已知$f(x)$在$[0,1]$上二阶可导，求证：$\displaystyle\int_0^1\left |f' (x)\right |\mathrm{d}x\le 9\int_0^1\left |f(x)\right |\mathrm{d}x +\int_0^1\left |f''(x)\right |\mathrm{d}x$
证明:
对任意$0<\xi<\dfrac{1}{3}$和$\dfrac{2}{3}<\eta<1$,则存在$\lambda\in(\xi,\eta)$,使得
\begin{equation*}
|f'(\lambda)|=\left|\frac{f(\eta)-f(\xi)}{\eta-\xi}\right|\leq 3|f(\xi)|+3|f(\eta)|
\end{equation*}
因此对任意的$x\in(0,1)$成立
\begin{equation*}
|f'(x)|=\left|f'(\lambda)+\int_\lambda^x f''(t)dt\right|\leq 3|f(\xi)|+3|f(\eta)|+\int_0^1|f''(t)|dt
\end{equation*}
分别对$\xi$在$(0,\dfrac{1}{3})$上和对$\eta$在$(\dfrac{2}{3},1)$上积分以上不等式，得
\begin{align*}
\frac{1}{9}|f'(x)|&\leq \int_0^{\frac{1}{3}}|f(\xi)|d\xi+\int_{\frac{2}{3}}^1 |f(\eta)|d\eta+
\frac{1}{9}\int_0^1|f''(t)|dt\\
&\leq \int_0^1|f(t)|dt+\frac{1}{9}\int_0^1|f''(t)|dt
\end{align*}
于是
\begin{equation*}
|f'(x)|\leq 9\int_0^1|f(t)|dt+\int_0^1|f''(t)|dt,\hspace{10mm}x\in[0,1]
\end{equation*}
对上式两边在$[0,1]$积分,得到
\begin{equation*}
\int_0^1\left |f' (x)\right |\mathrm{d}x\le 9\int_0^1\left |f(x)\right |\mathrm{d}x +\int_0^1\left |f''(x)\right |\mathrm{d}x
\end{equation*}
$\square$
问题似乎可以加强到
\[\displaystyle\int_0^1\left |f' (x)\right |\mathrm{d}x\le 4\int_0^1\left |f(x)\right |\mathrm{d}x +\int_0^1\left |f''(x)\right |\mathrm{d}x\]
求极限:
\[ \lim_{y\rightarrow+\infty}{\left(\ln^{2}{y}-2\int_{0}^{y}{\frac{\ln{y}}{\sqrt{x^2+1}}dx}\right)} \]
(Proposed by tian275461, Solution by sos440)
\[ \begin{align*} I&=\ln^{2}y-2\int_{0}^{y}{\frac{\ln{x}}{x}\cdot\frac{x}{\sqrt{x^2+1}}dx}\\
&=\ln^{2}{y}-\frac{y}{\sqrt{y^2+1}}\cdot\ln^{2}{y}+\int_{0}^{y}{\frac{\ln^{2}{x}}{(x^2+1)^{\frac{3}{2}}}dx} \end{align*}\]
notice that
\[ \lim_{y\rightarrow+\infty}{\left(\ln^{2}{y}-\frac{y}{\sqrt{y^2+1}}\cdot\ln^{2}{y}\right)}=0 \]
So,just find the value of the following integral
\[ \int_{0}^{\infty}{\frac{\ln^{2}{x}}{(x^2+1)^{\frac{3}{2}}}dx} \]
First note that
\[\begin{align*}
I(p, q) := \int_{0}^{\infty} \frac{x^{p}}{(1+x^2)^{q}} \, dx
&= \int_{0}^{\frac{\pi}{2}} \frac{\tan^{p} \theta}{\sec^{2q} \theta} \, \sec^2 \theta \, d\theta \qquad (x = \tan\theta)\\
&= \int_{0}^{\frac{\pi}{2}} \sin^{p} \theta \cos^{2q-p-2} \theta \, d\theta \\
&= \frac{1}{2}\beta\left( \frac{p+1}{2}, \frac{2q-p-1}{2} \right) \\
&= \frac{1}{2}\frac{\Gamma \left( \frac{p+1}{2} \right) \Gamma \left( \frac{2q-p-1}{2} \right)}{\Gamma(q)}.
