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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?

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  • $\begingroup$ I didn't know how to formulate the question. $\endgroup$ – user137654 Mar 28 '14 at 19:20
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    $\begingroup$ It smells like integration by parts (because of that x there). $\endgroup$ – orion Mar 28 '14 at 19:37
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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 $\dfrac{24}{5}$.

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  • $\begingroup$ 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) $\endgroup$ – abnry Jun 14 '14 at 4:16
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    $\begingroup$ 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. $\endgroup$ – robjohn Jun 14 '14 at 4:28
  • $\begingroup$ Can you, please, provide a solution with random limits of integration, say a and b, instead of 0 and 1. $\endgroup$ – George R. Mar 14 '17 at 6:23
  • $\begingroup$ @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 $t$. $\endgroup$ – robjohn Mar 14 '17 at 9:47

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