TITLE: Given $\int_{\frac13}^{\frac23}f(x)dx=0$, how to prove $4860(\int_0^1f(x)dx)^2\le 11\int_0^1|f''(x)|^2dx$?
QUESTION [7 upvotes]: Suppose $f\in C^2[0,1]$, and
  $\int_{\frac13}^{\frac23}f(x)dx=0$.
  Prove  that 
  $$\left(\int_0^1f(x)dx\right)^2\le \frac{11}{4860}\int_0^1|f''(x)|^2dx.$$

This problem is quite similar to Prove the following integral inequality: $\int_{0}^{1}(f''(x))^2dx\ge 1920\left(\int_{0}^{1}f(x)dx\right)^2$.
I have tried to write
$$\int_0^1f(x)dx=\int_0^{\frac13}f(x)dx+\lambda \int_{\frac13}^{\frac23}f(x)dx+\int_{\frac23}^{1}f(x)dx$$ for any $\lambda \in \mathbb{R}$, and pick a suitable $g$ such that
$$\int_0^1f(x)dx=\int_0^1g(x)f''(x)dx$$
then we can use Cauchy-Schwarz inequality to get what we want. How can I get the function $g$?

REPLY [2 votes]: By integration by parts, we have
$$
\begin{aligned}
\int_0^{\frac{1}{3}}g_1(x)f''(x)\mathrm{d}x
&=g_1(x)f'(x)\bigg|_0^{\frac{1}{3}}-\int_0^{\frac{1}{3}}g_1'(x)f'(x)\mathrm{d}x\\
&=g_1(x)f'(x)\bigg|_0^{\frac{1}{3}}-g_1'(x)f(x)\bigg|_0^{\frac{1}{3}}+\int_0^{\frac{1}{2}}g_1''(x)f(x)\mathrm{d}x
\end{aligned}\tag{1}
$$
$$
\begin{aligned}
\int_{\frac{1}{3}}^{\frac{2}{3}}g_2(x)f''(x)\mathrm{d}x
&=g_2(x)f'(x)\bigg|_{\frac{1}{3}}^{\frac{2}{3}}-\int_{\frac{1}{3}}^{\frac{2}{3}}g_2'(x)f'(x)\mathrm{d}x\\
&=g_2(x)f'(x)\bigg|_{\frac{1}{3}}^{\frac{2}{3}}-g_2'(x)f(x)\bigg|_{\frac{1}{3}}^{\frac{2}{3}}+\int_{\frac{1}{3}}^{\frac{2}{3}}g_2''(x)f(x)\mathrm{d}x
\end{aligned}\tag{2}
$$
$$
\begin{aligned}
\int_{\frac{2}{3}}^1g_3(x)f''(x)\mathrm{d}x
&=g_3(x)f'(x)\bigg|_{\frac{2}{3}}^1-\int_{\frac{2}{3}}^1g_3'(x)f'(x)\mathrm{d}x\\
&=g_3(x)f'(x)\bigg|_{\frac{2}{3}}^1-g_3'(x)f(x)\bigg|_{\frac{2}{3}}^1+\int_{\frac{2}{3}}^1g_3''(x)f(x)\mathrm{d}x
\end{aligned}\tag{3}
$$
Let $g(x)$ be the piecewise differentiable function, $g_1(x)$, $g_2(x)$ and $g_3(x)$ is polynomial.
$$g(x)=\begin{cases}
g_1(x),\quad x\in[0,\tfrac{1}{3})\\
g_2(x),\quad x\in[\tfrac{1}{3},\tfrac{2}{3})\\
g_3(x),\quad x\in[\tfrac{2}{3},1]
\end{cases}$$
Since $\textstyle\int_{\frac{1}{3}}^{\frac{2}{3}}f(x)\mathrm{d}x=0$, Let $g_1''(x)=g_3''(x)=c$, $g_2''(x)=k$, So $g_1(x),g_2(x),g_3(x)$ is quadratic polynomial.