\end{align*}\]
Thus
\[ \int_{0}^{\infty} \frac{\log^2 x}{(1+x^2)^{3/2}} \, dx = \frac{\partial^2 I}{\partial p^2}\left( 0, \tfrac{3}{2} \right) \]
and the rest is a mere calculation. Note first that
\[ I\left(p, \tfrac{3}{2} \right) = \frac{\Gamma \left(\frac{1+p}{2}\right) \Gamma \left(1 - \frac{p}{2}\right) }{\Gamma \left(\frac{1}{2}\right) }. \]
Differentiating, we have
\[ \frac{\partial I}{\partial p}\left(p, \tfrac{3}{2} \right) = \frac{\Gamma \left(\frac{1+p}{2}\right) \Gamma \left(1 - \frac{p}{2}\right) }{2\Gamma \left(\frac{1}{2}\right) } \left[ \psi_{0}\left( \frac{1+p}{2} \right) - \psi_{0}\left( 1 - \frac{p}{2} \right) \right]. \]
Differentiating once again, we have
\[ \frac{\partial^2 I}{\partial p^2} \left(p, \tfrac{3}{2} \right) = \frac{\Gamma \left(\frac{1+p}{2}\right) \Gamma \left(1 - \frac{p}{2}\right) }{4\Gamma \left(\frac{1}{2}\right) } \left[ \left\{\psi_{0}\left( \frac{1+p}{2} \right) - \psi_{0}\left( 1 - \frac{p}{2} \right) \right\}^2 + \left\{\psi_{1}\left( \frac{1+p}{2} \right) + \psi_{1}\left( 1 - \frac{p}{2} \right) \right\} \right] . \]
Plugging $p = 0$, we have
\[ \frac{\partial^2 I}{\partial p^2} \left(0, \tfrac{3}{2} \right) = \frac{1}{4} \left[ \left\{\psi_{0}\left( \tfrac{1}{2} \right) - \psi_{0}(1) \right\}^2 + \left\{\psi_{1}\left( \tfrac{1}{2} \right) + \psi_{1}(1) \right\} \right] . \]
In view of the expansion formula
\[ \psi_{0}(s) = -\gamma + \sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+s} \right), \]
we obtain
\[\begin{align*}
\psi_{0}\left( \tfrac{1}{2} \right) - \psi_{0}(1)
&= \sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+\frac{1}{2}} \right) \\
&= -2 \sum_{n=0}^{\infty} \left( \frac{1}{2n+1} - \frac{1}{2n+2} \right) \\
&= -2 \ln{2}
\end{align*}\]
and likewise
\[\begin{align*}
\psi_{1}\left( \tfrac{1}{2} \right) + \psi_{1}(1)
&= \sum_{n=0}^{\infty} \left( \frac{1}{\left( n+\frac{1}{2} \right)^2} + \frac{1}{(n+1)^2} \right) \\
&= 4 \sum_{n=0}^{\infty} \left( \frac{1}{(2n+1)^2} + \frac{1}{(2n+2)^2} \right) \\
&= 4 \zeta(2) .
