Eufisky - The lost book

## 级数方法解答的一道积分不等式题

$\int_0^{2\left( {k + 1} \right)} {\frac{{{x^k}}}{{k!}}{e^{ - x}}dx} = - \left. {{e^{ - x}}\sum\limits_{i = 0}^k {\frac{{{x^i}}}{i}} } \right|_0^{2\left( {k + 1} \right)} = 1 - {e^{ - 2\left( {k + 1} \right)}}\sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^i}}}{i}} .$

${e^{2k + 2}} > \left( {k + 1} \right)\sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} \Leftrightarrow \sum\limits_{i = 0}^\infty {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} > \left( {k + 1} \right)\sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} .$

$\sum\limits_{i = k + 1}^{2k + 1} {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} = \sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^{i + k + 1}}}}{{\left( {i + k + 1} \right)!}}} > k\sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} .$

${\left( {2k + 2} \right)^{k + 1}} > k\frac{{\left( {2k + 1} \right)!}}{{k!}} = k\left( {k + 1} \right) \cdots \left( {2k + 1} \right) > k\frac{{\left( {i + k + 1} \right)!}}{{i!}}$是显然的.

## 裴礼文上的一道积分不等式

Poof.For all $x$,

$$\left|\frac{\sin((2n+1)x)}{\sin(x)}\right|=\left|\sum_{k=-n}^{n}e^{i2kx}\right|\le2n+1$$
Note that $\displaystyle\sum_{k=-n}^{n}e^{i2kx}=1+2\sum_{k=1}^n\cos(2kx)$.

For $0\le x\le\pi/2$, we have $\sin(x)\ge2x/\pi$. Therefore,
$$\left|\frac{\sin((2n+1)x)}{\sin(x)}\right|\le\frac\pi{2x}$$
Thus,
\begin{align*}\int_0^{\pi/2}\left|\frac{\sin((2n+1)x)}{\sin(x)}\right|\,\mathrm{d}x&\le\int_0^{\pi/(4n+2)}(2n+1)\,\mathrm{d}x+\int_{\pi/(4n+2)}^{\pi/2}\frac\pi{2x}\,\mathrm{d}x\\&=\frac\pi2+\frac\pi2\log\left(2n+1\right)\end{align*}
For $n\ge3$, $2n+1\le\frac73n$. Therefore,
\begin{align*}\frac\pi2+\frac\pi2\log\left(2n+1\right)&\le\frac\pi2\left(1+\log\left(\frac73\right)+\log(n)\right)\\[6pt]&\le\pi\left(1+\frac{\log(n)}{2}\right).\end{align*}

Show that for $p>1$ and $x \ge 0$,$$\dfrac{2}{\pi}\int_{x}^{px}\left(\dfrac{\sin{t}}{t}\right)^2\,\mathrm dt\le 1-\dfrac{1}{p}$$

Proof.This is quite a difficult problem, and I found it very enjoyable.  Here is the solution I found:

First, we give some simple bounds when $x$ is large, or $px$ is small.  If $x\geq\frac{2}{\pi},$ then by using the bound $|\sin(t)|\leq1$,
we have that
$$\frac{2}{\pi}\int_{x}^{px}\left(\frac{\sin(t)}{t}\right)^{2}dt\leq\frac{2}{\pi}\int_{x}^{px}\frac{1}{t^{2}}dt=\frac{2}{\pi x}\left(1-\frac{1}{p}\right)\leq1-\frac{1}{p}.$$
Similarly, if $px\leq\frac{\pi}{2}$, then since $\frac{\text{sin}(t)}{t}\leq1$,
it follows that
$$\frac{2}{\pi}\int_{x}^{px}\left(\frac{\sin(t)}{t}\right)^{2}dt\leq\frac{2}{\pi}\left(px-x\right)=\frac{2xp}{\pi}\left(1-\frac{1}{p}\right)\leq\left(1-\frac{1}{p}\right).$$
Now, assume that $0\leq x\leq\frac{2}{\pi}$, and that $px\geq\frac{\pi}{2}$.
Then notice that
$$\frac{2}{\pi}\int_{x}^{px}\left(\frac{\sin(t)}{t}\right)^{2}dt=1-\frac{2}{\pi}\int_{px}^{\infty}\left(\frac{\sin(t)}{t}\right)^{2}dt-\frac{2}{\pi}\int_{0}^{x}\left(\frac{\sin(t)}{t}\right)^{2}dt$$
since $\int_{0}^{\infty}\left(\frac{\sin(t)}{t}\right)^{2}dt=\frac{\pi}{2}$.
We will now find a bound on the other two terms. Working over an interval
of length $\pi$, by pulling out a lower bound for $\frac{1}{t^{2}}$,
we have that for any $y$

