Integral inequality with a function twice differentiable

Question

Let $f:[0,1]\longrightarrow\mathbb{R}$ be a function twice differentiable with continous second derivative and $f(1)=f(0)$. The inequality: $$\int_{0}^{1}(f''(x))^2dx\geq 120\left(\int_{0}^{1}xf'(x)dx\right)^2$$ holds?

Answer 1

Let $$ A=\int_0^1xf'(x)\,\mathrm{d}x\tag{1} $$ Since $f(0)=f(1)$, we have $$ \int_0^1f'(x)\,\mathrm{d}x=0\tag{2} $$ $(1)$, $(2)$, and integration by parts gives $$ \begin{align} 2A &=\int_0^1(2x-1)f'(x)\,\mathrm{d}x\\ &=\int_0^1f'(x)\,\mathrm{d}x(x-1)\\ &=\int_0^1x(1-x)f''(x)\,\mathrm{d}x\tag{3} \end{align} $$ Apply Hölder to $(3)$: $$ \begin{align} 4A^2 &\le\int_0^1[x(1-x)]^2\,\mathrm{d}x\int_0^1f''(x)^2\,\mathrm{d}x\\ &=\frac1{30}\int_0^1f''(x)^2\,\mathrm{d}x\tag{4} \end{align} $$ Plugging $(1)$ into $(4)$ yields $$ 120\left(\int_0^1xf'(x)\,\mathrm{d}x\right)^2\le\int_0^1f''(x)^2\,\mathrm{d}x\tag{5} $$

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Using $f(x)=x(1-x)(1+x(1-x))$, we see that $(5)$ is sharp: both sides equal $\dfrac{24}{5}$.

Answer 2

Since $f(1)=f(0)\Rightarrow \int_0^1f'(x)\,\mathrm{d}x=0$, By integration by parts, \begin{equation}\label{MSE-730580} \int_0^1g(x)f''(x)\mathrm{d}x=g(x)f'(x)\bigg|_0^1-\int_0^1g'(x)f'(x)\mathrm{d}x \tag{1} \end{equation} Let $g(x)$ be a polynomial defined as, $$g(x)=ax^2+bx+c\Rightarrow g'(x)=2ax+b$$ Let $$g(x)f'(x)\bigg|_0^1=0\Rightarrow g(0)=g(1)=0\Rightarrow b=-a$$ we can get, $$g(x)=ax(x-1)\Rightarrow g'(x)=2ax-a \tag{2}$$ by $(1),(2)$, we have $$\int_0^1x(x-1)f''(x)\mathrm{d}x=\int_0^1(1-2x)f'(x)\mathrm{d}x=-2\int_0^1xf'(x)\mathrm{d}x$$ by the Cauchy-Schwarz inequality, \begin{align*} \left(\int_0^1xf'(x)\mathrm{d}x\right)^2 &=\left(\int_0^1\Big(-\frac{1}{2}x(x-1)\Big)f''(x)\mathrm{d}x\right)^2\\ &\leqslant\int_0^1\Big(-\frac{1}{2}x(x-1)\Big)^2\mathrm{d}x\int_0^1(f''(x))^2\mathrm{d}x \end{align*} Therefore, $$120\left(\int_0^1xf'(x)\,\mathrm{d}x\right)^2\le\int_0^1f''(x)^2\,\mathrm{d}x$$ because, $$\int_0^1\bigg(-\frac{1}{2}x(x-1)\bigg)^2\mathrm{d}x=\frac{1}{120}$$