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?

  • $\begingroup$ I didn't know how to formulate the question. $\endgroup$ – user137654 Mar 28 '14 at 19:20
  • 1
    $\begingroup$ It smells like integration by parts (because of that x there). $\endgroup$ – 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 $\dfrac{24}{5}$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.