$$\int_{y}^{y+\pi}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{1}{\left(y+\pi\right)^{2}}\int_{0}^{\pi}\sin^{2}(t)dt\geq\frac{\pi}{2}\int_{y+\pi}^{y+2\pi}\frac{1}{t^{2}}dt,$$
and so
$$\frac{2}{\pi}\int_{px}^{\infty}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\int_{px+\pi}^{\infty}\frac{1}{t^{2}}dt=\frac{1}{px+\pi}.$$
Since the function $\frac{\sin(t)}{t}$ is monotonically decreasing
on the interval $\left[0,\frac{2}{\pi}\right],$ it follows that for
$x\leq\frac{2}{\pi}$ we have

$$\frac{1}{x}\int_{0}^{x}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{\pi}{2}\int_{0}^{\frac{2}{\pi}}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{\pi}{2}\cdot\frac{5}{3\pi},$$

and hence

$$\frac{2}{\pi}\int_{0}^{x}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{5x}{3\pi}.$$

Now, notice that since $px\geq\frac{\pi}{2},$ and $p>1$, by plugging them in directly, we have that

$$\frac{5\left(xp\right)^{2}}{3\pi}+\frac{2}{3}px+p-\pi>\frac{5\pi}{12}+\frac{\pi}{3}+1-\pi=1-\frac{\pi}{4}>0.$$

Rearranging the above by dividing through by both $(px+\pi)$  and $p$, we obtain the inequality

$$\frac{5}{3\pi}x+\frac{1}{px+\pi}>\frac{1}{p},$$
for $px\geq\frac{\pi}{2}$, and $p>1$. It then follows that
$$\frac{2}{\pi}\int_{px}^{\infty}\left(\frac{\sin(t)}{t}\right)^{2}dt+\frac{2}{\pi}\int_{0}^{x}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{1}{p},$$
for $x\leq\frac{2}{\pi},$ and $px\geq\frac{\pi}{2}$, and hence
we have shown that for all $x\geq0$, and all $p>1$,
$$\int_{x}^{px}\left(\frac{\sin(t)}{t}\right)^{2}dt\leq1-\frac{1}{p},$$
as desired.

Show that if $f\in \mathcal C^{n+1}([a,b])$ and $f(a)=f^{'}(a)=\cdots=f^\left(n\right)(a)=0,$ then the following statements are ture:

$\mathbf a)$

$\forall r\in[1,\infty),$the inequality $$\left(\int_{a}^{b}|f(x)|^rdx\right)^{\frac{1}{r}} \leq \frac{(b-a)^{n+\frac{1}{r}}}{n!(nr+1)^{\frac{1}{r}}}\int_{a}^{b}|f^{(n+1)}(x)|dx$$holds.

$\mathbf b)$

$\forall r\in[1,\infty),$the inequality $$\left(\int_{a}^{b}|f(x)|^rdx\right)^{\frac{1}{r}} \leq \frac{2^{\frac{1}{r}}(b-a)^{n+\frac{1}{r}+\frac{1}{2}}}{n!\sqrt{2n+1}(2nr+r+1)^{\frac{1}{r}}}\left(\int_{a}^{b}|f^{(n+1)}(x)|^{2}dx\right)^{\frac{1}{2}}$$holds.

Proof.We have by Taylor's Theorem with Integral form of the Remainder

\begin{align*}f(x) = \int_a^x\dfrac{f^{(n+1)}(t)}{n!}(x-t)^ndt\end{align*}

Then we have
\begin{align*}\int_a^b |f(x)|^rdx &= \int_a^b \left|\int_a^x\dfrac{f^{(n+1)}(t)}{n!}(x-t)^ndt\right|^rdx \\&\leq \int_a^b \left(\int_a^x \left|\dfrac{f^{(n+1)}(t)}{n!}\right| \left|(x-t)^n \right|dt\right)^rdx \\&\leq \int_a^b \left(\int_a^x \left|\dfrac{f^{(n+1)}(t)}{n!}\right|dt (x-a)^n\right)^rdx \\&\leq \left(\int_a^b \left|\dfrac{f^{(n+1)}(t)}{n!}\right|dt\right)^r\int_a^b \left( (x-a)^n\right)^rdx \\& = \left(\int_a^b \left|\dfrac{f^{(n+1)}(t)}{n!}\right|dt\right)^r\frac{(b-a)^{nr+1}}{nr+1} .\end{align*}

So we get
\begin{align*}\left(\int_a^b |f(x)|^rdx\right)^{1/r} \leq \left(\frac{(b-a)^{nr+1}}{nr+1}\right)^{1/r} \int_a^b \left|\dfrac{f^{(n+1)}(x)}{n!}\right|dx\end{align*}
which is $\mathbf a)$