$$\begin{cases}
g_1(x)=cx^2+a_1x+a_0,& x\in[0,\tfrac{1}{3})\\
g_2(x)=kx^2+b_1x+b_0,& x\in[\tfrac{1}{3},\tfrac{2}{3})\\
g_3(x)=cx^2+c_1x+c_0,& x\in[\tfrac{2}{3},1]
\end{cases}$$
Let,
$$g_1(x)f'(x)\bigg|_0^{\frac{1}{3}}+g_3(x)f'(x)\bigg|_{\frac{2}{3}}^1+g_2(x)f'(x)\bigg|_{\frac{1}{3}}^{\frac{2}{3}}=0$$
we get,
$$\tag{4}
\begin{cases}
g(0)=g(1)=0\\
g_1(\tfrac{1}{3})-g_2(\tfrac{1}{3})=0\\
g_2(\tfrac{2}{3})-g_3(\tfrac{2}{3})=0
\end{cases}\Rightarrow\quad \begin{cases}
a_0=c+c_1+c_0=0\\
\tfrac{1}{9}c+\tfrac{1}{3}a_1+a_0=\tfrac{1}{9}k+\tfrac{1}{3}b_1+b_0\\
\tfrac{4}{9}k+\tfrac{2}{3}b_1+b_0=\tfrac{4}{9}c+\tfrac{2}{3}c_1+c_0
\end{cases}
$$
Let,
$$g_1'(x)f(x)\bigg|_0^{\frac{1}{3}}+g_3'(x)f(x)\bigg|_{\frac{2}{3}}^1+g_2'(x)f(x)\bigg|_{\frac{1}{3}}^{\frac{2}{3}}=0$$
we get,
$$\tag{5}
\begin{cases}
g'(0)=g'(1)=0\\
g_1'(\tfrac{1}{3})-g_2'(\tfrac{1}{3})=0\\
g_2'(\tfrac{2}{3})-g_3'(\tfrac{2}{3})=0
\end{cases}\Rightarrow\quad 
\begin{cases}
a_1=2c+c_1\\
\tfrac{2}{3}c+a_1=\tfrac{2}{3}k+b_1\\
\tfrac{4}{3}k+b_1=\tfrac{4}{3}c+c_1
\end{cases}
$$
by (4),(5),  it follows that
$$a_0=0,a_1=0,k=-2c,b_1=2c,b_0=-\frac{1}{3}c,c_0=c,c_1=-2c$$
Hence,
$$g(x)=\begin{cases}
g_1(x)=cx^2,& x\in[0,\tfrac{1}{3})\\
g_2(x)=-2cx^2+2cx-\frac{1}{3}c,& x\in[\tfrac{1}{3},\tfrac{2}{3})\\
g_3(x)=cx^2-2cx+c,& x\in[\tfrac{2}{3},1]
\end{cases}$$
by $(1)+(2)+(3)$, we have
\begin{align*}
\int_0^1g(x)f''(x)\mathrm{d}x
&=\int_0^{\frac{1}{3}}g_1(x)f''(x)\mathrm{d}x+\int_{\frac{1}{3}}^{\frac{2}{3}}g_2(x)f''(x)\mathrm{d}x+\int_{\frac{2}{3}}^1g_3(x)f''(x)\mathrm{d}x\\
&=\int_0^{\frac{1}{2}}g_1''(x)f(x)\mathrm{d}x+\underbrace{\int_{\frac{1}{3}}^{\frac{2}{3}}g''_2(x)f(x)\mathrm{d}x}_{=0}+\int_{\frac{2}{3}}^1g''_3(x)f(x)\mathrm{d}x\\
&=2c\int_0^1f(x)\mathrm{d}x
\end{align*}
By Cauchy-Schwarz inequality,
\begin{align*}
\left(2c\int_0^1f(x)\mathrm{d}x\right)^2&=\int_0^1g(x)f''(x)\mathrm{d}x\\
&\leqslant \int_0^1(g(x))^2\mathrm{d}x\int_0^1(f''(x))^2\mathrm{d}x
\end{align*}
After calculation, We finally get
$$\int_0^1(g(x))^2\mathrm{d}x=\frac{c^2}{1215}+\frac{c^2}{135}+\frac{c^2}{1215}=\frac{11c^2}{1215}$$
Therefore,
$$4860\left(\int_0^1f(x)\mathrm{d}x \right)^2\leqslant 11\int_0^1\big(f''(x)\big)^2\mathrm{d}x $$