X_i\sim U(0,\theta)$, then there exists UMP test for $H_0:\theta=\theta_0$ vs $H_1:\theta>\theta_0$
QUESTION [1 upvotes]: Let $X_{1},\dots,X_{n} \sim Unif(0,\theta)$ independently. Show there exists a UMP size $\alpha$ test for testing $H_{0}:\theta=\theta_{0}$ vs $H_{1}:\theta>\theta_{0}$.

My attempt:
I attempted to show that the distribution is of increasing monotone likelihood ratio by computing $$\frac{f(X_{1},\dots, X_{n}\mid\theta_{1})}{f(X_{1}\dots X_{n}\mid\theta_{2})}$$
This computation gave me: $$\frac{f(X_{1},\dots, X_{n}\mid\theta_{1})}{f(X_{1},\dots ,X_{n}\mid\theta_{2})}=\frac{\theta_{2}^{n}I(X_{(n)}<\theta_{1})}{\theta_{1}^{n}I(X_{(n)}<\theta_{2})}$$ This already confuses me because I'm not familiar with what happens when dividing indicator functions. Moreover, this does not seem to be a non-decreasing function of our test-statistic $t(x)=X_{(n)}$, which, according to my notes, means we cannot apply the Karlin-Rubin theorem for finding a UMP test. Since I get stuck here I cannot prove that there exists a UMP size $\alpha$ test and I can also not find its form.
Question: What is going wrong in my approach above and how can I solve this question?
Thanks!

REPLY [2 votes]: Just to elaborate on the division of indicator functions:
To test $H_0:\theta=\theta_0$ against $H_1:\theta=\theta_1(>\theta_0)$ according to Neyman-Pearson lemma, note that
the likelihood ratio $\lambda$ is of the form
\begin{align}
\lambda(x_1\ldots,x_n)&=\frac{f_{H_1}(x_1,\ldots,x_n)}{f_{H_0}(x_1,\ldots,x_n)}
\\\\&=\left(\frac{\theta_0}{\theta_1}\right)^n\frac{\mathbf1_{{x_{(n)}<\theta_1}}}{\mathbf1_{{x_{(n)}<\theta_0}}}
\\\\&=\begin{cases}\left(\frac{\theta_0}{\theta_1}\right)^n&,\text{ if }0<x_{(n)}<\theta_0\\\,\,\, \infty&,\text{ if }\theta_0< x_{(n)}<\theta_1\end{cases}
\end{align}
So I think it follows from here that $\lambda$ is a monotone non-decreasing function of $x_{(n)}$.
And by N-P lemma we know that an MP test of size $\alpha$ is given by $$\varphi(x_1,\ldots,x_n)=\mathbf1_{\lambda(x_1,\ldots,x_n)>k}$$, where $k$ is so chosen that $$E_{H_0}\varphi(X_1,\ldots,X_n)=\alpha$$
Now you can proceed with your proof.