To get $\mathbf b)$ we can proceed similarly using Holder's inequality

\begin{align*}\int_a^b |f(x)|^rdx &= \int_a^b \left|\int_a^x\dfrac{f^{(n+1)}(t)}{n!}(x-t)^ndt\right|^rdx \\&\leq \int_a^b \left(\int_a^x \left|\dfrac{f^{(n+1)}(t)}{n!}\right|^2dt \int_a^x \left|(x-t)^{2n} \right|dt\right)^{r/2} dx \\&\leq \left(\int_a^b \left|\dfrac{f^{(n+1)}(t)}{n!}\right|^2dt\right)^{r/2} \int_a^b  \left(\int_a^x \left|(x-t)^{2n} \right|dt\right)^{r/2} dx \\&= \left(\int_a^b \left|\dfrac{f^{(n+1)}(t)}{n!}\right|^2dt\right)^{r/2} \int_a^b  \left(\frac{(x-a)^{2n+1}}{2n+1}\right)^{r/2} dx\\& = \left(\int_a^b \left|\dfrac{f^{(n+1)}(t)}{n!}\right|^2dt\right)^{r/2} \left(\frac{1}{2n+1}\right)^{r/2} \left(\frac{(b-a)^{nr+\frac{r}{2} +1}}{nr+\frac{r}{2} +1}\right)\end{align*}

Let $f$ be a twice continuously differentiable function from $[0,1]$ into $R$,Give that
$$f(0)+2f(\frac{1}{2})+f(1)=0$$
show that
$$\int_{0}^{1}(f''(x))^2dx\ge 1920\left(\int_{0}^{1}f(x)dx\right)^2$$

Proof.1) Let $g_1(x)=x(x-1/2)$, $g_2(x)=(x-1)(x-1/2)$. By two integration by parts, we have

$$\int_0^{1/2}f^{\prime\prime}(x)g_1(x)dx=-\frac{f(1/2)+f(0)}{2}+2\int_0^{1/2}f(x)dx$$
and
$$\int_{1/2}^{1}f^{\prime\prime}(x)g_2(x)dx=-\frac{f(1/2)+f(1)}{2}+2\int_0^{1/2}f(x)dx$$
Hence
$$\int_0^{1/2}f^{\prime\prime}(x)g_1(x)dx+\int_{1/2}^{1}f^{\prime\prime}(x)g_2(x)dx=2\int_0^1f(x)dx$$

2) By Cauchy-Schwarz:

$$(\int_{0}^{1/2}f^{\prime\prime}(x)g_1(x)dx)^2\leq (\int_{0}^{1/2}f^{\prime\prime}(x)^2dx)\frac{1}{15.2^6}$$

$$(\int_{1/2}^{1}f^{\prime\prime}(x)g_2(x)dx)^2\leq (\int_{1/2}^{1}f^{\prime\prime}(x)^2dx)\frac{1}{15.2^6}$$

3) Use now $\sqrt{U}+\sqrt{V}\leq \sqrt{2}\sqrt{U+V}$ with $\displaystyle U=\int_{0}^{1/2}f^{\prime\prime}(x)^2dx$ and $\displaystyle V=\int_{1/2}^{1}f^{\prime\prime}(x)^2dx$ to finish the proof.

Let $f$ be a positive-valued,concave function on $[0,1]$,Prove that
$$6\left(\int_{0}^{1}f(x)dx\right)^2\le 1+ 8\int_{0}^{1}f^3(x)dx.$$

Proof.Let $c=\int_0^1 f(x)\,dx$ and $g=f/c$, so $\int_0^1 g(x)\,dx=1$. Then by Holder's inequality,

$$1\le \left(\int_0^1 g(x)^3\,dx\right)^{1/3}\left(\int_0^1 1^{3/2}\,dx\right)^{2/3} .$$
Therefore $\int_0^1 f(x)^3\,dx=c^3\int_0^1g(x)^3\,dx\ge c^3$, and
$$8\int_0^1 f(x)^3\,dx + 1-6\left(\int_0^1 f(x)\,dx\right)^2\ge 8c^3+1-6c^2 =: h(c).$$
For $c>0$ the right-hand side is minimized when $0=h'(c)=24c^2-12c$, meaning $c=1/2$ (noting $h'(c)<0$ for $c<1/2$ and $h'(c)>0$ for $c>1/2$). Thus $$h(1/2)=8(1/2)^3+1-6(1/2)^2=1/2\le h(c)$$ for all $c>0$.
Actually, then it follows
$$6\left(\int_0^1 f(x)\,dx\right)^2 \le \frac12 + 8\int_0^1f(x)^3\,dx.$$
Concavity of $f$ is not needed.

[1] 积分不等式的研究;

[2] 积分不等式的证明方法;