# Integral inequality with a function twice differentiable

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?

• I didn't know how to formulate the question. – user137654 Mar 28 '14 at 19:20
• It smells like integration by parts (because of that x there). – orion Mar 28 '14 at 19:37

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 $$120\left(\int_0^1xf'(x)\,\mathrm{d}x\right)^2\le\int_0^1f''(x)^2\,\mathrm{d}x\tag{5}$$
Using $f(x)=x(1-x)(1+x(1-x))$, we see that $(5)$ is sharp: both sides equal $\dfrac{24}{5}$.
• Okay, so using integration by parts and then Holder makes sense but one would first try it on $\int_0^1 x^2 f''(x)dx$. Why did it occur to you that the $f(0)=f(1)$ condition would allow for that $x^2$ to be changed to something else? (Nice answer, btw) – abnry Jun 14 '14 at 4:16
• What I wanted was to have $u=x(x-1)$ and $v=f'(x)$ so that the limit terms in the integration by parts would vanish, so I needed $(2x-1)\,\mathrm{d}x$ instead of $x\,\mathrm{d}x$ for $\mathrm{d}u$. I recognized that $f(0)=f(1)$ gives us that $\int_0^1f'(x)\,\mathrm{d}x=0$ and used it. – robjohn Jun 14 '14 at 4:28
• @GeorgeR.: Use $g(u)=f\left(\frac{u-a}{b-a}\right)$ or $f(t)=g(a+(b-a)t)$ and write everything in terms of $u$ instead of $t$. – robjohn Mar 14 '17 at 9:47