\end{align*}\]
Therefore we have
\[ \frac{\partial^2 I}{\partial p^2} \left(0, \tfrac{3}{2} \right) = \zeta(2) + \ln^{2}{2}.\]
$\square$
设$f:[0,1]\rightarrow R$是连续函数，且$\displaystyle \int_{0}^{1}{f^{3}(x)dx}=0$,求证:
\[ \int_{0}^{1}{f^{4}(x)dx}\geq \frac{27}{4}\left(\int_{0}^{1}{f(x)dx} \right)^{4} \]
(tian276461)
证明:令
\[ I_{n}=\int_{0}^{1}{f^{n}(x)dx} \]
由Cauchy-Schwarz得
\[ I_{2}\geq I_{1}^{2} \]
又由Cauchy-Schwarz得
\[ \left(\int_{0}^{1}{(r+f^{2}(x))\cdot f(x)dx} \right)^{2}\leq \int_{0}^{1}{(r+f^{2}(x))^{2}dx}\cdot\int_{0}^{1}{f^{2}(x)dx} \]
展开得
\[ (I_{2}-I^{2}_{1})r^2+2I_{2}^{2}r+I_{2}I_{4}\geq 0 \]
上式恒成立，所以 $ \Delta\leq 0 $
\[ \Rightarrow I_{4}\geq \frac{I^{3}_{2}}{I_{2}-I_{1}^{2}} \]
所以只要证明
\[ \frac{I^{3}_{2}}{I_{2}-I^{2}_{1}}\geq \frac{27}{4}I^{4}_{1} \]
由AM-GM
\[ (I_{2}-I_{1}^{2})I^{4}_{1}=\frac{1}{2}(2I_{2}-2I_{1}^{2})\cdot I^{2}_{1}\cdot I^{2}_{1}\leq \frac{4}{27}I_{2}^{3}\]
故有
\[ \int_{0}^{1}{f^{4}(x)dx}\geq \frac{27}{4}\left(\int_{0}^{1}{f(x)dx} \right)^{4} \]
$\square$
先看12的。
设$f:[0,1]\rightarrow R$是连续可导函数，若$\displaystyle \int_{0}^{\frac{1}{2}}{f(x)dx}=0 $,求证:
\[ \int_{0}^{1}{f'(x)^{2}dx}\geq 12\left(\int_{0}^{1}{f(x)dx} \right)^{2} \]
(tian 275461)
证明:
\[ \int_{0}^{\frac{1}{2}}{f(x)dx}=0\Rightarrow \int_{0}^{\frac{1}{2}}{xf'(x)dx}=\frac{1}{2}f\left(\frac{1}{2}\right)\]
\begin{align*}
\left(\int_{0}^{1}{f(x)dx} \right)^{2}&=\left[\int_{\frac{1}{2}}^{1}{(f(x)-f(\frac{1}{2}))dx}+\frac{1}{2}f(\frac{1}{2}) \right]^{2}\\
&=\left[\int_{\frac{1}{2}}^{1}{\int_{\frac{1}{2}}^{x}{f'(t) dt}dx}+\int_{0}^{\frac{1}{2}}{xf'(x)dx} \right]^{2}\\
&=\left[\int_{\frac{1}{2}}^{1}{(1-t)f'(t)dt}+\int_{0}^{\frac{1}{2}}{xf'(x)dx} \right]^{2}\\
&\leq 2\left[\int_{\frac{1}{2}}^{1}{(1-t)f'(t)dt}\right]^{2}+2\left[\int_{0}^{\frac{1}{2}}{xf'(x)dx} \right]^{2}\\
&\leq 2\left[\int_{\frac{1}{2}}^{1}{(1-t)^{2}dt}\int_{\frac{1}{2}}^{1}{f'(t)^{2}dt}+\int_{0}^{\frac{1}{2}}{x^2dx}\int_{0}^{\frac{1}{2}}{f'(t)^{2}dt} \right]\\
&=\frac{1}{12}\int_{0}^{1}{f'(x)^{2}dx}
\end{align*}
$\square$
再看27的。 
设$f(x)$在$[0,1]$连续可导且可积，若$\displaystyle \int_{\frac{1}{3}}^{\frac{2}{3}}{f(x)dx}=0 $
求证：
\[ \int_{0}^{1}{(f'(x))^{2}dx}\geq 27 \left(\int_{0}^{1}{f(x)dx}\right)^{2} \]
(tian275461)
证明
考虑
\[ G(x)= \left\{
\begin{array}{ll}
x, & \hbox{$x\in\left[0,\frac{1}{3}\right)$} \\
1-2x, & \hbox{$x\in\left[\frac{1}{3},\frac{2}{3}\right]$} \\
x-1, & \hbox{$x\in\left[\frac{2}{3},1\right)$}
\end{array}
\right. \]
由Cauchy-Schwarz容易证明.以下略
3个推广
1.若$f(x):[0,1]\rightarrow R$ 是连续可导函数，且$\displaystyle \int_{\frac{1}{2n}}^{\frac{1}{n}}{f(x)dx}=0 $,则有:
\[ \int_{0}^{1}{(f'(x))^{2}dx}\geq \frac{12n^{2}}{4n^2-10n+7}\left(\int_{0}^{1}{f(x)dx} \right)^{2} \]
提示：设
\[ g(x)=\left\{
\begin{array}{ll}
x, & \hbox{$x\in\left[0,\frac{1}{2n} \right]$} \\
1-(2n-1)x, & \hbox{$x\in\left[\frac{1}{2n},\frac{1}{n} \right]$} \\
x-1, & \hbox{$x\in\left[\frac{1}{n},1 \right]$.}
\end{array} \right. \]
然后仿照上面一样用Cauchy-Schwarz
2. 若$f(x):[a,b]\rightarrow R$ 是连续可导函数，且$\displaystyle \int_{a}^{b}{f(x)dx}=0 $,则有:
\[ \int_{a}^{2b-a}{(f'(x))^{2}dx}\geq \frac{3}{2(b-a)^{3}}\left(\int_{a}^{2b-a}{f(x)dx} \right)^{2} \]
提示：设
\[ g(x)=\left\{
\begin{array}{ll}
x-a, & \hbox{$x\in [a,b]$} \\
2b-a-x, & \hbox{$x\in [b,2b-a]$.}
\end{array} \right.\]
然后仿照上面一样用Cauchy-Schwarz
3. 若$f(x):[0,1]\rightarrow R$ 是连续可导函数，且$\displaystyle \int_{\frac{1}{2n+1}}^{\frac{2}{2n+1}}{f(x)dx}=0 $,则有:
\[ \int_{0}^{1}{(f'(x))^{2}dx}\geq \frac{3(2n+1)^{2}}{4n^2-6n+3}\left(\int_{0}^{1}{f(x)dx} \right)^{2} \]
提示：设
\[ g(x)=\left\{
\begin{array}{ll}
x, & \hbox{$x\in\left[0,\frac{1}{2n+1} \right]$} \\
1-2nx, & \hbox{$x\in\left[\frac{1}{2n+1},\frac{2}{2n+1} \right]$} \\
x-1, & \hbox{$x\in\left[\frac{2}{2n+1},1 \right]$.}\end{array}\right. \]
然后仿照上面一样用Cauchy-Schwarz
设$f(x)$是在$[0,1]$非负的连续的凹函数，且$f(0)=1$, 求证:
\[ 2\int_{0}^{1}{x^2f(x)dx}+\frac{1}{12}\leq \left(\int_{0}^{1}{f(x)dx} \right)^{2} \]
证明
设 \[ F(x)=\int_{0}^{x}{f(t)dt}\]
由于$f(x)$是凹函数，所以有
\[ \frac{f(t)-f(0)}{t-0}\geq \frac{f(x)-f(0)}{x-0}, \qquad t\in(0,x) \]
\[ \Rightarrow f(t)\geq \frac{t}{x}(f(x)-1)+1 \]
故
\[\begin{align*}
I=\int_{0}^{1}{x^2f(x)dx}&=\int_{0}^{1}{x^2dF(x)}\\
&=F(1)-2\int_{0}^{1}{x\int_{0}^{x}{f(t)dt}dx}\\
&\leq F(1)-I-\frac{1}{3}
\end{align*} \]
所以
\[ 2I\leq F(1)-\frac{1}{3} \]
只要证明
\[ F(1)-\frac{1}{3}+\frac{1}{12}\leq F^{2}(1) \]
\[ \Leftrightarrow \left(F(1)-\frac{1}{4}\right)^{2}\geq 0 \]
显然成立。
$\square$
下面的题和这个有着公共的内核。手段一样，所以在这里不证明了。
若$f:[0,1]\rightarrow \mathbf{R}$是连续凹函数，且满足$f(0)=1,$证明:
\[ \int_{0}^{1}{xf(x)dx}\leq \dfrac{2}{3}\left(\int_{0}^{1}{f(x)dx}\right)^{2} \]
